\(\left(2a+3\right)^2=\left(2a\right)^2+2.2a.3+3^2=4a^2+12a+9\)

\(A=\left(x+2\right)^2-\left(x+2\right)\left(x-2\right)+\left(x-4\right)^2\\ =x^2+4x+4-\left(x^2-4\right)+x^2-8x+16=x^2-4x+24\\ \cdot x=-2=>A=\left(-2\right)^2-4.\left(-2\right)+24=36\\ \cdot x=0=>A=0^2-4.0+24=24\\ \cdot x=2=>A=2^2-4.2+24=20\\ A=\left(x-2\right)^2+20>0\left(DPCM\right)\)

\(\left[{}\begin{matrix}x+3=0\\2x+4=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-2\\x=5\end{matrix}\right.\)

Ta có: \(\left(x+3\right)\left(2x+4\right)\left(x-5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\2x+4=0\\x-5=0\end{matrix}\right.\)

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

Vậy....

pp ẩn dụ thì mk ko bt chỉ bt đặt ẩn phụ thôi mà chỗ kia là đấu + ms đúng

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

đặt \(\left(x^2+x+2\right)^2=a\left(a>0\right)\)và \(x^2=b\)

\(=a^2-3ab+2b^2\)

\(=\left(a-b\right).\left(a-2b\right)\)

\(=\left[\left(x^2+x+2\right)^2-x^2\right].\left[\left(x^2+x+2\right)^2-2x^2\right]\)

bạn tự pt nốt nhé

Sửa đề: \(A=\left(x^2+x+2\right)^4-3x^2\left(x^2+x+2\right)^2+2x^4\)

- Đặt \(\left\{{}\begin{matrix}\left(x^2+x+2\right)^2=a\\x^2=b\end{matrix}\right.\)

\(\Rightarrow A=a^2-3ab+2b^2=a^2-ab-2ab+2b^2=a\left(a-b\right)-2b\left(a-b\right)=\left(a-b\right)\left(a-2b\right)\)

\(\Rightarrow A=\left[\left(x^2+x+2\right)^2-x^2\right]\left[\left(x^2+x+2\right)^2-2x^2\right]\)

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

a) \(x^2-6x+8=x^2-4x-2x+8=x\left(x-4\right)-2\left(x-4\right)=\left(x-2\right)\left(x-4\right)\)

b) \(x^2-8xy+12y^2=x^2-6xy-2xy+12y^2=x\left(x-6y\right)-2y\left(x-6y\right)=\left(x-6y\right)\left(x-2y\right)\)c) \(x^4+4x^2-5=x^4-x^3+x^3-x^2+5x^2-5\)

\(=x^3\left(x-1\right)+x^2\left(x-1\right)+5\left(x-1\right)\left(x+1\right)=\left(x-1\right)\left(x^3+x^2+5x+5\right)\)

\(=\left(x-1\right)\left[x^2\left(x+1\right)+5\left(x+1\right)\right]=\left(x-1\right)\left(x+1\right)\left(x^2+5\right)\)

d) \(x^4+x^2+1=x^4+x^3+x^2-x^3-x^2-x+x^2+x+1\)

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

a, \(x^2-2x-4x+8=x\left(x-2\right)-4\left(x-2\right)=\left(x-4\right)\left(x-2\right)\)

b, \(x^2-2.4xy+16y^2-4y^2=\left(x-4y\right)^2-4y^2=\left(x-6y\right)\left(x-2y\right)\)

c, \(x^4+4x^2+4-9=\left(x^2+2\right)^2-9=\left(x^2-1\right)\left(x^2+5\right)=\left(x-1\right)\left(x+1\right)\left(x^2+5\right)\)

d, \(x^4-2x^22y^2+4y^4+4x^2y^2=\left(x^2-2y^2\right)^2+4x^2y^2\)bạn xem lại đề nhé 

e, \(x^4+x^2+\dfrac{1}{4}-\dfrac{1}{4}+1=\left(x^2+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\)bạn xem lại đề nhé 

(a+b)²  = 100, 2ab = 8 ⇔ a² + b² = 100 - 8 = 92

(a² + b²)² =  92² = 8464 , ab = 4 ⇒ a²b² = 16 ⇒ 2a²b² = 32

a⁴ + b^4 ₌ 8464 - 32 = 8432

\(=\left(a^2+b^2\right)^2-2a^2b^2=\left[\left(a+b\right)^2-2ab\right]-2a^2b^2\)

-> \(\left(100-8\right)-2.16=92-32=60\)

`a)(3x+2)^2-2(x-1)^2-7x^2`

`=9x^2+12x+4-2(x^2-2x+1)-7x^2`

`=9x^2+12x+4-2x^2+4x-2-7x^2`

`=16x+2`

______________________________________________

`b)(x-2)(x+2)+(x-3)(x+3)-x(2x+1)-4`

`=x^2-4+x^2-9-2x^2-x-4`

`=-x-17`