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$ a/ 12x(x – 5) – 3x(4x - 10) = 120$

`<=>12x^2-60x-12x^2+30x=120`

`<=>-30x=120`

`<=>x=-4`

Vậy `x=-4`

$b/ 9x(x + 4) – 5x(3x + 2) = 112 - 2x(3x + 1)$

`<=>9x^2+36x-15x^2-10x=112-6x^2-2x`

`<=>-6x^2+26x=112-6x^2-2x`

`<=>28x=112`

`<=>x=4`

Vậy `x=4`

$c/ 3x(1 – x) - 5x(3x + 7) = 154 + 9x(5 – 2x)$

`<=>3x-3x^2-15x^2-35x=154+45x-18x^2`

`<=>-32x-18x^2=154+45x-18x^2`

`<=>77x=-154`

`<=>x=-2`

Vậy `x=-2`

\(\Leftrightarrow\left(x+1\right)\left(3x-5\right)=0\)

hay \(x\in\left\{-1;\dfrac{5}{3}\right\}\)

Thu gọn biểu thức:

\(9x\left(x+5\right)-\left(3x+2\right)\left(3x-2\right)\)

\(=9x^2+45x-\left(9x^2-4\right)\)

\(=45x+4\)

Tìm x. biết:

\(\left(3x-2\right)^2=49x^2\)

\(\Leftrightarrow\left[{}\begin{matrix}3x-2=7x\\3x-2=-7x\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=\dfrac{1}{5}\end{matrix}\right.\)

Vậy \(x\in\left\{-\dfrac{1}{2};\dfrac{1}{5}\right\}\)

a: Ta có: \(9x\cdot\left(x+5\right)-\left(3x+2\right)\left(3x-2\right)\)

\(=9x^2+45x-9x^2+4\)

=45x+4

b: Ta có: \(\left(3x-2\right)^2=49x^2\)

\(\Leftrightarrow\left(3x-2-7x\right)\left(3x-2+7x\right)=0\)

\(\Leftrightarrow\left(4x+2\right)\left(10x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=\dfrac{1}{5}\end{matrix}\right.\)

sai đề rồi bạn

đề đúng mà bạn

a) \(\left|2x-5\right|=x+1\)

<=> \(\orbr{\begin{cases}2x-5=x+1\left(x\ge\frac{5}{2}\right)\\5-2x=x+1\left(x< \frac{5}{2}\right)\end{cases}}\)

<=> \(\orbr{\begin{cases}x=-6\left(ktm\right)\\3x=4\end{cases}}\)

<=> \(x=\frac{4}{3}\left(tm\right)\)

b) \(\left|3x-2\right|-1=2x\) <=> \(\left|3x-2\right|=2x+1\)

<=> \(\orbr{\begin{cases}3x-2=2x+1\left(x\ge\frac{2}{3}\right)\\2-3x=2x+1\left(x< \frac{2}{3}\right)\end{cases}}\)

<=>  \(\orbr{\begin{cases}x=-3\left(ktm\right)\\5x=1\end{cases}}\) <=> \(x=\frac{1}{5}\left(tm\right)\)

c) \(\left|x-5\right|+5=x\) <=> \(\left|x-5\right|=x-5\)

<=> \(\orbr{\begin{cases}x-5=x-5\left(x\ge5\right)\\5-x=x-5\left(x< 5\right)\end{cases}}\)

<=> \(\orbr{\begin{cases}0x=0\\2x=10\end{cases}}\) <=> 0x = 0 (luôn đúng) hoặc x = 5 (ktm)

Vậy x \(\ge\)5

d) \(\left|3x-5\right|=3x-5\) <=> \(\orbr{\begin{cases}3x-5=3x-5\left(x\ge\frac{5}{3}\right)\\5-3x=3x-5\left(x< \frac{5}{3}\right)\end{cases}}\)

<=> \(\orbr{\begin{cases}0x=0\left(luônđúng\right)\\6x=10\end{cases}}\)

<=> \(\orbr{\begin{cases}x\ge\frac{5}{3}\\x=\frac{5}{3}\left(ktm\right)\end{cases}}\)Vậy x \(\ge\)5/3

x=0 nha bạn

|3x - 4| + |3x + 5| = 9

=> 3x. (-4 +5) = 9

=> 3x. 1 = 9

=> x = 3

Đoàn Thanh Bảo An làm kết quả đúng nhưng bn trình bày rõ ra mk tk cho