Question

[Julian Edward]

giai pt:

a) \(\frac{3x+\sqrt{x^2-x-1}}{x+1}=\frac{7}{3}\)

b) \(\frac{2}{2\sqrt{x^2-2x+1}}=\frac{1}{x-1}\)

c) \(\frac{6}{6-\sqrt{x}}+\frac{1}{\sqrt{x}}=1\)

d) \(\frac{2}{\sqrt{x-1}}+\sqrt{x-1}=\frac{3\sqrt{x-1}+1}{\sqrt{x-1}}-1\)

e) \(\sqrt{x+3-\sqrt{x-1}=2}\)

f) \(\sqrt{x^3+x^2+6x+28}=x+5\)

g) \(\sqrt{x^4-4x^3+14x-11}=1-x\)

Answer 1

[Nguyễn Thị Ngọc Thơ]

ĐK: \(x^4-4x^3+14x-11\ge0\) (*)

\(PT\Leftrightarrow\left\{{}\begin{matrix}x\le1\\x^4-4x^3+14x-11=x^2-2x+1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\le1\\x^4-4x^3-x^2+16x-12=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\le1\\\left(x-1\right)\left(x-2\right)\left(x-3\right)\left(x+2\right)=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)(tm)

Answer 2

[Nguyễn Việt Lâm]

e/ ĐKXĐ: \(x\ge1\)

\(\Leftrightarrow x+3-\sqrt{x-1}=4\)

\(\Leftrightarrow\sqrt{x-1}=x-1\)

\(\Leftrightarrow x-1=x^2-2x+1\)

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

f/ \(\Leftrightarrow\left\{{}\begin{matrix}x+5\ge0\\x^3+x^2+6x+28=\left(x+5\right)^2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-5\\x^3-4x+3=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-5\\\left(x-1\right)\left(x^2+x-3\right)=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=1\\x=\frac{-1\pm\sqrt{13}}{2}\\\end{matrix}\right.\)