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a) ĐK: \(x\geq \frac{1}{2}\)
Ta có: \(\sqrt{2x-1}-\sqrt{x+1}=2x-4\)
\(\Leftrightarrow \frac{(2x-1)-(x+1)}{\sqrt{2x-1}+\sqrt{x+1}}=2(x-2)\)
\(\Leftrightarrow \frac{x-2}{\sqrt{2x-1}+\sqrt{x+1}}=2(x-2)\)
\(\Leftrightarrow (x-2)\left(\frac{1}{\sqrt{2x-1}+\sqrt{x+1}}-2\right)=0\)
\(\Rightarrow \left[\begin{matrix} x-2=0\leftrightarrow x=2\\ \frac{1}{\sqrt{2x-1}+\sqrt{x+1}}=2(*)\end{matrix}\right.\)
Đối với $(*)$:
Vì \(x\geq \frac{1}{2}\Rightarrow \sqrt{2x-1}+\sqrt{x+1}\geq \sqrt{\frac{1}{2}+1}>1\)
\(\Rightarrow \frac{1}{\sqrt{2x-1}+\sqrt{x+1}}< 1\)
Do đó $(*)$ vô nghiệm
Vậy pt có nghiệm duy nhất $x=2$
b) ĐK:.....
\(\sqrt{2x^2-3x+10}+\sqrt{2x^2-5x+4}=x+3\)
TH1:
\(\sqrt{2x^2-3x+10}=\sqrt{2x^2-5x+4}\)
\(\Rightarrow 2x^2-3x+10=2x^2-5x+4\)
\(\Rightarrow 2x+6=0\Rightarrow x=-3\) (thử lại thấy không thỏa mãn)
TH2: \(\sqrt{2x^2-3x+10}\neq \sqrt{2x^2-5x+4}\), tức là \(x\neq -3\)
PT ban đầu tương đương với:
\(\frac{(2x^2-3x+10)-(2x^2-5x+4)}{\sqrt{2x^2-3x+10}-\sqrt{2x^2-5x+4}}=x+3\)
\(\Leftrightarrow \frac{2(x+3)}{\sqrt{2x^2-3x+10}-\sqrt{2x^2-5x+4}}=x+3\)
\(\Leftrightarrow \frac{2}{\sqrt{2x^2-3x+10}-\sqrt{2x^2-5x+4}}=1\) (do \(x\neq -3\) )
\(\Rightarrow \sqrt{2x^2-3x+10}-\sqrt{2x^2-5x+4}=2\)
\(\Rightarrow \sqrt{2x^2-3x+10}=2+\sqrt{2x^2-5x+4}\)
Bình phương 2 vế:
\(2x^2-3x+10=4+2x^2-5x+4+4\sqrt{2x^2-5x+4}\)
\(\Leftrightarrow x+1=2\sqrt{2x^2-5x+4}\)
\(\Rightarrow (x+1)^2=4(2x^2-5x+4)\)
\(\Rightarrow 7x^2-22x+15=0\Rightarrow \left[\begin{matrix} x=\frac{15}{7}\\ x=1\end{matrix}\right.\) (thử đều thấy t/m)
Vậy...........
2,\(pt\Leftrightarrow12\left(\sqrt{x+1}-2\right)+x^2+x-12=0\)
\(\Leftrightarrow12\cdot\frac{x-3}{\sqrt{x+1}+2}+\left(x-3\right)\left(x+4\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(\frac{12}{\sqrt{x+1}+2}+x+4\right)=0\)
Vì \(\left(\frac{12}{\sqrt{x+1}+2}+x+4\right)\ge0\left(\forall x>-1\right)\)
\(\Rightarrow x=3\)
a) ĐK: \(x\ge -1\)
Ta có: \(x^2+\sqrt{x+1}=1\)
\(\Leftrightarrow (x^2-1)+\sqrt{x+1}=0\)
\(\Leftrightarrow (x-1)(x+1)+\sqrt{x+1}=0\)
\(\Leftrightarrow \sqrt{x+1}[(x-1)\sqrt{x+1}+1]=0\)
\(\Rightarrow \left[\begin{matrix} \sqrt{x+1}=0(1)\\ (x-1)\sqrt{x+1}+1=0(2)\end{matrix}\right.\)
Với \((1)\Rightarrow x+1=0\Rightarrow x=-1\) (thỏa mãn)
Với \((2)\Rightarrow x\sqrt{x+1}-(\sqrt{x+1}-1)=0\)
\(\Leftrightarrow x\sqrt{x+1}-\frac{x}{\sqrt{x+1}+1}=0\)
\(\Leftrightarrow x\left(\sqrt{x+1}-\frac{1}{\sqrt{x+1}+1}\right)=0\)
\(\Leftrightarrow x.\frac{x+1+\sqrt{x+1}-1}{\sqrt{x+1}+1}=0\)
\(\Leftrightarrow x.\frac{x+\sqrt{x+1}}{\sqrt{x+1}+1}=0\)
\(\Rightarrow \left[\begin{matrix} x=0\\ x+\sqrt{x+1}=0\end{matrix}\right.\)
Với \(x+\sqrt{x+1}=0\Rightarrow x=-\sqrt{x+1}\Rightarrow \left\{\begin{matrix} x\leq 0\\ x^2=x+1\end{matrix}\right.\Rightarrow x=\frac{1-\sqrt{5}}{2}\)
Vậy \(x=\left\{-1; \frac{1-\sqrt{5}}{2}; 0\right\}\)
b) ĐK: \(-3\leq x\leq 6\)
Ta có: \((\sqrt{3+x}+\sqrt{6-x})^2=3+x+6-x+2\sqrt{(3+x)(6-x)}\)
\(=9+2\sqrt{(3+x)(6-x)}\geq 9\)
\(\Rightarrow \sqrt{3+x}+\sqrt{6-x}\geq 3\) do \(\sqrt{3+x}+\sqrt{6-x}\) không âm.
Dấu "=" xảy ra khi \(\sqrt{(3+x)(6-x)}=0\Leftrightarrow x=-3; x=6\)
Vậy \(x=-3\) or $x=6$
d, ĐKXĐ: \(x\ge-\frac{1}{4}\)
\(pt\Leftrightarrow4x^2+4x+2=2\sqrt{4x+1}\)
\(\Leftrightarrow4x^2+\left(4x+1-2\sqrt{4x+1}+1\right)=0\)
\(\Leftrightarrow4x^2+\left(\sqrt{4x+1}-1\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}4x^2=0\\\sqrt{4x+1}-1=0\end{matrix}\right.\Leftrightarrow x=0\left(tm\right)\)
a, ĐKXĐ: \(x\ge-1\)
\(pt\Leftrightarrow\sqrt{x+1}+\sqrt{x+8}=7\)
\(\Leftrightarrow\left(\sqrt{x+1}+\sqrt{x+8}\right)^2=49\)
\(\Leftrightarrow x+1+x+8+2\sqrt{\left(x+1\right)\left(x+8\right)}=49\)
\(\Leftrightarrow\sqrt{\left(x+1\right)\left(x+8\right)}=20-x\)
\(\Leftrightarrow\left\{{}\begin{matrix}20-x\ge0\\\left(x+1\right)\left(x+8\right)=\left(20-x\right)^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\le20\\49x=392\end{matrix}\right.\Leftrightarrow x=8\left(tm\right)\)
b, ĐKXĐ: \(x\ge-1\)
\(pt\Leftrightarrow\frac{x-3}{\sqrt[3]{\left(x-2\right)^2}+\sqrt[3]{x-2}+1}+\frac{x-3}{\sqrt{x+1}+2}=0\)
\(\Leftrightarrow\left(x-3\right)\left(\frac{1}{\sqrt[3]{\left(x-2\right)^2}+\sqrt[3]{x-2}+1}+\frac{1}{\sqrt{x+1}+2}\right)=0\)
Do \(\frac{1}{\sqrt[3]{\left(x-2\right)^2}+\sqrt[3]{x-2}+1}+\frac{1}{\sqrt{x+1}+2}>0,\forall x\ge-1\)
Nên \(x=3\left(tm\right)\)
c, ĐKXĐ: \(x\ge-\frac{3}{2}\)
\(pt\Leftrightarrow\left(x^2+2x+1\right)+\left(2x+3-2\sqrt{2x+3}+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)^2+\left(\sqrt{2x+3}-1\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+1=0\\\sqrt{2x+3}-1=0\end{matrix}\right.\Leftrightarrow x=-1\left(tm\right)\)
a.
ĐKXĐ: \(x\ge\dfrac{23}{4}\)
\(\sqrt{x+10}=\sqrt{x+3}+\sqrt{4x-23}\)
\(\Leftrightarrow x+10=5x-20+2\sqrt{\left(x+3\right)\left(4x-23\right)}\)
\(\Leftrightarrow\sqrt{4x^2-11x-69}=15-2x\) \(\left(x\le\dfrac{15}{2}\right)\)
\(\Leftrightarrow4x^2-11x-69=\left(15-2x\right)^2\)
\(\Leftrightarrow49x-294=0\)
\(\Leftrightarrow x=6\) (thỏa mãn)
b.
ĐKXĐ: \(x\ge-\dfrac{1}{2}\)
\(\sqrt{3x+4}=\sqrt{2x+1}+\sqrt{x+3}\)
\(\Leftrightarrow3x+4=3x+4+2\sqrt{\left(2x+1\right)\left(x+3\right)}\)
\(\Leftrightarrow\sqrt{\left(2x+1\right)\left(x+3\right)}=0\)
\(\Rightarrow\left[{}\begin{matrix}2x+1=0\\x+3=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=-3\left(loại\right)\end{matrix}\right.\)