1)

a) \(3\left(2x-1\right)\left(3x-1\right)-\left(2x-3\right)\left(9x-1\right)=0\)

\(\Leftrightarrow\left(6x-3\right)\left(3x-1\right)-\left(18x^2-2x-27x+3\right)=0\)

\(\Leftrightarrow18x^2-6x-9x+3-18x^2 +29x-3=0\)

\(\Leftrightarrow14x=0\)

\(\Rightarrow x=0\)

b) \(2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+10\)

\(\Leftrightarrow10x-16-12x+15=12x-16+10\)

\(\Leftrightarrow10x-12x+15=12x+10\)

\(\Leftrightarrow-2x+15=12x+10\)

\(\Leftrightarrow-2x-12x=10-15\)

\(\Leftrightarrow-14x=-5\)

\(\Rightarrow x=\dfrac{5}{14}\)

c) \(\left(3x-2\right)\left(2x+9\right)-\left(x+2\right)\left(6x+1\right)=0\)

\(\Leftrightarrow6x^2+27x-4x-18-\left(6x^2+x+12x+2\right)=0\)

\(\Leftrightarrow6x^2+27x-4x-18-6x^2-13x-2=0\)

\(\Leftrightarrow10x-20=0\)

\(\Leftrightarrow10x=20\)

\(\Rightarrow x=2\)

\(\left(x^2-4\right)\times\)\(\left(x^2-10\right)\subset0\)

\(\Rightarrow\)\(\left(x^2-4\right)< 0hoăc\left(x^2-10\right)< 0\)

còn đâu xét hai th trên thôi

\(x\cdot\left(x+2\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x+2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\)

Vậy: \(x\in\left\{0;-2\right\}\)/

x . (x + 2) = 0

x + 2 = 0 : x

x + 2 = 0

x = 0 - 2 

x = (-2)

\(\dfrac{x-1}{x}-\dfrac{1}{x+1}=\dfrac{2x-1}{x^2+x}\)

\(\Leftrightarrow\dfrac{x-1}{x}-\dfrac{1}{x+1}=\dfrac{2x-1}{x\left(x+1\right)}\)

ĐKXĐ : \(\left\{{}\begin{matrix}x\ne0\\x+1\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne0\\x\ne-1\end{matrix}\right.\)

Ta có : `(x-1)/x -1/(x+1) =(2x-1)/(x(x+1))`

\(\Leftrightarrow\dfrac{\left(x-1\right)\left(x+1\right)}{x\left(x+1\right)}-\dfrac{x}{x\left(x+1\right)}=\dfrac{2x-1}{x\left(x+1\right)}\)

`=> x^2 +x -x-1 -x-2x+1=0`

`<=> x^2 -3x =0`

`<=> x(x-3)=0`

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(ktm\right)\\x=3\end{matrix}\right.\)

__

`(x+2)(5-3x)=0`

\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\5-3x=0\end{matrix}\right.\)

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

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

__

\(\dfrac{5\left(1-2x\right)}{3}+\dfrac{x}{2}=\dfrac{3\left(x-5\right)}{4}-2\)

\(\Leftrightarrow\dfrac{20\left(1-2x\right)}{12}+\dfrac{6x}{12}=\dfrac{9\left(x-5\right)}{12}-\dfrac{24}{12}\)

`<=> 2x- 40x + 6x = 9x - 45 -24`

`<=>  2x- 40x + 6x-9x + 45 +24=0`

`<=>-41x+69=0`

`<=>-41x=-69`

`<=> x=69/41`

Cậu tách 2 câu 1 lượt mn trl nhanh hơn đó ạ

Bài 1 

ta có a+3+b-3 =a +b chia hết cho 4

nên (b-a )(a+b) cũng chia hết cho 4

bài 2.

ta có: \(M=6x^2-5x-6-12xy+6y^2+6y-3x+2y+2027\)

\(=6\left(x-y\right)^2-8\left(x-y\right)+2021=24-16+2021=2029\)

a: \(x\in\left\{0;25\right\}\)

c: \(x\in\left\{0;5\right\}\)

1 nghịch biến(a<0) 

2 đồng biến

3,4 thay các g trị tm đk vào

hojk tốt