cho biểu thức: A=\(\frac{\sqrt{x}}{\sqrt{x}-2}+\frac{3}{\sqrt{x}+2}-\frac{9\sqrt{x}-10}{x-4}\) ( x >= 0; x khác 4)
a, Rút gọn A
b, Tìm A biết x = 4 - 2√3
c, Tìm x thuộc Z để A thuộc Z
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B
29 tháng 7 2019
\(a,A=\frac{\sqrt{x}}{\sqrt{x}-2}+\frac{3}{\sqrt{x}+2}-\frac{9\sqrt{x}-10}{x-4}\left(x\ge0;x\ne16\right)\)
\(=\frac{\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\frac{3\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}-\frac{9\sqrt{x}-10}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{x+2\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\frac{3\sqrt{x}-6}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\frac{9\sqrt{x}-10}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{x+2\sqrt{x}+3\sqrt{x}-6-9\sqrt{x}+10}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{x-4\sqrt{x}+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\frac{\left(\sqrt{x}\right)^2-2.\sqrt{x}.2+2^2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\frac{\left(\sqrt{x}-2\right)^2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\frac{\sqrt{x}-2}{\sqrt{x}+2}\)
Vây...
\(b,\)Ta có:\(x=4-2\sqrt{3}=\left(1-\sqrt{3}\right)^2\)
Thay \(x=\left(1-\sqrt{3}\right)^2\)vào A ta được:
\(A=\frac{\sqrt{\left(1-\sqrt{3}\right)^2}-2}{\sqrt{\left(1-\sqrt{3}\right)^2}+2}=\frac{\sqrt{3}-1-2}{\sqrt{3}-1+2}=\frac{\sqrt{3}-3}{\sqrt{3}-1}=\frac{-\sqrt{3}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}=-\sqrt{3}\)
NC
29 tháng 8 2020
Bài làm:
Ta có:
\(P=\left(1-\frac{x-3\sqrt{x}}{x-9}\right)\div\left(\frac{\sqrt{x}-9}{2-\sqrt{x}}+\frac{\sqrt{x}-2}{3+\sqrt{x}}-\frac{9-x}{x+\sqrt{x}-6}\right)\)
\(P=\frac{x-9-x+3\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\div\left[\frac{\left(9-\sqrt{x}\right)\left(3+\sqrt{x}\right)+\left(\sqrt{x}-2\right)^2-9+x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\right]\)
\(P=\frac{3\sqrt{x}-9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\div\frac{-x+6\sqrt{x}+27+x-4\sqrt{x}+2-9+x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\)
\(P=\frac{3}{\sqrt{x}+3}\div\frac{x+2\sqrt{x}+20}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\)
\(P=\frac{3}{\sqrt{x}+3}\cdot\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}{x+2\sqrt{x}+20}\)
\(P=\frac{3\left(\sqrt{x}-2\right)}{x+2\sqrt{x}+20}=\frac{3\sqrt{x}-6}{x+2\sqrt{x}+20}\)
23 tháng 5 2021
Mình ghi nhầm. \(x=\frac{\sqrt{4+2\sqrt{3}}.\left(\sqrt{3}-1\right)}{\sqrt{6+2\sqrt{5}}-\sqrt{5}}\)nhé
VH
9 tháng 7 2019
a) \(A=\frac{\sqrt{4+2\sqrt{3}}}{\sqrt{3}+1}+\frac{5+3\sqrt{5}}{\sqrt{5}}-\left(\sqrt{5}+3\right)\)
\(A=\frac{\sqrt{\left(\sqrt{3}+1\right)^2}}{\sqrt{3}+1}+\frac{5+3\sqrt{5}}{\sqrt{5}}-\frac{\sqrt{5}\left(\sqrt{5}+3\right)}{\sqrt{5}}\)
\(A=\frac{\sqrt{3}+1}{\sqrt{3}+1}+\frac{5+3\sqrt{5}}{\sqrt{5}}-\frac{5+3\sqrt{5}}{\sqrt{5}}\)
\(A=1\)
b) Ta có:
\(B=\frac{1}{3-\sqrt{x}}+\frac{\sqrt{x}}{3+\sqrt{x}}-\frac{x+9}{x-9}\) ( x >= 0, x khác 9 )
\(B=\frac{3+\sqrt{x}}{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}+\frac{\sqrt{x}\left(3-\sqrt{x}\right)}{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}+\frac{x+9}{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}\)
\(B=\frac{3+\sqrt{x}+3\sqrt{x}-x+x+9}{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}\)
\(B=\frac{3+\sqrt{x}+3\sqrt{x}+9}{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}\)
\(B=\frac{\left(3+\sqrt{x}\right)+3\left(\sqrt{x}+3\right)}{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}\)
\(B=\frac{4\left(3+\sqrt{x}\right)}{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}\)
\(B=\frac{4}{3-\sqrt{x}}\)
Để B > A
\(\Rightarrow\frac{4}{3-\sqrt{x}}>1\)
\(\Rightarrow4>3-\sqrt{x}\)
\(\Rightarrow4-3+\sqrt{x}>0\)
\(\Rightarrow1+\sqrt{x}>0\)
\(\Rightarrow\sqrt{x}>-1\)
\(\Rightarrow x>1\)
WE
9 tháng 7 2019
a) A=\(\frac{\sqrt{4+2\sqrt{3}}}{\sqrt{3}+1}+\frac{5+3\sqrt{5}}{\sqrt{5}}-\left(\sqrt{5}+3\right)\)
\(=\frac{\sqrt{3+2\sqrt{3}+1}}{\sqrt{3}+1}+\frac{\sqrt{5}\cdot\left(\sqrt{5}+3\right)}{\sqrt{5}}\)
\(=\frac{\sqrt{\left(\sqrt{3}+1\right)^2}}{\sqrt{3}+1}+\left(\sqrt{5}+3\right)-\left(\sqrt{5}+3\right)\)
\(=\frac{\sqrt{3}+1}{\sqrt{3}+1}+0=1\)
b) B=\(\frac{1}{3-\sqrt{x}}+\frac{\sqrt{x}}{3+\sqrt{x}}-\frac{x+9}{x-9}\)
\(=\frac{3+\sqrt{x}+\sqrt{x}\left(3-\sqrt{x}\right)}{\left(3-\sqrt{x}\right)\cdot\left(3+\sqrt{x}\right)}+\frac{x+9}{9-x}\)
\(=\frac{3+\sqrt{x}+3\sqrt{x}-x}{\left(3-\sqrt{x}\right)\cdot\left(3+\sqrt{x}\right)}+\frac{x+9}{\left(3-\sqrt{x}\right)\cdot\left(3+\sqrt{x}\right)}\)
\(=\frac{4\text{}\sqrt{x}+12}{\left(3-\sqrt{x}\right)\cdot\left(3+\sqrt{x}\right)}\)
\(=\frac{4\left(\sqrt{x}+3\right)}{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}\)
\(=\frac{4}{3-\sqrt{x}}\)
\(B>A \Leftrightarrow\frac{4}{3-\sqrt{x}}>1\)
các giá trị của x là \(\left\{x\in R\backslash0\le x\le9\right\}\)
2 tháng 8 2019
\(a,đkxđ\Leftrightarrow\hept{\begin{cases}x\ge0\\x\ne4\end{cases}}\)
\(A=\frac{\sqrt{x}}{\sqrt{x}-2}+\frac{3}{\sqrt{x}+2}-\frac{9\sqrt{x}-10}{x-4}.\)
\(=\frac{\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\frac{3\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)\(-\frac{9\sqrt{x}-10}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{x+2\sqrt{x}+3\sqrt{x}-6-9\sqrt{x}+10}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{x-4\sqrt{x}-4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{\left(\sqrt{x}-2\right)^2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}=\frac{\sqrt{x}-2}{\sqrt{x}+2}\)
\(b,x=4-2\sqrt{3}=3-2\sqrt{3}+1=\left(\sqrt{3}-1\right)^2\)
\(\Rightarrow x=\sqrt{3}-1\)
\(\Rightarrow A=\frac{\sqrt{3}-1-2}{\sqrt{3}-1+2}=\frac{\sqrt{3}-3}{\sqrt{3}-1}\)
\(b,A=\frac{\sqrt{x}-2}{\sqrt{x}+2}=\frac{\sqrt{x}+2-4}{\sqrt{x}+2}\)\(=1-\frac{4}{\sqrt{x}+2}\)
\(A\in Z\Leftrightarrow1-\frac{4}{\sqrt{x}+2}\in Z\Rightarrow\frac{4}{\sqrt{x}+2}\in Z\)
\(\Rightarrow\sqrt{x}+2\inƯ_4\)
Mà \(Ư_4=\left\{\pm1;\pm2;\pm4\right\}\)Nhưng \(\sqrt{x}+2\ge2\)\(\Rightarrow\sqrt{x}+2\in\left\{2;4\right\}\)
\(Th1:\sqrt{x}+2=2\Rightarrow\sqrt{x}=0\Rightarrow x=0\)
\(Th2:\sqrt{x}+2=4\Rightarrow\sqrt{x}=2\Rightarrow x=4\)
\(KL:x\in\left\{0;4\right\}\)
31 tháng 5 2020
1) Ta có: \(A=\left(\frac{2}{\sqrt{x}-2}+\frac{3}{2\sqrt{x}+1}-\frac{5\sqrt{x}-7}{2x-3\sqrt{x}-2}\right):\frac{2\sqrt{x}+3}{5x-10\sqrt{x}}\)
\(=\left(\frac{2}{\sqrt{x}-2}+\frac{3}{2\sqrt{x}+1}-\frac{5\sqrt{x}-7}{2x-4\sqrt{x}+\sqrt{x}-2}\right)\cdot\frac{5\sqrt{x}\left(\sqrt{x}-2\right)}{2\sqrt{x}+3}\)
\(=\left(\frac{2\left(2\sqrt{x}+1\right)}{\left(\sqrt{x}-2\right)\left(2\sqrt{x}+1\right)}+\frac{3\left(\sqrt{x}-2\right)}{\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}-\frac{5\sqrt{x}-7}{\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\right)\cdot\frac{5\sqrt{x}\left(\sqrt{x}-2\right)}{2\sqrt{x}+3}\)
\(=\frac{4\sqrt{x}+2+3\sqrt{x}-6-5\sqrt{x}+7}{\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\cdot\frac{5\sqrt{x}\left(\sqrt{x}-2\right)}{2\sqrt{x}+3}\)
\(=\frac{2\sqrt{x}+3}{\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\cdot\frac{5\sqrt{x}\left(\sqrt{x}-2\right)}{2\sqrt{x}+3}\)
\(=\frac{5\sqrt{x}}{2\sqrt{x}+1}\)
a.
\(A=\frac{\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\frac{3\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\frac{9\sqrt{x}-10}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\\ =\frac{x+2\sqrt{x}+3\sqrt{x}-6-9\sqrt{x}+10}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\\ =\frac{x-4\sqrt{x}+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\\ =\frac{\left(\sqrt{x}-2\right)^2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\\ =\frac{\sqrt{x}-2}{\sqrt{x}+2}\)
b. Ta có
\(\sqrt{x}=\sqrt{4-2\sqrt{3}}=\sqrt{3-2\cdot\sqrt{3}\cdot1+1}=\sqrt{\left(\sqrt{3}-1\right)^2}=\sqrt{3}-1\)
Thay vào A ta được
\(A=\frac{\sqrt{x}-2}{\sqrt{x}+2}\\ =\frac{\sqrt{3}-1-2}{\sqrt{3}-1+2}\\ =\frac{\sqrt{3}-3}{\sqrt{3}+1}\\ =\frac{\left(\sqrt{3}-3\right)\left(\sqrt{3}-1\right)}{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}\\ =\frac{6-4\sqrt{3}}{2}=3-2\sqrt{3}\)
c. \(A=\frac{\sqrt{x}-2}{\sqrt{x}+2}=\frac{\sqrt{x}+2-4}{\sqrt{x}+2}=1-\frac{4}{\sqrt{x}+2}\)
Để \(A\in Z\Leftrightarrow4⋮\sqrt{x}+2\Leftrightarrow\sqrt{x}+2\inƯ\left(4\right)\)
Ta thấy \(\sqrt{x}\ge0\forall x\ge0\left(ĐK\right)\Leftrightarrow\sqrt{x}+2\ge2\)
Nên \(\sqrt{x}+2\in\left\{2;4\right\}\\ \Leftrightarrow\left[{}\begin{matrix}\sqrt{x}+2=2\\\sqrt{x}+2=4\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=0\\\sqrt{x}=2\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\left(tm\right)\\x=4\left(ktm\right)\end{matrix}\right.\)
Vậy x=0 thì A thuộc Z