Cho biểu thức: Q = x x 2 - y 2 - 1 + x x 2 - y 2 : y x - x 2 - y 2 với x > y > 0
A. x x + y
B. x + y x - y
C. x - y x + y
D. y x + y
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Ta có \(P=\frac{x^2+y\left(x+y\right)}{x^2-y^2}:\frac{\left(x-y\right)\left(x^2+xy+y^2\right)}{x^4\left(x-y\right)-y^4\left(x-y\right)}\)
\(=\frac{x^2+xy+y^2}{x^2-y^2}:\frac{\left(x-y\right)\left(x^2+xy+y^2\right)}{\left(x-y\right)\left(x^4-y^4\right)}\)\(=\frac{x^2+xy+y^2}{x^2-y^2}:\frac{\left(x-y\right)\left(x^2+xy+y^2\right)}{\left(x-y\right)\left(x^2-y^2\right)\left(x^2+y^2\right)}\)
\(=\frac{x^2+xy+y^2}{x^2-y^2}.\frac{\left(x-y\right)\left(x^2-y^2\right)\left(x^2+y^2\right)}{\left(x-y\right)\left(x^2+xy+y^2\right)}\)\(=x^2+y^2=\left(x+y\right)^2-2xy\)
Thay \(x+y=5;xy=-\frac{1}{2}\Rightarrow P=5^2-2.\left(-\frac{1}{2}\right)=26\)
Vậy P=26
\(A=\left(\dfrac{x+y}{y}+\dfrac{2y}{x-y}\right)\cdot\dfrac{-\left(x-y\right)}{x^2+y^2}+\left(\dfrac{2x^2+2-2x^2+x}{2\left(2x-1\right)}\right)\cdot\dfrac{1-2x}{x+2}\)
\(=\dfrac{x^2-y^2+2y^2}{y\left(x-y\right)}\cdot\dfrac{-\left(x-y\right)}{x^2+y^2}+\dfrac{x+2}{2\left(2x-1\right)}\cdot\dfrac{-\left(2x-1\right)}{x+2}\)
\(=\dfrac{-1}{y}+\dfrac{-1}{2}=\dfrac{-2-y}{2y}\)
P=(\(\dfrac{x^2}{x^2-y^2}+\dfrac{y\left(x+y\right)}{x^2-y^2}\)):\(\dfrac{\left(x-y\right)\left(x^2+xy+y^2\right)}{\left(x-y\right)\left(x^4-y^4\right)}\)
P=\(\dfrac{X^2+xy+y^2}{x^2-y^2}\).\(\dfrac{\left(x^2-y^2\right)\left(x^2+y^2\right)}{x^2+xy+y^2}\)
P=x^2+y^2=(x+y)^2-2xy=5^2-(-1)=26
Bài 1:
\(Q=\left(x+1\right)\left(x^2-x+1\right)-\left(x-1\right)\left(x^2+x+1\right)\)
\(Q=x^3+1-\left(x^3-1\right)=x^3+1-x^3+1=2\)
Vậy......................
Bài 2:
\(5x-\left(4-2x+x^2\right)\left(x+2\right)+x\left(x-1\right)\left(x+1\right)=0\)
\(\Rightarrow5x-\left(x^3+8\right)+x\left(x^2-1\right)=0\)
\(\Rightarrow5x-x^3-8+x^3-x=0\)
\(\Rightarrow4x=8\Rightarrow x=2\)
Chúc bạn học tốt!!!