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ĐKXĐ:...
\(\sqrt{3x^2-5x-1}-\sqrt{3x^2-7x+9}+\sqrt{x^2-2}-\sqrt{x^2-3x+13}=0\)
\(\Leftrightarrow\frac{2\left(x-5\right)}{\sqrt{3x^2-5x-1}+\sqrt{3x^2-7x+9}}+\frac{3\left(x-5\right)}{\sqrt{x^2-2}+\sqrt{x^2-3x+13}}=0\)
\(\Leftrightarrow\left(x-5\right)\left(\frac{2}{\sqrt{3x^2-5x-1}+\sqrt{3x^2-7x+9}}+\frac{3}{\sqrt{x^2-2}+\sqrt{x^2-3x+13}}\right)=0\)
\(\Leftrightarrow x-5=0\) (ngoặc to phía sau luôn dương)
\(\Rightarrow x=5\)
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\(\Leftrightarrow\sqrt{12-7x}-\sqrt{x^2-x}=\sqrt{3x^2-5x-1}-\sqrt{x^2-3x+4}\)
\(\Rightarrow-\sqrt{3x^2-5x-1}-\sqrt{x^2-x}+\sqrt{x^2-3x+4}+\sqrt{12-7x}=0\)
=>\(x\approx-3,4579061804411\)
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Giải phương trình sau:
√3x2−5x+1−√x2−2=√3(x2−x−1)−√x2−3x+4
ĐKXD: \(3x^2-7x+5\ge0;x^2-x+4\ge0;3x^2-5x+1\ge0\)
Phương trình tương đương
\(\sqrt{3x^2-7x+5}-\sqrt{3x^2-5x+1}=\sqrt{x^2-2}-\sqrt{x^2-x+4}\)
\(\left(=\right)\frac{-2\left(x-2\right)}{\sqrt{3x^2-7x+5}+\sqrt{3x^2-5x+1}}=\frac{x-2}{\sqrt{x^2+2}+\sqrt{x^2-x+4}}\)
\(\left(=\right)\left(x-2\right)\left(\frac{-2}{\sqrt{3x^2-7x+5}+\sqrt{3x^2-5x+1}}-\frac{1}{\sqrt{x^2+2}+\sqrt{x^2-x+4}}\right)=0\)
Dễ đàng đánh giá Trường hợp còn lại nhỏ hơn 0. Từ đó suy ra x=2(thỏa)
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ĐKXĐ: \(x\ge\frac{1}{2}\)
\(\Leftrightarrow7x+3-4\sqrt{x\left(x+3\right)}-2\sqrt{2x-1}=0\)
\(\Leftrightarrow2x-1-2\sqrt{2x-1}+1+4x-4\sqrt{x\left(x+3\right)}+x+3=0\)
\(\Leftrightarrow\left(\sqrt{2x-1}-1\right)^2+\left(2\sqrt{x}-\sqrt{x+3}\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{2x-1}-1=0\\2\sqrt{x}-\sqrt{x+3}=0\end{matrix}\right.\) \(\Rightarrow x=1\)
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Lời giải:
ĐKXĐ: \(x\leq \frac{2}{3}\)
Ta có: \(\sqrt{2-3x}=-3x^2+7x-1\)
\(\Leftrightarrow 3x^2-7x+1+\sqrt{2-3x}=0\)
\(\Leftrightarrow x(3x-1)-2(3x-1)+\sqrt{2-3x}-1=0\)
\(\Leftrightarrow x(3x-1)-2(3x-1)+\frac{2-3x-1}{\sqrt{2-3x}+1}=0\)
\(\Leftrightarrow (3x-1)\left(x-2-\frac{1}{\sqrt{2-3x}+1}\right)=0\)
Vì \(x\leq \frac{2}{3}; \frac{1}{\sqrt{2-3x}+1}>0\Rightarrow x-2-\frac{1}{\sqrt{2-3x}+1}< \frac{2}{3}-2-0<0\)
Tức là \(x-2-\frac{1}{\sqrt{2-3x}+1}\neq 0\Rightarrow 3x-1=0\Rightarrow x=\frac{1}{3}\) (t/m)
Vậy...........
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\(PT\Leftrightarrow3\left(x^2+7x+7\right)-3+2\sqrt{x^2+7x+7}-2=0.\)
\(\Leftrightarrow3\left(x^2+7x+7\right)+2\sqrt{x^2+7x+7}-5=0\)
Đặt \(a=\sqrt{x^2+7x+7}\)(a\(\ge\)0)
\(PT\Leftrightarrow3a^2+2a-5=0\)
\(\Leftrightarrow\left(a-1\right)\left(3a+5\right)=0\)
Vì a\(\ge\)0 nên a-1=0=> a=1
lúc đó x2+7x+7=1
<=> x2+7x+6=0
<=> (x+1)(x+6)=0
<=> \(\orbr{\begin{cases}x=-1\\x=-6\end{cases}}\)
Vậy.................................