a) rút gọn P
b)tính gt của p với x = 7 - 4 căn 3
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KT
12 tháng 7 2018
Bài 1:
a) \(B=\sqrt{1-4x+4x^2}\)
\(=\sqrt{\left(1-2x\right)^2}\)
\(=\left|1-2x\right|\)
Nếu \(x\le\frac{1}{2}\)thì: \(B=1-2x\)
Nếu \(x>\frac{1}{2}\)thì: \(B=2x-1\)
b) Tại \(x=-7\)thì: \(B=1-2.\left(-7\right)=15\)
15 tháng 6 2021
a, \(A=\left(\frac{1}{1-\sqrt{x}}+\frac{1}{1+\sqrt{x}}\right):\left(\frac{1}{1-\sqrt{x}}-\frac{1}{1+\sqrt{x}}\right)+\frac{1}{1-\sqrt{x}}\)ĐK : \(x>0;x\ne1\)
\(=\left(\frac{1+\sqrt{x}+1-\sqrt{x}}{1-x}\right):\left(\frac{1+\sqrt{x}-1+\sqrt{x}}{1-x}\right)+\frac{1}{1-\sqrt{x}}\)
\(=\frac{2}{1-x}.\frac{1-x}{2\sqrt{x}}+\frac{1}{1-\sqrt{x}}=\frac{1}{\sqrt{x}}+\frac{1}{1-\sqrt{x}}=\frac{1-\sqrt{x}+\sqrt{x}}{-x+\sqrt{x}}=\frac{1}{\sqrt{x}-x}\)
b, Ta có : \(x=7+4\sqrt{3}=7+2.2\sqrt{3}=\left(\sqrt{4}+\sqrt{3}\right)^2\)
\(A=\frac{1}{\sqrt{4}+\sqrt{3}-7+4\sqrt{3}}\)
13 tháng 6 2021
a, \(B=\frac{\sqrt{a}+3}{2\sqrt{a}-6}-\frac{3-\sqrt{a}}{2\sqrt{a}+6}=\frac{\left(2\sqrt{a}+6\right)\left(\sqrt{a}+3\right)+\left(2\sqrt{a}-6\right)\left(\sqrt{a}-3\right)}{4a-36}\)
\(=\frac{2a+12\sqrt{a}+18+2a-12\sqrt{a}+18}{4a-36}=\frac{4a+36}{4a-36}=\frac{a+9}{a-9}\)
b, Ta có : \(B>1\Rightarrow\frac{a+9}{a-9}>1\Leftrightarrow\frac{a+9}{a-9}-1>0\)
\(\Leftrightarrow\frac{a+9-a+9}{a-9}>0\Leftrightarrow\frac{18}{a-9}>0\Rightarrow a-9>0\Leftrightarrow a>9\)vì 18 > 0
\(B< 1\Rightarrow\frac{a+9}{a-9}< 1\Leftrightarrow\frac{a+9}{a-9}-1< 0\)
\(\Leftrightarrow\frac{a+9-a+9}{a-9}< 0\Leftrightarrow\frac{18}{a-9}< 0\Rightarrow a-9< 0\Leftrightarrow a< 9\)vì 18 > 0
c, Ta có : \(B=4\Rightarrow\frac{a+9}{a-9}=4\Rightarrow a+9=4a-36\Leftrightarrow3a=45\Leftrightarrow a=15\)
Vậy a = 15 thì B = 4
22 tháng 7 2020
1) Ta có: \(\frac{x+6\sqrt{x}+9}{x-9}=\frac{\left(\sqrt{x}+3\right)^2}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\frac{\sqrt{x}+3}{\sqrt{x}-3}\)
a: \(\frac{2x+\sqrt{x}-1}{1-x}+\frac{2x\sqrt{x}+x-\sqrt{x}}{1+x\sqrt{x}}\)
\(=\frac{2x+2\sqrt{x}-\sqrt{x}-1}{\left(1-\sqrt{x}\right)\left(1+\sqrt{x}\right)}+\frac{\sqrt{x}\left(2x+\sqrt{x}-1\right)}{\left(1+\sqrt{x}\right)\left(1-\sqrt{x}+x\right)}\)
\(=\left(\sqrt{x}+1\right)\left(2\sqrt{x}-1\right)\cdot\left(\frac{-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\frac{\sqrt{x}}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\right)\)
\(=\left(\sqrt{x}+1\right)\left(2\sqrt{x}-1\right)\cdot\frac{-x+\sqrt{x}-1+\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\)
\(=\left(2\sqrt{x}-1\right)\cdot\frac{-x+\sqrt{x}-1+x-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x-\sqrt{x}+1\right)}=\left(2\sqrt{x}-1\right)\cdot\frac{-1}{\left(\sqrt{x}-1\right)\left(x-\sqrt{x}+1\right)}=\frac{-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x-\sqrt{x}+1\right)}\)
Ta có: \(\frac{1}{1-\sqrt{x}}-\frac{1}{\sqrt{x}}\)
\(=\frac{-1}{\sqrt{x}-1}-\frac{1}{\sqrt{x}}\)
\(=\frac{-\sqrt{x}-\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}=\frac{-2\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}\)
Ta có: \(P=\left(\frac{1}{1-\sqrt{x}}-\frac{1}{\sqrt{x}}\right):\left(\frac{2x+\sqrt{x}-1}{1-x}+\frac{2x\sqrt{x}+x-\sqrt{x}}{1+x\sqrt{x}}\right)\)
\(=\frac{-2\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}:\frac{-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x-\sqrt{x}+1\right)}\)
\(=\frac{-2\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\frac{\left(\sqrt{x}-1\right)\left(x-\sqrt{x}+1\right)}{-2\sqrt{x}+1}=\frac{x-\sqrt{x}+1}{\sqrt{x}}\)
b: \(x=7-4\sqrt3\)
=>\(x=4-2\cdot2\cdot\sqrt3+3=\left(2-\sqrt3\right)^2\)
\(P=\frac{x-\sqrt{x}+1}{\sqrt{x}}\)
\(=\frac{7-4\sqrt3-\sqrt{\left(2-\sqrt3\right)^2}+1}{\sqrt{\left(2-\sqrt3\right)^2}}\)
\(=\frac{8-4\sqrt3-2+\sqrt3}{2-\sqrt3}=\frac{6-3\sqrt3}{2-\sqrt3}=\frac{3\left(2-\sqrt3\right)}{2-\sqrt3}\)
=3