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\(\sqrt{6+2\sqrt{2}.\sqrt{3-\sqrt{\sqrt{2}+2\sqrt{3}+\sqrt{18-8\sqrt{2}}}}}-\sqrt{3}\)\(=\sqrt{6+2.1,4.\sqrt{3-\sqrt{1,4+2.1,7+\sqrt{18-8.1,4\text{}}}}}-1,7\)
\(=\sqrt{6+2,8\sqrt{3-\sqrt{1,4+3,4+\sqrt{18-11,2}}}}-1,7\)
\(=\sqrt{8,8\sqrt{3-\sqrt{4,8+\sqrt{6,8}}}}-1,7\)
\(=\sqrt{8,8\sqrt{3-\sqrt{4,8+2,6}}}-1,7\)
\(=\sqrt{8,8\sqrt{3-\sqrt{7,4}}}-1,7\)
\(=\sqrt{8,8\sqrt{3-2,7}}-1,7\)
\(=\sqrt{88\sqrt{0,3}}-1,7\)
\(=\sqrt{88.0,54}-1,7\)
\(=\sqrt{47,52}-1,7\)
\(=6,9-1,7\)
\(=5,2\)
2,Mệt với câu 1 rồi nên câu 2 và câu 3 chịu
hình như sai rồi bạn ơi, lúc học thì thầy mình giải ra kết quả =1 và ko tính căn ra như thế
b/A=\(\frac{x-2\sqrt{x}-3-3\sqrt{x}+9}{x-2\sqrt{x}-3}=1-\frac{3\left(\sqrt{x}-3\right)}{\left(1+\sqrt{x}\right)\left(\sqrt{x}-3\right)}=1-\frac{3}{1+\sqrt{x}}\)
Vậy 1+ căn x thuốc Ư(3), mà \(\sqrt{x}\ge0\Rightarrow1+\sqrt{x}\ge1\)
Vậy \(1+\sqrt{x}=\left(1,3\right)\)
\(\Rightarrow\sqrt{x}=\left(0,2\right)\) Vì x nguyên nên x=0
\(\Leftrightarrow A=\frac{1+\sqrt{x}-\sqrt{x}}{1+\sqrt{x}}:\left(\frac{\sqrt{x}+3}{\sqrt{x}-2}-\frac{\sqrt{x}+2}{\sqrt{x}-3}+\frac{\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\right)\)
\(\Leftrightarrow\frac{1}{1+\sqrt{x}}:\frac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)-\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(\Leftrightarrow A=\frac{1}{1+\sqrt{x}}.\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}{x-9-x+4+\sqrt{x}+2}\)
\(\Leftrightarrow A=\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}{\left(1+\sqrt{x}\right)\left(\sqrt{x}-3\right)}\)
\(\Leftrightarrow A=\frac{x-5\sqrt{x}+6}{x-2\sqrt{x}-3}\)
Áp dụng bđt AM-GM cho 2 số dương ta có:
\(\left(\frac{1}{\sqrt{2x-3}}+\sqrt{2x-3}\right)+\left(\frac{4}{\sqrt{y-2}}+\sqrt{y-2}\right)\)\(+\left(\frac{16}{\sqrt{3z-1}}+\sqrt{3z-1}\right)\ge\)\(2\sqrt{\frac{1}{\sqrt{2x-3}}.\sqrt{2x-3}}+2\sqrt{\frac{4}{\sqrt{y-2}}.\sqrt{y-2}}\)\(+2\sqrt{\frac{16}{\sqrt{3z-1}}.\sqrt{3z-1}}=2.1+2.2+2.4=14\)
Dau "=" xay ra khi \(\left\{\begin{matrix}\frac{1}{\sqrt{2x-3}}=\sqrt{2x-3}\\\frac{4}{\sqrt{y-2}}=\sqrt{y-2}\\\frac{16}{\sqrt{3z-1}}=\sqrt{3z-1}\end{matrix}\right.\)\(\Rightarrow\left\{\begin{matrix}2x-3=1\\y-2=4\\3z-1=16\end{matrix}\right.\)=> \(\left\{\begin{matrix}x=1\\y=6\\z=\frac{17}{3}\end{matrix}\right.\) (không TM z nguyên dương)
Vay ...
Cái này là toán lớp 9 chứ.
a)
ĐKXĐ : \(x\ne\pm4\)
\(A=\left(\frac{x-\sqrt{x}+7}{x-4}+\frac{\sqrt{x}+2}{x-4}\right):\left(\frac{\left(\sqrt{x}+2\right)^2}{x-4}-\frac{\left(\sqrt{x}-2\right)^2}{x-4}-\frac{2\sqrt{x}}{x-4}\right)\)
\(=\left(\frac{x-\sqrt{x}+7+\sqrt{x}+2}{x-4}\right):\left(\frac{x+4\sqrt{x}+4-x+4\sqrt{x}-4-2\sqrt{x}}{x-4}\right)\)
\(=\frac{x+9}{x-4}\cdot\frac{x-4}{6\sqrt{x}}=\frac{x+9}{6\sqrt{x}}\)
b)
Ta có
\(x+9-6\sqrt{x}=\left(\sqrt{x}-3\right)^2\ge0\)
\(\Rightarrow x+9\ge6\sqrt{x}\)
\(\Rightarrow\frac{x+9}{6\sqrt{x}}\ge1\)
\(\Leftrightarrow A\ge1\)
\(\Leftrightarrow\frac{1}{A}\le1\)
\(\Rightarrow A\ge\frac{1}{A}\)
ĐKXĐ : \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)
Ta có :
\(A=\frac{\sqrt{x}+4}{\sqrt{x}+1}-\frac{3}{x-1}:\frac{1}{\sqrt{x}-1}\)
\(=\frac{\sqrt{x}+4}{\sqrt{x}+1}-\frac{3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}.\left(\sqrt{x}-1\right)\)
\(=\frac{\sqrt{x}+4}{\sqrt{x}+1}-\frac{3}{\sqrt{x}+1}\)
\(=\frac{\sqrt{x}+1}{\sqrt{x}+1}\)
\(=1\)
Vậy...
b/ ĐKXĐ : \(\left\{{}\begin{matrix}x\ge0\\x\ne4\end{matrix}\right.\)
Ta có :
\(B=\left(\frac{x-4\sqrt{x}+4}{\sqrt{x}-2}+6\right)\left(\frac{x\sqrt{x}-1}{x+\sqrt{x}+1}-3\right)\)
\(=\left(\frac{\left(\sqrt{x}-2\right)^2}{\sqrt{x}-2}+6\right)\left(\frac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{x+\sqrt{x}+1}-3\right)\)
\(=\left(\sqrt{x}-2+6\right)\left(\sqrt{x}-1-3\right)\)
\(=\left(\sqrt{x}+4\right)\left(\sqrt{x}-4\right)\)
\(=x-16\)
Vậy..
c/ ĐKXĐ : \(\left\{{}\begin{matrix}x>0\\x\ne1\end{matrix}\right.\)
Ta có :
\(C=\frac{2\sqrt{x}}{x-1}+\frac{1}{x+\sqrt{x}}+\frac{1}{\sqrt{x}-x}\)
\(=\frac{2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\frac{1}{\sqrt{x}\left(\sqrt{x}+1\right)}-\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}\)
\(=\frac{2x}{\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\frac{\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{2x+\sqrt{x}-1-\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{2x-2}{\sqrt{x}\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(=\frac{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(=\frac{2}{\sqrt{x}}\)
Vậy..
Ta phải có m , n > 0 để m/n > 0 và n/m > 0 ta được:
\(\sqrt{x^2-4}=\sqrt{\frac{\left(m-n\right)^2}{mn}}=\frac{|m-n|}{\sqrt{mn}}\)
\(A=\frac{2n.\frac{|m-n|}{\sqrt{mn}}}{\left(\sqrt{\frac{m}{n}}+\sqrt{\frac{n}{m}}\right)-\frac{|m-n|}{\sqrt{mn}}}\)
\(=\frac{2n|m-n|}{\sqrt{mn}\left(\sqrt{\frac{m}{n}}+\sqrt{\frac{n}{m}}\right)-|m-n|}\)
\(=\frac{2n|m-n|}{\left(\sqrt{m^2}+\sqrt{n^2}\right)-|m-n|}\)
Đến đây ta xét hai trường hợp:
+ TH1: m > 0 và n > 0
Khi đó \(\sqrt{m^2}+\sqrt{n^2}=m+n\)
và \(A=\frac{2n.|m-n|}{m+n-|m-n|}\)
Nếu \(m\ge n>0\Rightarrow|m-n|=n-m\) do đó: A = m - n
Nếu \(0< m< n\Rightarrow|m-n|=n-m\) do đó\(A=\frac{n\left(n-m\right)}{m}\)
Còn TH2: m < 0 ; n < 0 bạn tự giải nốt:vv
Bé Mon: Giải hết luôn trường hợp 2 cho mình đi