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Bài 1:
\(A=\sqrt{8}-2\sqrt{2}+\sqrt{20}-2\sqrt{5}-2=2\sqrt{2}-2\sqrt{2}+2\sqrt{5}-2\sqrt{5}-2=-2\)\(B=\frac{\sqrt{x}}{\sqrt{x}-1}-\frac{2\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-1\right)}=\frac{x-2\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}=\frac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}\left(\sqrt{x}-1\right)}=\frac{\sqrt{x}-1}{\sqrt{x}}\)
Rút gọn bt:
Câu 1: a, \(\left(\sqrt{50}+\sqrt{48}-\sqrt{72}\right)2\sqrt{3}\)
b, \(\sqrt{25a}+2\sqrt{45a}-3\sqrt{80a}+2\sqrt{16a}\left(a\ge0\right)\)ư
Câu 2: Cho bt: P =\(\left(1+\frac{\sqrt{a}}{a+1}\right):\left(\frac{1}{\sqrt{a}-1}-\frac{2\sqrt{a}}{a\sqrt{a}+\sqrt{a}-a-1}\right)\)
a, Tìm ĐKXĐ . Rút gọn P
B, Tìm x nguyên để P có gt nguyên
c, Tìm GTNN của P với a >1
Câu 3: Giair các pt
a, \(\sqrt{\left(2x-1\right)^2}=4\)
b, \(\sqrt{4x+4}+\sqrt{9x+9}-8\sqrt{\frac{x+1}{16}}=5\)
`a)(2sqrtx-9)/(x-5sqrtx+6)-(sqrtx+3)/(sqrtx-2)-(2sqrtx+1)/(3-sqrtx)(x>=0,x ne 4,x ne 9)`
`=(2sqrtx-9)/(x-5sqrtx+6)-(sqrtx+3)/(sqrtx-2)+(2sqrtx+1)/(sqrtx-3)`
`=(2sqrtx-9+(sqrtx-3)(sqrtx+3)+(2sqrtx+1)(sqrtx-2))/(x-5sqrtx+6)`
`=(2sqrtx-9+x-9+2x-3sqrtx-2)/(x-5sqrtx+6)`
`=(3x-sqrtx-20)/
a) Ta có: \(P=\left(\frac{\sqrt{a}}{\sqrt{a}-1}+\frac{\sqrt{a}}{a-1}\right):\left(\frac{2}{a}-\frac{2-a}{a\sqrt{a}+a}\right)\)
\(=\left(\frac{\sqrt{a}\left(\sqrt{a}+1\right)}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}+\frac{\sqrt{a}}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\right):\left(\frac{2\left(\sqrt{a}+1\right)}{a\left(\sqrt{a}+1\right)}-\frac{2-a}{a\left(\sqrt{a}+1\right)}\right)\)
\(=\frac{a+\sqrt{a}+\sqrt{a}}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}:\frac{2\sqrt{a}+2-2+a}{a\left(\sqrt{a}+1\right)}\)
\(=\frac{a+2\sqrt{a}}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\cdot\frac{a\left(\sqrt{a}+1\right)}{a+2\sqrt{a}}\)
\(=\frac{a}{\sqrt{a}-1}\)
b)
ĐKXĐ: \(a\notin\left\{1;0\right\}\)
Để P-2 là số dương thì P-2>0
⇔\(\frac{a}{\sqrt{a}-1}-2>0\)
\(\Leftrightarrow\frac{a}{\sqrt{a}-1}-\frac{2\left(\sqrt{a}-1\right)}{\sqrt{a}-1}>0\)
\(\Leftrightarrow\frac{a-2\sqrt{a}+2}{\sqrt{a}-1}>0\)
mà \(a-2\sqrt{a}+2=\left(\sqrt{a}-1\right)^2+1>0\forall a\)
nên \(\sqrt{a}-1>0\)
\(\Leftrightarrow\sqrt{a}>1\)
\(\Leftrightarrow a>1\)(tm)
Vậy: Khi a>1 thì P-2 là số dương
A=\((\frac{\sqrt{a}\left(\sqrt{a}+1\right)+\sqrt{a}}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}):\left(\frac{2\left(\sqrt{a}+1\right)-\left(2-a\right)}{a\left(\sqrt{a}+1\right)}\right)\)
\(A=\left(\frac{a+\sqrt{a}+\sqrt{a}}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\right):\left(\frac{2\sqrt{a}+2-2+a}{a\left(\sqrt{a}+1\right)}\right)\)
\(A=\frac{a+2\sqrt{a}}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}.\frac{a\left(\sqrt{a}+1\right)}{2\sqrt{a}-a}\)
\(A=\frac{a}{\sqrt{a}-1}\)
\(a,\left(\sqrt{50}+\sqrt{48}-\sqrt{72}\right)2\sqrt{3}\)
\(=\left(5\sqrt{2}+4\sqrt{3}-6\sqrt{2}\right)2\sqrt{3}\)
\(=\left(4\sqrt{3}-\sqrt{2}\right)2\sqrt{3}\)
\(=24-2\sqrt{6}\)
Bài 1 :
a, ĐKXĐ : \(\left\{{}\begin{matrix}x\ge0\\\sqrt{x}-1\ne0\end{matrix}\right.\) => \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)
b, ĐKXĐ : \(-x^2+10x-25\ge0\)
=> \(x^2-10x+25\le0\)
=> \(\left(x-5\right)^2\le0\)
=> \(x-5\le0\)
=> \(x\le5\)
Bài 2 :
a, Ta có : \(A=\sqrt{\left(2\sqrt{2}-5\right)^2}+\sqrt{\left(2-\sqrt{5}\right)^2}\)
=> \(A=5-2\sqrt{2}+\sqrt{5}-2=3-2\sqrt{2}+\sqrt{5}\)
b, Ta có : \(B=\sqrt{9+4\sqrt{5}}-\sqrt{6-2\sqrt{5}}\)
=> \(B=\sqrt{4+2.2\sqrt{5}+5}-\sqrt{1-2\sqrt{5}+5}\)
=> \(B=\sqrt{\left(2+\sqrt{5}\right)^2}-\sqrt{\left(1-\sqrt{5}\right)^2}\)
=> \(B=2+\sqrt{5}-\sqrt{5}+1=3\)
c, Ta có : \(C=\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}\)
=> \(C=\frac{\sqrt{4+2\sqrt{3}}}{\sqrt{2}}+\frac{\sqrt{4-2\sqrt{3}}}{\sqrt{2}}\)
=> \(C=\frac{\sqrt{1+2\sqrt{3}+3}}{\sqrt{2}}+\frac{\sqrt{1-2\sqrt{3}+3}}{\sqrt{2}}\)
=> \(C=\frac{\sqrt{\left(1+\sqrt{3}\right)^2}}{\sqrt{2}}+\frac{\sqrt{\left(1-\sqrt{3}\right)^2}}{\sqrt{2}}\)
=> \(C=\frac{1+\sqrt{3}}{\sqrt{2}}+\frac{\sqrt{3}-1}{\sqrt{2}}=\frac{2\sqrt{3}}{\sqrt{2}}=\sqrt{6}\)