\(x^4+2x^3+8x^2+10x+15=0\)

\(\Leftrightarrow\left(x^4+5x^2\right)+\left(2x^3+10x\right)+\left(3x^2+15\right)=0\)

\(\Leftrightarrow x^2\left(x^2+5\right)+2x\left(x^2+5\right)+3\left(x^2+5\right)=0\)

\(\Leftrightarrow\left(x^2+5\right)\left(x^2+2x+3\right)=0\)

mà ta có: \(x^2+5\ge5>0;x^2+2x+3=\left(x+1\right)^2+1\ge1>0\)

nên suy ra phương trình vô nghiệm.

\(x^4+2x^3+8x^2+10x+15=\left(x^4+2x^3+x^2\right)+\left(7x^2+10x+15\right)\)

\(\Leftrightarrow\left(x^2+x\right)^2+2.4.\left(x^2+x\right)+16=x^2-2x+1\\ \)

\(\left(x^2+x+4\right)^2=\left(x-1\right)^2\)

\(\left[\begin{matrix}x^2+x+4=x-1\left(1\right)\\x^2+x+4=1-x\left(2\right)\end{matrix}\right.\)

\(\left[\begin{matrix}\left(1\right)\Leftrightarrow x^2=-5\\\left(x+1\right)^2=-3\end{matrix}\right.\)Vo. N_o

(x^4+2x^3+3x^2)+(5x^2+10x+15)=0

x^2(x^2+2x+3)+5(x^2+2x+3)=0

(x^2+2x+3)(x^2+5)=0

x^2+2x+3=0 hoặc x^2+5=0

Mà:x^2+2x^3+3=(x+1)^2+2>0 suy ra pt vô nghiệm.

x^2+5>0 suy ra pt vô nghiệm.

Vậy pt đã cho vô nghiệm.

Nhớ chọn đúng nha chó yến như :p :p :p

k,\(\dfrac{x}{3}-\dfrac{2x+1}{2}=\dfrac{x}{6}-x\)

giúp mk câu k nhé đề bài như trên

b: \(\Leftrightarrow4x+8-9=4x-4\)

=>-1=-4(loại)

d: \(\Leftrightarrow3\left(x-2\right)+2\left(x+1\right)=8x\)

=>8x=3x-6+2x+2=5x-4

=>3x=-4

=>x=-4/3

f: \(\Leftrightarrow3\left(x+2\right)+4\left(2x-3\right)=2\left(x-12\right)\)

=>3x+6+8x-12=2x-24

=>11x-6=2x-24

=>9x=-18

=>x=-2

`b,2(x+1)=5x-7`

`=>2x+2=5x-7`

`=>3x=9`

`=>x=3`

`c,3-4x(25-2x)=8x^2+x-300`

`<=>3-100x+8x^2=8x^2+x-300`

`<=>101x=303`

`<=>x=3`

`d,(10x+3)/12=1+(6+8x)/9`

`<=>(10x+3)/12=(8x+15)/9`

`<=>30x+9=32x+60`

`<=>2x=-51`

`<=>x=-51/2`

1:

a: =>3x=6

=>x=2

b: =>4x=16

=>x=4

c: =>4x-6=9-x

=>5x=15

=>x=3

d: =>7x-12=x+6

=>6x=18

=>x=3

2:

a: =>2x<=-8

=>x<=-4

b: =>x+5<0

=>x<-5

c: =>2x>8

=>x>4

ĐKXĐ: \(x\ne0\)

Phương trình tương đương:

\(\dfrac{4}{4x-8+\dfrac{7}{x}}+\dfrac{3}{4x-10+\dfrac{7}{x}}=1\)

Đặt \(4x-10+\dfrac{7}{x}=t\)

\(\Rightarrow\dfrac{4}{t+2}+\dfrac{3}{t}=1\)

\(\Rightarrow4t+3\left(t+2\right)=t\left(t+2\right)\)

\(\Leftrightarrow t^2-5t-6=0\Rightarrow\left[{}\begin{matrix}t=-1\\t=6\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}4x-10+\dfrac{7}{x}=-1\\4x-10+\dfrac{7}{x}=6\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}4x^2-9x+7=0\left(vn\right)\\4x^2-16x+7=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=\dfrac{1}{2}\end{matrix}\right.\)