Cho \(x=\frac{2\sqrt[3]{3}}{\sqrt[3]{9}+\sqrt[3]{3}+1};y=\frac{4\sqrt[3]{3}}{9-\sqrt[3]{3}+1}\)
Tính:\(P=\frac{x^3y+xy^3+6\sqrt[3]{3}}{\sqrt[3]{9}}\)
Những câu hỏi liên quan
Cho biểu thức: \(P=\left(\frac{\sqrt{x}-3}{2-\sqrt{x}}+\frac{\sqrt{x}+2}{3+\sqrt{x}}-\frac{9-x}{x+\sqrt{x}-6}\right):\left(1-\frac{3\sqrt{x}-9}{x-9}\right)\)
a)Rút gọn biểu thức
b)Tính P với \(x=\frac{\sqrt{4+2\sqrt{3}}\left(\sqrt{x}-1\right)}{\sqrt{6+2\sqrt{5}-\sqrt{5}}}\)
Mình ghi nhầm. \(x=\frac{\sqrt{4+2\sqrt{3}}.\left(\sqrt{3}-1\right)}{\sqrt{6+2\sqrt{5}}-\sqrt{5}}\)nhé
rút gọn
P=\(\frac{15\sqrt{x}-11}{x+2\sqrt{x}-3}+\frac{3\sqrt{x}-2}{1-\sqrt{x}}-\frac{2\sqrt{x}+3}{\sqrt{x}+3}\)
N= \(\left(\frac{x-3\sqrt{x}}{x-9}-1\right):\left(\frac{9-x}{x+\sqrt{x}-6}-\frac{\sqrt{x}-3}{2-\sqrt{x}}-\frac{\sqrt{x}-2}{\sqrt{x}+3}\right)\)
\(P=\frac{15\sqrt{x}-11}{x+2\sqrt{x}-3}+\frac{3\sqrt{x}-2}{1-\sqrt{x}}-\frac{2\sqrt{x}+3}{\sqrt{x}+3}\)
\(=\frac{15\sqrt{x}-11}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}+\frac{3\sqrt{x}-2}{1-\sqrt{x}}-\frac{2\sqrt{x}+3}{\sqrt{x}+3}\)
\(=\frac{15\sqrt{x}-11-\left(3\sqrt{x}-2\right)\left(\sqrt{x}+3\right)-\left(2\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{-5x+7\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\frac{\left(1-\sqrt{x}\right)\left(5\sqrt{x}-2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\frac{2-5\sqrt{x}}{3+\sqrt{x}}\)
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\(N=\left(\frac{x-3\sqrt{x}}{x-9}-1\right):\left(\frac{9-x}{x+\sqrt{x}-6}-\frac{\sqrt{x}-3}{2-\sqrt{x}}-\frac{\sqrt{x}-2}{\sqrt{x}+3}\right)\)
\(=\left(\frac{\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}-1\right):\left(\frac{9-x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}-\frac{\sqrt{x}-3}{2-\sqrt{x}}-\frac{\sqrt{x}-2}{\sqrt{x}+3}\right)\)
\(=\left(\frac{\sqrt{x}}{\left(\sqrt{x}+3\right)}-1\right):\left(\frac{9-x+\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)-\left(\sqrt{x}-2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\right)\)
\(=-3:\left(\frac{-\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+3\right)}\right)\)
\(=3.\left(\frac{\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-2\right)}\right)\)
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Cái đó chỗ số 3 phải có mẫu nữa mà mình quên mất mẫu
\(\frac{3}{\sqrt{x}+3}:\frac{\sqrt{x}-2}{\sqrt{x}+3}=\frac{3}{\sqrt{x}-2}\)
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Rút gọn
\(A=\left(\frac{\sqrt{x}-1}{3\sqrt{x}-1}-\frac{1}{3\sqrt{x}+1}+\frac{8\sqrt{x}}{9x-1}\right):\left(1-\frac{3\sqrt{x}-2}{3\sqrt{x}+1}\right)\)
\(B=\left(\frac{x-3\sqrt{x}}{x-9}-1\right):\left(\frac{9-x}{x+\sqrt{x}-6}+\frac{\sqrt{x}-3}{\sqrt{x}-2}-\frac{\sqrt{x}-2}{\sqrt{x}+3}\right)\)
cho D= \(\left(\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}}{\sqrt{x}-3}+\frac{3x+3}{9-x}\right):\left(\frac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\)
rút gọn D.
tìm x để D=x/9
Tìm ĐKXĐ và rút gọn biểu thứcAfrac{left(sqrt{a}+sqrt{b}right)^2-4sqrt{ab}}{sqrt{a}-sqrt{b}}-frac{asqrt{b}+bsqrt{a}}{sqrt{ab}}Bleft(frac{2sqrt{x}-x}{xsqrt{x}-1}-frac{1}{sqrt{x}-1}right):frac{x-1}{x+sqrt{x}+1}Cleft(1-frac{x-3sqrt{x}}{x-9}right):left(frac{sqrt{x}-3}{2-sqrt{x}}+frac{sqrt{x}-2}{3+sqrt{x}}-frac{9-x}{x+sqrt{x}-6}right)Dleft(frac{sqrt{x}}{3+sqrt{x}}+frac{x+9}{9-x}right):left(frac{3sqrt{x}+1}{x-3sqrt{x}}-frac{1}{sqrt{x}}right)CM rằng GT của bthức A ko phụ thuộc vào aTìm x để C 4Tìm x sa...
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Tìm ĐKXĐ và rút gọn biểu thức
\(A=\frac{\left(\sqrt{a}+\sqrt{b}\right)^2-4\sqrt{ab}}{\sqrt{a}-\sqrt{b}}-\frac{a\sqrt{b}+b\sqrt{a}}{\sqrt{ab}}\)
\(B=\left(\frac{2\sqrt{x}-x}{x\sqrt{x}-1}-\frac{1}{\sqrt{x}-1}\right):\frac{x-1}{x+\sqrt{x}+1}\)
\(C=\left(1-\frac{x-3\sqrt{x}}{x-9}\right):\left(\frac{\sqrt{x}-3}{2-\sqrt{x}}+\frac{\sqrt{x}-2}{3+\sqrt{x}}-\frac{9-x}{x+\sqrt{x}-6}\right)\)
\(D=\left(\frac{\sqrt{x}}{3+\sqrt{x}}+\frac{x+9}{9-x}\right):\left(\frac{3\sqrt{x}+1}{x-3\sqrt{x}}-\frac{1}{\sqrt{x}}\right)\)
CM rằng GT của bthức A ko phụ thuộc vào a
Tìm x để C = 4
Tìm x sao cho D < -1
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: \(C=\dfrac{x-9-x+3\sqrt{x}}{x-9}:\left(\dfrac{3-\sqrt{x}}{\sqrt{x}-2}+\dfrac{\sqrt{x}-2}{\sqrt{x}+3}+\dfrac{x-9}{x+\sqrt{x}-6}\right)\)
\(=\dfrac{3\left(\sqrt{x}-3\right)}{x-9}:\dfrac{9-x+x-4\sqrt{x}+4+x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{3}{\sqrt{x}+3}\cdot\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}{x-4\sqrt{x}+4}\)
\(=\dfrac{3}{\sqrt{x}-2}\)
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tính giá trị biểu thức
1) A frac{15sqrt{x}-11}{x-2sqrt{x}-3}+frac{3sqrt{x}-2}{1-sqrt{x}}-frac{2sqrt{x}+3}{sqrt{x}+3} tại x3-2sqrt{2}
2) Bfrac{x+2}{xsqrt{x}-1}+frac{sqrt{x}+1}{x+sqrt{x}+1}-frac{sqrt{x}+1}{x-1} tại x7-2sqrt{6}
3) Cleft(frac{sqrt{x}}{3+sqrt{x}}+frac{x+9}{9-x}right):left(frac{3sqrt{x}+1}{x-3sqrt{x}}-frac{1}{sqrt{x}}right) tại x7-4sqrt{3}
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tính giá trị biểu thức
1) A = \(\frac{15\sqrt{x}-11}{x-2\sqrt{x}-3}+\frac{3\sqrt{x}-2}{1-\sqrt{x}}-\frac{2\sqrt{x}+3}{\sqrt{x}+3}\) tại \(x=3-2\sqrt{2}\)
2) \(B=\frac{x+2}{x\sqrt{x}-1}+\frac{\sqrt{x}+1}{x+\sqrt{x}+1}-\frac{\sqrt{x}+1}{x-1}\) tại \(x=7-2\sqrt{6}\)
3) \(C=\left(\frac{\sqrt{x}}{3+\sqrt{x}}+\frac{x+9}{9-x}\right):\left(\frac{3\sqrt{x}+1}{x-3\sqrt{x}}-\frac{1}{\sqrt{x}}\right)\) tại \(x=7-4\sqrt{3}\)
1) Sửa đề: \(A=\frac{15\sqrt{x}-11}{x+2\sqrt{x}-3}+\frac{3\sqrt{x}-2}{1-\sqrt{x}}-\frac{2\sqrt{x}+3}{\sqrt{x}+3}\)
Ta có: \(A=\frac{15\sqrt{x}-11}{x+2\sqrt{x}-3}+\frac{3\sqrt{x}-2}{1-\sqrt{x}}-\frac{2\sqrt{x}+3}{\sqrt{x}+3}\)
\(=\frac{15\sqrt{x}-11}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}-\frac{\left(3\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}-\frac{\left(2\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\frac{15\sqrt{x}-11-\left(3x+7\sqrt{x}-6\right)-\left(2x+\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\frac{15\sqrt{x}-11-3x-7\sqrt{x}+6-2x-\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\frac{-5x+7\sqrt{x}-2}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\frac{-5x+5\sqrt{x}+2\sqrt{x}-2}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\frac{-5\sqrt{x}\left(\sqrt{x}-1\right)+2\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\frac{\left(\sqrt{x}-1\right)\left(-5\sqrt{x}+2\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\frac{-5\sqrt{x}+2}{\sqrt{x}+3}\)
Ta có: \(x=3-2\sqrt{2}\)
\(=2-2\cdot\sqrt{2}\cdot1+1\)
\(=\left(\sqrt{2}-1\right)^2\)
Thay \(x=\left(\sqrt{2}-1\right)^2\) vào biểu thức \(A=\frac{-5\sqrt{x}+2}{\sqrt{x}+3}\), ta được:
\(A=\frac{-5\cdot\sqrt{\left(\sqrt{2}-1\right)^2}+2}{\sqrt{\left(\sqrt{2}-1\right)^2}+3}\)
\(=\frac{-5\cdot\left(\sqrt{2}-1\right)+2}{\sqrt{2}-1+3}\)
\(=\frac{-5\sqrt{2}+5+2}{\sqrt{2}+2}\)
\(=\frac{-5\sqrt{2}+7}{\sqrt{2}+2}\)
Vậy: Khi \(x=3-2\sqrt{2}\) thì \(A=\frac{-5\sqrt{2}+7}{\sqrt{2}+2}\)
2) Ta có: \(B=\frac{x+2}{x\sqrt{x}-1}+\frac{\sqrt{x}+1}{x+\sqrt{x}+1}-\frac{\sqrt{x}+1}{x-1}\)
\(=\frac{\left(x+2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\left(x+\sqrt{x}+1\right)\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}-\frac{\left(\sqrt{x}+1\right)\left(x+\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\frac{x\sqrt{x}+x+2\sqrt{x}+2+x+x\sqrt{x}-\sqrt{x}-1-\left(2x+2\sqrt{x}+x\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\frac{2x+2x\sqrt{x}+\sqrt{x}+1-2x-2\sqrt{x}-x\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\frac{x\sqrt{x}-\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\frac{\sqrt{x}\left(x-1\right)}{\left(x-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\frac{\sqrt{x}}{x+\sqrt{x}+1}\)
Ta có: \(x=7-2\sqrt{6}\)
\(=6-2\sqrt{6}\cdot1+1\)
\(=\left(\sqrt{6}-1\right)^2\)
Thay \(x=\left(\sqrt{6}-1\right)^2\) vào biểu thức \(B=\frac{\sqrt{x}}{x+\sqrt{x}+1}\), ta được:
\(B=\frac{\sqrt{\left(\sqrt{6}-1\right)^2}}{\left(\sqrt{6}-1\right)^2+\sqrt{\left(\sqrt{6}-1\right)^2}+1}\)
\(=\frac{\sqrt{6}-1}{7-2\sqrt{6}+\sqrt{6}-1+1}\)
\(=\frac{\sqrt{6}-1}{7-\sqrt{6}}\)
Vậy: Khi \(x=7-2\sqrt{6}\) thì \(B=\frac{\sqrt{6}-1}{7-\sqrt{6}}\)
3) Ta có: \(C=\left(\frac{\sqrt{x}}{3+\sqrt{x}}+\frac{x+9}{9-x}\right):\left(\frac{3\sqrt{x}+1}{x-3\sqrt{x}}-\frac{1}{\sqrt{x}}\right)\)
\(=\left(\frac{\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}-\frac{x+9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\right):\left(\frac{3\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-3\right)}-\frac{\sqrt{x}-3}{\sqrt{x}\left(\sqrt{x}-3\right)}\right)\)
\(=\frac{x-3\sqrt{x}-x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}:\frac{3\sqrt{x}+1-\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}-3\right)}\)
\(=\frac{x-3\sqrt{x}-x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\cdot\frac{\sqrt{x}\left(\sqrt{x}-3\right)}{2\sqrt{x}+4}\)
\(=\frac{\sqrt{x}\left(x-3\sqrt{x}-x-9\right)}{\left(\sqrt{x}+3\right)\left(2\sqrt{x}+4\right)}\)
\(=\frac{\sqrt{x}\left(-3\sqrt{x}-9\right)}{\left(\sqrt{x}+3\right)\cdot2\cdot\left(\sqrt{x}+2\right)}\)
\(=\frac{-3\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(2\sqrt{x}+4\right)}\)
\(=\frac{-3\sqrt{x}}{2\sqrt{x}+4}\)
Ta có: \(x=7-4\sqrt{3}\)
\(=4-2\cdot2\cdot\sqrt{3}+3\)
\(=\left(2-\sqrt{3}\right)^2\)
Thay \(x=\left(2-\sqrt{3}\right)^2\) vào biểu thức \(C=\frac{-3\sqrt{x}}{2\sqrt{x}+4}\), ta được:
\(C=\frac{-3\cdot\sqrt{\left(2-\sqrt{3}\right)^2}}{2\cdot\sqrt{\left(2-\sqrt{3}\right)^2}+4}\)
\(=\frac{-3\cdot\left(2-\sqrt{3}\right)}{2\cdot\left(2-\sqrt{3}\right)+4}\)
\(=\frac{-6+3\sqrt{3}}{4-2\sqrt{3}+4}\)
\(=\frac{-6+3\sqrt{3}}{8-2\sqrt{3}}\)
Vậy: Khi \(x=7-4\sqrt{3}\) thì \(C=\frac{-6+3\sqrt{3}}{8-2\sqrt{3}}\)
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1. sqrt{x-2sqrt{x-1}}+sqrt{x+3-4sqrt{x-1}}left(2 x 5right)2. frac{6}{1-sqrt{3}}-frac{3sqrt{3}-1}{sqrt{3}+1}+sqrt{3}3. sqrt{29-12sqrt{5}+sqrt{24-8sqrt{3}}}4. sqrt{frac{4}{9-4sqrt{5}}}-sqrt{frac{4}{9+4sqrt{5}}}5. 5sqrt{frac{1}{5}}+frac{1}{2}sqrt{x}-frac{5}{4}sqrt{frac{4}{5}+sqrt{5}}6. frac{6-sqrt{6}}{sqrt{6}-1}-9sqrt{frac{2}{3}}-frac{4}{2-sqrt{6}}7. left(frac{sqrt{x}-2}{x-1}-frac{sqrt{x}+2}{x+2sqrt{x}+1}right)frac{left(sqrt{x}-1right)^2}{2}left(xge0,xne1right)
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1. \(\sqrt{x-2\sqrt{x-1}}+\sqrt{x+3-4\sqrt{x-1}}\left(2< x< 5\right)\)
2. \(\frac{6}{1-\sqrt{3}}-\frac{3\sqrt{3}-1}{\sqrt{3}+1}+\sqrt{3}\)
3. \(\sqrt{29-12\sqrt{5}+\sqrt{24-8\sqrt{3}}}\)
4. \(\sqrt{\frac{4}{9-4\sqrt{5}}}-\sqrt{\frac{4}{9+4\sqrt{5}}}\)
5. \(5\sqrt{\frac{1}{5}}+\frac{1}{2}\sqrt{x}-\frac{5}{4}\sqrt{\frac{4}{5}+\sqrt{5}}\)
6. \(\frac{6-\sqrt{6}}{\sqrt{6}-1}-9\sqrt{\frac{2}{3}}-\frac{4}{2-\sqrt{6}}\)
7. \(\left(\frac{\sqrt{x}-2}{x-1}-\frac{\sqrt{x}+2}{x+2\sqrt{x}+1}\right)\frac{\left(\sqrt{x}-1\right)^2}{2}\left(x\ge0,x\ne1\right)\)
Trả lời nhanh giúp mình với mình cần gấp lắm
Cho biểu thức : R = \(\left(\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}}{\sqrt{x}-3}-\frac{3.\left(\sqrt{x}+3\right)}{x-9}\right):\left(\frac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\)
a. Rút gọn biểu thức.
b. Tìm x để R < -1
\(a)\)\(R=\left(\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}}{\sqrt{x}-3}-\frac{3\left(\sqrt{x}+3\right)}{x-9}\right):\left(\frac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\)
\(R=\left(\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}}{\sqrt{x}-3}-\frac{3\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\right):\frac{\sqrt{x}+1}{\sqrt{x}-3}\)
\(R=\left(\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}}{\sqrt{x}-3}-\frac{3}{\sqrt{x-3}}\right):\frac{\sqrt{x}+1}{\sqrt{x}-3}\)
\(R=\left(\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}-3}{\sqrt{x}-3}\right):\frac{\sqrt{x}+1}{\sqrt{x}-3}\)
\(R=\left(\frac{2\sqrt{x}}{\sqrt{x}+3}+1\right):\frac{\sqrt{x}+1}{\sqrt{x}-3}\)
\(R=\frac{3\sqrt{x}+3}{\sqrt{x}+3}.\frac{\sqrt{x}-3}{\sqrt{x+1}}\)
\(R=\frac{3\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}+1\right)}\)
\(R=\frac{3\left(\sqrt{x}-3\right)}{\sqrt{x}+3}\)
\(b)\) Ta có : \(R< -1\)
\(\Leftrightarrow\)\(\frac{3\left(\sqrt{x}-3\right)}{\sqrt{x}+3}< -1\)
\(\Leftrightarrow\)\(\frac{\sqrt{x}-3}{\sqrt{x}+3}< \frac{-1}{3}\)
\(\Leftrightarrow\)\(3\sqrt{x}-9< -\sqrt{x}-3\)
\(\Leftrightarrow\)\(4\sqrt{x}< 6\)
\(\Leftrightarrow\)\(\sqrt{x}< \frac{3}{2}\)
\(\Leftrightarrow\)\(x< \frac{9}{4}\)
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a) Rút gọn biểu thức sau A=\(\sqrt{3+2\sqrt{2}}-\frac{1}{1+\sqrt{2}}\)
b)Chứng minh rằng:\(\left(\frac{\sqrt{x}}{\sqrt{x}+3}+\frac{3}{\sqrt{x}-3}\right).\frac{\sqrt{x}+3}{x+9}=\frac{1}{\sqrt{x}-3}\)với x≥0 và x ≠ 9
a) Ta có: \(A=\sqrt{3+2\sqrt{2}}-\frac{1}{1+\sqrt{2}}\)
\(=\sqrt{1+2\cdot1\cdot\sqrt{2}+2}-\frac{1}{1+\sqrt{2}}\)
\(=\sqrt{\left(1+\sqrt{2}\right)^2}-\frac{1}{1+\sqrt{2}}\)
\(=1+\sqrt{2}-\frac{1}{1+\sqrt{2}}\)
\(=\frac{\left(1+\sqrt{2}\right)^2}{1+\sqrt{2}}-\frac{1}{1+\sqrt{2}}\)
\(=\frac{1+2\sqrt{2}+2-1}{1+\sqrt{2}}\)
\(=\frac{2\sqrt{2}+2}{1+\sqrt{2}}\)
\(=\frac{2\left(\sqrt{2}+1\right)}{\sqrt{2}+1}=2\)
b) Ta có: \(\left(\frac{\sqrt{x}}{\sqrt{x}+3}+\frac{3}{\sqrt{x}-3}\right)\cdot\frac{\sqrt{x}+3}{x+9}\)
\(=\left(\frac{\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\frac{3\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\right)\cdot\frac{1}{\sqrt{x}-3}\)
\(=\frac{x-3\sqrt{x}+3\sqrt{x}+9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\cdot\frac{1}{\sqrt{x}-3}\)
\(=\frac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\cdot\frac{1}{\sqrt{x}-3}\)
\(=\frac{1}{\sqrt{x}-3}\)(đpcm)
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