Đa thức này không phân tích được thành nhân tử.

Bạn coi lại đề.

Ta có: \(4\left(x+5\right)\left(x+6\right)\left(x+10\right)\left(x+12\right)+3x^2\)

\(=4\left(x^2+60+17x\right)\left(x^2+60x+16x\right)+3x^2\)

\(=4\left[\left(x^2+60\right)^2+33x\left(x^2+60\right)+272x^2\right]+3x^2\)

\(=4\left(x^2+60\right)^2+132x\left(x^2+60\right)+1091x^2\)

Lần sau đừng tự tiện xếp vào phần bất pt bạn nhé :(

Ta có : \(4\left(x+5\right)\left(x+6\right)\left(x+10\right)\left(x+12\right)=3x^2\)

\(\Leftrightarrow4\left(x+5\right)\left(x+12\right)\left(x+6\right)\left(x+10\right)=3x^2\)

\(\Leftrightarrow4\left(x^2+17x+60\right)\left(x^2+16x+60\right)=3x^2\)(1)

Đặt \(x^2+16x+60=a\)

Pt (1) \(\Leftrightarrow4\left(a+x\right)a=3x^2\)

         \(\Leftrightarrow4\left(a^2+ax\right)=3x^2\)

          \(\Leftrightarrow4a^2+4ax=3x^2\)

          \(\Leftrightarrow4a^2+4ax+x^2=4x^2\)

         \(\Leftrightarrow\left(2a+x\right)^2=4x^2\)

          \(\Leftrightarrow\orbr{\begin{cases}2a+x=2x\\2a+x=-2x\end{cases}}\)

*Nếu \(2a+x=2x\)

\(\Leftrightarrow2a=x\)

\(\Leftrightarrow x^2+16x+60=x\)

\(\Leftrightarrow x^2+15x+60=0\)

\(\Leftrightarrow x^2+2.\frac{15}{2}.x+\frac{225}{4}+\frac{15}{4}=0\)

\(\Leftrightarrow\left(x+\frac{15}{2}\right)^2+\frac{15}{4}=0\)

Pt vô nghiệm

*Nếu \(2a+x=-2x\)

\(\Leftrightarrow2a+3x=0\)

\(\Leftrightarrow2\left(x^2-16x+60\right)+3x=0\)

\(\Leftrightarrow2x^2-32x+120+3x=0\)

\(\Leftrightarrow2x^2-29x+120=0\)

\(\Leftrightarrow x^2-\frac{29}{2}x+60=0\)

\(\Leftrightarrow x^2-2.\frac{29}{4}.x+\frac{841}{16}+\frac{119}{16}=0\)

\(\Leftrightarrow\left(x-\frac{29}{4}\right)^2+\frac{119}{16}=0\)

Pt vô nghiệm

Vậy pt vô nghiệm

a) ĐKXĐ: \(x\notin\left\{2;-2\right\}\)

Ta có: \(\dfrac{x+1}{x-2}-\dfrac{5}{x+2}=\dfrac{12}{x^2-4}+1\)

\(\Leftrightarrow\dfrac{\left(x+1\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\dfrac{5\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}=\dfrac{12}{\left(x-2\right)\left(x+2\right)}+\dfrac{x^2-4}{\left(x-2\right)\left(x+2\right)}\)

Suy ra: \(x^2+3x+2-5x+10=12+x^2-4\)

\(\Leftrightarrow x^2-2x+12-8-x^2=0\)

\(\Leftrightarrow-2x+4=0\)

\(\Leftrightarrow-2x=-4\)

hay x=2(loại)

Vậy: \(S=\varnothing\)

b) Ta có: \(\left|2x+6\right|-x=3\)

\(\Leftrightarrow\left|2x+6\right|=x+3\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+6=x+3\left(x\ge-3\right)\\-2x-6=x+3\left(x< -3\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-x=3-6\\-2x-x=3+6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\left(nhận\right)\\x=-3\left(loại\right)\end{matrix}\right.\)

Vậy: S={-3}

\(\dfrac{1}{\left(x-1\right)\left(x-2\right)}+\dfrac{1}{\left(x-2\right)\left(x-3\right)}+\dfrac{1}{\left(x-3\right)\left(x-4\right)}+\dfrac{1}{\left(x-4\right)\left(x-5\right)}+\dfrac{1}{\left(x-5\right)\left(x-6\right)}=\dfrac{1}{10}\)

\(\Leftrightarrow\dfrac{1}{x-1}-\dfrac{1}{x-2}+\dfrac{1}{x-2}-\dfrac{1}{x-3}+\dfrac{1}{x-3}-....+\dfrac{1}{x-5}-\dfrac{1}{x-6}=\dfrac{1}{10}\)

\(\Leftrightarrow\dfrac{1}{x-1}-\dfrac{1}{x-6}=\dfrac{1}{10}\Leftrightarrow\dfrac{x-6-x+1}{\left(x-1\right)\left(x-6\right)}=\dfrac{1}{10}\)

\(\Leftrightarrow x^2-7x+56=0\Leftrightarrow x^2-2.\dfrac{7}{2}x+\dfrac{49}{4}+\dfrac{175}{4}=\left(x-\dfrac{7}{2}\right)^2+\dfrac{175}{4}>0\)

Vậy phương trình vô nghiệm 

ĐKXĐ: \(x\notin\left\{1;2;3;4;5;6\right\}\)

Ta có: \(\dfrac{1}{\left(x-1\right)\left(x-2\right)}+\dfrac{1}{\left(x-2\right)\left(x-3\right)}+\dfrac{1}{\left(x-3\right)\left(x-4\right)}+\dfrac{1}{\left(x-4\right)\left(x-5\right)}+\dfrac{1}{\left(x-5\right)\left(x-6\right)}=\dfrac{1}{10}\)

\(\Leftrightarrow\dfrac{1}{x-2}-\dfrac{1}{x-1}+\dfrac{1}{x-3}-\dfrac{1}{x-2}+\dfrac{1}{x-4}+\dfrac{1}{x-3}+\dfrac{1}{x-5}-\dfrac{1}{x-4}+\dfrac{1}{x-6}-\dfrac{1}{x-5}=\dfrac{1}{10}\)

\(\Leftrightarrow\dfrac{1}{x-6}-\dfrac{1}{x-1}=\dfrac{1}{10}\)

\(\Leftrightarrow\dfrac{10\left(x-1\right)}{10\left(x-6\right)\left(x-1\right)}-\dfrac{10\left(x-6\right)}{10\left(x-1\right)\left(x-6\right)}=\dfrac{\left(x-1\right)\left(x-6\right)}{10\left(x-1\right)\left(x-6\right)}\)

Suy ra: \(x^2-7x+6=10x-10-10x+60\)

\(\Leftrightarrow x^2-7x+6=50\)

\(\Leftrightarrow x^2-7x-44=0\)

\(\Leftrightarrow x^2-11x+4x-44=0\)

\(\Leftrightarrow x\left(x-11\right)+4\left(x-11\right)=0\)

\(\Leftrightarrow\left(x-11\right)\left(x+4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-11=0\\x+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=11\left(nhận\right)\\x=-4\left(nhận\right)\end{matrix}\right.\)

Vậy: S={11;-4}

ĐKXĐ : \(x\notin\left\{1;2;...;6\right\}\)

\(\dfrac{1}{\left(x-1\right)\left(x-2\right)}+\dfrac{1}{\left(x-2\right)\left(x-3\right)}+...+\dfrac{1}{\left(x-5\right)\left(x-6\right)}=\dfrac{1}{10}\\ \Leftrightarrow\dfrac{\left(x-1\right)-\left(x-2\right)}{\left(x-1\right)\left(x-2\right)}+\dfrac{\left(x-2\right)-\left(x-3\right)}{\left(x-2\right)\left(x-3\right)}+...+\dfrac{\left(x-5\right)-\left(x-6\right)}{\left(x-5\right)\left(x-6\right)}=\dfrac{1}{10}\\ \Leftrightarrow\dfrac{1}{x-2}-\dfrac{1}{x-1}+\dfrac{1}{x-3}-\dfrac{1}{x-2}+...+\dfrac{1}{x-6}-\dfrac{1}{x-5}=\dfrac{1}{10}\\ \Leftrightarrow\dfrac{1}{x-6}-\dfrac{1}{x-1}=\dfrac{1}{10}\\ \Leftrightarrow\dfrac{5}{\left(x-1\right)\left(x-6\right)}=\dfrac{5}{50}\\ \Rightarrow\left(x-1\right)\left(x-6\right)=50\\ \Leftrightarrow x^2-7x-44=0\\ \Leftrightarrow\left(x-11\right)\left(x+4\right)=0\\ \Leftrightarrow\begin{matrix}x=-4\\x=11\end{matrix}\left(t.m\right)\)

Bài 2:

a: =>2x^2-4x+1=x^2+x+5

=>x^2-5x-4=0

=>\(x=\dfrac{5\pm\sqrt{41}}{2}\)

b: =>11x^2-14x-12=3x^2+4x-7

=>8x^2-18x-5=0

=>x=5/2 hoặc x=-1/4

1: =>(x+2)^2-3|x+2|=0

=>|x+2|(|x+2|-3)=0

=>x+2=0 hoặc x+2=3 hoặc x+2=-3

=>x=-2; x=1; x=-5