\(=\left[{}\begin{matrix}\dfrac{5}{8}-x+\dfrac{1}{4}-\dfrac{3}{2}\\x-\dfrac{5}{8}+\dfrac{1}{4}-\dfrac{3}{2}\end{matrix}\right.=\left[{}\begin{matrix}-x-\dfrac{5}{8}\\x-\dfrac{15}{8}\end{matrix}\right.\)

\(2.\left|\dfrac{5}{8}-x\right|=\dfrac{5}{4}\\\left|\dfrac{5}{8}-x\right|=\dfrac{5}{8} \\ \left[{}\begin{matrix}\dfrac{5}{8}-x=\dfrac{5}{8}\\\dfrac{5}{8}-x=-\dfrac{5}{8}\end{matrix}\right.\left[{}\begin{matrix}x=0\\x=\dfrac{5}{4}\end{matrix}\right.\)

\(|\dfrac{5}{8}-x|+\dfrac{1}{4}=\dfrac{3}{2}\)

\(|\dfrac{5}{8}-x|=\dfrac{3}{2}-\dfrac{1}{4}\)

\(|\dfrac{5}{8}-x|=\dfrac{5}{4}\)

\(\Rightarrow\dfrac{5}{8}-x=\pm\dfrac{5}{4}\)

\(\dfrac{5}{8}-x=\dfrac{5}{4}\)                                        \(\dfrac{5}{8}-x=-\dfrac{5}{4}\)

\(x=\dfrac{5}{8}-\dfrac{5}{4}\)                                         \(x=\dfrac{5}{8}+\dfrac{5}{4}\)

\(x=-\dfrac{5}{8}\)                                              \(x=\dfrac{15}{8}\)

\(\Rightarrow x=\left\{-\dfrac{5}{8};\dfrac{15}{8}\right\}\)

\(\left[{}\begin{matrix}x-1,7=2,3\\x-1,7=-2,3\end{matrix}\right.\left[{}\begin{matrix}x=4\\x\neg-\dfrac{3}{5}\end{matrix}\right.\)

\(\left[{}\begin{matrix}x+\dfrac{3}{4}=\dfrac{1}{3}\\x+\dfrac{3}{4}=-\dfrac{1}{3}\end{matrix}\right.\left[{}\begin{matrix}x=-\dfrac{5}{12}\\x=-\dfrac{13}{12}\end{matrix}\right.\)

ĐK: `x \ne kπ`

`cot(x-π/4)+cot(π/2-x)=0`

`<=>cot(x-π/4)=-cot(π/2-x)`

`<=>cot(x-π/4)=cot(x-π/2)`

`<=> x-π/4=x-π/2+kπ`

`<=>0x=-π/4+kπ` (VN)

Vậy PTVN.

hahihihihi

\(\dfrac{1}{x}+\dfrac{3}{2x}-\dfrac{5}{24}=0\)

\(\Leftrightarrow24+36-5x=0\)

\(\Leftrightarrow5x=60\)

\(\Leftrightarrow x=12\)

\(\dfrac{x-5}{3}+\dfrac{x+1}{2}>3\)

\(\Leftrightarrow2x-10+3x+3>18\)

\(\Leftrightarrow5x>25\)

\(\Leftrightarrow x>5\)

\(\dfrac{2}{x-1}+\dfrac{1}{x+2}=\dfrac{2x}{\left(x+2\right)\left(x-1\right)}\)

\(\Leftrightarrow2x+4+x-1=2x\)

\(\Leftrightarrow x=-3\)

a)Đk:\(x\ne4\)

\(\dfrac{x^4}{4-x}+x^3+1=\dfrac{x^4+\left(x^3+1\right)\left(4-x\right)}{4-x}\)\(=\dfrac{x^4+\left(-x^4+4x^3+4-x\right)}{4-x}=\dfrac{4x^3-x+4}{4-x}\)

b) Đk: \(x\ne0;x\ne1\)

\(\dfrac{1}{x^2-x}+\dfrac{2x}{x-1}=\dfrac{1}{x\left(x-1\right)}+\dfrac{2x^2}{x\left(x-1\right)}=\dfrac{1+2x^2}{x\left(x-1\right)}\)

\(1,ĐK:x\ge2\\ PT\Leftrightarrow\sqrt{3x-6}+x-2-\left(\sqrt{2x-3}-1\right)=0\\ \Leftrightarrow\dfrac{3\left(x-2\right)}{\sqrt{3x-6}}+\left(x-2\right)-\dfrac{2\left(x-2\right)}{\sqrt{2x-3}+1}=0\\ \Leftrightarrow\left(x-2\right)\left(\dfrac{3}{\sqrt{3x-6}}-\dfrac{2}{\sqrt{2x-3}+1}+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\left(tm\right)\\\dfrac{3}{\sqrt{3x-6}}-\dfrac{2}{\sqrt{2x-3}+1}+1=0\left(1\right)\end{matrix}\right.\)

Với \(x>2\Leftrightarrow-\dfrac{2}{\sqrt{2x-3}+1}>-\dfrac{2}{1+1}=-1\left(3x-6\ne0\right)\)

\(\Leftrightarrow\left(1\right)>0-1+1=0\left(vn\right)\)

Vậy \(x=2\)

\(2,ĐK:x\ge-1\)

Đặt \(\left\{{}\begin{matrix}\sqrt{x+1}=a\\\sqrt{x^2-x+1}=b\end{matrix}\right.\left(a,b\ge0\right)\Leftrightarrow a^2+b^2=x^2+2\)

\(PT\Leftrightarrow2a^2+2b^2-5ab=0\\ \Leftrightarrow\left(a-2b\right)\left(2a-b\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}a=2b\\b=2a\end{matrix}\right.\)

Với \(a=2b\Leftrightarrow x+1=4x^2-4x+4\left(vn\right)\)

Với \(b=2a\Leftrightarrow4x+4=x^2-x+1\Leftrightarrow x^2-5x-3=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5+\sqrt{37}}{2}\left(tm\right)\\x=\dfrac{5-\sqrt{37}}{2}\left(tm\right)\end{matrix}\right.\)

Vậy ...

\(3,ĐK:x\ge-1\\ PT\Leftrightarrow3\left(x^2-x+1\right)-2\left(x+1\right)=5\sqrt{x^3+1}\) 

Đặt \(\left\{{}\begin{matrix}\sqrt{x+1}=a\\\sqrt{x^2-x+1}=b\end{matrix}\right.\left(a,b\ge0\right)\)

\(PT\Leftrightarrow3b^2-2a^2=5ab\\ \Leftrightarrow2a^2+5ab-3b^2=0\\ \Leftrightarrow\left[{}\begin{matrix}a=2b\\a=-3b\left(vn\right)\end{matrix}\right.\Leftrightarrow a=2b\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5+\sqrt{37}}{2}\\x=\dfrac{5-\sqrt{37}}{2}\end{matrix}\right.\left(\text{giống bài 2}\right)\)

E tk nha: