Các bn giúp mk vs mk đang vội mang mk phải kiểm tra rùi!!!
\(\left(\sqrt{2}+1\right)^3-\left(\sqrt{2}-1\right)^3\)
K
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19 tháng 10 2021
\(a,=\sqrt{5}\left(2\sqrt{5}-3\right)+3\sqrt{5}=10-3\sqrt{5}+3\sqrt{5}=10\\ b,=5-\sqrt{3}-\left(2-\sqrt{3}\right)=3\\ c,=\dfrac{2\left(\sqrt{5}-1\right)}{4}-\dfrac{2\left(3+\sqrt{5}\right)}{4}=\dfrac{2\sqrt{5}-2-6-2\sqrt{5}}{4}=\dfrac{-8}{4}=-2\)
S
22 tháng 8 2021
a. \(P=\left(\dfrac{\sqrt{x}}{\sqrt{x}-2}-\dfrac{4}{x-2\sqrt{x}}\right)\cdot\left(\dfrac{1}{\sqrt{x}+2}+\dfrac{4}{x-4}\right)\)
<=> \(P=\left(\dfrac{\sqrt{x}}{\sqrt{x}-2}-\dfrac{4}{\sqrt{x}\left(\sqrt{x}-2\right)}\right)\cdot\dfrac{\sqrt{x}-2+4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
<=> \(P=\dfrac{x-4}{\sqrt{x}\left(\sqrt{x}-2\right)}\cdot\dfrac{\sqrt{x}+2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
<=> \(P=\dfrac{\sqrt{x}+2}{x-2\sqrt{x}}\)
b. Khi \(x=7+4\sqrt{3}=\left(2+\sqrt{3}\right)^2\) => \(\sqrt{x}=2+\sqrt{3}\)
=> \(P=\dfrac{2+\sqrt{3}+2}{7+4\sqrt{3}-2\left(2+\sqrt{3}\right)}=\dfrac{4+\sqrt{3}}{7+4\sqrt{3}-4-2\sqrt{3}}=\dfrac{4+\sqrt{3}}{3+2\sqrt{3}}=\dfrac{5\sqrt{3}-6}{3}\)
check giùm mik
NC
29 tháng 9 2020
Ta có: \(\frac{1}{3}\left(\sqrt{6}+\sqrt{5}\right)^2-\frac{1}{4}\sqrt{120}-2\sqrt{\frac{15}{2}}\)
\(=\frac{1}{3}\left(11+2\sqrt{30}\right)-\frac{\sqrt{30}}{2}-\sqrt{30}\)
\(=\frac{11}{3}+\frac{2}{3}\sqrt{30}-\frac{\sqrt{30}}{2}-\sqrt{30}\)
\(=\frac{11}{3}-\frac{5}{6}\sqrt{30}\)
\(=\frac{22-5\sqrt{30}}{6}\)
NC
29 tháng 9 2020
Ta có: \(\left(\frac{1}{2}\sqrt{\frac{2}{3}}-\frac{3}{4}\sqrt{54}+\frac{1}{3}\sqrt{\frac{8}{3}}\right)\div\sqrt{\frac{81}{6}}\)
\(=\left(\frac{\sqrt{6}}{6}-\frac{9\sqrt{6}}{4}+\frac{2\sqrt{6}}{9}\right)\div\frac{3\sqrt{6}}{2}\)
\(=-\frac{67\sqrt{6}}{36}\cdot\frac{2}{3\sqrt{6}}\)
\(=-\frac{67}{54}\)
5 tháng 10 2021
Sửa đề: \(C=\left(\dfrac{1}{\sqrt{a}-1}-\dfrac{1}{\sqrt{a}}\right):\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-1}\right)\)
\(a,C=\dfrac{\sqrt{a}-\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\dfrac{a-1-a+4}{\left(\sqrt{a}-1\right)\left(\sqrt{a}-2\right)}\left(a>0;a\ne1;a\ne4\right)\\ C=\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{\left(\sqrt{a}-1\right)\left(\sqrt{a}-2\right)}{3}=\dfrac{\sqrt{a}-2}{3\sqrt{a}}\\ b,C\ge\dfrac{1}{6}\Leftrightarrow\dfrac{\sqrt{a}-2}{3\sqrt{a}}-\dfrac{1}{6}\ge0\Leftrightarrow\dfrac{\sqrt{a}-4}{6\sqrt{a}}\ge0\\ \Leftrightarrow\sqrt{a}-4\ge0\left(6\sqrt{a}>0\right)\\ \Leftrightarrow a\ge16\)
AH
Akai Haruma
Giáo viên
26 tháng 8 2021
Lời giải:
ĐKXĐ: $x>0$
a. \(P=\frac{x-1}{\sqrt{x}}:\left[\frac{(\sqrt{x}-1)(\sqrt{x}+1)}{\sqrt{x}(\sqrt{x}+1)}+\frac{1-\sqrt{x}}{\sqrt{x}(\sqrt{x}+1)}\right]\)
\(=\frac{x-1}{\sqrt{x}}:\frac{x-1+1-\sqrt{x}}{\sqrt{x}(\sqrt{x}+1)}=\frac{x-1}{\sqrt{x}}:\frac{\sqrt{x}(\sqrt{x}-1)}{\sqrt{x}(\sqrt{x}+1)}=\frac{(\sqrt{x}-1)(\sqrt{x}+1)}{\sqrt{x}}.\frac{\sqrt{x}+1}{\sqrt{x}-1}\)
\(=\frac{(\sqrt{x}+1)^2}{\sqrt{x}}\)
b.
\(x=\frac{4}{4+2\sqrt{3}}=(\frac{2}{\sqrt{3}+1})^2\Rightarrow \sqrt{x}=\frac{2}{\sqrt{3}+1}\)
\(P=\frac{(\frac{2}{\sqrt{3}+1}+1)^2}{\frac{2}{\sqrt{3}+1}}=\frac{3+3\sqrt{3}}{2}\)
26 tháng 8 2021
a: Ta có: \(P=\left(\sqrt{x}-\dfrac{1}{\sqrt{x}}\right):\left(\dfrac{\sqrt{x}-1}{\sqrt{x}}+\dfrac{1-\sqrt{x}}{x+\sqrt{x}}\right)\)
\(=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}}:\dfrac{x-1+1-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}\)
\(=\dfrac{x+2\sqrt{x}+1}{\sqrt{x}}\)
25 tháng 2 2022
\(A=\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}+\dfrac{1}{x-\sqrt{x}}\right):\left(\dfrac{1}{\sqrt{x}+1}+\dfrac{2}{x-1}\right)\)
\(\Rightarrow A=\left(\dfrac{x}{\sqrt{x}\left(\sqrt{x}-1\right)}+\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}\right):\left(\dfrac{\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\dfrac{2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\)
\(\Rightarrow A=\dfrac{x+1}{\sqrt{x}\left(\sqrt{x}-1\right)}:\dfrac{\sqrt{x}-1+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(\Rightarrow A=\dfrac{x+1}{\sqrt{x}\left(\sqrt{x}-1\right)}:\dfrac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(\Rightarrow A=\dfrac{x+1}{\sqrt{x}\left(\sqrt{x}-1\right)}.\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}+1}\)
\(\Rightarrow A=\dfrac{x+1}{\sqrt{x}}\)
30 tháng 9 2016
a)\(\left(\frac{1}{5}\right)^{3n-1}=\frac{1}{25}\)
\(\Leftrightarrow\left(\frac{1}{5}\right)^{3n-1}=\left(\frac{1}{5}\right)^2\)
\(\Leftrightarrow3n-1=2\)
\(\Leftrightarrow3n=3\)
\(\Leftrightarrow n=1\)
b)\(\left(\frac{4}{7}\right)^{n+2}=\frac{7}{4}\)
\(\Leftrightarrow\left(\frac{4}{7}\right)^{n+2}=\left(\frac{4}{7}\right)^{-1}\)
\(\Leftrightarrow n+2=-1\)
\(\Leftrightarrow n=-3\)
c)\(\left(\frac{2}{3}\right)^{-n+1}=\frac{3^3}{2^3}\)
\(\Leftrightarrow\left(\frac{2}{3}\right)^{-n+1}=\left(\frac{3}{2}\right)^3\)
\(\Leftrightarrow\left(\frac{2}{3}\right)^{-n+1}=\left(\frac{2}{3}\right)^{-3}\)
\(\Leftrightarrow-n+1=-3\)
\(\Leftrightarrow n=-4\)
c)\(\left(0,7\right)^{3n+1}=10^3:7^3\)
\(\Leftrightarrow\left(\frac{7}{10}\right)^{3n+1}=\left(\frac{10}{7}\right)^3\)
\(\Leftrightarrow\left(\frac{7}{10}\right)^{3n+1}=\left(\frac{7}{10}\right)^{-3}\)
\(\Leftrightarrow3n+1=-3\)
\(\Leftrightarrow3n=-4\)
\(\Leftrightarrow n=-\frac{4}{3}\)
\(\left(\sqrt{2}+1\right)^3-\left(\sqrt{2}-1\right)^3\)
\(=\left(\sqrt{2}+1-\sqrt{2}+1\right)\left[\left(\sqrt{2}+1\right)^2+\left(\sqrt{2}-1\right)^2+\left(\sqrt{2}+1\right)\left(\sqrt{2}-1\right)\right]\)
\(=2\cdot\left(2+2\sqrt{2}+1+2-2\sqrt{2}+1+2-1\right)\)
\(=2\cdot7\)
\(=14\)
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