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a,
\(\left(x+y\right)^2+\left(x-y\right)^2\\ =x^2+2xy+y^2+x^2-2xy+y^2\\ =2\left(x^2+y^2\right)\)
b,
\(2\left(x-y\right)\left(x+y\right)+\left(x+y\right)^2+\left(x-y\right)^2\\ =2\left(x^2-y^2\right)+x^2+2xy+y^2+x^2-2xy+y^2\\ =2x^2-2y^2+2x^2+2y^2\\ =4x^2\)
c,
\(\left(x-y+z\right)^2+\left(z-y\right)^2+2\left(x-y+z\right)\left(y-z\right)\\ =\left(x-y+z\right)^2+2\left(x-y+z\right)\left(y-z\right)+\left(y-z\right)^2\\ =\left(x-y+z+y-z\right)^2=x^2\)
\(\left(x+y\right)^2+\left(x-y\right)^2=x^2+2xy+y^2+x^2-2xy+y^2=2x^2+2y^2=2\left(x^2+y^2\right)\)\(b,2\left(x-y\right)\left(x+y\right)+\left(x+y\right)^2+\left(x-y\right)^2=2x^2-2y^2+x^2+2xy+y^2+x^2-2xy+y^2=3\left(x^2-y^2\right)\)\(c,\left(x-y+z\right)^2+\left(z-y\right)^2+2\left(x-y+z\right)\left(y-z\right)=\left(x-y+z\right)^2-2\left(x-y+z\right)\left(z-y\right)+\left(z-y\right)^2=\left[\left(x-y+z\right)-\left(z-y\right)\right]^2=x^2\)
Từ \(x\in\left[-1,2\right]\Rightarrow\left(x+1\right)\left(x-2\right)\le0\)
\(\Rightarrow x^2\le x+6\)
Tương tự \(y^2\le y+6\);\(z^2\le z+6\)
Suy ra \(x^2+y^2+z^2\le x+y+z+6=6\)
Dấu "=" xảy ra <=> \(\left(x,y,z\right)=\left(-1,-1,2\right)\) và các hoán vị của nó
(x⁴ - x²y + x²y² - xy²)(y² + x²)
=[x²(x² + y²) - xy(x + y)](x² + y²)
=(x² + y²)(x + y)(x² - xy)(x² + y²)
=x(x² + y²)(x + y)(x + y)
=x(x² + y²)(x + y)²
Chắc bây giờ bạn cũng chả cần nữa :)
a, \(\left(x+2\right)\left(2-x\right)-\left(2x-1\right)\left(x+3\right)=4-x^2-\left(2x^2+5x-3\right)\)
\(=4-x^2-2x^2-5x+3=-3x^2-5x+7\)
b, \(\left(x+1\right)^2-2\left(x^2-1\right)+\left(x-1\right)^2=\left(x+1\right)^2-2\left(x+1\right)\left(x-1\right)+\left(x-1\right)^2\)
\(=\left(x+1-x+1\right)^2=2^2=4\)
c, \(\left(x+y\right)^3-\left(x-y\right)^3-6x^2y\)
\(=\left(x+y-x+y\right)\left[\left(x+y\right)^2+\left(x+y\right)\left(x-y\right)+\left(x-y\right)^2\right]-6x^2y\)
\(=2y\left(x^2+2xy+y^2+x^2-y^2+x^2-2xy+y^2\right)-6x^2y\)
\(=2y\left(3x^2+y^2\right)-6x^2y=6x^2y+2y^3-6x^2y=2y^3\)
a)(x+2y-3z-t)(x+2y+3z+t)
=\(\left(x+2y\right)^2-\left(3z+t\right)^2\)
Ta có:
\(x+y+z=\frac{a-b}{a+b}+\frac{b-c}{b+c}+\frac{c-a}{c+a}\).
\(x+y+z=\frac{\left(a-b\right)\left(b+c\right)\left(c+a\right)+\left(b-c\right)\left(a+b\right)\left(c+a\right)+\left(c-a\right)\left(a+b\right)\left(b+c\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)
Ta có:
\(\left(a-b\right)\left(b+c\right)\left(c+a\right)+\left(b-c\right)\left(c+a\right)\left(a+b\right)+\left(c-a\right)\left(a+b\right)\left(b+c\right)\).
\(=\left(c+a\right)\left[\left(a-b\right)\left(b+c\right)+\left(b-c\right)\left(a+b\right)\right]+\left(c-a\right)\left(a+b\right)\left(b+c\right)\).
\(=\left(c+a\right)\left(ab+ac-b^2-bc+ab+b^2-ac-bc\right)\)\(+\left(c-a\right)\left(ab+ac+b^2+bc\right)\).
\(=\left(c+a\right)\left(2ab-2bc\right)-\left(a-c\right)\left(ab+ac+b^2+bc\right)\).
\(=2b\left(c+a\right)\left(a-c\right)-\left(a-c\right)\left(ab+ac+b^2+bc\right)\).
\(=\left(2bc+2ab\right)\left(a-c\right)-\left(a-c\right)\left(ab+ac+b^2+bc\right)\).
\(=\left(a-c\right)\left(2ab+2bc-ab-ac-b^2-bc\right)\).
\(=\left(a-c\right)\left(ab+bc-b^2-ac\right)=\left(a-c\right)\left[\left(ab-b^2\right)-\left(ac-bc\right)\right]\).
\(=\left(a-c\right)\left[b\left(a-b\right)-c\left(a-b\right)\right]=\left(a-c\right)\left(a-b\right)\left(b-c\right)\).
Do đó\(x+y+z=\frac{\left(a-c\right)\left(a-b\right)\left(b-c\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}=\frac{-\left(a-b\right)\left(b-c\right)\left(c-a\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\).
Mà \(xyz=\frac{\left(a-b\right)\left(b-c\right)\left(c-a\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)nên:
\(x+y+z=-xyz\).
\(\Rightarrow x+y+z+xyz=0\)(điều phải chứng minh).
Ta có: \(x^2+y^2-z^2\)
\(=\left(x+y\right)^2-z^2-2xy\)
\(=\left(x+y+z\right)\left(x+y-z\right)-2xy\)
\(=-2xy\)
Ta có: \(x^2+z^2-y^2\)
\(=\left(x+z\right)^2-y^2-2xz\)
\(=\left(x+y+z\right)\left(x+z-y\right)-2xz\)
\(=-2xz\)
Ta có: \(y^2+z^2-x^2\)
\(=\left(y+z\right)^2-x^2-2yz\)
\(=\left(x+y+z\right)\left(y+z-x\right)-2yz\)
\(=-2yz\)
Ta có: \(\dfrac{xy}{x^2+y^2-z^2}+\dfrac{xz}{x^2+z^2-y^2}+\dfrac{yz}{y^2+z^2-x^2}\)
\(=\dfrac{xy}{-2xy}+\dfrac{xz}{-2xz}+\dfrac{yz}{-2yz}\)
\(=\dfrac{1}{-2}+\dfrac{1}{-2}+\dfrac{1}{-2}\)
\(=\dfrac{-3}{2}\)