Đề bạn có mấy chỗ thiếu mk bổ sung nha

\(a,2^3+4^2+6x=8+16+6x=6x+24=x\left(x+4\right)\\ b,x^2-4=\left(x-2\right)\left(x+2\right)\\ c,x^2-10x+25=\left(x-5\right)^2\\ d,x^3-4x=x\left(x^2-4\right)=x\left(x-2\right)\left(x+2\right)\\ e,x^2+xy-3x-3y=x\left(x+y\right)-3\left(x+y\right)=\left(x-3\right)\left(x+y\right)\\ g,x^2-y^2-4x+4=\left(x-2\right)^2-y^2=\left(x-y-2\right)\left(x+y-2\right)\)

Tick plzz

a: Ta có: \(2x^3+4x^2+6x\)

\(=2x\left(x^2+2x+3\right)\)

b: \(x^2-4=\left(x-2\right)\left(x+2\right)\)

c: \(x^2-10x+25=\left(x-5\right)^2\)

d: \(x^3-4x=x\left(x-2\right)\left(x+2\right)\)

e: \(x^2+xy-3x-3y\)

\(=x\left(x+y\right)-3\left(x+y\right)\)

\(=\left(x+y\right)\left(x-3\right)\)

g: \(x^2-4x+4-y^2\)

\(=\left(x-2\right)^2-y^2\)

\(=\left(x-y-2\right)\left(x+y-2\right)\)

`@` `\text {Ans}`

`\downarrow`

`(8x-3)(3x+2)-(4x+7)(x+4)=(2x+1)(5x-1)-33`

`\Leftrightarrow 8x(3x+2) -3(3x+2) - 4x(x+4) + 7(x+4) = 2x(5x-1) + 5x-1 - 33`

`\Leftrightarrow 24x^2 + 16x - 9x - 6 - 4x^2 - 16x - 7x - 28 = 10x^2 - 2x + 5x - 1 - 33`

`\Leftrightarrow 20x^2 -16x - 34 = 10x^2 + 3x - 34`

`\Leftrightarrow 20x^2 - 16x - 34 - 10x^2 - 3x + 34 = 0`

`\Leftrightarrow 10x^2 - 19x = 0`

`\Leftrightarrow x(10x - 19)=0`

`\Leftrightarrow `\(\left[{}\begin{matrix}x=0\\10x-19=0\end{matrix}\right.\)

`\Leftrightarrow `\(\left[{}\begin{matrix}x=0\\10x=19\end{matrix}\right.\)

`\Leftrightarrow `\(\left[{}\begin{matrix}x=0\\x=\dfrac{19}{10}\end{matrix}\right.\)

Vậy, `x={0; 19/10}.`

Với bài này bn áp dụng bài phần tử của tập hợp nhé! 

(8-4):6=129

Gọi 129 là x

X-7=59

Gọi  59 làc

Vậy phần bài này là phần tử

Đs 78/9

a: \(x-2-\frac{x^2-10}{x+2}\)

\(=\frac{\left(x-2\right)\left(x+2\right)-x^2+10}{x+2}\)

\(=\frac{x^2-4-x^2+10}{x+2}=\frac{6}{x+2}\)

b: \(\frac{x}{y^2-xy}-\frac{y}{xy-x^2}\)

\(=\frac{-x}{y\left(x-y\right)}+\frac{y}{x\left(x-y\right)}=\frac{-x^2+y^2}{xy\left(x-y\right)}=\frac{-\left(x-y\right)\left(x+y\right)}{xy\left(x-y\right)}\)

\(=\frac{-x-y}{xy}\)

c: \(\frac{1}{x\left(x-1\right)}+\frac{1}{\left(x-1\right)\left(x-2\right)}+\frac{1}{\left(x-2\right)\left(x-3\right)}+\frac{1}{\left(x-3\right)\left(x-4\right)}+\frac{1}{\left(x-4\right)\left(x-5\right)}\)

\(=-\frac{1}{x}+\frac{1}{x-1}-\frac{1}{x-1}+\frac{1}{x-2}-\frac{1}{x-2}+\frac{1}{x-3}-\frac{1}{x-3}+\frac{1}{x-4}-\frac{1}{x-4}+\frac{1}{x-5}\)

\(=\frac{1}{x-5}-\frac{1}{x}=\frac{x-\left(x-5\right)}{x\left(x-5\right)}=\frac{5}{x\left(x-5\right)}\)

\(\Leftrightarrow x^2+6x+8-x^2=7\\ \Leftrightarrow6x=-1\Leftrightarrow x=-\dfrac{1}{6}\)

(x + 4)(x+2) - x² =7

x²+ 2x + 4x + 8 - x² = 7

6x + 8 = 7

6x = 7 - 8 = -1

=> x = \(\dfrac{-1}{6}\)

Đề bài sai nhé, từ giả thiết chỉ xác định được \(x+y=0\Rightarrow y=-x\)

\(\Rightarrow A=4x^2-x^2+x^2+15=4x^2+15\) ko rút gọn được