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Ta có: \(x+y=7\Rightarrow\left(x+y\right)^2=49\Rightarrow x^2+y^2+2xy=49\)
Mà: \(x^2+y^2=25\Rightarrow2xy=24\Rightarrow xy=12\)
\(x^3+y^3=\left(x+y\right)\left(x^2-xy+y^2\right)=7\left(25-12\right)=91\)
(Vì\(x+y=7;x^2+y^2=25;xy=12\))
`B = x^2- 2xy + y^2 + 2x - 10y + 17
`2B = 2x^2 - 4xy + 2y^2 + 4x - 20y + 34`
`= (x-y)^2 + (x+2)^2 + (y-5)^2 + 5 >= 5`.
\(=\dfrac{2x^2-x-x-1+2-x^2}{x-1}=\dfrac{x^2-2x+1}{x-1}=\dfrac{\left(x-1\right)^2}{x-1}=x-1\)
\(a,x^2-x+1\)
\(x^2-x+\left(\frac{1}{2}\right)^2+\frac{3}{4}\)
\(\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\)
\(< =>MIN=\frac{3}{4}\)dấu"=" xảy ra khi \(x=\frac{1}{2}\)
\(b,x^2+y^2-4\left(x+y\right)+16\)
\(x^2+y^2-4x-4y+16\)
\(\left(x^2-4x+4\right)+\left(y^2-4y+4\right)+8\)
\(\left(x-2\right)^2+\left(y-2\right)^2+8\ge8\)
\(MIN=8\)dấu "=" xảy ra khi \(x=y=2\)
\(2x^2+8x+9\)
\(\left(x^2+8x+16\right)+x^2-7\)
\(\left(x+4\right)^2+x^2-7\ge-7\)
\(< =>MIN=-7\)dấu "=" xảy ra khi \(x=-4\)
m) \(\dfrac{1}{4}x^2-4x^2=\left(\dfrac{1}{2}x-2x\right)\left(\dfrac{1}{2}x+2x\right)\)
n) \(\dfrac{4}{49}-4x^2=\left(\dfrac{2}{7}-2x\right)\left(\dfrac{2}{7}+2x\right)\)
o) \(\left(x-3\right)\left(x+3\right)=x^2-9\)
\(2,=\left(x-y\right)^2-2\left(x-y\right)=\left(x-y\right)\left(x-y-2\right)\\ 3,=\left(3x-5\right)\left(x+1\right)\\ 4,sai.đề\\ 5,=\left(x-1\right)^2-y^2=\left(x-y-1\right)\left(x+y-1\right)\\ 6,=\left(x+3\right)\left(x+5\right)\)
\(25\left(x+y\right)^2-16\left(x-y\right)^2\)
\(=\left(5x+5y\right)^2-\left(4x-4y\right)^2\)
\(=\left(5x+5y+4x-4y\right)\left(5x+5y-4x+4y\right)\)
\(=\left(9x+y\right)\left(x+9y\right)\)
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