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Đk:\(-4\le x\le1.\)
Đặt \(\sqrt{1-x}=a,\sqrt{4+x}=b.\)
\(\Rightarrow\hept{\begin{cases}a+b=3\\a^2+b^2=5\end{cases}\Leftrightarrow\hept{\begin{cases}\left(a+b\right)^2=9\\a^2+b^2=5\end{cases}\Rightarrow}ab=2\Rightarrow\left(a-b\right)^2=1.\Rightarrow\orbr{\begin{cases}a-b=1\\a-b=-1\end{cases}\Rightarrow}\orbr{\begin{cases}a=2,b=1\\a=1,b=2\end{cases}}.}\)
Từ đó suy ra x=-3,x=0
\(C=\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{6}+2\sqrt{2}+4}{\sqrt{2}+\sqrt{3}+2}\)
\(=\dfrac{3\sqrt{2}+\sqrt{3}+\sqrt{6}+4}{\sqrt{2}+\sqrt{3}+2}\)
\(=\dfrac{\left(3\sqrt{2}+\sqrt{3}+\sqrt{6}+4\right)\left(\sqrt{2}+\sqrt{3}-2\right)}{1+2\sqrt{6}}\)
\(=\dfrac{6+3\sqrt{6}+6\sqrt{2}+\sqrt{6}+3-2\sqrt{3}+\sqrt{12}+\sqrt{18}-2\sqrt{6}+4\sqrt{2}+4\sqrt{3}-8}{1+2\sqrt{6}}\)
\(=\dfrac{6+3\sqrt{6}-6\sqrt{2}+\sqrt{6}+3-2\sqrt{3}+2\sqrt{3}+\sqrt{18}+2\sqrt{6}+4\sqrt{2}+4\sqrt{3}-8}{1+2\sqrt{6}}\)
\(=\dfrac{6+3\sqrt{6}-6\sqrt{2}+\sqrt{6}+3+3\sqrt{2}-2\sqrt{6}+4\sqrt{2}+4\sqrt{3}-8}{1+2\sqrt{6}}\)
\(=\dfrac{1+2\sqrt{6}+\sqrt{2}+4\sqrt{3}}{1+2\sqrt{6}}\)
\(=1+\sqrt{2}\)
\(C=\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}{\sqrt{2}+\sqrt{3}+\sqrt{4}}=\dfrac{\sqrt{2}+\sqrt{3}+2+2+\sqrt{6}+\sqrt{8}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(C=\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{2}\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}=\dfrac{\left(1+\sqrt{2}\right)\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(C=1+\sqrt{2}\)
a) ĐKXĐ: \(2x^2-9\ge0\Leftrightarrow2x^2\ge9\Leftrightarrow x^2\ge\frac{9}{2}\Leftrightarrow\left[{}\begin{matrix}x\ge\frac{3}{\sqrt{2}}\\x\le\frac{-3}{\sqrt{2}}\end{matrix}\right.\)
Ta có: \(\sqrt{2x^2-9}=x\)
\(\Leftrightarrow2x^2-9=x^2\)
\(\Leftrightarrow2x^2-9-x^2=0\)
\(\Leftrightarrow x^2-9=0\)
\(\Leftrightarrow x^2=9\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\left(nhận\right)\\x=-3\left(nhận\right)\end{matrix}\right.\)
Vậy: S={3;-3}
b) ĐKXĐ: \(x\in R\)
Ta có: \(\sqrt{x^2-8x+16}=4\)
\(\Leftrightarrow\sqrt{\left(x-4\right)^2}=4\)
\(\Leftrightarrow\left|x-4\right|=4\)
\(\Leftrightarrow\left[{}\begin{matrix}x-4=-4\\x-4=4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(nhận\right)\\x=8\left(nhận\right)\end{matrix}\right.\)
Vậy: S={0;8}
c) ĐKXĐ: \(x\ge0\)
Ta có: \(\sqrt{4x}=\sqrt{5}\)
\(\Leftrightarrow4x=5\)
hay \(x=\frac{5}{4}\)(nhận)
Vậy: \(S=\left\{\frac{5}{4}\right\}\)
a/ \(\sqrt{2x^2-9}=x\)
\(\Leftrightarrow2x^2-9=x^2\)
\(\Leftrightarrow2x^2-x^2-9=0\)
\(\Leftrightarrow x^2-9=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x+3=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)
Vậy...
b/ \(\sqrt{x^2-8x+16}=4\)
\(\Leftrightarrow\sqrt{\left(x-4\right)^2}=4\)
\(\Leftrightarrow\left(x-4\right)^2=4\)
\(\Leftrightarrow\left(x-4\right)^2-4=0\)
\(\Leftrightarrow\left(x-4-2\right)\left(x-4+2\right)=0\)
\(\Leftrightarrow\left(x-6\right)\left(x-2\right)=0\)
\(\Leftrightarrow\)\(\left[{}\begin{matrix}x-6=0\\x-2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=6\\x=2\end{matrix}\right.\)
Vậy....
c/ ĐK : \(x\ge0\)
Ta có :
\(\sqrt{4x}=\sqrt{5x}\)
\(\Leftrightarrow4x=5x\)
\(\Leftrightarrow5x-4x=0\)
\(\Leftrightarrow x=0\)
Vậy....
Bạn tự xét ĐKXĐ nhé ^^
Ta có : \(\sqrt{3x^2-5x+1}-\sqrt{x^2-2}=\sqrt{3\left(x^2-x-1\right)}-\sqrt{x^2-3x+4}\)
\(\Leftrightarrow\left(\sqrt{3x^2-5x+1}-\sqrt{3}\right)-\left(\sqrt{x^2-2}-\sqrt{2}\right)-\left[\sqrt{3\left(x^2-x-1\right)}-\sqrt{3}\right]+\left(\sqrt{x^2-3x+4}-\sqrt{2}\right)=0\)
\(\Leftrightarrow\frac{3x^2-5x+1-3}{\sqrt{3x^2-5x+1}+\sqrt{3}}-\frac{x^2-2-2}{\sqrt{x^2-2}+\sqrt{2}}-\frac{3x^2-3x-3-3}{\sqrt{3\left(x^2-x-1\right)}+\sqrt{3}}+\frac{x^2-3x+4-2}{\sqrt{x^2-3x+4}+\sqrt{2}}=0\)
\(\Leftrightarrow\frac{\left(x-2\right)\left(3x+1\right)}{\sqrt{3x^2-5x+1}+\sqrt{3}}-\frac{\left(x-2\right)\left(x+2\right)}{\sqrt{x^2-2}+\sqrt{2}}-\frac{3\left(x-2\right)\left(x+1\right)}{\sqrt{3\left(x^2-x-1\right)}+\sqrt{3}}+\frac{\left(x-2\right)\left(x-1\right)}{\sqrt{x^2-3x+4}+\sqrt{2}}=0\)
\(\Leftrightarrow\left(x-2\right)\left(\frac{3x+1}{\sqrt{3x^2-5x+1}+\sqrt{3}}-\frac{x+2}{\sqrt{x^2-2}+\sqrt{2}}-\frac{3x+3}{\sqrt{3\left(x^2-x-1\right)}+\sqrt{3}}+\frac{x-1}{\sqrt{x^2-3x+4}+\sqrt{2}}\right)=0\)Tới đây bạn tự làm tiếp ^^
Dài quá ^^
ĐKXĐ: \(x\ge1\)
\(\Leftrightarrow x-\sqrt{x^2-1}+x+\sqrt{x^2-1}+2\sqrt{x^2-\left(x^2-1\right)}=4\)
\(\Leftrightarrow2x+2=4\)
\(\Leftrightarrow x=1\)
ĐKXĐ: \(x\ge0;x\ne4.\)
\(A=\frac{x}{x-4}+\frac{1}{\sqrt{x}-2}+\frac{1}{\sqrt{x}+2}.\)
\(=\frac{x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\frac{\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\frac{\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{x+\sqrt{x}+2+\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}=\frac{x+2\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}.\)
\(=\frac{\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}=\frac{\sqrt{x}}{\sqrt{x}-2}.\)
b) Để \(A=\frac{5}{4}\)\(\Leftrightarrow\frac{\sqrt{x}}{\sqrt{x}-2}=\frac{5}{4}\Leftrightarrow\frac{4\sqrt{x}}{4\left(\sqrt{x}-2\right)}-\frac{5\left(\sqrt{x}-2\right)}{4\left(\sqrt{x}-2\right)}=0\)
\(\Leftrightarrow\frac{4\sqrt{x}-5\sqrt{x}+10}{4\left(\sqrt{x}-2\right)}=0\Leftrightarrow-\sqrt{x}+10=0\)
\(\Leftrightarrow\sqrt{x}=10\Leftrightarrow x=100\left(tmđk\right).\)
Vậy để A=5/4 thì x=100
Tự tìm ĐK nha
a) \(A=\frac{x}{x-4}+\frac{1}{\sqrt{x}-2}+\frac{1}{\sqrt{x}+2}\)
\(A=\frac{x}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}+\frac{\sqrt{x}+2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}+\frac{\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(A=\frac{x+\sqrt{x}+2+\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(A=\frac{x+2\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(A=\frac{\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(A=\frac{\sqrt{x}}{\sqrt{x}-2}\)
b) \(A=\frac{5}{4}\Leftrightarrow\frac{\sqrt{x}}{\sqrt{x}-2}=\frac{5}{4}\)
\(\Leftrightarrow4\sqrt{x}=5\left(\sqrt{x}-2\right)\)
\(\Leftrightarrow4\sqrt{x}=5\sqrt{x}-10\)
\(\Leftrightarrow\sqrt{x}=10\)
\(\Leftrightarrow x=100\)( thỏa mãn )
Vậy...
Lời giải:
a)
\(M=\left[\frac{\sqrt{x}(\sqrt{x}-1)}{\sqrt{x}-1}-\frac{\sqrt{x}+1}{\sqrt{x}(\sqrt{x}+1)}\right].\frac{\sqrt{x}+1}{x}\)
\(=\left(\sqrt{x}-\frac{1}{\sqrt{x}}\right).\frac{\sqrt{x}+1}{x}=\frac{x-1}{\sqrt{x}}.\frac{\sqrt{x}+1}{x}=\frac{(x-1)(\sqrt{x}+1)}{x\sqrt{x}}\)
b)
$M< 0\Leftrightarrow \frac{(x-1)\sqrt{x}+1)}{x\sqrt{x}}< 0$
$\Leftrightarrow x-1<0$ (do $\frac{\sqrt{x}+1}{x\sqrt{x}}>0$ với mọi $x>0; x\neq 1$)
$\Leftrightarrow x< 1$
Kết hợp với ĐKXĐ suy ra để $M< 0$ thì $0< x< 1$