\(\sqrt{\frac{m}{1-2x+x^2}}.\sqrt{\frac{4m-8mx+4mx^2}{81}}\) voi \(m>0;x\ne1\)
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M
30 tháng 6 2019
\(M=\sqrt{\frac{m}{1-2x+x^2}}\times\sqrt{\frac{4m-8mx+4mx^2}{81}}\)
\(=\frac{\sqrt{m}}{\sqrt{1-2x+x^2}}\times\frac{\sqrt{4m\times\left(1-2x+x^2\right)}}{\sqrt{81}}\)
\(=\frac{\sqrt{m}}{\sqrt{1-2x+x^2}}\times\frac{\sqrt{4m}\times\sqrt{1-2x+x^2}}{9}\)
\(=\frac{\sqrt{m}\times\sqrt{4m}}{9}\)
\(=\frac{2m}{9}\)
vậy . . .
HH
18 tháng 11 2020
\(M=\sqrt{\frac{m}{1-2x+x^2}}.\sqrt{\frac{4m-8mx+4mx^2}{81}}\)
\(=\sqrt{\frac{m}{\left(1-x\right)^2}}.\sqrt{\frac{4m\left(1-2x+x^2\right)}{81}}\)
\(=\sqrt{\frac{m}{\left(1-x\right)^2}.\frac{4m\left(1-x\right)^2}{81}}\)
\(=\frac{\sqrt{4m^2}}{81}\)
\(=\frac{\sqrt{4m^2}}{\sqrt{81}}=\frac{2m}{9}\)
Vậy : \(M=\frac{2m}{9}\)
19 tháng 7 2017
\(\sqrt{\dfrac{m}{1-2x+x^2}}+\sqrt{\dfrac{4m-8mx+4mx^2}{81}}=\sqrt{\dfrac{m}{\left(1-x\right)^2}}+\sqrt{\dfrac{4m\left(1-2x+x^2\right)}{81}}\\ =\dfrac{\sqrt{m}}{\left|1-x\right|}+\dfrac{2\sqrt{m}\left|1-x\right|}{9}=\dfrac{9\sqrt{m}+2\sqrt{m}\left(1-x\right)^2}{\left|1-x\right|.9}\\ =\dfrac{\sqrt{m}\left(9+2-4x+2x^2\right)}{\left(x-1\right).9}\)
tới đây thì hết bt rồi
A
19 tháng 6 2019
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: \(\Rightarrow\frac{a+b}{b^2}\cdot\frac{\left|a\right|b^2}{a+b}=\left|a\right|\)
A
19 tháng 6 2019
2/\(\left(\frac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right)\left(\frac{1-\sqrt{a}}{1-a}\right)^2\)
\(=\left(\frac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right)\cdot\frac{\left(1-\sqrt{a}\right)^2}{\left(1-a\right)^2}\)
\(=\left(\frac{1-a\sqrt{a}}{1-\sqrt{a}}+\frac{\sqrt{a}\left(1-\sqrt{a}\right)}{1-\sqrt{a}}\right)\cdot\frac{\left(1-\sqrt{a}\right)^2}{\left(1-a\right)^2}\)
\(=\left(\frac{1-a\sqrt{a}}{1-\sqrt{a}}+\frac{\sqrt{a}-a}{1-\sqrt{a}}\right)\cdot\frac{\left(1-\sqrt{a}\right)^2}{\left(1-a\right)^2}\)
\(=\frac{1-a\sqrt{a}+\sqrt{a}-a}{1-\sqrt{a}}\cdot\frac{\left(1-\sqrt{a}\right)^2}{\left(1-a\right)^2}\)
\(=\frac{1-a\sqrt{a}+\sqrt{a}-a}{1}\cdot\frac{1-\sqrt{a}}{\left(1-a\right)^2}\)
\(=\frac{\left(1-a\sqrt{a}+\sqrt{a}-a\right)\cdot\left(1-\sqrt{a}\right)}{\left(1-a\right)^2}\)
\(=\frac{1-a\sqrt{a}+\sqrt{a}-a-\sqrt{a}+a^2-a+a\sqrt{a}}{\left(1-a\right)^2}\)
\(=\frac{a^2-2a+1}{\left(1-a\right)^2}\)
\(=\frac{\left(a-1\right)^2}{\left(1-a\right)^2}=\frac{-\left(1-a\right)^2}{\left(1-a\right)^2}=-1\)
QD
22 tháng 4 2017
a) \(\sqrt{\dfrac{a}{b}}+\sqrt{ab}+\dfrac{a}{b}\sqrt{\dfrac{a}{b}}\) với a>0 và b>0
b) \(\sqrt{\dfrac{m}{1-2x+x^2}}.\sqrt{\dfrac{4m-8mx+4mx^2}{81}}=\sqrt{\dfrac{m}{1-2x+x^2}}.\sqrt{\dfrac{4m\left(2-2x+x^2\right)}{81}}\)
\(=\sqrt{\dfrac{4m^2\left(1-2x+x^2\right)}{81\left(1-2x+x^2\right)}}=\sqrt{\dfrac{4m^2}{81}}=\sqrt{\dfrac{2m}{9}}\)
KL
8 tháng 8 2016
\(A=\left(\sqrt{6\left(x^2-2xy^2+y^3\right)}+\sqrt{6.4x^2y}\right).\frac{1}{\sqrt{6y}}\)
\(=\left(\sqrt{6\left(x^2-xy^2+y^3\right)}+2x\sqrt{6y}\right).\frac{1}{\sqrt{6y}}\)
\(=\left[\sqrt{6}\left(\sqrt{x^2-xy^2+y^3}+2x\sqrt{y}\right)\right].\frac{1}{\sqrt{6y}}=\sqrt{6}\left(\sqrt{x^2-xy^2+y^3}-2x\sqrt{y}\right).\frac{1}{\sqrt{6}\sqrt{y}}\)
\(=\frac{x^2-xy^2+y^3}{\sqrt{y}}-\frac{2x\sqrt{y}}{\sqrt{y}}=\frac{x^2-xy^2+y^3}{\sqrt{y}}-2x\)
mik chỉ lm đến đây đc thui
16 tháng 10 2020
Cho biểu thức:
\(P=\frac{2\sqrt{x}}{\sqrt{x}+m}+\frac{\sqrt{x}}{\sqrt{x}-m}-\frac{m^2}{4x-4m^2}\) với \(x>0\)
a) Rút gọn P
\(P=\frac{2\sqrt{x}}{\sqrt{x}+m}+\frac{\sqrt{x}}{\sqrt{x}-m}-\frac{m^2}{4x-4m^2}\)
\(P=\frac{2\sqrt{x}\left(\sqrt{x}-m\right)}{\left(\sqrt{x}\right)^2-m^2}+\frac{\sqrt{x}\left(\sqrt{x}+m\right)}{\left(\sqrt{x}\right)^2-m^2}-\frac{m^2}{4\left(x-m^2\right)}\)
\(P=\frac{8\sqrt{x}\left(\sqrt{x}-m\right)+4\sqrt{x}\left(\sqrt{x}+m\right)-m^2}{4\left(x-m^2\right)}\)
\(P=\frac{12\sqrt{x}\left(\sqrt{x}-m\right)-m^2}{4\left(x-m^2\right)}\)
\(\)
5 tháng 4 2020
\(\left(\frac{\sqrt{x}-2}{x-1}-\frac{\sqrt{x}+2}{x+2\sqrt{x}+1}\right).\frac{x^2-2x+1}{2}\)
a)
Đkxđ:\(\left\{{}\begin{matrix}x-1\ne0\\x\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne1\\x\ge0\end{matrix}\right.\)
\(=\)\(\left(\frac{\sqrt{x}-2}{x-1}-\frac{\sqrt{x}+2}{x+2\sqrt{x}+1}\right).\frac{\left(x-1\right)^2}{2}\)
\(=\frac{x\sqrt{x}+2x+\sqrt{x}-2x-4\sqrt{x}-2-x\sqrt{x}+\sqrt{x}-2x+2}{\left(x-1\right)\left(x+2\sqrt{x}+1\right)}.\frac{\left(x-1\right)^2}{2}\)
\(=\frac{-2\sqrt{x}-2x}{\left(x-1\right)\left(\sqrt{x}+1\right)^2}.\frac{\left(x-1\right)^2}{2}\)
\(=\frac{-2\sqrt{x}\left(1+\sqrt{x}\right)}{\left(x-1\right)\left(\sqrt{x}+1\right)^2}.\frac{\left(x-1\right)^2}{2}\)
\(=\frac{-2\sqrt{x}\left(x-1\right)}{2\left(\sqrt{x}+1\right)}=\frac{-2\sqrt{x}\left(x-1\right)}{2\sqrt{x}+2}\)
RR
13 tháng 5 2018
a) Với x = 25 thì \(N=\frac{\sqrt{25}+1}{\sqrt{25}}=\frac{6}{5}\)
b) Ta có \(M=\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)^2.\left(\sqrt{x}-1\right)}-\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)^2.\left(\sqrt{x}-1\right)}\)
\(M=\frac{2\sqrt{x}}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}\)
Suy ra \(S=M.N=\frac{2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(\sqrt{\frac{m}{1-2x+x^2}}.\sqrt{\frac{4m-8mx+4mx^2}{81}}=\sqrt{\frac{m}{\left(x-1\right)^2}}.\sqrt{\frac{4m\left(1-2x+x^2\right)}{81}}\)
\(=\sqrt{\frac{m}{\left(x-1\right)^2}}.\sqrt{\frac{4m\left(x-1\right)^2}{81}}=\frac{\sqrt{m}}{\left|x-1\right|}.\frac{2\sqrt{m}.\left|x-1\right|}{9}=\frac{2m}{9}\)
\(x\ne1\) chứ không phải x>1 nên không thể ghi |x-1|=x-1 nhé Despacito
A..mk vua nghi ra bai nay
\(\sqrt{\frac{m}{x^2-2x+1}}.\sqrt{\frac{4m-8mx+4mx^2}{81}}\)
\(=\sqrt{\frac{m}{\left(x-1\right)^2}}.\sqrt{\frac{4m\left(1-2x+x^2\right)}{81}}\)
\(=\frac{\sqrt{m}}{\left|x-1\right|}.\frac{\sqrt{4m\left(x-1\right)^2}}{9}\) ( Thoa man DKXD \(m>0;x\ne1\)
\(=\frac{\sqrt{m}}{x-1}.\frac{2\left(x-1\right).\sqrt{m}}{9}\)
\(=\frac{2m}{9}\)
ko biet co dung ko nua