\(5\sqrt{\dfrac{1}{5}}+\dfrac{1}{2}\sqrt{20}+\sqrt{5}\)
Những câu hỏi liên quan
dfrac{2sqrt{3}-3sqrt{2}}{sqrt{6}}-dfrac{2}{1-sqrt{3}}dfrac{4}{sqrt{6}+sqrt{2}}-dfrac{sqrt{54}+sqrt{2}}{sqrt{3}+1}dfrac{5+2sqrt{5}}{sqrt{5}}-dfrac{20}{5+sqrt{5}}-sqrt{20}Bài 2sqrt{25x^2-10x+1}sqrt{4x^2+8x+4}sqrt{x^2-3}+1xsqrt{7-2x}sqrt{x^2+7}sqrt{9x-27}+dfrac{1}{2}sqrt{4x-12}-9sqrt{dfrac{x-3}{9}}2
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\(\dfrac{2\sqrt{3}-3\sqrt{2}}{\sqrt{6}}-\dfrac{2}{1-\sqrt{3}}\)
\(\dfrac{4}{\sqrt{6}+\sqrt{2}}-\dfrac{\sqrt{54}+\sqrt{2}}{\sqrt{3}+1}\)
\(\dfrac{5+2\sqrt{5}}{\sqrt{5}}-\dfrac{20}{5+\sqrt{5}}-\sqrt{20}\)
Bài 2
\(\sqrt{25x^2-10x+1}=\sqrt{4x^2+8x+4}\)
\(\sqrt{x^2-3}+1=x\)
\(\sqrt{7-2x}=\sqrt{x^2+7}\)
\(\sqrt{9x-27}+\dfrac{1}{2}\sqrt{4x-12}-9\sqrt{\dfrac{x-3}{9}}=2\)
\(2,\\ a,PT\Leftrightarrow\sqrt{\left(5x-1\right)^2}=\sqrt{4\left(x+1\right)^2}\\ \Leftrightarrow\left|5x-1\right|=2\left|x+1\right|\\ \Leftrightarrow\left[{}\begin{matrix}5x-1=2\left(x+1\right)\\1-5x=2\left(x+1\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=3\\7x=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{1}{7}\end{matrix}\right.\)
\(b,ĐK:x^2-3\ge0\\ PT\Leftrightarrow\sqrt{x^2-3}=x-1\\ \Leftrightarrow x^2-3=x^2-2x+1\\ \Leftrightarrow2x=4\Leftrightarrow x=2\left(tm\right)\\ c,ĐK:x\le\dfrac{7}{2}\\ PT\Leftrightarrow7-2x=x^2+7\\ \Leftrightarrow x^2+2x=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\left(tm\right)\\x=2\left(tm\right)\end{matrix}\right.\\ d,ĐK:x\ge3\\ PT\Leftrightarrow3\sqrt{x-3}+\dfrac{1}{2}\cdot2\sqrt{x-3}-9\cdot\dfrac{1}{3}\sqrt{x-3}=2\\ \Leftrightarrow\sqrt{x-3}=2\\ \Leftrightarrow x-3=4\Leftrightarrow x=7\left(tm\right)\)
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Bài 1:
d: Ta có: \(\dfrac{5+2\sqrt{5}}{\sqrt{5}}-\dfrac{20}{5+\sqrt{5}}-\sqrt{20}\)
\(=\sqrt{5}+2-5+\sqrt{5}-2\sqrt{5}\)
=-3
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1. dfrac{-2}{sqrt{3}-1}2. dfrac{5}{1-sqrt{6}}3. dfrac{2+sqrt{5}}{2-sqrt{5}}4. dfrac{1}{5+2sqrt{6}}5. dfrac{sqrt{5}+2}{sqrt{5}-2}6. dfrac{5sqrt{2}-2sqrt{5}}{sqrt{2}-sqrt{5}}7. dfrac{sqrt{20}-3sqrt{10}}{3-sqrt{2}}8. dfrac{6-2sqrt{5}}{3+sqrt{5}}9. dfrac{9+4sqrt{5}}{sqrt{5}+2}
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1. \(\dfrac{-2}{\sqrt{3}-1}\)
2. \(\dfrac{5}{1-\sqrt{6}}\)
3. \(\dfrac{2+\sqrt{5}}{2-\sqrt{5}}\)
4. \(\dfrac{1}{5+2\sqrt{6}}\)
5. \(\dfrac{\sqrt{5}+2}{\sqrt{5}-2}\)
6. \(\dfrac{5\sqrt{2}-2\sqrt{5}}{\sqrt{2}-\sqrt{5}}\)
7. \(\dfrac{\sqrt{20}-3\sqrt{10}}{3-\sqrt{2}}\)
8. \(\dfrac{6-2\sqrt{5}}{3+\sqrt{5}}\)
9. \(\dfrac{9+4\sqrt{5}}{\sqrt{5}+2}\)
Rút gọn biểu thức :
\((5\sqrt{\dfrac{1}{5}}+\dfrac{1}{2}\sqrt{20}-\dfrac{5}{4}\sqrt{\dfrac{4}{5}+\sqrt{5}}):2\sqrt{5}\) và \(\dfrac{1}{3}\sqrt{48}+3\sqrt{75}-\sqrt{27}-10\sqrt{1\dfrac{1}{3}}\)
`(5sqrt{1/5}+1/2sqrt{20}-5/4sqrt{4/5}+sqrt{5}):2/5
`=(sqrt5+1/2*2sqrt5-sqrt{5/4}+sqrt5):2/5`
`=(sqrt5+sqrt5+sqrt5-sqrt5/2):2/5`
`=(5/2*sqrt5):2/5`
`=25/4sqrt5`
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`1/3sqrt{48}+3sqrt{75}-sqrt{27}-10sqrt{1 1/3}`
`=1/3*4sqrt3+3*5sqrt3-3sqrt3-10sqrt{4/3}`
`=4/sqrt3+15sqrt3-3sqrt3-20/sqrt3`
`=12sqrt3-16/sqrt3`
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5 câu:1) dfrac{sqrt{3}+sqrt{2}}{sqrt{6}+2}-dfrac{sqrt{3}-sqrt{2}}{sqrt{6}-2}2) dfrac{3}{sqrt{5}-sqrt{2}}-dfrac{2}{2-sqrt{2}}+dfrac{1}{sqrt{3}+sqrt{2}}3) dfrac{12}{sqrt{5}+1}-dfrac{4}{sqrt{5}+2}+dfrac{20}{3+sqrt{5}}4) dfrac{5}{3-sqrt{7}}-dfrac{2}{sqrt{2}+sqrt{3}}-dfrac{1}{sqrt{2}-1}5) dfrac{sqrt{12}-6}{sqrt{8}-sqrt{24}}-dfrac{3+sqrt{3}}{sqrt{3}}-dfrac{4}{sqrt{7}-1}
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5 câu:
1) \(\dfrac{\sqrt{3}+\sqrt{2}}{\sqrt{6}+2}-\dfrac{\sqrt{3}-\sqrt{2}}{\sqrt{6}-2}\)
2) \(\dfrac{3}{\sqrt{5}-\sqrt{2}}-\dfrac{2}{2-\sqrt{2}}+\dfrac{1}{\sqrt{3}+\sqrt{2}}\)
3) \(\dfrac{12}{\sqrt{5}+1}-\dfrac{4}{\sqrt{5}+2}+\dfrac{20}{3+\sqrt{5}}\)
4) \(\dfrac{5}{3-\sqrt{7}}-\dfrac{2}{\sqrt{2}+\sqrt{3}}-\dfrac{1}{\sqrt{2}-1}\)
5) \(\dfrac{\sqrt{12}-6}{\sqrt{8}-\sqrt{24}}-\dfrac{3+\sqrt{3}}{\sqrt{3}}-\dfrac{4}{\sqrt{7}-1}\)
Bài 1: Rút gọn biểu thức:a) left(5sqrt{dfrac{1}{5}}+dfrac{1}{2}+sqrt{20}-dfrac{5}{4}sqrt{dfrac{4}{5}+sqrt{5}}right)b) dfrac{1}{3}sqrt{48}+3sqrt{75}-sqrt{27}-10sqrt{1dfrac{1}{3}}c) dfrac{5sqrt{7}-7sqrt{5}+2sqrt{70}}{sqrt{35}}d) sqrt{dfrac{3}{4}}+sqrt{dfrac{1}{3}}+sqrt{dfrac{1}{12}}Bài 2: Giải các phương trình sau:a) x^2+4x+52sqrt{2x+3}b) x^2+9x+202sqrt{3x+10}c) x^2+7x+142sqrt{x+4}d) 4sqrt{x+1}x^2-5x+14e) sqrt{6-x}3x-4f) sqrt{5x-9}9-2xMọi người làm ơn giúp mình với. Mình đang cần gấp ạ. Cảm ơn mọi...
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Bài 1: Rút gọn biểu thức:
a) \(\left(5\sqrt{\dfrac{1}{5}}+\dfrac{1}{2}+\sqrt{20}-\dfrac{5}{4}\sqrt{\dfrac{4}{5}+\sqrt{5}}\right)\)
b) \(\dfrac{1}{3}\sqrt{48}+3\sqrt{75}-\sqrt{27}-10\sqrt{1\dfrac{1}{3}}\)
c) \(\dfrac{5\sqrt{7}-7\sqrt{5}+2\sqrt{70}}{\sqrt{35}}\)
d) \(\sqrt{\dfrac{3}{4}}+\sqrt{\dfrac{1}{3}}+\sqrt{\dfrac{1}{12}}\)
Bài 2: Giải các phương trình sau:
a) \(x^2+4x+5=2\sqrt{2x+3}\)
b) \(x^2+9x+20=2\sqrt{3x+10}\)
c) \(x^2+7x+14=2\sqrt{x+4}\)
d) \(4\sqrt{x+1}=x^2-5x+14\)
e) \(\sqrt{6-x}=3x-4\)
f) \(\sqrt{5x-9}=9-2x\)
Mọi người làm ơn giúp mình với. Mình đang cần gấp ạ. Cảm ơn mọi người rất nhiều
Bài 1:
a) Ta có: \(\left(5\sqrt{\dfrac{1}{5}}+\dfrac{1}{2}\sqrt{20}-\dfrac{5}{4}\sqrt{\dfrac{4}{5}}+\sqrt{5}\right)\)
\(=\left(\sqrt{5}+\sqrt{5}-\dfrac{5}{4}\cdot\dfrac{2}{\sqrt{5}}+\sqrt{5}\right)\)
\(=3\sqrt{5}-\dfrac{1}{2}\sqrt{5}\)
\(=\dfrac{5}{2}\sqrt{5}\)
c) Ta có: \(\dfrac{5\sqrt{7}-7\sqrt{5}+2\sqrt{70}}{\sqrt{35}}\)
\(=\dfrac{\sqrt{35}\left(\sqrt{5}-\sqrt{7}+2\sqrt{2}\right)}{\sqrt{35}}\)
\(=2\sqrt{2}+\sqrt{5}-\sqrt{7}\)
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Bài 2:
e) ĐKXĐ: \(\dfrac{4}{3}\le x\le6\)
Ta có: \(\sqrt{6-x}=3x-4\)
\(\Leftrightarrow6-x=\left(3x-4\right)^2\)
\(\Leftrightarrow9x^2-24x+16+6-x=0\)
\(\Leftrightarrow9x^2-25x+22=0\)
\(\Delta=\left(-25\right)^2-4\cdot9\cdot22=625-792< 0\)
Vậy: Phương trình vô nghiệm
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Tính:
\(A=3\sqrt{20}-\sqrt{45}+2\sqrt{18}+\sqrt{72}\)
\(B=\dfrac{12}{3-\sqrt{5}}-\dfrac{16}{\sqrt{5}+1}\)
\(C=10\sqrt{\dfrac{1}{5}}+\dfrac{1}{5}\sqrt{125}-2\sqrt{20}\)
\(E=\sqrt{\left(\sqrt{3}+1\right)^2}-\sqrt{\left(\sqrt{3}-2\right)^2}\)
\(F=\sqrt{6+2\sqrt{5}}-\sqrt{9-4\sqrt{5}}\)
a) Ta có: \(A=3\sqrt{20}-\sqrt{45}+2\sqrt{18}+\sqrt{72}\)
\(=6\sqrt{5}-3\sqrt{5}+6\sqrt{2}+6\sqrt{2}\)
\(=3\sqrt{5}+12\sqrt{2}\)
b) Ta có: \(B=\dfrac{12}{3-\sqrt{5}}-\dfrac{16}{\sqrt{5}+1}\)
\(=\dfrac{12\left(3+\sqrt{5}\right)}{4}-\dfrac{16\left(\sqrt{5}-1\right)}{4}\)
\(=3\left(3+\sqrt{5}\right)-4\left(\sqrt{5}-1\right)\)
\(=9+3\sqrt{5}-4\sqrt{5}+4\)
\(=13-\sqrt{5}\)
c) Ta có: \(C=10\sqrt{\dfrac{1}{5}}+\dfrac{1}{5}\sqrt{125}-2\sqrt{20}\)
\(=\dfrac{10}{\sqrt{5}}+\dfrac{1}{5}\cdot5\sqrt{5}-2\cdot2\sqrt{5}\)
\(=2\sqrt{5}+\sqrt{5}-4\sqrt{5}\)
\(=-\sqrt{5}\)
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e) Ta có: \(E=\sqrt{\left(\sqrt{3}+1\right)^2}-\sqrt{\left(\sqrt{3}-2\right)^2}\)
\(=\sqrt{3}+1-2+\sqrt{3}\)
\(=2\sqrt{3}-1\)
f) Ta có: \(F=\sqrt{6+2\sqrt{5}}-\sqrt{9-4\sqrt{5}}\)
\(=\sqrt{5}+1-\sqrt{5}+2\)
=3
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Tính:
\(A=3\sqrt{20}-\sqrt{45}+2\sqrt{18}+\sqrt{72}\)
\(B=\dfrac{12}{3-\sqrt{5}}-\dfrac{16}{\sqrt{5}+1}\)
\(C=10\sqrt{\dfrac{1}{5}}+\dfrac{1}{5}\sqrt{125}-2\sqrt{20}\)
\(E=\sqrt{\left(\sqrt{3}+1\right)^2}-\sqrt{\left(\sqrt{3}-2\right)^2}\)
\(F=\sqrt{6+2\sqrt{5}}-\sqrt{9-4\sqrt{5}}\)
e) Ta có: \(E=\sqrt{\left(\sqrt{3}+1\right)^2}-\sqrt{\left(\sqrt{3}-2\right)^2}\)
\(=\sqrt{3}+1-2+\sqrt{3}\)
\(=2\sqrt{3}-1\)
f) Ta có: \(F=\sqrt{6+2\sqrt{5}}-\sqrt{9-4\sqrt{5}}\)
\(=\sqrt{5}+1-\sqrt{5}+2\)
=3
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a) Ta có: \(A=3\sqrt{20}-\sqrt{45}+2\sqrt{18}+\sqrt{72}\)
\(=6\sqrt{5}-3\sqrt{5}+6\sqrt{2}+6\sqrt{2}\)
\(=3\sqrt{5}+12\sqrt{2}\)
b) Ta có: \(B=\dfrac{12}{3-\sqrt{5}}-\dfrac{16}{\sqrt{5}+1}\)
\(=\dfrac{12\left(3+\sqrt{5}\right)}{4}-\dfrac{16\left(\sqrt{5}-1\right)}{4}\)
\(=3\left(3+\sqrt{5}\right)-4\left(\sqrt{5}-1\right)\)
\(=9+3\sqrt{5}-4\sqrt{5}+4\)
\(=13-\sqrt{5}\)
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Tính: a, \(\left(4\sqrt{2}-\dfrac{11}{2}\sqrt{8}-\dfrac{1}{3}\sqrt{288}+\sqrt{50}\right)\left(\dfrac{1}{2}\sqrt{2}\right)\)
b, \(\left(\dfrac{4}{5}\sqrt{5}-\dfrac{1}{3}\sqrt{\dfrac{1}{5}}+3\sqrt{20}+\dfrac{1}{2}\sqrt{245}\right)\div\sqrt{5}\)
a: Ta có: \(\left(4\sqrt{2}-\dfrac{11}{2}\sqrt{8}-\dfrac{1}{3}\sqrt{288}+\sqrt{50}\right)\cdot\left(\dfrac{1}{2}\sqrt{2}\right)\)
\(=\dfrac{1}{2}\sqrt{2}\cdot\left(4\sqrt{2}-11\sqrt{2}-4\sqrt{2}+5\sqrt{2}\right)\)
\(=\dfrac{1}{2}\sqrt{2}\cdot6\sqrt{2}=3\)
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Giải 5 câu sau:
1. \(\dfrac{\sqrt{5}+2}{\sqrt{5}-2}\)
2. \(\dfrac{5\sqrt{2}-2\sqrt{5}}{\sqrt{2}-\sqrt{5}}\)
3. \(\dfrac{\sqrt{20}-3\sqrt{10}}{3-\sqrt{5}}\)
4. \(\dfrac{6-2\sqrt{5}}{3+\sqrt{5}}\)
5. \(\dfrac{9+4\sqrt{5}}{\sqrt{5}+2}\)
1) \(\dfrac{\sqrt{5}+2}{\sqrt{5}-2}=9+4\sqrt{5}\)
2) \(\dfrac{5\sqrt{2}-2\sqrt{5}}{\sqrt{2}-\sqrt{5}}=\dfrac{\sqrt{10}\left(\sqrt{5}-\sqrt{2}\right)}{-\left(\sqrt{5}-\sqrt{2}\right)}=-\sqrt{10}\)
3) \(\dfrac{\sqrt{20}-3\sqrt{10}}{3-\sqrt{5}}=\dfrac{\sqrt{10}\left(\sqrt{5}-3\right)}{-\left(\sqrt{5}-3\right)}=-\sqrt{10}\)
4) \(\dfrac{6-2\sqrt{5}}{3+\sqrt{5}}=\dfrac{\left(6-2\sqrt{5}\right)\left(3-\sqrt{5}\right)}{4}=\dfrac{18-6\sqrt{5}-6\sqrt{5}+10}{4}=\dfrac{28-12\sqrt{5}}{4}=7-3\sqrt{5}\)
5)\(\dfrac{9+4\sqrt{5}}{\sqrt{5}+2}=\sqrt{5}+2\)
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Tính
\(A=\sqrt{20}-3\sqrt{8}+5\sqrt{45}\)
\(B=\dfrac{30}{\sqrt{7}-1}+\dfrac{15}{\sqrt{7}+2}\)
\(C=\left(3-\dfrac{5-\sqrt{5}}{\sqrt{5}-1}\right)\left(3+\dfrac{5+\sqrt{5}}{\sqrt{5}+1}\right)\)
\(D=\sqrt{\left(3-\sqrt{2}\right)^2}-\sqrt{\left(1-\sqrt{2}\right)^2}\)
\(E=\sqrt{7-4\sqrt{3}}-\sqrt{3+2\sqrt{3}}\)
1) \(A=2\sqrt{5}-6\sqrt{2}+3\sqrt{5}=5\sqrt{5}-6\sqrt{2}\)
2) \(B=\dfrac{30\left(\sqrt{7}+1\right)}{7-1}+\dfrac{15\left(\sqrt{7}-2\right)}{7-4}=5\sqrt{7}+5+5\sqrt{7}-10=-5+10\sqrt{7}\)
3) \(C=\left(3-\dfrac{\sqrt{5}\left(\sqrt{5}-1\right)}{\sqrt{5}-1}\right)\left(3+\dfrac{\sqrt{5}\left(\sqrt{5}+1\right)}{\sqrt{5}+1}\right)=\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)=9-5=4\)
4) \(D=3-\sqrt{2}+1-\sqrt{2}=4-2\sqrt{2}\)
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