a: \(x^2-2xy+y^2+3x-3y-4\)

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

\(=\left(x-y+4\right)\left(x-y-1\right)\)

Đặt: \(x^2-6x+1=a;x^2+1=b\)

Khi đó đa thức này có dạng:

\(2a^2+5ab+2b^2=2a^2+4ab+ab+2b^2\)

\(=2a\left(a+2b\right)+b\left(a+2b\right)=\left(a+2b\right)\left(2a+b\right)\)

Thay lại a và b thì được:

\(\left(a+2b\right)\left(2a+b\right)=\left(x^2-6x+1+2x^2+2\right)\left(2x^2-12x+2+x^2+1\right)\)

\(=\left(3x^2-6x+3\right)\left(3x^2-12x+3\right)\)

\(=9\left(x-1\right)^2\left(x^2-4x+1\right)\)

Vậy ...

Hình như bạn ghi thiếu dấu + đó

Bạn áp dụng hằng đẳng thức \(a^2+2ab+b^2=\left(a+b\right)^2\)

Khi đó\(a=x^2+1\)

         \(b=x^2+6x-1\)

1) Ta có: \(\left(3-x^2\right)+6-2x=0\)

\(\Leftrightarrow3-x^2+6-2x=0\)

\(\Leftrightarrow-x^2-2x+9=0\)

\(\Leftrightarrow x^2+2x-9=0\)

\(\Leftrightarrow x^2+2x+1=10\)

\(\Leftrightarrow\left(x+1\right)^2=10\)

\(\Leftrightarrow\left[{}\begin{matrix}x+1=\sqrt{10}\\x+1=-\sqrt{10}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{10}-1\\x=-\sqrt{10}-1\end{matrix}\right.\)

Vậy: \(S=\left\{\sqrt{10}-1;-\sqrt{10}-1\right\}\)

2) Ta có: \(5\left(2x-1\right)+7=4\left(2-x\right)+2\)

\(\Leftrightarrow10x-5+7=8-4x+2\)

\(\Leftrightarrow10x+4x=8+2+5-7\)

\(\Leftrightarrow14x=8\)

\(\Leftrightarrow x=\dfrac{4}{7}\)

Vậy: \(S=\left\{\dfrac{4}{7}\right\}\)

3) Ta có: \(x^2-6x+4\left(x-6\right)=0\)

\(\Leftrightarrow x\left(x-6\right)+4\left(x-6\right)=0\)

\(\Leftrightarrow\left(x-6\right)\left(x+4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-6=0\\x+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-4\end{matrix}\right.\)

Vậy: S={6;-4}

a: \(=\dfrac{x-z}{2}\)

b: \(=\dfrac{3x}{4y^3}\)

1: Ta có: \(4x^2-36=0\)

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

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

2: Ta có: \(\left(x-1\right)^2+x\left(4-x\right)=11\)

\(\Leftrightarrow x^2-2x+1+4x-x^2=11\)

\(\Leftrightarrow2x=10\)

hay x=5

a) \(\left(x+2\right)\left(x^2-4x+4\right)-\left(x^3+2x^2\right)=5\)

\(\Leftrightarrow\left(x+2\right)\left(x^2-4x+4\right)-x^2\left(x+2\right)=5\)

\(\Leftrightarrow\left(x+2\right)\left(x^2-4x+4-x^2\right)=5\)

\(\Leftrightarrow\left(x+2\right)\left(4-4x\right)=5\)

\(\Leftrightarrow4x-4x^2+8-8x=5\)

\(\Leftrightarrow-4x^2-4x+3=0\)

\(\Leftrightarrow4x^2+4x-3=0\)

\(\Leftrightarrow4x^2-2x+6x-3=0\)

\(\Leftrightarrow2x\left(2x-1\right)+3\left(2x-1\right)=0\)

\(\Leftrightarrow\left(2x-1\right)\left(2x+3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\2x+3=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{2}\\x=-\frac{3}{2}\end{matrix}\right.\)

Vậy \(x=\left\{\frac{1}{2};-\frac{3}{2}\right\}\)

b) \(6x^2-6x\left(-2+x\right)=36\)

\(\Leftrightarrow6x^2+12x-6x^2=36\)

\(\Leftrightarrow12x=36\)

\(\Leftrightarrow x=3\)

Vậy x = 3

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

\(\Leftrightarrow x^2+4x+4+x^2-6x+9-2\left(x^2-1\right)=9\)

\(\Leftrightarrow2x^2-2x+13-2x^2+2=9\)

\(\Leftrightarrow15-2x=9\)

\(\Leftrightarrow2x=6\)

\(\Leftrightarrow x=3\)

Vậy x = 3

d) \(\left(x+5\right)^2-9=0\)

\(\Leftrightarrow\left(x+5\right)^2=9\)

\(\Leftrightarrow\left[{}\begin{matrix}\left(x+5\right)^2=3^2\\\left(x+5\right)^2=\left(-3\right)^2\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x+5=3\\x+5=-3\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=-8\end{matrix}\right.\)

Vậy x ={-2; -8}

e) \(\left(x-2\right)^3=x^3+6x^2=7\) (Câu này sai đề thì phải! Mình sửa lại đề, có gì không giống với đề của bạn thì ib mình sửa nha!)

\(\left(x-2\right)^3-x^3+6x^2=7\)

\(\Leftrightarrow x^3-6x^2+12x-8-x^3+6x^2=7\)

\(\Leftrightarrow12x-8=7\)

\(\Leftrightarrow12x=15\)

\(\Leftrightarrow x=\frac{5}{4}\)

Vậy \(x=\frac{5}{4}\)

#Chúc bạn học tốt!

\(\left(3x-1\right)^2-5\left(2x+1\right)^2+\left(6x-3\right)\left(2x+1\right)=\left(x-1\right)^2\\ \Leftrightarrow\left(9x^2-6x+1\right)-\left(20x^2+20x+5\right)+\left(12x^2-3\right)-\left(x^2-2x+1\right)=0\\ \Leftrightarrow12x-8=0\\ \Leftrightarrow x=\dfrac{2}{3}\)