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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: \(B=1-2.\left(-7\right)=15\)
Lời giải :
a) \(\sqrt{x^2\left(x-1\right)^2}=\left|x\right|\cdot\left|x-1\right|=-x\left(1-x\right)=x^2-x\)
b) \(\sqrt{13x}\cdot\sqrt{\frac{52}{x}}=\sqrt{\frac{13x\cdot52}{x}}=\sqrt{676}=26\)
c) \(5xy\cdot\sqrt{\frac{25x^2}{y^6}}=5xy\cdot\sqrt{\left(\frac{5x}{y^3}\right)^2}=5xy\cdot\frac{-5x}{y^3}=\frac{-25x^2}{y^2}\)
d) \(\sqrt{\frac{9+12x+4x^2}{y^2}}=\sqrt{\frac{\left(2x+3\right)^2}{y^2}}=\frac{2x+3}{-y}=\frac{-2x-3}{y}\)
a) \(ĐKXĐ:\hept{\begin{cases}x\ge0\\x\ne25\end{cases}}\)
\(A=\frac{x+3\sqrt{x}}{x-25}+\frac{1}{\sqrt{x}+5}\)
\(=\frac{x+3\sqrt{x}+\sqrt{x}-5}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}\)
\(=\frac{x+4\sqrt{x}-5}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}\)
\(=\frac{\sqrt{x}-1}{\sqrt{x}-5}\)
\(\Rightarrow P=\frac{\sqrt{x}-1}{\sqrt{x}-5}:\frac{\sqrt{x}+2}{\sqrt{x}-5}=\frac{\sqrt{x}-1}{\sqrt{x}+2}\)
b) Để P nguyên
\(\Leftrightarrow\sqrt{x}-1⋮\sqrt{x}+2\)
\(\Leftrightarrow3⋮\sqrt{x}+2\)
\(\Leftrightarrow\sqrt{x}+2\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)
\(\Leftrightarrow\sqrt{x}\in\left\{-3;-1;-5;1\right\}\)
Mà \(\sqrt{x}\ge0,\forall x\)
\(\Leftrightarrow\sqrt{x}=1\)
\(\Leftrightarrow x=1\)
Vậy để P nguyên \(\Leftrightarrow x=1\)
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