Question

Giải phương trình:a) \(\sqrt{x+1}=7-\sqrt{x+8}\)

b)\(\sqrt[3]{x-2}+\sqrt{x+1}=3\)

c)\(x^2+4x+5=2\sqrt{2x+3}\)

d)2\(x^2+2x+1=\sqrt{4x+1}\)

e)\(\sqrt{5-x^6}-2\sqrt{3x^4-2}=1\)

f)\(\sqrt[5]{x^2+28}+2\sqrt[3]{x^2+23}+\sqrt{x-1}+\sqrt{x}=\sqrt{2}+9\)

h)\(\sqrt{x-1}+\sqrt{x^3+x^2+x-1}=1+\sqrt{x^4-1}\)

Answer 1

d, ĐKXĐ: \(x\ge-\frac{1}{4}\)

\(pt\Leftrightarrow4x^2+4x+2=2\sqrt{4x+1}\)

\(\Leftrightarrow4x^2+\left(4x+1-2\sqrt{4x+1}+1\right)=0\)

\(\Leftrightarrow4x^2+\left(\sqrt{4x+1}-1\right)^2=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}4x^2=0\\\sqrt{4x+1}-1=0\end{matrix}\right.\Leftrightarrow x=0\left(tm\right)\)

Answer 2

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: \(x\ge-\frac{3}{2}\)

\(pt\Leftrightarrow\left(x^2+2x+1\right)+\left(2x+3-2\sqrt{2x+3}+1\right)=0\)

\(\Leftrightarrow\left(x+1\right)^2+\left(\sqrt{2x+3}-1\right)^2=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x+1=0\\\sqrt{2x+3}-1=0\end{matrix}\right.\Leftrightarrow x=-1\left(tm\right)\)