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a, \(\frac{2+\sqrt{3}}{2+\sqrt{4+2\sqrt{3}}}+\frac{2-\sqrt{3}}{2-\sqrt{4-2\sqrt{3}}}\)
\(=\frac{2+\sqrt{3}}{2+\sqrt{\left(\sqrt{3}+1\right)^2}}+\frac{2-\sqrt{3}}{2-\sqrt{\left(\sqrt{3}-1\right)^2}}\)
\(=\frac{2+\sqrt{3}}{2+\sqrt{3}+1}+\frac{2-\sqrt{3}}{2-\sqrt{3}+1}\)
\(=\frac{2+\sqrt{3}}{3+\sqrt{3}}+\frac{2-\sqrt{3}}{3-\sqrt{3}}\)
\(=\frac{\left(2+\sqrt{3}\right)\left(3-\sqrt{3}\right)+\left(2-\sqrt{3}\right)\left(3+\sqrt{3}\right)}{\left(3+\sqrt{3}\right)\left(3-\sqrt{3}\right)}\)
\(=\frac{6+\sqrt{3}-3+6-\sqrt{3}-3}{9-3}=\frac{6}{6}=1\)
b, \(\frac{1}{x+\sqrt{x}}+\frac{2\sqrt{x}}{x-1}-\frac{1}{x-\sqrt{x}}\)
\(=\frac{1}{\sqrt{x}\left(\sqrt{x}+1\right)}+\frac{2\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x-1}\right)}-\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}\)
\(=\frac{\sqrt{x}-1+2x-\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}=\frac{2\left(x-1\right)}{\sqrt{x}\left(x-1\right)}=\frac{2}{\sqrt{x}}\)
Giải pt :
1
a. ĐKXĐ : \(x\ge4\)
Ta có :
\(\sqrt{x+3}-\sqrt{x-4}=1\\ \Leftrightarrow\sqrt{x+3}=1+\sqrt{x-4}\\ \Leftrightarrow x+3=x-3+2\sqrt{x-4}\\ \Leftrightarrow6=2\sqrt{x-4}\)
\(\Leftrightarrow3=\sqrt{x-4}\\ \Leftrightarrow x-4=9\)
\(\Leftrightarrow x=13\) (TM ĐKXĐ)
Vậy \(S=\left\{13\right\}\)
b.ĐKXĐ : \(-3\le x\le10\)
Ta có :
\(\sqrt{10-x}+\sqrt{x+3}=5\\ \Leftrightarrow13+2\sqrt{-x^2+7x+30}=25\\ \Leftrightarrow\sqrt{-x^2+7x+30}=6\\ \Leftrightarrow-x^2+7x+30=36\\ \Leftrightarrow-x^2+7x-6=0\\ \Leftrightarrow-x^2+x+6x-6=0\\ \Leftrightarrow-x\left(x-1\right)+6\left(x-1\right)=0\\ \Leftrightarrow\left(x-1\right)\left(6-x\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\left(TMĐKXĐ\right)\\x=6\left(TMĐKXĐ\right)\end{matrix}\right.\)
Vậy \(S=\left\{1;6\right\}\)
mầy câu 1;3;;4;5 cách làm nhu nhau(nhân liên hop hoac bình phuong lên)
1.
\(DK:x\in\left[-4;5\right]\)
\(\Leftrightarrow\sqrt{x-5}+\left(\sqrt{x+4}-3\right)=0\)
\(\Leftrightarrow\sqrt{x-5}+\frac{x-5}{\sqrt{x+4}+3}=0\)
\(\Leftrightarrow\sqrt{x-5}\left(1+\frac{\sqrt{x-5}}{\sqrt{x+4}+3}\right)=0\)
Vi \(1+\frac{\sqrt{x-5}}{\sqrt{x+4}+3}>0\)
\(\Rightarrow\sqrt{x-5}=0\)
\(x=5\left(n\right)\)
Vay nghiem cua PT la \(x=5\)
2.
\(DK:x\ge0\)
\(\Leftrightarrow\sqrt{\left(\sqrt{x}-2\right)^2}+\sqrt{\left(\sqrt{x}-3\right)^2}=1\)
\(\Leftrightarrow|\sqrt{x}-2|+|\sqrt{x}-3|=1\)
Ta co:
\(|\sqrt{x}-2|+|\sqrt{x}-3|=|\sqrt{x}-2|+|3-\sqrt{x}|\ge|\sqrt{x}-2+3-\sqrt{x}|=1\)
Dau '=' xay ra khi \(\left(\sqrt{x}-2\right)\left(3-\sqrt{x}\right)\ge0\)
TH1:
\(\hept{\begin{cases}\sqrt{x}-2\ge0\\3-\sqrt{x}\ge0\end{cases}\Leftrightarrow4\le x\le9\left(n\right)}\)
TH2:(loai)
Vay nghiem cua PT la \(x\in\left[4;9\right]\)
b) Nhẩm thấy \(x=-2\) là nghiệm, ta xét trường hợp:
* Với \(x>-2\) thì
\(\sqrt[3]{x+1}+\sqrt[3]{x+2}+\sqrt[3]{x+3}>-1+0+1=0=VP\)
* Với \(x< -2\) thì
\(\sqrt[3]{x+1}+\sqrt[3]{x+2}+\sqrt[3]{x+3}< -1+0+1=0=VP\)
Do đó pt có nghiệm duy nhất \(x=-2\)
c) Đặt \(\sqrt[4]{1-x}=a;\sqrt[4]{1+x}=b\)
\(\Rightarrow a^4+b^4=2\)
Theo đề bài \(a+b+ab=3\Rightarrow a+b=3-ab\)
Cần giải cái hệ (đợi một xíu em ăn xong em làm tiếp hoặc là nếu bận thì thứ 6 tuần này em làm):v \(\left\{{}\begin{matrix}a^4+b^4=3\\a+b=3-ab\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left(a^2+b^2\right)^2=3+2a^2b^2\\ab=3-a-b\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left[\left(a+b\right)^2-2ab\right]^2=3+2\left(3-a-b\right)^2\\ab=3-a-b\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left[\left(a+b\right)^2-2\left(3-a-b\right)\right]^2=3+2\left(3-a-b\right)^2\\ab=3-a-b\end{matrix}\right.\)
b) \(< =>\sqrt{x+1}\left(\sqrt{x^2-x+1}+\sqrt{x+3}\right)=0\)
<=> x=-1
hoặc \(x^2-x+1=x+3\) => \(x^2-2x-2=0...\)
a,
\(\sqrt{\sqrt{3}+2\sqrt{\sqrt{3}-1}}+\sqrt{\sqrt{3}-2\sqrt{\sqrt{3}-1}}\\ =\sqrt{\sqrt{3}-1+2\sqrt{\sqrt{3}-1}+1}+\sqrt{\sqrt{3}-1-2\sqrt{\sqrt{3}-1}+1}\\ =\sqrt{\left(\sqrt{\sqrt{3}-1}+1\right)^2}+\sqrt{\left(1-\sqrt{\sqrt{3}-1}\right)^2}\\ =\sqrt{\sqrt{3}-1}+1+1-\sqrt{\sqrt{3}-1}\\ =2\)
b.
\(\sqrt{x-3-2\sqrt{x-4}}-\sqrt{x-4\sqrt{x-4}}\\ =\sqrt{x-4-2\sqrt{x-4}+1}-\sqrt{x-4-4\sqrt{x-4}+4}\\ =\sqrt{\left(\sqrt{x-4}-1\right)^2}-\sqrt{\left(\sqrt{x-4}-2\right)^2}\\ =\sqrt{x-4}-1-\sqrt{x-4}+2\\ =1\left(đpcm\right)\)\