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<=> 11 - 15 - x = x - 25 + 9
<=> -x - x = - 25 + 9 - 11 + 15
<=> - 2x = -12
<=> x = -12 : (- 2)
<=> x = 6
Tìm số nguyên x biết:
\(11-\left(15+x\right)=x-\left(25-9\right)\)
\(11-\left(15+x\right)=x-16\)
\(11-15-x=x-16\)
\(-x-x=-16-11+15\)
\(-2x=-16-11+15\)
\(-2x=-27+15\)
\(-2x=-12\)
\(-x=-12:2\)
\(-x=-6\)
\(=>x=6\)
Vậy \(x=6\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Lâu rồi mình ko giải, sai thì thôi nhé!
a) \(\left(10-2x\right)^2=25-\left(-11\right)\)'=
\(\Leftrightarrow\left(10-2x\right)^2=36\)
\(\Leftrightarrow\left(10-2x\right)^2=6^2\)
\(\Leftrightarrow\orbr{\begin{cases}10-2x=6\\10-2x=-6\end{cases}\Leftrightarrow\orbr{\begin{cases}2x=10-6\\2x=10-\left(-6\right)\end{cases}\Leftrightarrow}\orbr{\begin{cases}2x=4\\2x=16\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=2\\x=8\end{cases}}}\)
Vậy \(x\in\left\{2;8\right\}\)
b) \(-2\left(-x+5\right)-3\left(5-x\right)=4\left(2+x\right)\)
\(\Leftrightarrow-2\left(5-x\right)-3\left(5-x\right)=4\left(2+x\right)\)
\(\Leftrightarrow-5\left(5-x\right)=4\left(2+x\right)\)
\(\Leftrightarrow-25+5x=8+4x\)
\(\Leftrightarrow5x-25=8+4x\)
\(\Leftrightarrow5x=8+4x+25\)
\(\Leftrightarrow5x=4x+33\)
\(\Leftrightarrow5x-4x=33\)
\(\Leftrightarrow1x=33\)
\(\Leftrightarrow x=33\)
Vậy \(x=33\)
a) (10 - 2x)2 = 25 - (-11)
(10 - 2x)2 = 36
(10 - 2x)2 = 62
=> 10 - 2x = 6
2x = 10 - 6
2x = 4
x =4:2
x=2
Vậy x = 2
b)-2(-x+5) - 3(5 - x) = 4(2+x)
2x - 10 - 15 +3x = 8 + 4x
2x - 25 + 3x = 8 +4x
2x + 3x - 4x = 8 + 25
5x - 4x = 33
x= 33
Vậy x = 33
![](https://rs.olm.vn/images/avt/0.png?1311)
ta có
\(\frac{x+1\text{0}}{8}+\frac{x+11}{7}=-2\)
\(\Leftrightarrow\frac{7\left(x+1\text{0}\right)}{7\cdot8}+\frac{8\left(x+11\right)}{8\cdot7}=-\frac{112}{56}\)
\(\Leftrightarrow\frac{7x+7\text{0}+8x+88}{56}=-\frac{112}{56}\)
\(\Leftrightarrow15x+158=-112\)
\(\Leftrightarrow15x=-27\text{0}\)
\(\Leftrightarrow x=-18\)
Vậy ...........
chúc bn học tốt
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{1}{2}\) \(\times\) ( \(x\) - \(\dfrac{2}{3}\)) - \(\dfrac{1}{3}\) \(\times\) ( 2\(x\) - 3) = \(x\)
\(\dfrac{1}{2}\) \(\times\) \(\dfrac{3x-2}{3}\) - \(\dfrac{2x-3}{3}\) = \(x\)
\(\dfrac{3x-2}{6}\) - \(\dfrac{4x-6}{6}\) = \(\dfrac{6x}{6}\)
3\(x-2-4x\) + 6 = 6\(x\)
-\(x\) + 4 - 6\(x\) = 0
7\(x\) = 4
\(x\) = \(\dfrac{4}{7}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
X - { [ -x + (x+3) ] } - [ (x+3) - (x-2)] = 0
X - { -x + x + 3 } - [ x +3 - x +2] = 0
X - 3 - 5 = 0
x - 8 = 0
x = 8
Vì x^2-3 > x^2-11
Mà (x^2-3).(x^2-11) < 0
=> x^2-3 >0 ; x^2-11 < 0
=> x^2 > 3 ; x^2 < 11
=> 3 < x^2 < 11
=> \(-\sqrt{11}< x< -\sqrt{3}\) hoặc \(\sqrt{3}< x< \sqrt{11}\)
Vậy ............
Tk mk nha