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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 ^^
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....
a,\(\sqrt{1-x}=\sqrt[3]{27}\left(đk:x\le1\right)\Leftrightarrow\sqrt{1-x}=3\)
\(< =>\sqrt{1-x}^2=9< =>1-x=9< =>x=-8\)tm
b,\(\sqrt{x^2-10x+25}=x+1\)
\(< =>\sqrt{\left(x-5\right)^2}=x+1\)
\(< =>|x-5|=x+1\)
\(< =>\orbr{\begin{cases}-x+5=x+1\left(x< 5\right)\\x-5=x+1\left(x\ge5\right)\end{cases}}\)
\(< =>\orbr{\begin{cases}2x=4< =>x=2\left(tm\right)\\-5-1=0\left(vo-li\right)\end{cases}}\)
c, Đặt \(\sqrt{x}=t\left(t\ge0\right)\)khi đó pt tương đương
\(t^2+t-6=0< =>t^2-2t+3t-6=0\)
<\(< =>t\left(t-2\right)+3\left(t-2\right)=0< =>\left(t+3\right)\left(t-2\right)=0\)
\(< =>\orbr{\begin{cases}t+3=0\\t-2=0\end{cases}}< =>\orbr{\begin{cases}t=-3\left(ktm\right)\\t=2\left(tm\right)\end{cases}}\)
khi đó ta được \(\sqrt{x}=t< =>x=4\)
a) \(\sqrt{1-x}=\sqrt[3]{27}\)
\(\Leftrightarrow\sqrt{1-x}=3\)
\(\Leftrightarrow1-x=9\)
\(\Rightarrow x=-8\)
b) \(\sqrt{x^2-10x+25}=x+1\)
\(\Leftrightarrow\sqrt{\left(x-5\right)^2}=x+1\)
\(\Leftrightarrow\left|x-5\right|=x+1\)
\(\Leftrightarrow\orbr{\begin{cases}x-5=x+1\\x-5=-x-1\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}0=6\left(vl\right)\\2x=4\end{cases}}\Rightarrow x=2\)
c) \(x+\sqrt{x}-6=0\)
\(\Leftrightarrow\left(x+3\sqrt{x}\right)-\left(2\sqrt{x}+6\right)=0\)
\(\Leftrightarrow\sqrt{x}\left(\sqrt{x}+3\right)-2\left(\sqrt{x}+3\right)=0\)
\(\Leftrightarrow\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x}-2=0\\\sqrt{x}+3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}\sqrt{x}=2\\\sqrt{x}=-3\left(vl\right)\end{cases}}\Rightarrow x=4\)
cho mình hỏi hai ý đầu thôi, hai ý sau mình giải ra rồi. Thanks Zero ~
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...
cần gấp thì mình làm cho
\(\sqrt{x^2+2x+1}=\sqrt{x+1}\left(đk:x\ge1\right)\)
\(< =>\sqrt{\left(x+1\right)^2}=\sqrt{x+1}\)
\(< =>x+1=\sqrt{x+1}\)
\(< =>\frac{x+1}{\sqrt{x+1}}=1\)
\(< =>\sqrt{x+1}=1< =>x=0\left(ktm\right)\)
ĐKXĐ : \(x\ge-1\)
Bình phương 2 vế , ta có :
\(x^2+2x+1=x+1\)
\(\Leftrightarrow x^2+2x+1-x-1=0\)
\(\Leftrightarrow x^2+x=0\)
\(\Leftrightarrow x\left(x+1\right)=0\Leftrightarrow\orbr{\begin{cases}x=0\\x=-1\end{cases}\left(TM\right)}\)\
Vậy ...............................
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