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מתמטיקה תיכונית/אלגברה תיכונית/משוואות/משוואות עם שברים/תרגילים/רמה א
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עריכה
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מתמטיקה תיכונית
|
אלגברה תיכונית
|
משוואות
|
משוואות עם שברים
|
תרגילים
פתרו את המשוואות הבאות. יש למצוא את
כל
הפתרונות:
1
2
x
−
1
4
+
−
19
2
−
2
x
−
2
=
x
−
3
4
+
−
3
2
−
2
x
−
2
{\displaystyle {\frac {1}{2}}x-{\frac {1}{4}}+{\frac {-{\frac {19}{2}}}{-2x-2}}=x-{\frac {3}{4}}+{\frac {-{\frac {3}{2}}}{-2x-2}}}
3
x
−
3
+
1
x
=
2
x
+
1
x
{\displaystyle 3x-3+{\frac {1}{x}}=2x+{\frac {1}{x}}}
−
5
+
−
6
−
1
2
x
−
1
=
2
x
−
3
+
−
4
−
1
2
x
−
1
{\displaystyle -5+{\frac {-6}{-{\frac {1}{2}}x-1}}=2x-3+{\frac {-4}{-{\frac {1}{2}}x-1}}}
3
2
x
−
5
2
+
3
x
−
2
=
1
2
x
+
3
2
+
2
x
−
2
{\displaystyle {\frac {3}{2}}x-{\frac {5}{2}}+{\frac {3}{x-2}}={\frac {1}{2}}x+{\frac {3}{2}}+{\frac {2}{x-2}}}
−
1
2
x
+
−
9
−
x
−
2
=
1
2
x
−
2
+
−
4
−
x
−
2
{\displaystyle -{\frac {1}{2}}x+{\frac {-9}{-x-2}}={\frac {1}{2}}x-2+{\frac {-4}{-x-2}}}
1
2
x
+
7
4
+
15
8
x
−
1
2
=
−
1
2
x
+
1
4
+
9
8
x
−
1
2
{\displaystyle {\frac {1}{2}}x+{\frac {7}{4}}+{\frac {\frac {15}{8}}{x-{\frac {1}{2}}}}=-{\frac {1}{2}}x+{\frac {1}{4}}+{\frac {\frac {9}{8}}{x-{\frac {1}{2}}}}}
−
x
+
6
+
−
3
2
−
1
2
x
+
1
=
x
+
2
+
−
3
2
−
1
2
x
+
1
{\displaystyle -x+6+{\frac {-{\frac {3}{2}}}{-{\frac {1}{2}}x+1}}=x+2+{\frac {-{\frac {3}{2}}}{-{\frac {1}{2}}x+1}}}
−
x
−
3
2
+
3
2
−
2
x
+
1
=
−
1
2
x
+
1
4
+
−
9
4
−
2
x
+
1
{\displaystyle -x-{\frac {3}{2}}+{\frac {\frac {3}{2}}{-2x+1}}=-{\frac {1}{2}}x+{\frac {1}{4}}+{\frac {-{\frac {9}{4}}}{-2x+1}}}
−
3
x
−
5
2
+
1
−
x
=
−
2
x
+
1
2
+
−
1
−
x
{\displaystyle -3x-{\frac {5}{2}}+{\frac {1}{-x}}=-2x+{\frac {1}{2}}+{\frac {-1}{-x}}}
−
1
2
x
+
5
+
13
−
x
−
2
=
1
2
x
+
1
−
x
−
2
{\displaystyle -{\frac {1}{2}}x+5+{\frac {13}{-x-2}}={\frac {1}{2}}x+{\frac {1}{-x-2}}}
1
2
x
+
1
2
+
−
5
x
−
1
=
−
1
2
x
+
1
2
+
1
x
−
1
{\displaystyle {\frac {1}{2}}x+{\frac {1}{2}}+{\frac {-5}{x-1}}=-{\frac {1}{2}}x+{\frac {1}{2}}+{\frac {1}{x-1}}}
1
2
x
+
3
2
+
−
3
2
x
+
1
=
−
1
2
x
−
1
2
+
1
2
x
+
1
{\displaystyle {\frac {1}{2}}x+{\frac {3}{2}}+{\frac {-{\frac {3}{2}}}{x+1}}=-{\frac {1}{2}}x-{\frac {1}{2}}+{\frac {\frac {1}{2}}{x+1}}}
10
+
31
−
1
2
x
−
2
=
2
x
−
10
+
−
18
−
1
2
x
−
2
{\displaystyle 10+{\frac {31}{-{\frac {1}{2}}x-2}}=2x-10+{\frac {-18}{-{\frac {1}{2}}x-2}}}
4
x
+
−
11
2
1
2
x
=
2
x
−
2
+
1
2
1
2
x
{\displaystyle 4x+{\frac {-{\frac {11}{2}}}{{\frac {1}{2}}x}}=2x-2+{\frac {\frac {1}{2}}{{\frac {1}{2}}x}}}
1
4
x
+
3
16
+
61
32
2
x
+
1
2
=
−
1
4
x
−
3
16
+
67
32
2
x
+
1
2
{\displaystyle {\frac {1}{4}}x+{\frac {3}{16}}+{\frac {\frac {61}{32}}{2x+{\frac {1}{2}}}}=-{\frac {1}{4}}x-{\frac {3}{16}}+{\frac {\frac {67}{32}}{2x+{\frac {1}{2}}}}}
−
2
x
+
3
2
+
−
1
4
−
x
−
1
2
=
−
x
+
1
−
x
−
1
2
{\displaystyle -2x+{\frac {3}{2}}+{\frac {-{\frac {1}{4}}}{-x-{\frac {1}{2}}}}=-x+{\frac {1}{-x-{\frac {1}{2}}}}}
2
+
−
12
1
2
x
+
2
=
−
2
x
+
10
+
−
19
1
2
x
+
2
{\displaystyle 2+{\frac {-12}{{\frac {1}{2}}x+2}}=-2x+10+{\frac {-19}{{\frac {1}{2}}x+2}}}
−
3
4
+
−
2
2
x
=
−
1
2
x
+
1
4
+
−
2
2
x
{\displaystyle -{\frac {3}{4}}+{\frac {-2}{2x}}=-{\frac {1}{2}}x+{\frac {1}{4}}+{\frac {-2}{2x}}}
x
−
1
+
−
5
−
x
−
1
=
2
x
−
4
+
−
5
−
x
−
1
{\displaystyle x-1+{\frac {-5}{-x-1}}=2x-4+{\frac {-5}{-x-1}}}
1
+
4
−
2
x
=
1
2
x
−
1
2
+
2
−
2
x
{\displaystyle 1+{\frac {4}{-2x}}={\frac {1}{2}}x-{\frac {1}{2}}+{\frac {2}{-2x}}}
x
−
1
4
+
−
7
8
2
x
+
1
2
=
1
2
x
+
3
8
+
13
16
2
x
+
1
2
{\displaystyle x-{\frac {1}{4}}+{\frac {-{\frac {7}{8}}}{2x+{\frac {1}{2}}}}={\frac {1}{2}}x+{\frac {3}{8}}+{\frac {\frac {13}{16}}{2x+{\frac {1}{2}}}}}
3
2
x
+
7
+
18
x
−
2
=
1
2
x
+
1
+
3
x
−
2
{\displaystyle {\frac {3}{2}}x+7+{\frac {18}{x-2}}={\frac {1}{2}}x+1+{\frac {3}{x-2}}}
2
x
−
5
+
9
x
+
2
=
x
−
1
+
1
x
+
2
{\displaystyle 2x-5+{\frac {9}{x+2}}=x-1+{\frac {1}{x+2}}}
14
+
−
8
1
2
x
+
1
=
−
2
x
+
8
+
−
8
1
2
x
+
1
{\displaystyle 14+{\frac {-8}{{\frac {1}{2}}x+1}}=-2x+8+{\frac {-8}{{\frac {1}{2}}x+1}}}
1
4
x
−
1
16
+
−
95
32
2
x
+
1
2
=
−
1
4
x
−
7
16
+
−
25
32
2
x
+
1
2
{\displaystyle {\frac {1}{4}}x-{\frac {1}{16}}+{\frac {-{\frac {95}{32}}}{2x+{\frac {1}{2}}}}=-{\frac {1}{4}}x-{\frac {7}{16}}+{\frac {-{\frac {25}{32}}}{2x+{\frac {1}{2}}}}}
1
2
x
−
1
+
3
x
+
2
=
−
1
2
x
+
1
x
+
2
{\displaystyle {\frac {1}{2}}x-1+{\frac {3}{x+2}}=-{\frac {1}{2}}x+{\frac {1}{x+2}}}
3
4
x
−
1
16
+
−
289
32
2
x
−
1
2
=
1
4
x
−
3
16
+
−
3
32
2
x
−
1
2
{\displaystyle {\frac {3}{4}}x-{\frac {1}{16}}+{\frac {-{\frac {289}{32}}}{2x-{\frac {1}{2}}}}={\frac {1}{4}}x-{\frac {3}{16}}+{\frac {-{\frac {3}{32}}}{2x-{\frac {1}{2}}}}}
−
1
+
−
1
−
x
−
1
=
x
−
1
+
−
1
−
x
−
1
{\displaystyle -1+{\frac {-1}{-x-1}}=x-1+{\frac {-1}{-x-1}}}
−
1
4
x
+
25
16
+
57
32
−
2
x
−
1
2
=
1
4
x
−
1
16
+
31
32
−
2
x
−
1
2
{\displaystyle -{\frac {1}{4}}x+{\frac {25}{16}}+{\frac {\frac {57}{32}}{-2x-{\frac {1}{2}}}}={\frac {1}{4}}x-{\frac {1}{16}}+{\frac {\frac {31}{32}}{-2x-{\frac {1}{2}}}}}
x
+
3
2
+
9
2
2
x
−
1
=
1
2
x
+
1
4
+
9
4
2
x
−
1
{\displaystyle x+{\frac {3}{2}}+{\frac {\frac {9}{2}}{2x-1}}={\frac {1}{2}}x+{\frac {1}{4}}+{\frac {\frac {9}{4}}{2x-1}}}
−
1
2
x
+
7
2
+
5
−
x
=
1
2
x
−
1
2
+
2
−
x
{\displaystyle -{\frac {1}{2}}x+{\frac {7}{2}}+{\frac {5}{-x}}={\frac {1}{2}}x-{\frac {1}{2}}+{\frac {2}{-x}}}
1
2
x
−
1
2
+
−
3
−
2
x
−
2
=
x
−
2
+
−
3
−
2
x
−
2
{\displaystyle {\frac {1}{2}}x-{\frac {1}{2}}+{\frac {-3}{-2x-2}}=x-2+{\frac {-3}{-2x-2}}}
−
5
+
17
2
−
x
+
1
=
x
+
−
1
2
−
x
+
1
{\displaystyle -5+{\frac {\frac {17}{2}}{-x+1}}=x+{\frac {-{\frac {1}{2}}}{-x+1}}}
−
3
x
+
3
+
3
−
x
−
1
=
−
2
x
+
1
+
3
−
x
−
1
{\displaystyle -3x+3+{\frac {3}{-x-1}}=-2x+1+{\frac {3}{-x-1}}}
−
1
2
x
+
11
2
+
−
5
2
−
x
+
1
=
1
2
x
+
5
2
+
−
5
2
−
x
+
1
{\displaystyle -{\frac {1}{2}}x+{\frac {11}{2}}+{\frac {-{\frac {5}{2}}}{-x+1}}={\frac {1}{2}}x+{\frac {5}{2}}+{\frac {-{\frac {5}{2}}}{-x+1}}}
−
4
x
−
10
+
7
−
1
2
x
+
1
=
−
2
x
−
6
+
7
−
1
2
x
+
1
{\displaystyle -4x-10+{\frac {7}{-{\frac {1}{2}}x+1}}=-2x-6+{\frac {7}{-{\frac {1}{2}}x+1}}}
3
2
x
−
1
+
−
5
x
=
1
2
x
+
1
+
−
2
x
{\displaystyle {\frac {3}{2}}x-1+{\frac {-5}{x}}={\frac {1}{2}}x+1+{\frac {-2}{x}}}
−
x
+
2
−
2
x
−
2
=
−
1
2
x
+
1
2
+
3
−
2
x
−
2
{\displaystyle -x+{\frac {2}{-2x-2}}=-{\frac {1}{2}}x+{\frac {1}{2}}+{\frac {3}{-2x-2}}}
−
2
+
7
2
x
+
2
=
−
x
+
1
+
−
5
2
x
+
2
{\displaystyle -2+{\frac {\frac {7}{2}}{x+2}}=-x+1+{\frac {-{\frac {5}{2}}}{x+2}}}
6
x
+
30
+
59
1
2
x
−
2
=
4
x
+
18
+
38
1
2
x
−
2
{\displaystyle 6x+30+{\frac {59}{{\frac {1}{2}}x-2}}=4x+18+{\frac {38}{{\frac {1}{2}}x-2}}}
−
3
2
x
−
11
2
+
15
2
−
x
+
1
=
−
1
2
x
−
1
2
+
−
1
2
−
x
+
1
{\displaystyle -{\frac {3}{2}}x-{\frac {11}{2}}+{\frac {\frac {15}{2}}{-x+1}}=-{\frac {1}{2}}x-{\frac {1}{2}}+{\frac {-{\frac {1}{2}}}{-x+1}}}
−
x
+
−
4
−
1
2
x
+
1
=
x
+
4
+
−
4
−
1
2
x
+
1
{\displaystyle -x+{\frac {-4}{-{\frac {1}{2}}x+1}}=x+4+{\frac {-4}{-{\frac {1}{2}}x+1}}}
−
1
2
x
−
5
4
+
−
33
4
2
x
−
1
=
−
x
−
3
2
+
1
2
2
x
−
1
{\displaystyle -{\frac {1}{2}}x-{\frac {5}{4}}+{\frac {-{\frac {33}{4}}}{2x-1}}=-x-{\frac {3}{2}}+{\frac {\frac {1}{2}}{2x-1}}}
−
x
+
1
2
+
−
13
2
x
=
−
2
x
−
1
2
+
−
1
2
x
{\displaystyle -x+{\frac {1}{2}}+{\frac {-{\frac {13}{2}}}{x}}=-2x-{\frac {1}{2}}+{\frac {-{\frac {1}{2}}}{x}}}
−
3
4
x
−
15
8
+
47
8
−
2
x
+
1
=
−
1
4
x
+
3
8
+
−
3
8
−
2
x
+
1
{\displaystyle -{\frac {3}{4}}x-{\frac {15}{8}}+{\frac {\frac {47}{8}}{-2x+1}}=-{\frac {1}{4}}x+{\frac {3}{8}}+{\frac {-{\frac {3}{8}}}{-2x+1}}}
2
+
−
1
−
x
−
1
2
=
x
+
1
2
+
−
7
4
−
x
−
1
2
{\displaystyle 2+{\frac {-1}{-x-{\frac {1}{2}}}}=x+{\frac {1}{2}}+{\frac {-{\frac {7}{4}}}{-x-{\frac {1}{2}}}}}
0
+
−
5
1
2
x
+
2
=
−
2
x
+
6
+
−
11
1
2
x
+
2
{\displaystyle 0+{\frac {-5}{{\frac {1}{2}}x+2}}=-2x+6+{\frac {-11}{{\frac {1}{2}}x+2}}}
−
1
+
−
6
x
−
1
2
=
−
x
−
1
2
+
1
4
x
−
1
2
{\displaystyle -1+{\frac {-6}{x-{\frac {1}{2}}}}=-x-{\frac {1}{2}}+{\frac {\frac {1}{4}}{x-{\frac {1}{2}}}}}
1
2
x
−
7
4
+
−
7
8
x
−
1
2
=
−
1
2
x
+
3
4
+
3
8
x
−
1
2
{\displaystyle {\frac {1}{2}}x-{\frac {7}{4}}+{\frac {-{\frac {7}{8}}}{x-{\frac {1}{2}}}}=-{\frac {1}{2}}x+{\frac {3}{4}}+{\frac {\frac {3}{8}}{x-{\frac {1}{2}}}}}
−
6
x
−
12
+
13
2
−
1
2
x
+
1
=
−
4
x
−
10
+
21
2
−
1
2
x
+
1
{\displaystyle -6x-12+{\frac {\frac {13}{2}}{-{\frac {1}{2}}x+1}}=-4x-10+{\frac {\frac {21}{2}}{-{\frac {1}{2}}x+1}}}
תשובות סופיות
עריכה
{
3
,
−
3
}
{\displaystyle \{3,-3\}}
{
3
}
{\displaystyle \{3\}}
{
−
3
,
0
}
{\displaystyle \{-3,0\}}
{
3
,
3
}
{\displaystyle \{3,3\}}
{
−
3
,
3
}
{\displaystyle \{-3,3\}}
{
−
1
,
0
}
{\displaystyle \{-1,0\}}
{
2
,
2
}
{\displaystyle \{2,2\}}
{
−
2
,
−
1
}
{\displaystyle \{-2,-1\}}
{
−
1
,
−
2
}
{\displaystyle \{-1,-2\}}
{
1
,
2
}
{\displaystyle \{1,2\}}
{
3
,
−
2
}
{\displaystyle \{3,-2\}}
{
0
,
−
3
}
{\displaystyle \{0,-3\}}
{
3
,
3
}
{\displaystyle \{3,3\}}
{
−
3
,
2
}
{\displaystyle \{-3,2\}}
{
0
,
−
1
}
{\displaystyle \{0,-1\}}
{
−
1
,
2
}
{\displaystyle \{-1,2\}}
{
3
,
−
3
}
{\displaystyle \{3,-3\}}
{
2
,
0
}
{\displaystyle \{2,0\}}
{
−
1
,
3
}
{\displaystyle \{-1,3\}}
{
2
,
1
}
{\displaystyle \{2,1\}}
{
2
,
−
1
}
{\displaystyle \{2,-1\}}
{
−
1
,
−
3
}
{\displaystyle \{-1,-3\}}
{
2
,
0
}
{\displaystyle \{2,0\}}
{
−
2
,
−
3
}
{\displaystyle \{-2,-3\}}
{
−
2
,
1
}
{\displaystyle \{-2,1\}}
{
−
1
,
0
}
{\displaystyle \{-1,0\}}
{
−
3
,
3
}
{\displaystyle \{-3,3\}}
{
−
1
,
0
}
{\displaystyle \{-1,0\}}
{
3
,
0
}
{\displaystyle \{3,0\}}
{
−
1
,
−
1
}
{\displaystyle \{-1,-1\}}
{
3
,
1
}
{\displaystyle \{3,1\}}
{
−
1
,
3
}
{\displaystyle \{-1,3\}}
{
−
2
,
−
2
}
{\displaystyle \{-2,-2\}}
{
2
,
−
1
}
{\displaystyle \{2,-1\}}
{
1
,
3
}
{\displaystyle \{1,3\}}
{
2
,
−
2
}
{\displaystyle \{2,-2\}}
{
3
,
−
1
}
{\displaystyle \{3,-1\}}
{
−
2
,
0
}
{\displaystyle \{-2,0\}}
{
0
,
1
}
{\displaystyle \{0,1\}}
{
1
,
−
3
}
{\displaystyle \{1,-3\}}
{
−
1
,
−
3
}
{\displaystyle \{-1,-3\}}
{
−
2
,
2
}
{\displaystyle \{-2,2\}}
{
−
3
,
3
}
{\displaystyle \{-3,3\}}
{
2
,
−
3
}
{\displaystyle \{2,-3\}}
{
−
2
,
−
2
}
{\displaystyle \{-2,-2\}}
{
0
,
1
}
{\displaystyle \{0,1\}}
{
2
,
−
3
}
{\displaystyle \{2,-3\}}
{
−
2
,
3
}
{\displaystyle \{-2,3\}}
{
0
,
3
}
{\displaystyle \{0,3\}}
{
−
2
,
3
}
{\displaystyle \{-2,3\}}
