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a: \(\dfrac{x^2-3x+2}{x^2-1}=\dfrac{\left(x-2\right)\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}=\dfrac{x-2}{x+1}\)
\(a,VP=\dfrac{\left(x-1\right)\left(x-2\right)}{\left(x-1\right)\left(x+1\right)}=\dfrac{x-2}{x+1}=VP\\ b,VT=\dfrac{u\left(4u^2-1\right)}{5\left(1-2u\right)}=\dfrac{-u\left(1-2u\right)\left(1+2u\right)}{5\left(1-2u\right)}=\dfrac{-u\left(1+2u\right)}{5}=-\dfrac{2u^2+u}{5}=VP\)
\(\dfrac{x^2+3x-4}{x-1}=\dfrac{x^2+4x-x-4}{\left(x-1\right)}=\dfrac{\left(x+4\right)\left(x-1\right)}{x-1}=x+4\)
\(B=\left(x+\frac{1}{x}\right)^2+\left(y+\frac{1}{y}\right)^2+\left(xy+\frac{1}{xy}\right)^2\)
\(-\left(x+\frac{1}{x}\right)\left(y+\frac{1}{y}\right)\left(xy+\frac{1}{xy}\right)\)
\(\Rightarrow B=x^2+2+\frac{1}{x^2}+y^2+2+\frac{1}{y^2}+x^2y^2+2+\frac{1}{x^2y^2}-x^2y^2\)
\(-2-x^2-y^2-\frac{1}{y^2}-\frac{1}{x^2}-\frac{1}{x^2y^2}\)
\(\Rightarrow B=x^2y^2-x^2y^2+x^2-x^2+1.\frac{1}{x^2}+1.\frac{1}{x^2y^2}-1.\frac{1}{x^2}-1\)
\(.\frac{1}{x^2y^2}+1.\frac{1}{y^2}-1.\frac{1}{y^2}+y^2-y^2+2+2+2-2\)
\(\Rightarrow B=4\)
1: \(B=\dfrac{2x+1-x^2+2x^2-3x-1}{x\left(2x+1\right)}=\dfrac{x^2-x}{x\left(2x+1\right)}=\dfrac{x-1}{2x+1}\)
2: \(C=A:B\)
\(=\dfrac{x-1}{x^2}:\dfrac{x-1}{2x+1}=\dfrac{2x+1}{x^2}\)
\(C+1=\dfrac{2x+1+x^2}{x^2}=\dfrac{\left(x+1\right)^2}{x^2}>=0\)
=>C>=-1
Câu 1:
b: ĐKXĐ: \(x\notin\left\{3;-3\right\}\)
\(\dfrac{1}{x-3}-\dfrac{1}{x+3}+\dfrac{2x}{9-x^2}\)
\(=\dfrac{1}{x-3}-\dfrac{1}{x+3}-\dfrac{2x}{\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{x+3-x+3-2x}{\left(x-3\right)\left(x+3\right)}=\dfrac{-2x+6}{\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{-2\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}=-\dfrac{2}{x+3}\)
c: ĐKXĐ: \(x\notin\left\{2;0\right\}\)
Sửa đề: \(\dfrac{x+1}{x-2}+\dfrac{4-5x}{x^3+4x}:\dfrac{x-2}{x^2+4}\)
\(=\dfrac{x+1}{x-2}+\dfrac{4-5x}{x\left(x^2+4\right)}\cdot\dfrac{x^2+4}{x-2}\)
\(=\dfrac{x+1}{x-2}+\dfrac{4-5x}{x\left(x-2\right)}\)
\(=\dfrac{x\left(x+1\right)+4-5x}{x\left(x-2\right)}=\dfrac{x^2+x-5x+4}{x\left(x-2\right)}\)
\(=\dfrac{x^2-4x+4}{x\left(x-2\right)}=\dfrac{\left(x-2\right)^2}{x\left(x-2\right)}=\dfrac{x-2}{x}\)
\(\left(x+y\right)\left(\frac{1}{x}+\frac{1}{y}\right)=1+\frac{x}{y}+1+\frac{y}{x}=2+\frac{x}{y}+\frac{y}{x}\)
Áp dụng BĐT cô si ,ta có:
\(\frac{x}{y}+\frac{y}{x}\ge2\sqrt{\frac{x\cdot y}{y\cdot x}}=2\)
Vậy ta được đpcm
ta có:
\(a+\frac{1}{a}-2=\left(\sqrt{a}\right)^2+\left(\frac{1}{\sqrt{a}}\right)^2-2\sqrt{a\cdot\frac{1}{a}}=\left(\sqrt{a}+\frac{1}{\sqrt{a}}\right)^2\ge0\Rightarrow a+\frac{1}{a}\ge2\)
Vì a và 1/a cùng dấu nên 2 căn (a*1/a) lớn hơn 0 nha