Question

Rút Gọn Biểu Thức Có sử dụng hàng đẳng thức

(x-y+z)²+(z-y)²+2(x-y+z)(y-z)

Answer 1

,[x-y+z]^2+[z-y]^2+2.[x-y+z][y-z] (x - y + z)² + (z - y)² + 2(x - y + z)(y - z)
= (x - y + z)² + 2(x - y + z)(y - z) + (y - z)²
= (x - y + z + y - z)²
= x²

Answer 2

Ta có:

\(\left(x-y+z\right)^2+\left(z-y\right)^2+2.\left(x-y+z\right).\left(y-z\right)\)

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

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

\(=x^2\)

Học tốt nhé 

Answer 3

\(\left(x-y+z\right)^2+2\left(x-y+z\right)\left(y-z\right)+\left(z-y\right)^2\)

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

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

Answer 4

Ta có : \(z-y=-\left(y-z\right)\)

\(\Rightarrow\left(z-y\right)^2=\left[-\left(y-z\right)\right]^2=\left(y-z\right)^2\)

\(\Rightarrow\left(x-y+z\right)^2+\left(z-y\right)^2+2\left(x-y+z\right)\left(y-z\right)\)

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

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

\(=x^2\)

Answer 5

Ta có:\(\left(z-y\right)=-\left(y-z\right)\)

\(\left(x-y+z\right)^2+\left(z-y\right)^2+2\left(x-y+z\right)\left(y-z\right)\)

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

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

\(=x^2\)

Answer 6

\(\left(x-y+z\right)^2+\left(z-y\right)^2+2\left(x-y+z\right)\left(y-z\right)\)

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

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

\(=x^2\)