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

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

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

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

Vậy phương trình vô nghiệm 

oke cảm ơn bn nhìu :)))

bai 1

1 thay k=0 vao pt ta co 4x^2-25+0^2+4*0*x=0

<=>(2x)^2-5^2=0

<=>(2x+5)*(2x-5)=0

<=>2x+5=0 hoăc 2x-5 =0 tiếp tục giải ý 2 tương tự

2:

a: =>x-1=0 hoặc 3x+1=0

=>x=1 hoặc x=-1/3

b: =>x-5=0 hoặc 7-x=0

=>x=5 hoặc x=7

c: =>\(\left[{}\begin{matrix}x-1=0\\x+5=0\\3x-8=0\end{matrix}\right.\Leftrightarrow x\in\left\{1;-5;\dfrac{8}{3}\right\}\)

d: =>x=0 hoặc x^2-1=0

=>\(x\in\left\{0;1;-1\right\}\)

Bạn tách ra từng câu thoi nhe .

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}

Ta có : |1 - 5x| - 1 = 3

=> |1 - 5x| = 4

\(\Leftrightarrow\orbr{\begin{cases}1-5x=4\\1-5x=-4\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}5x=1-4\\5x=1+4\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}5x=3\\5x=5\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{3}{5}\\x=1\end{cases}}\)

2. \(x\left(x+2\right)\left(x+3\right)\left(x+5\right)=280\)

\(\Leftrightarrow x\left(x+5\right)\left(x+2\right)\left(x+3\right)=280\)

\(\Leftrightarrow\left(x^2+5x\right)\left(x^2+5x+6\right)=280\)

Đặt \(x^2+5x+3=t\)

\(\Rightarrow\left(t-3\right)\left(t+3\right)=280\)

\(\Leftrightarrow t^2-9=280\)

\(\Leftrightarrow t^2=289\Leftrightarrow\left[{}\begin{matrix}t=17\\t=-17\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2+5x+3=17\\x^2+5x+3=-17\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2+5x-14=0\\x^2+5x+20=0\end{matrix}\right.\)

\(\Leftrightarrow x^2+5x-14=0\text{(vì }x^2+5x+20=\left(x+\dfrac{5}{2}\right)^2+\dfrac{55}{4}>0\forall x\text{)}\)

\(\Leftrightarrow x^2-2x+7x-14=0\)

\(\Leftrightarrow x\left(x-2\right)+7\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x+7\right)=0\)

\(\Leftrightarrow\) x - 2 = 0 hoặc x + 7 = 0

\(\Leftrightarrow\) x = 2 hoặc x = - 7

Vậy x = 2 hoặc x = -7.

3. \(\left(x+3\right)\left(x+4\right)\left(x+5\right)=x\)

\(\Leftrightarrow\left(x+3\right)\left(x+4\right)\left(x+5\right)-x=0\)

\(\Leftrightarrow x^3+12x^2+47x+60-x=0\)

\(\Leftrightarrow x^3+12x^2+46x+60=0\)

\(\Leftrightarrow x^3+6x^2+6x^2+36x+10x+60=0\)

\(\Leftrightarrow x^2\left(x+6\right)+6x\left(x+6\right)+10\left(x+6\right)=0\)

\(\Leftrightarrow\left(x+6\right)\left(x^2+6x+10\right)=0\)

\(\Leftrightarrow x+6=0\text{(vì }x^2+6x+10=\left(x+3\right)^2+1>0\forall x\text{)}\)

\(\Leftrightarrow x=-6\)

Vậy x = -6.

thansk you