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bài 2 : ĐKXĐ : \(x\ge0\) và \(x\ne1\)
Rút gọn :\(B=\frac{\sqrt{x}+1}{\sqrt{x}-1}-\frac{\sqrt{x}-1}{\sqrt{x}+1}-\frac{5\sqrt{x}-1}{x-1}\)
\(B=\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{5\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(B=\frac{x+2\sqrt{x}+1-x+2\sqrt{x}-1-5\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(B=\frac{-\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(B=\frac{-1}{\sqrt{x}+1}\)
Bài 1:
a, \(4\sqrt{3+2\sqrt{2}}-\sqrt{57+40\sqrt{2}}\)
\(=4\sqrt{\left(\sqrt{2}+1\right)^2}-\sqrt{\left(4\sqrt{2}+5\right)^2}\)
\(=4\left(\sqrt{2}+1\right)-4\sqrt{2}-5\)
\(=4\sqrt{2}+4-4\sqrt{2}-5=-1\)
b, \(B=\sqrt{1100}-7\sqrt{44}+2\sqrt{176}-\sqrt{1331}\)
\(=10\sqrt{11}-14\sqrt{11}+8\sqrt{11}-11\sqrt{11}=-7\sqrt{11}\)
c, \(C=\sqrt{\left(1-\sqrt{2002}\right)^2}.\sqrt{2003+2\sqrt{2002}}\)
\(=\left(1-\sqrt{2002}\right).\sqrt{\left(\sqrt{2002}+1\right)^2}\)
\(=\left(1-\sqrt{2002}\right).\left(\sqrt{2002}+1\right)=-2001\)
Câu d bạn kiểm tra lại đề bài nhé.
Bài 2:
\(A=\frac{1}{2\sqrt{x}-2}-\frac{1}{2\sqrt{2}+2}+\frac{\sqrt{x}}{1-x}\)
a, ĐK: \(x\ge0,x\ne1\)
b, ĐK: \(x\ge0,x\ne1\)
\(A=\frac{1}{2\sqrt{x}-2}-\frac{1}{2\sqrt{2}+2}+\frac{\sqrt{x}}{1-x}\)
\(=\frac{1}{2\sqrt{x}-2}-\frac{1}{2\sqrt{2}+2}-\frac{\sqrt{x}}{x-1}\)
\(=\frac{1}{2\left(\sqrt{x}-1\right)}-\frac{1}{2\left(\sqrt{x}+1\right)}-\frac{\sqrt{x}}{x-1}\)
\(=\frac{2\sqrt{x}+2-2\sqrt{x}+2}{4\left(x-1\right)}-\frac{\sqrt{x}}{x-1}\)
\(=\frac{4-4\sqrt{x}}{4\left(x-1\right)}=\frac{4\left(1-\sqrt{x}\right)}{4\left(1-x\right)}=\frac{1-\sqrt{x}}{1-x}\)
Thay \(x=3\left(TM\right)\)vào A ta có: \(A=\frac{1-\sqrt{3}}{3-1}=\frac{1-\sqrt{3}}{2}\)
Vậy với \(x=3\)thì \(A=\frac{1-\sqrt{3}}{2}\)
c, \(\left|A\right|=\frac{1}{2}\Leftrightarrow\orbr{\begin{cases}A=\frac{1}{2}\\A=-\frac{1}{2}\end{cases}}\)
TH1: \(A=\frac{1}{2}\)\(\Leftrightarrow\frac{1-\sqrt{x}}{x-1}=\frac{1}{2}\Leftrightarrow2-2\sqrt{x}=x-1\)\(\Leftrightarrow x-1-2+2\sqrt{x}=0\)\(\Leftrightarrow x+2\sqrt{x}-3=0\Leftrightarrow\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)=0\)\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x}-1=0\\\sqrt{x}+3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1\left(TM\right)\\\sqrt{x}=-3\left(L\right)\end{cases}}}\)
TH2: \(A=-\frac{1}{2}\Leftrightarrow\frac{1-\sqrt{x}}{x-1}=-\frac{1}{2}\)\(\Leftrightarrow2-2\sqrt{x}=1-x\Leftrightarrow-x+1-2+2\sqrt{x}=0\)\(\Leftrightarrow-x-1+2\sqrt{x}=0\Leftrightarrow x-2\sqrt{x}+1=0\)\(\Leftrightarrow\left(\sqrt{x}+1\right)^2=0\Leftrightarrow\sqrt{x}=-1\left(L\right)\)
Vậy với \(x=1\)thì \(\left|A\right|=\frac{1}{2}\)
a: \(A=\dfrac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\sqrt{a}-\sqrt{b}}-\dfrac{\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{ab}}\)
\(=\sqrt{a}-\sqrt{b}-\sqrt{a}-\sqrt{b}=-2\sqrt{b}\)
b: \(B=\dfrac{2\sqrt{x}-x-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\dfrac{x+\sqrt{x}+1}{x-1}\)
\(=\dfrac{-2x+\sqrt{x}-1}{\sqrt{x}-1}\cdot\dfrac{1}{x-1}\)
c: \(C=\dfrac{x-9-x+3\sqrt{x}}{x-9}:\left(\dfrac{3-\sqrt{x}}{\sqrt{x}-2}+\dfrac{\sqrt{x}-2}{\sqrt{x}+3}+\dfrac{x-9}{x+\sqrt{x}-6}\right)\)
\(=\dfrac{3\left(\sqrt{x}-3\right)}{x-9}:\dfrac{9-x+x-4\sqrt{x}+4+x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{3}{\sqrt{x}+3}\cdot\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}{x-4\sqrt{x}+4}\)
\(=\dfrac{3}{\sqrt{x}-2}\)
mình giúp bài 3 cho
\(\sqrt{25x-125}-3\sqrt{\frac{x-5}{9}}-\frac{1}{3}\sqrt{9x-45}=6\left(ĐKXĐ:x\ge5\right)\)
\(< =>\sqrt{25\left(x-5\right)}-3\sqrt{\frac{x-5}{9}}-\frac{1}{3}\sqrt{9\left(x-5\right)}=6\)
\(< =>\sqrt{25}.\sqrt{x-5}-3\frac{\sqrt{x-5}}{\sqrt{9}}-\frac{1}{3}\sqrt{9}.\sqrt{x-5}=6\)
\(< =>5.\sqrt{x-5}-3.\frac{\sqrt{x-5}}{3}-\frac{1}{3}.3.\sqrt{x-5}=6\)
\(< =>5.\sqrt{x-5}-\sqrt{x-5}-\sqrt{x-5}=6\)
\(< =>3\sqrt{x-5}=6< =>\sqrt{x-5}=2\)
\(< =>x-5=4< =>x=4+5=9\left(tmđk\right)\)
ĐK \(\hept{\begin{cases}x\ge0\\x\ne4;x\ne9\end{cases}}\)
a. P=\(\left(\frac{\sqrt{x}+2}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}+\frac{\sqrt{x}+3}{\sqrt{x}-2}-\frac{\sqrt{x}+2}{\sqrt{x}-3}\right):\frac{2\sqrt{x}+2-\sqrt{x}}{\sqrt{x}+1}\)
\(=\frac{\sqrt{x}+2+\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)-\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}:\frac{\sqrt{x}+2}{\sqrt{x}+1}\)
\(=\frac{\sqrt{x}+2+x-9-x+4}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}.\frac{\sqrt{x}+1}{\sqrt{x}+2}=\frac{\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}.\frac{\sqrt{x}+1}{\sqrt{x}+2}\)
\(=\frac{\sqrt{x}+1}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
b. Với \(x=4-2\sqrt{3}\Rightarrow P=\frac{\sqrt{4-2\sqrt{3}}+1}{4-2\sqrt{3}-4}=\frac{\sqrt{\left(\sqrt{3}-1\right)^2}+1}{-2\sqrt{3}}\)
\(=\frac{\sqrt{3}-1+1}{-2\sqrt{3}}=-\frac{1}{2}\)
c. Để \(\frac{1}{P}\le\frac{-5}{2}\Leftrightarrow\frac{x-4}{\sqrt{x}+1}+\frac{5}{2}\le0\Leftrightarrow\frac{2x-8+5\sqrt{x}+5}{2\left(\sqrt{x}+1\right)}\le0\)
\(\Leftrightarrow\frac{2x+5\sqrt{x}-3}{2\left(\sqrt{x}+1\right)}\le0\Leftrightarrow2x+5\sqrt{x}-3\le0\)vì \(2\left(\sqrt{x}+1\right)\ge0\forall x\ge0\)
\(\Leftrightarrow\left(\sqrt{x}+3\right)\left(2\sqrt{x}-1\right)\le0\Leftrightarrow2\sqrt{x}-1\le0\Leftrightarrow0\le x\le\frac{1}{4}\left(tm\right)\)
Vậy với \(0\le x\le\frac{1}{4}\)thì \(\frac{1}{P}\le-\frac{5}{2}\)
d. Ta có \(B=P\left(\sqrt{x}-2\right)=\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\frac{\sqrt{x}+1}{\sqrt{x}+2}=1-\frac{1}{\sqrt{x}+2}\)
Gỉa sử \(B\in Z\Leftrightarrow\sqrt{x}+2\inƯ\left(1\right)\Leftrightarrow\sqrt{x}+2\in\left\{-1;1\right\}\Leftrightarrow x\in\left\{\phi\right\}\)
Vậy B không nhận giá trị nguyên với mọi x để P có nghĩa
Bài 2 :
a) Sửa đề :
\(A=\sqrt{\left(\sqrt{3}-1\right)^2}-\sqrt{3}\)
\(A=\sqrt{3}-1-\sqrt{3}\)
\(A=-1\)
b) \(B=\sqrt{3+2\sqrt{2}}-\sqrt{3-2\sqrt{2}}\)
\(B=\sqrt{\left(\sqrt{2}+1\right)^2}-\sqrt{\left(\sqrt{2}-1\right)^2}\)
\(B=\sqrt{2}+1-\sqrt{2}+1\)
\(B=2\)
c) \(C=\sqrt{7-4\sqrt{3}}+\sqrt{7+4\sqrt{3}}\)
\(C=\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{\left(2+\sqrt{3}\right)^2}\)
\(C=2-\sqrt{3}+2+\sqrt{3}\)
\(C=4\)
d) \(D=\sqrt{23+8\sqrt{7}}-\sqrt{7}\)
\(D=\sqrt{\left(4+\sqrt{7}\right)^2}-\sqrt{7}\)
\(D=4+\sqrt{7}-\sqrt{7}\)
\(D=4\)
Bài 1 :
a) Để \(\sqrt{\left(x-1\right)\left(x-3\right)}\) có nghĩa
\(\Leftrightarrow\left(x-1\right)\left(x-3\right)\ge0\)
TH1 :\(\hept{\begin{cases}x-1\ge0\\x-3\ge0\end{cases}\Leftrightarrow}\hept{\begin{cases}x\ge1\\x\ge3\end{cases}\Leftrightarrow x\ge3}\)
TH2 : \(\hept{\begin{cases}x-1\le0\\x-3\le0\end{cases}\Leftrightarrow\hept{\begin{cases}x\le1\\x\le3\end{cases}\Leftrightarrow}x\le1}\)
Vậy để biểu thức có nghĩa thì \(\orbr{\begin{cases}x\ge3\\x\le1\end{cases}}\)
b) Để \(\sqrt{\frac{1-x}{x+2}}\)có nghĩa
\(\Leftrightarrow\frac{1-x}{x+2}\ge0\)
TH1 : \(\hept{\begin{cases}1-x\ge0\\x+2\ge0\end{cases}\Leftrightarrow\hept{\begin{cases}x\le1\\x\ge-2\end{cases}\Leftrightarrow}-2\le x\le1}\)
TH2 : \(\hept{\begin{cases}1-x\le0\\x+2\le0\end{cases}\Leftrightarrow}\hept{\begin{cases}x\ge1\\x\le-2\end{cases}\Leftrightarrow x\in\varnothing}\)
Vậy để biểu thức có nghĩa thì \(-2\le x\le1\)
\(đkxđ\Leftrightarrow\hept{\begin{cases}x\ge0\\x\ne49\end{cases}}\)
\(B=\left(\frac{\sqrt{x}}{x-49}-\frac{\sqrt{x}-7}{x+7\sqrt{x}}\right):\)\(\frac{2\sqrt{x}-7}{x+7\sqrt{x}}+\frac{\sqrt{x}}{7-\sqrt{x}}\)
\(=\left(\frac{\sqrt{x}}{\left(\sqrt{x}-7\right)\left(\sqrt{x}+7\right)}-\frac{\left(\sqrt{x}-7\right)^2}{\sqrt{x}\left(\sqrt{x}+7\right)\left(\sqrt{x}-7\right)}\right)\)\(:\frac{2\sqrt{x}-7}{\sqrt{x}\left(\sqrt{x}+7\right)}-\frac{\sqrt{x}}{\sqrt{x}-7}\)
\(\frac{x-x+14\sqrt{x}-49}{\sqrt{x}\left(\sqrt{x}-7\right)\left(\sqrt{x}+7\right)}:\frac{2\sqrt{x}-7}{\sqrt{x}\left(\sqrt{x}+7\right)}\)\(-\frac{\sqrt{x}}{\sqrt{x}-7}\)
\(=\frac{7\left(2\sqrt{x}-7\right)\sqrt{x}\left(\sqrt{x}+7\right)}{\sqrt{x}\left(\sqrt{x}+7\right)\left(\sqrt{x}-7\right)\left(2\sqrt{x}-7\right)}\)\(-\frac{\sqrt{x}}{\sqrt{x}-7}\)
\(=\frac{7}{\sqrt{x}-7}-\frac{\sqrt{x}}{\sqrt{x}-7}=\frac{7-\sqrt{x}}{\sqrt{x}-7}=-1\)