\(\frac{2+\sqrt{2}}{1+\sqrt{2}}+\sqrt{\left(2-\sqrt{2}\right)^2}\)
Tính giúp mk vs ạ
K
Khách
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28 tháng 11 2016
Ta có
\(x=\frac{\sqrt{4+2\sqrt{3}}-\sqrt{3}}{\left(\sqrt{5}+2\right)\sqrt[3]{17\sqrt{5}-38}-2}\)
\(=\frac{\sqrt{3+2\sqrt{3}+1}-\sqrt{3}}{\left(\sqrt{5}+2\right)\sqrt[3]{5\sqrt{5}-3.5.2+3.4.\sqrt{5}-8}-2}\)
\(=\frac{\sqrt{3}+1-\sqrt{3}}{\left(\sqrt{5}+2\right)\left(\sqrt{5}-2\right)-2}=\frac{1}{5-4-2}=-1\)
Thế vào ta được
\(P=\left(x^2+x+1\right)^{2013}+\left(x^2+x-1\right)^{2013}\)
\(=\left(1-1+1\right)^{2013}+\left(1-1-1\right)^{2013}=1-1=0\)
19 tháng 10 2021
\(a,=\sqrt{5}\left(2\sqrt{5}-3\right)+3\sqrt{5}=10-3\sqrt{5}+3\sqrt{5}=10\\ b,=5-\sqrt{3}-\left(2-\sqrt{3}\right)=3\\ c,=\dfrac{2\left(\sqrt{5}-1\right)}{4}-\dfrac{2\left(3+\sqrt{5}\right)}{4}=\dfrac{2\sqrt{5}-2-6-2\sqrt{5}}{4}=\dfrac{-8}{4}=-2\)
1 tháng 12 2021
a,ĐKXĐ:\(x\ge2\)
\(4\sqrt{x-2}+\sqrt{9x-18}-\sqrt{\dfrac{x-2}{4}}=26\\ \Leftrightarrow4\sqrt{x-2}+3\sqrt{x-2}-\dfrac{\sqrt{x-2}}{2}=26\\ \Leftrightarrow8\sqrt{x-2}+6\sqrt{x-2}-\sqrt{x-2}=52\\ \Leftrightarrow13\sqrt{x-2}=52\\ \Leftrightarrow\sqrt{x-2}=4\\ \Leftrightarrow x-2=16\\ \Leftrightarrow x=18\left(tm\right)\)
b,ĐKXĐ:\(x\in R\)
\(3x+\sqrt{4x^2-8x+4}=1\\ \Leftrightarrow2\sqrt{x^2-2x+1}=1-3x\\ \Leftrightarrow\left|x-1\right|=\dfrac{1-3x}{2}\\ \Leftrightarrow\left[{}\begin{matrix}x-1=\dfrac{1-3x}{2}\\x-1=\dfrac{3x-1}{2}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}2x-2=1-3x\\2x-2=3x-1\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{5}\left(tm\right)\\x=-1\left(tm\right)\end{matrix}\right.\)
c, ĐKXĐ:\(x\ge0\)
\(\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)=7\\ \Leftrightarrow\sqrt{x}\left(2\sqrt{x}+1\right)-2\left(2\sqrt{x}+1\right)=7\\ \Leftrightarrow2x+\sqrt{x}-4\sqrt{x}-2=7\\ \Leftrightarrow2x-3\sqrt{x}-9=0\\ \Leftrightarrow\left(2x+3\sqrt{x}\right)-\left(6\sqrt{x}+9\right)=0\\ \Leftrightarrow\sqrt{x}\left(2\sqrt{x}+3\right)-3\left(2\sqrt{x}+3\right)=0\\ \Leftrightarrow\left(\sqrt{x}-3\right)\left(2\sqrt{x}+3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=3\\2\sqrt{x}=-3\left(vô.lí\right)\end{matrix}\right.\\ \Leftrightarrow x=9\left(tm\right)\)
MA
1
AH
Akai Haruma
Giáo viên
9 tháng 5 2021
Lời giải:
\(K=\sqrt{4+|1-\sqrt{5}|}.(\sqrt{10}-\sqrt{2})=\sqrt{4+\sqrt{5}-1}.\sqrt{2}(\sqrt{5}-1)\)
\(=\sqrt{3+\sqrt{5}}.\sqrt{2}.(\sqrt{5}-1)=\sqrt{6+2\sqrt{5}}.(\sqrt{5}-1)\)
\(=\sqrt{(\sqrt{5}+1)^2}(\sqrt{5}-1)=(\sqrt{5}+1)(\sqrt{5}-1)=4\)
28 tháng 12 2019
- ĐKXĐ : \(\left\{{}\begin{matrix}a\ge0\\\sqrt{a}-1\ne0\\\sqrt{a}+1\ne0\\2\sqrt{a}\ne0\end{matrix}\right.\)=> \(\left\{{}\begin{matrix}a\ne0\\a\ge0\\a\ne1\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}a>0\\a\ne1\end{matrix}\right.\)
- Ta có phương trình : \(\left(\frac{\sqrt{a}}{2}-\frac{1}{2\sqrt{a}}\right)\left(\frac{a-\sqrt{a}}{\sqrt{a}+1}-\frac{a+\sqrt{a}}{\sqrt{a}-1}\right)\)
=\(\left(\frac{a}{2\sqrt{a}}-\frac{1}{2\sqrt{a}}\right)\left(\frac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}+1}-\frac{\sqrt{a}\left(\sqrt{a}+1\right)}{\sqrt{a}-1}\right)\)
= \(\left(\frac{a-1}{2\sqrt{a}}\right)\left(\frac{\sqrt{a}\left(\sqrt{a}-1\right)^2}{a-1}-\frac{\sqrt{a}\left(\sqrt{a}+1\right)^2}{a-1}\right)\)
= \(\left(\frac{a-1}{2\sqrt{a}}\right)\left(\frac{\sqrt{a}\left(\sqrt{a}-1\right)^2-\sqrt{a}\left(\sqrt{a}+1\right)^2}{a-1}\right)\)
= \(\left(\frac{a-1}{2\sqrt{a}}\right)\left(\frac{\sqrt{a}\left(\left(\sqrt{a}-1\right)^2-\left(\sqrt{a}+1\right)^2\right)}{a-1}\right)\)
= \(\left(\frac{a-1}{2\sqrt{a}}\right)\left(\frac{\sqrt{a}\left(a-2\sqrt{a}+1-a-2\sqrt{a}-1\right)}{a-1}\right)\)
= \(\left(\frac{a-1}{2\sqrt{a}}\right)\left(\frac{\sqrt{a}\left(-4\sqrt{a}\right)}{a-1}\right)\)
= \(\left(\frac{a-1}{2\sqrt{a}}\right)\left(\frac{-4a}{a-1}\right)\)= \(\frac{-4a\left(a-1\right)}{2\sqrt{a}\left(a-1\right)}\) = \(\frac{-4a}{2\sqrt{a}}\)
= \(\frac{-4\sqrt{a}\sqrt{a}}{2\sqrt{a}}\) = \(-2\sqrt{a}\)
NV
Nguyễn Việt Lâm
Giáo viên
9 tháng 10 2019
ĐKXĐ: \(y\ge0;y\ne4;9\)
\(A=\left(\frac{8\sqrt{y}-4y+8y}{\left(2+\sqrt{y}\right)\left(2-\sqrt{y}\right)}\right):\left(\frac{\sqrt{y}-1}{\sqrt{y}\left(\sqrt{y}-2\right)}-\frac{2\left(\sqrt{y}-2\right)}{\sqrt{y}\left(\sqrt{y}-2\right)}\right)\)
\(=\left(\frac{4\sqrt{y}\left(2+\sqrt{y}\right)}{\left(2+\sqrt{y}\right)\left(2-\sqrt{y}\right)}\right):\left(\frac{-\sqrt{y}+3}{\sqrt{y}\left(\sqrt{y}-2\right)}\right)\)
\(=\left(\frac{4\sqrt{y}}{2-\sqrt{y}}\right):\left(\frac{\sqrt{y}-3}{\sqrt{y}\left(2-\sqrt{y}\right)}\right)\)
\(=\frac{4\sqrt{y}}{\left(2-\sqrt{y}\right)}.\frac{\sqrt{y}\left(2-\sqrt{y}\right)}{\left(\sqrt{y}-3\right)}=\frac{4y}{\sqrt{y}-3}\)
\(A=-2\Leftrightarrow\frac{4y}{\sqrt{y}-3}=-2\)
\(\Rightarrow2y=-\sqrt{y}+3\Rightarrow2y+\sqrt{y}-3=0\)
\(\Rightarrow\left[{}\begin{matrix}\sqrt{y}=1\\\sqrt{y}=-\frac{3}{2}< 0\left(l\right)\end{matrix}\right.\) \(\Rightarrow y=1\)
16 tháng 9 2021
\(A=\left(\dfrac{1}{\sqrt{a}-1}-\dfrac{1}{\sqrt{a}}\right):\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-1}\right)\left(đk:a>0,a\ne1\right)\)
\(=\dfrac{\sqrt{a}-\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\dfrac{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)-\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\)
\(=\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}.\dfrac{\left(\sqrt{a}-1\right)\left(\sqrt{a}-2\right)}{a-1-a+2}=\dfrac{1}{\sqrt{a}}.\dfrac{\sqrt{a}-2}{1}=\dfrac{\sqrt{a}-2}{\sqrt{a}}\)
Để A nguyên
\(\Leftrightarrow A=\dfrac{\sqrt{a}-2}{\sqrt{a}}=1-\dfrac{2}{\sqrt{a}}\in Z\)
Do \(\sqrt{a}>0,\sqrt{a}\ne1\)
\(\Leftrightarrow\sqrt{a}\inƯ\left(2\right)=\left\{2\right\}\)
\(\Leftrightarrow a=4\)
\(\frac{2+\sqrt{2}}{1+\sqrt{2}}+\sqrt{\left(2-\sqrt{2}\right)^2}\)
\(=\frac{\sqrt{2}\left(\sqrt{2}+1\right)}{1+\sqrt{2}}+2-\sqrt{2}\)do \(2-\sqrt{2}>0\)
\(=\sqrt{2}+2-\sqrt{2}=2\)