ĐKXĐ: ...

Đặt \(\sqrt{x+\frac{3}{4}}=a\ge0\Rightarrow x=a^2-\frac{3}{4}\)

\(\sqrt{a^2-\frac{3}{4}+1+a}+a^2-\frac{3}{4}=-\frac{1}{4}\)

\(\Leftrightarrow\sqrt{a^2+a+\frac{1}{4}}+a^2-\frac{1}{2}=0\)

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

\(\Leftrightarrow a^2+a=0\Rightarrow\left[{}\begin{matrix}a=0\\a=-1\left(l\right)\end{matrix}\right.\) \(\Rightarrow x=-\frac{3}{4}\)

em ms hok lớp 1

a/ ĐKXĐ: \(-\frac{3}{2}\le x\le4\)

\(\sqrt{2x+3}+\sqrt{4-x}=6x-3\left(x+7-2\sqrt{\left(2x+3\right)\left(4-x\right)}\right)-10\)

\(\Leftrightarrow\sqrt{2x+3}+\sqrt{4-x}=3\left(x+7+2\sqrt{\left(2x+3\right)\left(4-x\right)}\right)-52\)

Đặt \(\sqrt{2x+3}+\sqrt{4-x}=a>0\Rightarrow a^2=x+7+2\sqrt{\left(2x+3\right)\left(4-x\right)}\)

Phương trình trở thành:

\(a=3a^2-52\Leftrightarrow3a^2-a-52=0\Rightarrow\left[{}\begin{matrix}a=-4\left(l\right)\\a=\frac{13}{3}\end{matrix}\right.\)

\(\sqrt{2x+3}+\sqrt{4-x}=\frac{13}{3}\)

Phương trình này vô nghiệm nên ko muốn giải tiếp, bạn bình phương lên và chuyển vế thôi :(

b/ ĐKXĐ: \(-\frac{1}{4}\le x\le1\)

Đặt \(\sqrt{4x+1}+2\sqrt{1-x}=a>0\Rightarrow a^2=5+4\sqrt{-4x^2+3x+1}\)

\(\Rightarrow\sqrt{-4x^2+3x+1}=\frac{a^2-5}{4}\)

Pt trở thành:

\(a+10\left(\frac{a^2-5}{4}\right)=13\)

\(\Leftrightarrow5a^2+2a-51=0\Rightarrow\left[{}\begin{matrix}a=3\\a=-\frac{17}{5}\left(l\right)\end{matrix}\right.\)

\(\Rightarrow\sqrt{-4x^2+3x+1}=\frac{a^2-5}{4}=1\)

\(\Leftrightarrow-4x^2+3x=0\Rightarrow\left[{}\begin{matrix}x=0\\x=\frac{3}{4}\end{matrix}\right.\)

c/ \(\Leftrightarrow x^2\left(x^2+2\right)=12-x\sqrt{2x^2+4}\)

\(\Leftrightarrow x^2\left(2x^2+4\right)=24-2x\sqrt{2x^2+4}\)

Đặt \(x\sqrt{2x^2+4}=a\) ta được:

\(a^2=24-2a\Leftrightarrow a^2+2a-24=0\Leftrightarrow\left[{}\begin{matrix}a=4\\a=-6\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x\sqrt{2x^2+4}=4\left(x>0\right)\\x\sqrt{2x^2+4}=-6\left(x< 0\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2\left(2x^2+4\right)=16\\x^2\left(2x^2+4\right)=36\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^4+2x^2-8=0\\x^4+2x^2-18=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x^2=2\\x^2=-4\left(l\right)\\x^2=\sqrt{19}-1\\x^2=-\sqrt{19}-1\left(l\right)\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\sqrt{2}\\x=-\sqrt{2}< 0\left(l\right)\\x=-\sqrt{\sqrt{19}-1}\\x=\sqrt{\sqrt{19}-1}>0\left(l\right)\end{matrix}\right.\)

d/ ĐKXĐ: \(x\ge1\)

Nhân cả tử và mẫu của vế phải với liên hợp của nó ta được:

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

Đặt \(\sqrt{x+1}+\sqrt{x-1}=a>0\)

\(\Rightarrow a^2-3=\frac{a}{2}\Rightarrow2a^2-a-6=0\Rightarrow\left[{}\begin{matrix}a=2\\a=-\frac{3}{2}\left(l\right)\end{matrix}\right.\)

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

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

\(\Leftrightarrow\sqrt{x^2-1}=2-x\) (\(x\le2\))

\(\Leftrightarrow x^2-1=x^2-4x+4\)

\(\Rightarrow x=\frac{5}{4}\)

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)

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

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

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

a/ ĐKXĐ: ...

\(\Leftrightarrow9x+3\sqrt{x^2-x-1}=7x+7\)

\(\Leftrightarrow3\sqrt{x^2-x-1}=7-2x\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\le\frac{7}{2}\\9\left(x^2-x-1\right)=\left(7-2x\right)^2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\le\frac{7}{2}\\5x^2+19x-58=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=2\\x=-\frac{29}{5}\end{matrix}\right.\)

b/ ĐKXĐ: \(x\ne1\)

\(\Leftrightarrow\frac{1}{\sqrt{\left(x-1\right)^2}}=\frac{1}{x-1}\)

\(\Leftrightarrow\frac{1}{\left|x-1\right|}=\frac{1}{x-1}\)

\(\Rightarrow x-1>0\Rightarrow x>1\)

điều kiện: \(-2\le x\le2\)

pt\(\Leftrightarrow\frac{\left(2x+4\right)-4\left(2-x\right)}{\sqrt{2x+4}+2\sqrt{2-x}}=\frac{6x-4}{\sqrt{x^2+4}}\)

\(\Leftrightarrow6x-4=0\Leftrightarrow x=\frac{2}{3}\)(t/m)