a)

\(\dfrac{1}{x+1}+\dfrac{2}{x^3-x^2-x+1}+\dfrac{3}{x^2-1}=0\) (\(x\ne\pm1\))

\(\Rightarrow\dfrac{\left(x-1\right)^2}{\left(x+1\right)\left(x-1\right)^2}+\dfrac{2}{\left(x+1\right)\left(x-1\right)^2}+\dfrac{3\left(x-1\right)}{\left(x+1\right)\left(x-1\right)^2}=0\)

\(\Rightarrow\dfrac{x^2-2x+1+2+3x-3}{\left(x+1\right)\left(x-1\right)^2}=0\)

\(\Rightarrow\dfrac{x^2+x-2}{\left(x+1\right)\left(x-1\right)^2}=0\)

\(\Rightarrow x^2-x+2=0\)

\(\Rightarrow\left(x-1\right)\left(x+2\right)=0\)

=> Th1 :

x- 1 =0

=> x = 1 ( hư cấu vì không thỏa mãn ĐK )

Th2 :

x+2 = 0

=> x = -2 ( hợp lí )

Vậy nghiệm của phương trình là x = -2

a.

\(\dfrac{1}{x^2-7x+12}+\dfrac{1}{x^2-11x+28}+\dfrac{1}{x^2-19x+84}=\dfrac{1}{4}\\ \Rightarrow\dfrac{1}{\left(x-3\right)\left(x-4\right)}+\dfrac{1}{\left(x-4\right)\left(x-7\right)}+\dfrac{1}{\left(x-7\right)\left(x-12\right)}=\dfrac{1}{4}\\ \Rightarrow\dfrac{1}{x-3}-\dfrac{1}{x-12}=\dfrac{1}{4}\\ \Rightarrow\dfrac{-9}{\left(x-3\right)\left(x-12\right)}=\dfrac{1}{4}\\ \Rightarrow x^2-15x+36=-36\\ \)

Tự giải tiếp

các bn giúp mik với!! vài câu cx được

a: \(\Leftrightarrow-12x-4=8x-2-8-6x\)

=>-12x-4=2x-10

=>-14x=-6

hay x=3/7

b: \(\Leftrightarrow3\left(5x-3\right)-2\left(5x-1\right)=-4\)

=>15x-9-10x+2=-4

=>5x-7=-4

=>5x=3

hay x=3/5(loại)

c: \(\Leftrightarrow x^2-4+3x+3=3+x^2-x-2\)

\(\Leftrightarrow x^2+3x-1=x^2-x+1\)

=>4x=2

hay x=1/2(nhận)

\(a.\Leftrightarrow\frac{5x^2+16}{\left(x+4\right)\left(x-4\right)}=\frac{\left(2x-1\right)\left(x-4\right)+\left(3x-1\right)\left(x+4\right)}{\left(x+4\right)\left(x-4\right)}DKXD:x\ne4;-4\)

\(\Rightarrow5x^2+16=2x^2-8x-x+4+3x^2+12x-x-4\)

\(\Leftrightarrow2x=16\)

\(\Leftrightarrow x=8\)

\(b.\Leftrightarrow\frac{\left(y+1\right)\left(y+2\right)-5\left(y-2\right)}{\left(y-2\right)\left(y+2\right)}=\frac{12+\left(y-2\right)\left(y+2\right)}{\left(y-2\right)\left(y+2\right)}.DKXD:y\ne2;-2\)

\(\Rightarrow y^2+2y+y+2-5y+10=12+y^2-4\)

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

\(\Leftrightarrow y=2\)

a: \(\Leftrightarrow20x^2-12x+15x+5< 10x\left(2x+1\right)-30\)

\(\Leftrightarrow20x^2+3x+5< 20x^2+10x-30\)

=>3x+5<10x-30

=>-7x<-35

hay x>5

b: \(\Leftrightarrow4\left(5x-20\right)-6\left(2x^2+x\right)>4x\left(1-3x\right)-15x\)

\(\Leftrightarrow20x-80-12x^2-6x>4x-12x^2-15x\)

=>14x-80>-11x

=>25x>80

hay x>16/5

Nhật Linh Liệu có cách khác nhẹ nhàng hơn không bạn

a: \(\dfrac{1}{x^2+9x+20}+\dfrac{1}{x^2+11x+30}+\dfrac{1}{x^2+13x+42}=\dfrac{1}{18}\)

\(\Leftrightarrow\dfrac{1}{x+4}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+6}+\dfrac{1}{x+6}-\dfrac{1}{x+7}=\dfrac{1}{18}\)

=>\(\dfrac{x+7-x-4}{\left(x+4\right)\left(x+7\right)}=\dfrac{1}{18}\)

=>(x+4)(x+7)=54

=>x^2+11x+28-54=0

=>(x+13)(x-2)=0

=>x=-13 hoặc x=2

b: \(\Leftrightarrow\dfrac{1}{x+1}-\dfrac{1}{x+2}+\dfrac{1}{x+2}-...+\dfrac{1}{x+4}-\dfrac{1}{x+5}=\dfrac{1}{3}\)

=>\(\dfrac{x+5-x-1}{\left(x+5\right)\left(x+1\right)}=\dfrac{1}{3}\)

=>x^2+6x+5=12

=>x^2+6x-7=0

=>(x+7)(x-1)=0

=>x=-7 hoặc x=1

1) điều kiện xác định : \(x\notin\left\{-1;-2;-3;-4\right\}\)

ta có : \(\dfrac{1}{x^2+3x+2}+\dfrac{1}{x^2+5x+6}+\dfrac{1}{x^2+7x+12}=\dfrac{1}{6}\)

\(\Leftrightarrow\dfrac{x^2+7x+12+x^2+5x+4+x^2+3x+2}{\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)}=\dfrac{1}{6}\)

\(\Leftrightarrow\dfrac{3x^2+15x+18}{\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)}=\dfrac{1}{6}\)

\(\Leftrightarrow6\left(3x^2+15x+18\right)=\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)\)

\(\Leftrightarrow18\left(x^2+5x+6\right)=\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)\)

\(\Leftrightarrow18\left(x+2\right)\left(x+3\right)=\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)\)

\(\Leftrightarrow18=\left(x+1\right)\left(x+4\right)\) ( vì điều kiện xác định )

\(\Leftrightarrow18=x^2+5x+4\Leftrightarrow x^2+5x-14=0\)

\(\Leftrightarrow\left(x-2\right)\left(x+7\right)=0\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+7=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-7\end{matrix}\right.\left(tmđk\right)\)

vậy \(x=2\) hoặc \(x=-7\) mấy câu kia lm tương tự nha bn