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Vì: |2\(x\) - 1| = |1 - 2\(x\)|
Nên: |2\(x\) - 1| + |1 - 2\(x\)| = 8
⇒ |2\(x\) - 1| + |2\(x\) - 1| = 8
2.|2\(x\) - 1| = 8
|2\(x\) - 1| = 8:2
|2\(x\) - 1| = 4
\(\left[{}\begin{matrix}2x-1=-4\\2x-1=4\end{matrix}\right.\)
\(\left[{}\begin{matrix}2x=-4+1\\2x=4+1\end{matrix}\right.\)
\(\left[{}\begin{matrix}2x=-3\\2x=5\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{5}{2}\end{matrix}\right.\)
Vậy \(x\) \(\in\){- \(\dfrac{3}{2}\); \(\dfrac{5}{2}\)}
![](https://rs.olm.vn/images/avt/0.png?1311)
Toán này toán 8 Đức em à :
Ta có A = (2 + 1)(22 + 1)(24 + 1)(28 + 1)
=> A = (2 - 1)(2 + 1)(22 + 1)(24 + 1)(28 + 1)
=> A = (22 - 1)(22 + 1)(24 + 1)(28 + 1)
=> A = (2x4 - 1)(24 + 1)(28 + 1)
=> A = (28 - 1)(28 + 1)
=> A = 216 - 1
B = {[(22)2]2}2 = 22.2.2.2 = 216
Vậy A < B
![](https://rs.olm.vn/images/avt/0.png?1311)
* Trả lời:
\(\left(1\right)\) \(-3\left(1-2x\right)-4\left(1+3x\right)=-5x+5\)
\(\Leftrightarrow-3+6x-4-12x=-5x+5\)
\(\Leftrightarrow6x-12x+5x=3+4+5\)
\(\Leftrightarrow x=12\)
\(\left(2\right)\) \(3\left(2x-5\right)-6\left(1-4x\right)=-3x+7\)
\(\Leftrightarrow6x-15-6+24x=-3x+7\)
\(\Leftrightarrow6x+24x+3x=15+6+7\)
\(\Leftrightarrow33x=28\)
\(\Leftrightarrow x=\dfrac{28}{33}\)
\(\left(3\right)\) \(\left(1-3x\right)-2\left(3x-6\right)=-4x-5\)
\(\Leftrightarrow1-3x-6x+12=-4x-5\)
\(\Leftrightarrow-3x-6x+4x=-1-12-5\)
\(\Leftrightarrow-5x=-18\)
\(\Leftrightarrow x=\dfrac{18}{5}\)
\(\left(4\right)\) \(x\left(4x-3\right)-2x\left(2x-1\right)=5x-7\)
\(\Leftrightarrow4x^2-3x-4x^2+2x=5x-7\)
\(\Leftrightarrow-x-5x=-7\)
\(\Leftrightarrow-6x=-7\)
\(\Leftrightarrow x=\dfrac{7}{6}\)
\(\left(5\right)\) \(3x\left(2x-1\right)-6x\left(x+2\right)=-3x+4\)
\(\Leftrightarrow6x^2-3x-6x^2-12x=-3x+4\)
\(\Leftrightarrow-15x+3x=4\)
\(\Leftrightarrow-12x=4\)
\(\Leftrightarrow x=-\dfrac{1}{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(2x+\frac{3}{15}=\frac{7}{5}\)
=> \(2x=\frac{7}{5}-\frac{3}{15}=\frac{21}{15}-\frac{3}{15}=\frac{18}{15}\)
=> \(x=\frac{18}{15}:2=\frac{18}{15}\cdot\frac{1}{2}=\frac{9}{15}\cdot\frac{1}{1}=\frac{9}{15}\)
b) \(x-\frac{2}{9}=\frac{8}{3}\)
=> \(x=\frac{8}{3}+\frac{2}{9}\)
=> \(x=\frac{24}{9}+\frac{2}{9}=\frac{26}{9}\)
c) \(\frac{-8}{x}=\frac{-x}{18}\)
=> x(-x) = (-8).18
=> -x2 = -144
=> x2 = 144(bỏ dấu âm)
=> x = \(\pm\)12
d) \(\frac{2x+3}{6}=\frac{x-2}{5}\)
=> 5(2x + 3) = 6(x - 2)
=> 10x + 15 = 6x - 12
=> 10x + 15 - 6x + 12 = 0
=> 4x + 27 = 0
=> 4x = -27
=> x = -27/4
e) \(\frac{x+1}{22}=\frac{6}{x}\)
=> x(x + 1) = 132
=> x(x + 1) = 11.12
=> x = 11
f) \(\frac{2x-1}{2}=\frac{5}{x}\)
=> x(2x - 1) = 10
=> 2x2 - x = 10
=> 2x2 - x - 10 = 0
tới đây tự làm đi nhé
g) \(\frac{2x-1}{21}=\frac{3}{2x+1}\)
=> (2x - 1)(2x + 1) = 63
=> 4x2 - 1 = 63
=> 4x2 = 64
=> x2 = 16
=> x = \(\pm\)4
h) Tương tự
a) \(\frac{2x+3}{15}=\frac{7}{5}\Leftrightarrow10x+15=105\Leftrightarrow10x=90\Rightarrow x=9\)
b) \(\frac{x-2}{9}=\frac{8}{3}\Leftrightarrow3x-6=72\Leftrightarrow3x=78\Rightarrow x=26\)
c) \(\frac{-8}{x}=\frac{-x}{18}\Leftrightarrow x^2=144\Leftrightarrow\orbr{\begin{cases}x=12\\x=-12\end{cases}}\)
d) \(\frac{2x+3}{6}=\frac{x-2}{5}\Leftrightarrow10x+15=12x-12\Leftrightarrow2x=27\Rightarrow x=\frac{27}{2}\)
e) \(\frac{x+1}{22}=\frac{6}{x}\Leftrightarrow x^2+x-132=0\Leftrightarrow\left(x-11\right)\left(x+12\right)=0\Leftrightarrow\orbr{\begin{cases}x=11\\x=-12\end{cases}}\)
f) \(\frac{2x-1}{2}=\frac{5}{x}\Leftrightarrow2x^2-x-10=0\Leftrightarrow\left(x-2\right)\left(2x+5\right)=0\Leftrightarrow\orbr{\begin{cases}x=2\\x=-\frac{5}{2}\end{cases}}\)
g) \(\frac{2x-1}{21}=\frac{3}{2x+1}\Leftrightarrow4x^2=64\Leftrightarrow x^2=16\Rightarrow\orbr{\begin{cases}x=4\\x=-4\end{cases}}\)
h) \(\frac{10x+5}{6}=\frac{5}{x+1}\Leftrightarrow10x^2+15x-25=0\Leftrightarrow5\left(x-1\right)\left(2x+5\right)=0\Leftrightarrow\orbr{\begin{cases}x=1\\x=-\frac{5}{2}\end{cases}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{x+1}{16}-\dfrac{2x-1}{8}=\dfrac{-x-8}{2}\)
\(\Rightarrow\dfrac{x+1}{16}-\dfrac{4x-2}{16}+\dfrac{8\left(x+8\right)}{16}=0\)
\(\Rightarrow x+1-4x+2+8x+64=0\)
\(\Rightarrow5x+67=0\)
\(\Rightarrow5x=-67\)
\(\Rightarrow x=-\dfrac{67}{5}\)
`(x+1)/16-(2x-1)/8=(-x-8)/2`
`=>x+1-2(2x-1)=8(-x-8)`
`=>x+1-4x+2=-8x-64`
`=>-3x+3=-8x-64`
`=>5x=67`
`=>x=67/5`
Vậy `x=67/5`
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{2x-1}{2}=\frac{8}{2x-1}\)
<=> \(\left(2x-1\right)\times\left(2x-1\right)=2\times8\)
<=> \(\left(2x-1\right)^2=16\)
<=> \(\left(2x-1\right)^2=4^2\)
<=> \(2x-1=4\)
<=> \(2x=4+1\)
<=> \(2x=5\)
<=> \(x=\frac{5}{2}\)
*Lưu ý: Đối với số nguyên dương, khi nâng lên mũ 2 (bình phương) thì thỏa mãn 2 trường hợp: số dương, âm*
Ta có: \(\frac{2x-1}{2}=\frac{8}{2x-1}\)
\(\Rightarrow\left(2x-1\right)\left(2x-1\right)=2.8\)
\(\Rightarrow\left(2x-1\right)^2=16\)
\(\Rightarrow\left(2x-1\right)^2=\left(\pm4\right)^2\)
\(\Rightarrow\orbr{\begin{cases}2x-1=-4\\2x-1=4\end{cases}\Rightarrow\orbr{\begin{cases}2x=-3\\2x=5\end{cases}}\Rightarrow\orbr{\begin{cases}x=-\frac{3}{2}\\x=\frac{5}{2}\end{cases}}}\)
\(\Rightarrow x\in\left\{-\frac{3}{2};\frac{5}{2}\right\}\)
vi 1/8=1/2^3=>2x-1=3
=>2x=4=>x=2
\(\frac{1}{2}^{2x-1}=\frac{1}{8}\)
\(=>\left(\frac{1}{2}^2\right)^x:\frac{1}{2}=\frac{1}{8}\)
\(=>\frac{1}{4}^x=\frac{1}{8}\cdot\frac{1}{2}=\frac{1}{16}\)
\(=>\frac{1}{4}^x=\left(\frac{1}{4}\right)^2\)
\(=>x=2\)