Cho a,b thỏa ab=1; a+b\(\ne\) 0 Tính
\(P=\frac{1}{\left(a+b\right)^3}\left(\frac{1}{a^3}+\frac{1}{b^3}\right)+\frac{1}{\left(a+b\right)^4}\left(\frac{1}{a^2}+\frac{1}{b^2}\right)+\frac{6}{\left(a+b\right)^5}\left(\frac{1}{a}+\frac{1}{b}\right)\)
1) 0<a,b,c<1 và ab+bc+ca=1.find Min of:
\(M=\frac{a^2\left(1-2b\right)}{b}+\frac{b^2\left(1-2c\right)}{c}+\frac{c^2\left(1-2a\right)}{a}\)
2) a,b,c>0.CMR:
\(\frac{1}{\left(2a+b\right)^2}+\frac{1}{\left(2b+c\right)^2}+\frac{1}{\left(2c+a\right)^2}\ge\frac{1}{ab+bc+ca}\)
3)a,b,c>0 CMR:
\(\left(\frac{a}{a+b}\right)^2+\left(\frac{b}{b+c}\right)^2+\left(\frac{c}{c+a}\right)^2\ge\frac{1}{2}\left(\frac{b}{a+b}+\frac{c}{b+c}+\frac{a}{c+a}\right)\)
a) gpt \(\left(x-1\right)\left(x+2\right)+4\left(x-1\right)\sqrt{\frac{x+2}{x-1}}=12\)
b) ghpt \(\left\{\begin{matrix}2\sqrt{x}\left(1+\frac{1}{x+y}\right)=3\\2\sqrt{y}\left(1-\frac{1}{x+y}\right)=1\end{matrix}\right.\)
Giải hệ:
\(\left\{\begin{matrix}\frac{\left(a+5\right)\left(b+2\right)}{2}=\frac{ab}{2}+45\\\frac{\left(a-1\right)\left(b-1\right)}{2}=\frac{ab}{2}-15\end{matrix}\right.\)
\(\frac{\left[\left(e-m\right)^2-\left(e+m\right)^2\right]\left[\left(y-1\right)^2-\left(y+1\right)^2\right]}{a.16.n.h}\) . \(\frac{e}{u^{-1}}\)
cho bt p=\(\left(\frac{x+3\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}-\frac{x+\sqrt{x}}{x-1}\right):\left(\frac{1}{\sqrt{x}+1}+\frac{1}{\sqrt{x}-1}\right)\)
a) rg p
b) tìm x để \(\frac{1}{p}-\frac{\sqrt{x}+1}{8}>\)hoặc bằng 1
Tìm ĐKXĐ và rút gọn biểu thức
\(A=\frac{\left(\sqrt{a}+\sqrt{b}\right)^2-4\sqrt{ab}}{\sqrt{a}-\sqrt{b}}-\frac{a\sqrt{b}+b\sqrt{a}}{\sqrt{ab}}\)
\(B=\left(\frac{2\sqrt{x}-x}{x\sqrt{x}-1}-\frac{1}{\sqrt{x}-1}\right):\frac{x-1}{x+\sqrt{x}+1}\)
\(C=\left(1-\frac{x-3\sqrt{x}}{x-9}\right):\left(\frac{\sqrt{x}-3}{2-\sqrt{x}}+\frac{\sqrt{x}-2}{3+\sqrt{x}}-\frac{9-x}{x+\sqrt{x}-6}\right)\)
\(D=\left(\frac{\sqrt{x}}{3+\sqrt{x}}+\frac{x+9}{9-x}\right):\left(\frac{3\sqrt{x}+1}{x-3\sqrt{x}}-\frac{1}{\sqrt{x}}\right)\)
CM rằng GT của bthức A ko phụ thuộc vào a
Tìm x để C = 4
Tìm x sao cho D < -1
Chứng minh với mọi a , b , c > 0 ta luôn có :
\(\frac{1}{a\left(a+b\right)}+\frac{1}{b\left(b+c\right)}+\frac{1}{c\left(c+a\right)}\ge\frac{27}{2\left(a+b+c\right)^2}\)
rút gọn biểu thức:
A=\(\left(\frac{\sqrt{x}-2}{x-1}-\frac{\sqrt{x}+2}{x+2\sqrt{2}+1}\right).\frac{\left(1-x\right)^2}{2}\)
ai trả lời đúng yêu lun ý!!!
