giải phương trình: \(\sqrt{x}+\sqrt{x+7}+2\sqrt{x^2+7x}=35-2x\)
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Giải phương trình: \(\sqrt{x^2+x+19}+\sqrt{7x^2-2x+4}+\sqrt{13x^2+19x+7}=\sqrt{3}.\left(x+5\right)\)
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a:
ĐKXĐ: x>=5/2
\(\sqrt{x-2+\sqrt{2x-5}}+\sqrt{x+2+3\sqrt{2x-5}}=7\sqrt{2}\)
=>\(\sqrt{2x-4+2\sqrt{2x-5}}+\sqrt{2x+4+6\cdot\sqrt{2x-5}}=14\)
=>\(\sqrt{\left(\sqrt{2x-5}+1\right)^2}+\sqrt{\left(\sqrt{2x-5}+3\right)^2}=14\)
=>\(\sqrt{2x-5}+1+\sqrt{2x-5}+3=14\)
=>\(2\sqrt{2x-5}+4=14\)
=>\(\sqrt{2x-5}=5\)
=>2x-5=25
=>2x=30
=>x=15
b: \(x^2-4x=\sqrt{x+2}\)
=>\(x+2=\left(x^2-4x\right)^2\) và x^2-4x>=0
=>x^4-8x^3+16x^2-x-2=0 và x^2-4x>=0
=>(x^2-5x+2)(x^2-3x-1)=0 và x^2-4x>=0
=>\(\left[{}\begin{matrix}x=\dfrac{5+\sqrt{17}}{2}\\x=\dfrac{3-\sqrt{13}}{2}\end{matrix}\right.\)
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1.
\(\Leftrightarrow\left(2x+1\right)\sqrt{2x^2+4x+5}-\left(2x+1\right)\left(x+3\right)+x^2-2x-4=0\)
\(\Leftrightarrow\left(2x+1\right)\left(\sqrt{2x^2+4x+5}-\left(x+3\right)\right)+x^2-2x-4=0\)
\(\Leftrightarrow\dfrac{\left(2x+1\right)\left(x^2-2x-4\right)}{\sqrt{2x^2+4x+5}+x+3}+x^2-2x-4=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-2x-4=0\\\dfrac{2x+1}{\sqrt{2x^2+4x+5}+x+3}+1=0\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow2x+1+\sqrt{2x^2+4x+5}+x+3=0\)
\(\Leftrightarrow\sqrt{2x^2+4x+5}=-3x-4\) \(\left(x\le-\dfrac{4}{3}\right)\)
\(\Leftrightarrow2x^2+4x+5=9x^2+24x+16\)
\(\Leftrightarrow7x^2+20x+11=0\)
2.
ĐKXĐ: ...
\(\Leftrightarrow2x\sqrt{2x+7}+7\sqrt{2x+7}=x^2+2x+7+7x\)
\(\Leftrightarrow\left(x^2-2x\sqrt{2x+7}+2x+7\right)+7\left(x-\sqrt{2x+7}\right)=0\)
\(\Leftrightarrow\left(x-\sqrt{2x+7}\right)^2+7\left(x-\sqrt{2x+7}\right)=0\)
\(\Leftrightarrow\left(x-\sqrt{2x+7}\right)\left(x+7-\sqrt{2x+7}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{2x+7}\\x+7=\sqrt{2x+7}\end{matrix}\right.\)
\(\Leftrightarrow...\)
ĐKXĐ: \(x\ge0\)
\(\Leftrightarrow2x+7+2\sqrt{x^2+7x}+\sqrt{x}+\sqrt{x+7}-42=0\)
Đặt \(\sqrt{x}+\sqrt{x+7}=t>0\)
\(\Rightarrow2x+7+2\sqrt{x^2+7x}=t^2\)
Pt trở thành:
\(t^2+t-42=0\Rightarrow\left[{}\begin{matrix}t=6\\t=-7\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{x}+\sqrt{x+7}=6\)
\(\Leftrightarrow2x+7+2\sqrt{x^2+7x}=36\)
\(\Leftrightarrow2\sqrt{x^2+7x}=29-2x\) (\(x\le\frac{29}{2}\))
\(\Leftrightarrow4\left(x^2+7x\right)=\left(29-2x\right)^2\)
\(\Leftrightarrow144x-841=0\Rightarrow x=\frac{841}{144}\)