Ta có: 5x=6y

nên \(\dfrac{x}{6}=\dfrac{y}{5}\)

Đặt \(\dfrac{x}{6}=\dfrac{y}{5}=k\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=6k\\y=5k\end{matrix}\right.\)

Ta có: xy=120

\(\Leftrightarrow30k^2=120\)

\(\Leftrightarrow k^2=4\)

Trường hợp 1: k=2

\(\Leftrightarrow\left\{{}\begin{matrix}x=6k=12\\y=5k=10\end{matrix}\right.\)

Trường hợp 2: k=-2

\(\Leftrightarrow\left\{{}\begin{matrix}x=6k=-12\\y=5k=-10\end{matrix}\right.\)

c: Ta có: \(x^3+3x^2+3x-7=0\)

\(\Leftrightarrow x+1=2\)

hay x=1

b: Ta có: \(x\left(x-3\right)-4x+12=0\)

\(\Leftrightarrow\left(x-3\right)\left(x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=4\end{matrix}\right.\)

\(a)2x(x+1)-2x^2=0 <=> 2x^2+2x-2x^2=0 \\<=>2x=0<=>x=0 \\b)3(x-7)+4x(x-7)=0<=>(4x+3)(x-7)=0 \\<=>4x+3=0\ hoặc\ x+7=0 \\<=>x=\dfrac{-3}{4}\ hoặc\ x=-7 \\c)x^2-x=12<=>x^2-x-12=0 \\<=>(x+3)(x-4)=0 \\<=>x+3=0\ hoặc\ x-4=0 \\<=>x=-3\ hoặc\ x=4\)

\(d,=24x^2-38x+3\\ e,=x^2-12x+35\\ f,=\left(x^2-144\right)\left(4x-1\right)=4x^3-x^2-576x+144\)

Tính giá trị biểu thức

ý d, e, f

d: \(\left(12x-1\right)\left(2x-3\right)\)

\(=24x^2-36x-2x+3\)

\(=24x^2-38x+3\)

`a)`

`x-12+4x=25+2x-1`

`<=>5x-12=24+2x`

`<=>5x-2x=12+24`

`<=>3x=36`

`<=>x=12`

`b)`

`7-(2x+4)=-(x+4)`

`<=>7-2x-4=-x-4`

`<=>3-2x=-x-4`

`<=>2x-x=4+3`

`<=>x=7`

a: ĐKXĐ: \(x\in R\)

\(\sqrt{x^2-4x+4}=7\)

=>\(\sqrt{\left(x-2\right)^2}=7\)

=>|x-2|=7

=>\(\left[{}\begin{matrix}x-2=7\\x-2=-7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=9\\x=-5\end{matrix}\right.\)

b: ĐKXĐ: x>=-3

\(\sqrt{4x+12}-3\sqrt{x+3}+\dfrac{4}{3}\cdot\sqrt{9x+27}=6\)

=>\(2\sqrt{x+3}-3\sqrt{x+3}+\dfrac{4}{3}\cdot3\sqrt{x+3}=6\)

=>\(3\sqrt{x+3}=6\)

=>\(\sqrt{x+3}=2\)

=>x+3=4

=>x=1(nhận)

a, 4x(x-5)-x(4x-7)=12

<=>4x²-20x-4x+7x=12

<=>-13x=12

<=>x=-12/13

b,(2-3x)x+6x(x-7)=4

<=>2x-3x²+6x²-42x=4

<=>3x²-40x=4

Đến đây thì mid