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Đặt \(\left\{{}\begin{matrix}\sqrt[3]{x^2+3x+1}=a\\\sqrt[3]{5x+1}=b\end{matrix}\right.\)
\(\Rightarrow a+a^3-b^3=b\)
\(\Leftrightarrow a-b+\left(a-b\right)\left(a^2+ab+b^2\right)=0\)
\(\Leftrightarrow\left(a-b\right)\left(a^2+ab+b^2+1\right)=0\)
\(\Leftrightarrow a=b\)
\(\Leftrightarrow\sqrt[3]{x^2+3x+1}=\sqrt[3]{5x+1}\)
\(\Leftrightarrow x^2+3x+1=5x+1\)
\(\Leftrightarrow...\)
a.
ĐKXĐ: \(x^2+2x-1\ge0\)
\(x^2+2x-1+2\left(x-1\right)\sqrt{x^2+2x-1}-4x=0\)
Đặt \(\sqrt{x^2+2x-1}=t\ge0\)
\(\Rightarrow t^2+2\left(x-1\right)t-4x=0\)
\(\Delta'=\left(x-1\right)^2+4x=\left(x+1\right)^2\)
\(\Rightarrow\left[{}\begin{matrix}t=1-x+x+1=2\\t=1-x-x-1=-2x\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\sqrt{x^2+2x-1}=2\\\sqrt{x^2+2x-1}=-2x\left(x\le0\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+2x-5=0\\3x^2-2x+1=0\left(vn\right)\end{matrix}\right.\)
\(\Rightarrow x=-1\pm\sqrt{6}\)
b.
ĐKXĐ: \(x\ge\dfrac{1}{5}\)
\(2x^2+x-3+2x-\sqrt{5x-1}+2-\sqrt[3]{9-x}=0\)
\(\Leftrightarrow\left(x-1\right)\left(2x+3\right)+\dfrac{\left(x-1\right)\left(4x-1\right)}{2x+\sqrt[]{5x-1}}+\dfrac{x-1}{4+2\sqrt[3]{9-x}+\sqrt[3]{\left(9-x\right)^2}}=0\)
\(\Leftrightarrow\left(x-1\right)\left(2x+3+\dfrac{4x-1}{2x+\sqrt[]{5x-1}}+\dfrac{1}{4+2\sqrt[3]{9-x}+\sqrt[3]{\left(9-x\right)^2}}\right)=0\)
\(\Leftrightarrow x=1\) (ngoặc đằng sau luôn dương)
\(\sqrt{3x+1}+2\sqrt{x+3}=3\sqrt{5x-1}\)
=>\(\sqrt{3x+1}-2+2\sqrt{x+3}-4=3\sqrt{5x-1}-6\)
=>\(\dfrac{3x+1-4}{\sqrt{3x+1}+2}+2\left(\sqrt{x+3}-2\right)-3\left(\sqrt{5x-1}-2\right)=0\)
=>\(\dfrac{3\left(x-1\right)}{\sqrt{3x+1}+2}+2\cdot\dfrac{x+3-4}{\sqrt{x+3}+2}-3\cdot\dfrac{5x-1-4}{\sqrt{5x-1}+2}=0\)
=>\(\left(x-1\right)\left(\dfrac{3}{\sqrt{3x+1}+2}+\dfrac{2}{\sqrt{x+3}+2}-\dfrac{15}{\sqrt{5x-1}+2}\right)=0\)
=>x-1=0
=>x=1
1 ĐKXD \(x\ge1\)
.\(2x^2+5x-1=7\sqrt{\left(x-1\right)\left(x^2+x+1\right)}\)
Đặt \(\sqrt{x-1}=a;\sqrt{x^2+x+1}=b\left(a,b\ge0\right)\)
=> \(2b^2+3a^2=2x^2+5x-1\)
=> \(2b^2+3a^2-7ab=0\)
<=> \(\orbr{\begin{cases}a=2b\\a=\frac{1}{3}b\end{cases}}\)
+ \(a=2b\)
=> \(2\sqrt{x^2+x+1}=\sqrt{x-1}\)
=> \(4x^2+3x+5=0\)vô nghiệm
+ \(a=\frac{1}{3}b\)
=> \(\sqrt{x^2+x+1}=3\sqrt{x-1}\)
=> \(x^2-8x+10=0\)
<=> \(\orbr{\begin{cases}x=4+\sqrt{6}\left(tmĐK\right)\\x=4-\sqrt{6}\left(kotmĐK\right)\end{cases}}\)
Vậy \(x=4+\sqrt{6}\)
ĐKXĐ:\(2x^2-1\ge0;x^2-3x-2\ge0;2x^2+2x+3\ge0;x^2-x+2\ge0\)
\(\sqrt{2x^2-1}+\sqrt{x^2-3x-2}=\sqrt{2x^2+2x+3}+\sqrt{x^2-x+2}\)
<=> \(\left(\sqrt{2x^2+2x+3}-\sqrt{2x^2-1}\right)+\left(\sqrt{x^2-x+2}-\sqrt{x^2-3x-2}\right)=0\)
\(\Leftrightarrow\frac{2x+4}{\sqrt{2x^2+2x+3}+\sqrt{2x^2-1}}+\frac{2x+4}{\sqrt{x^2-x+2}+\sqrt{x^2-3x-2}}=0\)
<=> \(\left(2x+4\right)\left(\frac{1}{\sqrt{2x^2+2x+3}+\sqrt{2x^2-1}}+\frac{1}{\sqrt{x^2-x+2}+\sqrt{x^2-3x-2}}\right)=0\)(1)
Vì \(\frac{1}{\sqrt{2x^2+2x+3}+\sqrt{2x^2-1}}+\frac{1}{\sqrt{x^2-x+2}+\sqrt{x^2-3x-2}}>0\)
nên pt(1) <=> \(2x+4=0\Leftrightarrow x=-2\)(tmđk)
Vậy x=-2
Em kiểm tra lại đề bài câu trên nhé