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\(a,\Rightarrow x+35=103\\ \Rightarrow x=68\\ b,\Rightarrow x+7=42\\ \Rightarrow x=35\\ c,\Rightarrow x-15=9\\ \Rightarrow x=24\\ d,\Rightarrow x+23=31\\ \Rightarrow x=8\\ e,\Rightarrow18-x=6\\ \Rightarrow x=12\\ f,\Rightarrow x+14=34\\ \Rightarrow x=20\\ g,\Rightarrow2x+1=21\\ \Rightarrow x=11\\ h,\Rightarrow3\left(x+1\right)=99\\ \Rightarrow x+1=33\Rightarrow x=32\\ i,5\left(x-3\right)=25\\ \Rightarrow x-3=5\\ \Rightarrow x=8\\ j,\Rightarrow8\left(2x+7\right)=88\\ \Rightarrow2x+7=11\Rightarrow x=2\\ k,\Rightarrow5\left(x+4\right)=85\\ \Rightarrow x+4=17\\ \Rightarrow x=13\)
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17x + 3. ( -16x – 37) = 2x + 43 - 4x
<=>17x-48x-111=-2x+43
<=>-29x=154
<=> \(x=-\frac{154}{29}\)
-3. (2x + 5) -16 < -4. (3 – 2x)
\(\Leftrightarrow-6x-31< -12+8x.\)
\(\Leftrightarrow-14x< 19\Rightarrow x< -\frac{19}{14}\)
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x(x+2)=0
suy ra x=0 hoặc x+2=0
5-2x=-7
2x=-7+5
2x=-(7-5)
2x=-2
x=-2:2
x=-1
Vậy x=-1
NHỚ TÍCH MK NHA
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Jim Rohn – Triết lý cuộc đời
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\(1\)) \(5-\left(10-x\right)=7\)
\(10-x=5-7\)
\(10-x=-2\)
\(x=10-\left(-2\right)\)
\(x=12\)
\(2\)) \(-32-\left(x-5\right)=0\)
\(x-5=-32-0\)
\(x-5=-32\)
\(x=-32+5\)
\(x=-27\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1) Ta có: \(\left(-86-x\right)-\left(3+2x\right)=-4-15\)
\(\Leftrightarrow-86-x-3-2x+4+15=0\)
\(\Leftrightarrow-3x-70=0\)
\(\Leftrightarrow-3x=70\)
hay \(x=-\dfrac{70}{3}\)
Vậy: \(x=-\dfrac{70}{3}\)
2) Ta có: \(18+\left(-x\right)-\left(40-28\right)=-32-\left(-18\right)\)
\(\Leftrightarrow18-x-40+28+32-18=0\)
\(\Leftrightarrow-x+20=0\)
\(\Leftrightarrow-x=-20\)
hay x=20
Vậy: x=20
3) Ta có: \(-27-\left(-31+x\right)-25=-5-17\)
\(\Leftrightarrow-27+31-x-25+5+17=0\)
\(\Leftrightarrow-x+1=0\)
\(\Leftrightarrow-x=-1\)
hay x=1
Vậy: x=1
4) Ta có: \(-9-14-x+42-38=-5+13\)
\(\Leftrightarrow-x-19=8\)
\(\Leftrightarrow-x=27\)
hay x=-27
Vậy: x=-27
![](https://rs.olm.vn/images/avt/0.png?1311)
a: =1/8-1/8+2/5=2/5
b: =(-1/15-14/15)+(23/27+31/27)=2-1=1
c: \(=\dfrac{3}{7}\left(\dfrac{22}{21}+\dfrac{5}{21}+\dfrac{15}{21}\right)=\dfrac{3}{7}\cdot2=\dfrac{6}{7}\)
d: \(=\dfrac{-8}{9}\cdot\dfrac{3}{2}+\dfrac{1}{9}\cdot\dfrac{-3}{2}=-\dfrac{3}{2}\)
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a) \(\dfrac{2}{3}x-\dfrac{1}{2}=\dfrac{1}{10}\)
\(\dfrac{2}{3}x=\dfrac{1}{10}+\dfrac{1}{2}=\dfrac{3}{5}\)
\(x=\dfrac{3}{5}:\dfrac{2}{3}=\dfrac{9}{10}\)
b) \(\dfrac{39}{7}:x=13\)
\(x=\dfrac{\dfrac{39}{7}}{13}=\dfrac{3}{7}\)
c) \(\left(\dfrac{14}{5}x-50\right):\dfrac{2}{3}=51\)
\(\dfrac{14}{5}x-50=51\cdot\dfrac{2}{3}=34\)
\(\dfrac{14}{5}x=34+50=84\)
\(x=\dfrac{84}{\dfrac{14}{5}}=30\)
d) \(\left(x+\dfrac{1}{2}\right)\left(\dfrac{2}{3}-2x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=0\\\dfrac{2}{3}-2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=\dfrac{1}{3}\end{matrix}\right.\)
e) \(\dfrac{2}{3}x-\dfrac{1}{2}x=\dfrac{5}{12}\)
\(\dfrac{1}{6}x=\dfrac{5}{12}\)
\(x=\dfrac{5}{12}:\dfrac{1}{6}=\dfrac{5}{2}\)
g) \(\left(x\cdot\dfrac{44}{7}+\dfrac{3}{7}\right)\dfrac{11}{5}-\dfrac{3}{7}=-2\)
\(\left(x\cdot\dfrac{44}{7}+\dfrac{3}{7}\right)\cdot\dfrac{11}{5}=-2+\dfrac{3}{7}=-\dfrac{11}{7}\)
\(x\cdot\dfrac{44}{7}+\dfrac{3}{7}=-\dfrac{11}{7}:\dfrac{11}{5}=-\dfrac{5}{7}\)
\(\dfrac{44}{7}x=-\dfrac{5}{7}-\dfrac{3}{7}=-\dfrac{8}{7}\)
\(x=-\dfrac{8}{7}:\dfrac{44}{7}=-\dfrac{2}{11}\)
h) \(\dfrac{13}{4}x+\left(-\dfrac{7}{6}\right)x-\dfrac{5}{3}=\dfrac{5}{12}\)
\(\dfrac{25}{12}x-\dfrac{5}{3}=\dfrac{5}{12}\)
\(\dfrac{25}{12}x=\dfrac{5}{12}+\dfrac{5}{3}=\dfrac{25}{12}\)
\(x=1\)
Mỏi tay woa bn làm nốt nha!!
Ta có: \(4.\left(5-x\right)-3.\left(1-x\right)=-3^2\)
\(\Leftrightarrow20-4x-3+3x=9\)
\(\Leftrightarrow-4x+3x=9-20+3\)
\(\Leftrightarrow-x=-8\)
\(\Leftrightarrow x=8\)
Vậy: \(x=8\)
Ta có: \(3.\left|2x-3\right|=7-\left(-14\right)\)
\(\Leftrightarrow3.\left|2x-3\right|=7+14=21\)
\(\Leftrightarrow\left|2x-3\right|=\dfrac{21}{3}=7\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-3=7\\2x-3=-7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=10\\2x=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)
Vậy: \(x=-2;5\)
Ta có: \(2^{2x-1}-3^2=23\)
\(\Leftrightarrow2^{2x-1}-9=23\)
\(\Leftrightarrow2^{2x-1}=32\)
\(\Leftrightarrow2^{2x-1}=2^5\)
\(\Leftrightarrow2x-1=5\)
\(\Leftrightarrow2x=6\)
\(\Leftrightarrow x=3\)
Vậy: \(x=3\)
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