bạn ơi cái chỗ 21x^2 ? y^2 là dấu cộng hay trừ vậy?

Nó là 1 á

\(=4x^4+21x^2y^2+y^4-25x^2y^2\)

\(\left(2x^2+y^2\right)-\left(5xy\right)^2\)

\(\left(2x^2+y^2-5xy\right)\left(2x^2+y^2+5xy\right)\)

$=4{}^{}+21{}^{}{}^{}+{}^{}-25{}^{}{}^{}$

$\left(2{}^{}+{}^{}\right)-{}^{}$

$\left(2{}^{}+{}^{}-5xy\right)\left(2{}^{}+{}^{}+5xy\right)$

4x⁴ - 21 x²y² + y⁴ 

= (4x⁴ + 4x²y² + y⁴) - 25x²y² 

=  [(2x²)² + 2x² . 2 . y² + (y²)²] - 25x²y²

= (2x² + y²) - 25x²y² 

= (2x² + y² - 5xy) (2x² + y² + 5xy)

c: =5a(a-b)+7(a-b)

=(a-b)(5a+7)

a) \(4x^4-21x^2y^2+y^4\)

Ấn nhầm :v

a) \(4x^4-21x^2y^2+y^4\)

\(=\left(2x^2\right)^2-2\cdot2x^2\cdot y^2+y^2-25x^2y^2\)

\(=\left(2x^2-y^2\right)^2-\left(5xy\right)^2\)

\(=\left(2x^2-5xy-y^2\right)\left(2x^2+5xy-y^2\right)\)

b) \(x^5-5x^3+4x\)

\(=x^5-4x^3-x^3+4x\)

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

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

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

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

        \(4x^4-21x^2y^2+y^4\)

\(=\left(2x^2\right)^2+2.2x^2.y^2+\left(y^2\right)^2-25x^2y^2\)

\(=\left(2x^2+y^2\right)^2-\left(5xy\right)^2\)

\(=\left(2x^2-5xy+y^2\right)\left(2x^2+5xy+y^2\right)\)

       \(x^5-5x^3+4x\)

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

\(=x\left[x^4-x^2-4x^2+4\right]\)

\(=x\left[x^2\left(x-1\right)-4\left(x^2-1\right)\right]\)

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

\(1,=x\left(x^2-2x+1-y^2\right)=x\left[\left(x-1\right)^2-y^2\right]=x\left(x-y-1\right)\left(x+y-1\right)\\ 2,=\left(x+y\right)^3\\ 3,=\left(2y-z\right)\left(4x+7y\right)\\ 4,=\left(x+2\right)^2\\ 5,Sửa:x\left(x-2\right)-x+2=0\\ \Leftrightarrow\left(x-2\right)\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)

4x^4+4x^2y^2+y^4-25x^2y^2

=(2x^2+y^2)^2-(5xy)^2

=(2x^2+y^2-5xy)(2x^2+y^2+5xy)

1D  2C

Câu 1: D

Câu 2: C