\(\left(\dfrac{7}{2}x-7\right)\left(21x-63\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}\dfrac{7}{2}x-7=0\\21x-63=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}\dfrac{7}{2}x=7\\21x=63\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=3\end{matrix}\right.\)

TH1: 2

TH2: 3

c) \(\dfrac{x-1}{2}+\dfrac{x-1}{3}+\dfrac{x-1}{2016}=0\\ \Leftrightarrow\dfrac{1008\left(x-1\right)}{2016}+\dfrac{672\left(x-1\right)}{2016}+\dfrac{x-1}{2016}=0\\ \Leftrightarrow1008x-1008+672x-672+x-1=0\\ \Leftrightarrow1008x+672x+x=0+1+672+1008\\ \Leftrightarrow1681x=1681\\ \Leftrightarrow x=1\)

d) \(\dfrac{x-4}{5}+\dfrac{3x-4}{10}-x=\dfrac{2x-5}{3}-\dfrac{7x+3}{6}\\ \Leftrightarrow\dfrac{6\left(x-4\right)}{30}+\dfrac{3\left(3x-4\right)}{30}-\dfrac{30x}{30}=\dfrac{10\left(2x-5\right)}{30}-\dfrac{5\left(7x+3\right)}{30}\\ \Leftrightarrow6x-24+9x-12-30x=20x-50-35x+15\\ \Leftrightarrow6x+9x-20x+35x-30x=-50+15+24+12\\ \Leftrightarrow0x=1\left(vô.lí\right)\)

c. \(\dfrac{x-1}{x-2}+\dfrac{5}{x+2}=\dfrac{12}{x^2-4}+1\\ ĐKXĐ:x\ne2;x\ne-2\\ \Leftrightarrow\dfrac{\left(x-1\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\dfrac{5\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{12}{\left(x-2\right)\left(x+2\right)}+\dfrac{\left(x-2\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\\ \Rightarrow x^2+2x-x-2+5x-10=12+x^2-4\\ \Leftrightarrow x^2+2x-x+5x-x^2=12-4+2+10\\ \Leftrightarrow6x=20\\ \Leftrightarrow x=\dfrac{20}{6}\\ \Leftrightarrow x=\dfrac{10}{3}\left(nhận\right)\)

đk : x khác -2 ;2 

\(x^2+x-2+5x-10=12+x^2-4\Leftrightarrow6x-12=8\Leftrightarrow x=\dfrac{10}{3}\left(tm\right)\)

ĐKXĐ: \(\hept{\begin{cases}x\ne0\\x\ne-5\end{cases}}\)

pt đã cho \(\Leftrightarrow\frac{90\left(x+5\right)-90x}{x\left(x+5\right)}=\frac{1}{4}\)\(\Leftrightarrow\frac{90x+450-90x}{x^2+5x}=\frac{1}{4}\)\(\Leftrightarrow\frac{450}{x^2+5x}=\frac{1}{4}\)\(\Rightarrow x^2+5x=1800\)\(\Leftrightarrow x^2+5x-1800=0\)\(\Leftrightarrow x^2+45x-40x-1800=0\)\(\Leftrightarrow x\left(x+45\right)-40\left(x+45\right)=0\)\(\Leftrightarrow\left(x+45\right)\left(x-40\right)=0\)\(\Leftrightarrow\orbr{\begin{cases}x=-45\left(nhận\right)\\x=40\left(nhận\right)\end{cases}}\)

Vậy tập nghiệm của pt đã cho là \(S=\left\{-45;40\right\}\)

\(f\left(x\right)=\left(x+1\right)\left[\left(x+2\right)\left(x+5\right)\right]\left[\left(x+3\right)\left(x+4\right)\right]+2014\)

\(=\left(x+1\right)\left(x^2+7x+10\right)\left(x^2+7x+12\right)+2014\)

\(x^2+7x+10\)chia \(g\left(x\right)\)dư 2

\(x^2+7x+12\)chia \(g\left(x\right)\)dư 4

Vậy \(f\left(x\right)\)chia \(g\left(x\right)\)dư \(\left(x+1\right).4.2+2014=8x+2022\)

Nhận thấy \(x=0\)không phải là nghiệm của pt đã cho nên ta có thể chia cả 2 vế của pt này cho \(x^2\). Khi đó, ta được:

\(\frac{x^4-2x^3+3x^2-2x+1}{x^2}=\frac{0}{x^2}\)\(\Leftrightarrow x^2-2x+3-\frac{2}{x}+\frac{1}{x^2}=0\)\(\Leftrightarrow\left(x^2+\frac{1}{x^2}\right)-2\left(x+\frac{1}{x}\right)+3=0\)\(\Leftrightarrow\left(x^2+2+\frac{1}{x^2}\right)-2\left(x+\frac{1}{x}\right)+1=0\)\(\Leftrightarrow\left(x^2+2x.\frac{1}{x}+\frac{1}{x^2}\right)-2\left(x+\frac{1}{x}\right)+1=0\)\(\Leftrightarrow\left(x+\frac{1}{x}\right)^2-2\left(x+\frac{1}{x}\right)+1=0\)\(\Leftrightarrow\left(x+\frac{1}{x}-1\right)^2=0\)\(\Leftrightarrow x+\frac{1}{x}-1=0\)\(\Leftrightarrow x^2+1-x=0\)(*)

Mà \(x^2-x+1=\left(x^2-2x.\frac{1}{2}+\frac{1}{4}\right)+\frac{3}{4}=\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}>0\)nên (*) vô nghiệm

Vậy pt đã cho vô nghiệm.

TL

=> 143547

~HT~