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5 . ( x + 2 ) . ( x - 2 ) - ( 3 . 4x )2 .
= 5( x\(^2\) - 4) - 12x\(^2\) = 5x\(^2\) - 20 - 12x\(^2\) = -7x\(^2\) - 20
2 . ( x - y ) . ( x + y ) + ( x + y )2 + ( x - y )2
= 2( x\(^2\) - y\(^2\)) + ( x\(^2\) + 2xy + y\(^2\)) + ( x\(^2\) - 2xy + y\(^2\))
= 2x\(^2\) - 2y\(^2\) + x\(^2\) + 2xy + y\(^2\) + x\(^2\) - 2xy + y\(^2\)
= 4x\(^2\)
Ta có: x+y+z=0
\(\Leftrightarrow\left(x+y+z\right)^2=0\)
\(\Leftrightarrow x^2+y^2+z^2+2xy+2yz+2xz=0\)(1)
Ta có: \(K=\dfrac{x^2+y^2+z^2}{\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2}\)
\(=\dfrac{x^2+y^2+z^2}{x^2-2xy+y^2+y^2-2yz+z^2+z^2-2xz+x^2}\)
\(=\dfrac{x^2+y^2+z^2}{3x^2+3y^2+3z^2-x^2-y^2-z^2-2xy-2yz-2xz}\)
\(=\dfrac{x^2+y^2+z^2}{3\left(x^2+y^2+z^2\right)-\left(x^2+y^2+z^2+2xy+2yz-2xz\right)}\)
\(=\dfrac{x^2+y^2+z^2}{3\left(x^2+y^2+z^2\right)}=\dfrac{1}{3}\)
Vậy: \(K=\dfrac{1}{3}\)
\(K=\dfrac{x^2+y^2+z^2}{2\left(x^2+y^2+z^2\right)-2\left(xy+yz+zx\right)}\)
\(K=\dfrac{x^2+y^2+z^2}{3\left(x^2+y^2+z^2\right)-\left(x+y+z\right)^2}=\dfrac{1}{3}\)
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 = x 2
(x + y + z)2 – 2.(x + y + z).(x + y) + (x + y)2
= [(x + y + z) – (x + y)]2 (Áp dụng HĐT (2) với A = x + y + z ; B = x + y)
= z2.
\(\left(x+y-z\right)^2+2.\left(x+y-z\right).\left(z-y\right)+\left(y-z\right)^2=\left[\left(x+y-z\right)+\left(z-y\right)\right]^2=x^2\)
Sai đề.
(x-y+z)2 + (z-y)2 + 2.(x-y+z).(y-z)
= (x-y+z)2 + (y-z)2 + 2.(x-y+z).(y-z)
=[(x-y+z)+(y-z)]2
=(x-y+z+y-z)2
=x2
hằng đẳng thức nha đổi vị trí tth]s 2 xuoong3 và 3 lên 2 ra rồi tự làm nha
(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
,[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
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