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1/ \(\sqrt{\frac{m}{1-2x+x^2}}\cdot\sqrt{\frac{4m-8mx+4mx^2}{81}}\)
\(=\sqrt{\frac{m}{\left(1-x\right)^2}}\cdot\sqrt{\frac{4m\left(1-2x+x^2\right)}{81}}\)
\(=\sqrt{\frac{m}{\left(1-x\right)^2}}\cdot\sqrt{\frac{4m\left(1-x\right)^2}{81}}\)
\(=\sqrt{\frac{m}{\left(1-x\right)^2}\cdot\frac{4m\left(1-x\right)^2}{81}}\)
\(=\sqrt{\frac{4m^2}{81}}=\sqrt{\frac{\left(2m\right)^2}{9^2}}=\frac{2\left|m\right|}{9}\)
3/\(\frac{a+b}{b^2}\sqrt{\frac{a^2b^4}{a^2+2ab+b^2}}\)
\(=\frac{a+b}{b^2}\sqrt{\frac{\left(ab^2\right)^2}{\left(a+b\right)^2}}\)
\(=\frac{a+b}{b^2}\cdot\frac{\left|a\right|b^2}{\left|a+b\right|}\)
TH1: \(\Rightarrow\frac{a+b}{b^2}\cdot\frac{\left|a\right|b^2}{-\left(a+b\right)}=-\left|a\right|\)
TH2: \(\Rightarrow\frac{a+b}{b^2}\cdot\frac{\left|a\right|b^2}{a+b}=\left|a\right|\)
2/\(\left(\frac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right)\left(\frac{1-\sqrt{a}}{1-a}\right)^2\)
\(=\left(\frac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right)\cdot\frac{\left(1-\sqrt{a}\right)^2}{\left(1-a\right)^2}\)
\(=\left(\frac{1-a\sqrt{a}}{1-\sqrt{a}}+\frac{\sqrt{a}\left(1-\sqrt{a}\right)}{1-\sqrt{a}}\right)\cdot\frac{\left(1-\sqrt{a}\right)^2}{\left(1-a\right)^2}\)
\(=\left(\frac{1-a\sqrt{a}}{1-\sqrt{a}}+\frac{\sqrt{a}-a}{1-\sqrt{a}}\right)\cdot\frac{\left(1-\sqrt{a}\right)^2}{\left(1-a\right)^2}\)
\(=\frac{1-a\sqrt{a}+\sqrt{a}-a}{1-\sqrt{a}}\cdot\frac{\left(1-\sqrt{a}\right)^2}{\left(1-a\right)^2}\)
\(=\frac{1-a\sqrt{a}+\sqrt{a}-a}{1}\cdot\frac{1-\sqrt{a}}{\left(1-a\right)^2}\)
\(=\frac{\left(1-a\sqrt{a}+\sqrt{a}-a\right)\cdot\left(1-\sqrt{a}\right)}{\left(1-a\right)^2}\)
\(=\frac{1-a\sqrt{a}+\sqrt{a}-a-\sqrt{a}+a^2-a+a\sqrt{a}}{\left(1-a\right)^2}\)
\(=\frac{a^2-2a+1}{\left(1-a\right)^2}\)
\(=\frac{\left(a-1\right)^2}{\left(1-a\right)^2}=\frac{-\left(1-a\right)^2}{\left(1-a\right)^2}=-1\)
a) \(\sqrt{7-4\sqrt{3}}-\sqrt{7+4\sqrt{3}}\)
\(=\sqrt{4-4\sqrt{3}+3}-\sqrt{4+4\sqrt{3}+3}\)
\(=\sqrt{\left(2-\sqrt{3}\right)^2}-\sqrt{\left(2+\sqrt{3}\right)^2}\)
\(=\left|2-\sqrt{3}\right|-\left|2+\sqrt{3}\right|\)
\(=2-\sqrt{3}-2-\sqrt{3}\)
\(=-2\sqrt{3}\)
ĐKXĐ : \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)
Ta có :
\(A=\frac{\sqrt{x}+4}{\sqrt{x}+1}-\frac{3}{x-1}:\frac{1}{\sqrt{x}-1}\)
\(=\frac{\sqrt{x}+4}{\sqrt{x}+1}-\frac{3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}.\left(\sqrt{x}-1\right)\)
\(=\frac{\sqrt{x}+4}{\sqrt{x}+1}-\frac{3}{\sqrt{x}+1}\)
\(=\frac{\sqrt{x}+1}{\sqrt{x}+1}\)
\(=1\)
Vậy...
b/ ĐKXĐ : \(\left\{{}\begin{matrix}x\ge0\\x\ne4\end{matrix}\right.\)
Ta có :
\(B=\left(\frac{x-4\sqrt{x}+4}{\sqrt{x}-2}+6\right)\left(\frac{x\sqrt{x}-1}{x+\sqrt{x}+1}-3\right)\)
\(=\left(\frac{\left(\sqrt{x}-2\right)^2}{\sqrt{x}-2}+6\right)\left(\frac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{x+\sqrt{x}+1}-3\right)\)
\(=\left(\sqrt{x}-2+6\right)\left(\sqrt{x}-1-3\right)\)
\(=\left(\sqrt{x}+4\right)\left(\sqrt{x}-4\right)\)
\(=x-16\)
Vậy..
c/ ĐKXĐ : \(\left\{{}\begin{matrix}x>0\\x\ne1\end{matrix}\right.\)
Ta có :
\(C=\frac{2\sqrt{x}}{x-1}+\frac{1}{x+\sqrt{x}}+\frac{1}{\sqrt{x}-x}\)
\(=\frac{2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\frac{1}{\sqrt{x}\left(\sqrt{x}+1\right)}-\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}\)
\(=\frac{2x}{\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\frac{\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{2x+\sqrt{x}-1-\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{2x-2}{\sqrt{x}\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(=\frac{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(=\frac{2}{\sqrt{x}}\)
Vậy..
Bài 1: diendantoanhoc.net
Đặt \(a=\frac{1}{x};b=\frac{1}{y};c=\frac{1}{z}\) BĐT cần chứng minh trở thành
\(\frac{x}{\sqrt{3zx+2yz}}+\frac{x}{\sqrt{3xy+2xz}}+\frac{x}{\sqrt{3yz+2xy}}\ge\frac{3}{\sqrt{5}}\)
\(\Leftrightarrow\frac{x}{\sqrt{5z}\cdot\sqrt{3x+2y}}+\frac{y}{\sqrt{5x}\cdot\sqrt{3y+2z}}+\frac{z}{\sqrt{5y}\cdot\sqrt{3z+2x}}\ge\frac{3}{5}\)
Theo BĐT AM-GM và Cauchy-Schwarz ta có:
\( {\displaystyle \displaystyle \sum }\)\(_{cyc}\frac{x}{\sqrt{5z}\cdot\sqrt{3x+2y}}\ge2\)\( {\displaystyle \displaystyle \sum }\)\(\frac{x}{3x+2y+5z}\ge\frac{2\left(x+y+z\right)^2}{x\left(3x+2y+5z\right)+y\left(5x+3y+2z\right)+z\left(2x+5y+3z\right)}\)
\(=\frac{2\left(x+y+z\right)^2}{3\left(x^2+y^2+z^2\right)+7\left(xy+yz+zx\right)}\)
\(=\frac{2\left(x+y+z\right)^2}{3\left(x^2+y^2+z^2\right)+\frac{1}{3}\left(xy+yz+zx\right)+\frac{20}{3}\left(xy+yz+zx\right)}\)
\(\ge\frac{2\left(x+y+z\right)^2}{3\left(x^2+y^2+z^2\right)+\frac{1}{3}\left(x^2+y^2+z^2\right)+\frac{20}{3}\left(xy+yz+zx\right)}\)
\(=\frac{2\left(x^2+y^2+z^2\right)}{5\left[x^2+y^2+z^2+2\left(xy+yz+zx\right)\right]}=\frac{3}{5}\)
Bổ sung bài 1:
BĐT được chứng minh
Đẳng thức xảy ra <=> a=b=c
\(\sqrt{6+2\sqrt{2}.\sqrt{3-\sqrt{\sqrt{2}+2\sqrt{3}+\sqrt{18-8\sqrt{2}}}}}-\sqrt{3}\)\(=\sqrt{6+2.1,4.\sqrt{3-\sqrt{1,4+2.1,7+\sqrt{18-8.1,4\text{}}}}}-1,7\)
\(=\sqrt{6+2,8\sqrt{3-\sqrt{1,4+3,4+\sqrt{18-11,2}}}}-1,7\)
\(=\sqrt{8,8\sqrt{3-\sqrt{4,8+\sqrt{6,8}}}}-1,7\)
\(=\sqrt{8,8\sqrt{3-\sqrt{4,8+2,6}}}-1,7\)
\(=\sqrt{8,8\sqrt{3-\sqrt{7,4}}}-1,7\)
\(=\sqrt{8,8\sqrt{3-2,7}}-1,7\)
\(=\sqrt{88\sqrt{0,3}}-1,7\)
\(=\sqrt{88.0,54}-1,7\)
\(=\sqrt{47,52}-1,7\)
\(=6,9-1,7\)
\(=5,2\)
2,Mệt với câu 1 rồi nên câu 2 và câu 3 chịu
hình như sai rồi bạn ơi, lúc học thì thầy mình giải ra kết quả =1 và ko tính căn ra như thế