`(x+2)(x^2 -2x+4) -x(x^2-2)=15`

`<=> x^3 +8 - x^3 + 2x-15=0`

`<=> 2x-7=0`

`<=> 2x=7`

`<=>x=7/2`

__

`(x-4)^2 -(x-2)(x+2)=6`

`<=>x^2 - 8x+16- x^2 +4-6=0`

`<=> -8x+14=0`

`<=> -8x=-14`

`<=>x=14/8= 7/4`

__

`x^4 -2x^3 +x^2-2x=0`

`<=>x(x^3-2x^2+x-2)=0`

`<=> x(x^3+x-2x^2-2)=0`

`<=>x(x(x^2+1) -2(x^2+1))=0`

`<=> x(x^2+1)(x-2)=0`

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x^2+1=0\\x-2=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)

a) \(\left(x+2\right)\left(x^2-2x+4\right)-x\left(x^2-2\right)=15\)

\(\Leftrightarrow\left(x^3+2^3\right)-\left(x^3-2x\right)=15\)

\(\Leftrightarrow x^3+8-x^3+2x=15\)

\(\Leftrightarrow2x+8=15\) 

\(\Leftrightarrow2x=15-8\)

\(\Leftrightarrow2x=7\)

\(\Leftrightarrow x=\dfrac{7}{2}\)

b) \(\left(x-4\right)^2-\left(x+2\right)\left(x-2\right)=6\)

\(\Leftrightarrow x^2-8x+16-\left(x^2-4\right)=6\)

\(\Leftrightarrow x^2-8x+16-x^2+4=6\)

\(\Leftrightarrow-8x+20=6\)

\(\Leftrightarrow-8x=6-20\)

\(\Leftrightarrow-8x=-14\)

\(\Leftrightarrow x=\dfrac{7}{4}\) 

c) \(x^4-2x^3+x^2-2x=0\)

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

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

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

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

|2x+6|+ |x+2| + |2x+2| =4

=>2x+6+ x+2 + 2x+2 =4 hoặc 2x+6+ x+2 + 2x+2 =-4

=>2x+6+ x+2 + 2x+2 =4         2x+6+ x+2 + 2x+2 =-4

=>(2x+x+2x)+(6+2+2)=4         (2x+x+2x)+(6+2+2)=-4

=> 5x+10=4                              5x+10=-4   

=>5x=4-10                                 5x=(-4)-10     

=>5x=-6                                     5x=-14

=>x=(-6):5                                     x= (-14):5

=>x=-6/5                                         x=-14/5

vậy x∈{-6,5;-14/5}

\(\left|2x+6\right|+\left|x+2\right|+\left|2x+2\right|=4\)

\(\Leftrightarrow2\left|x+3\right|+\left|x+2\right|+2\left|x+1\right|=4\) (1)

+, Với \(x< -3\) thì (1) trờ thành:

\(2(-x-3)+(-x-2)+2(-x-1)=4\)

\(\Rightarrow-2x-6-x-2-2x-2=4\)

\(\Rightarrow-5x-10=4\)

\(\Rightarrow-5x=14\)

\(\Rightarrow x=-\dfrac{14}{5}\left(ktm\right)\)

+, Với \(-3\le x< -2\), thì (1) trở thành:

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

\(\Rightarrow2x+6-x-2-2x-2=4\)

\(\Rightarrow-x+2=4\)

\(\Rightarrow x=-2\left(ktm\right)\)

+, Với \(-2\le x< -1\), thì (1) trở thành:

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

\(\Rightarrow2x+6+x+2-2x-2=4\)

\(\Rightarrow x+6=4\)

\(\Rightarrow x=-2\left(tm\right)\)

+, Với \(x\ge-1\), thì (1) trở thành:

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

\(\Rightarrow2x+6+x+2+2x+2=4\)

\(\Rightarrow5x+10=4\)

\(\Rightarrow5x=-6\)

\(\Rightarrow x=-\dfrac{6}{5}\left(ktm\right)\)

Vậy \(x=-2\)

#Urushi☕

1: =>(x+3)(x-5)=0

=>x=5 hoặc x=-3

2: =>(x-1)(5x-1)=0

=>x=1/5 hoặc x=1

5: =>(x-4)*x=0

=>x=0 hoặc x=4

10: =>(x+5)(x-3)=0

=>x=3 hoặc x=-5

9: =>(x-2)(x-4)=0

=>x=2 hoặc x=4

7: =>(x-6)(2x-1)=0

=>x=1/2 hoặc x=6

8: =>(2x-1)(3x-12)=0

=>x=4 hoặc x=1/2

\(a,3\left(2x-3\right)+2\left(2-x\right)=-3\\ \Leftrightarrow6x-9+4-2x=-3\\ \Leftrightarrow4x=2\\ \Leftrightarrow x=\dfrac{1}{2}\\ b,x\left(5-2x\right)+2x\left(x-1\right)=13\\ \Leftrightarrow5x-2x^2+2x^2-2x=13\\ \Leftrightarrow3x=13\\ \Leftrightarrow x=\dfrac{13}{3}\\ c,5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\\ \Leftrightarrow5x^2-5x-5x^2-3x+14=6\\ \Leftrightarrow-8x=-8\\ \Leftrightarrow x=1\\ d,3x\left(2x+3\right)-\left(2x+5\right)\left(3x-2\right)=8\\ \Leftrightarrow6x^2+9x-6x^2-11x+10=8\\ \Leftrightarrow-2x=-2\\ \Leftrightarrow x=1\)

\(e,2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\\ \Leftrightarrow10x-16-12x+15=12x-16+11\\ \Leftrightarrow-14x=-4\\ \Leftrightarrow x=\dfrac{2}{7}\\ f,2x\left(6x-2x^2\right)+3x^2\left(x-4\right)=8\\ \Leftrightarrow12x^2-4x^3+3x^3-12x^2=8\\ \Leftrightarrow-x^3-8=0\\ \Leftrightarrow-\left(x^3+8\right)=0\\ \Leftrightarrow-\left(x+2\right)\left(x^2-2x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-2\\x\in\varnothing\left(x^2-2x+4=\left(x-1\right)^2+3>0\right)\end{matrix}\right.\)

Bài 4:

a: Ta có: \(3\left(2x-3\right)-2\left(x-2\right)=-3\)

\(\Leftrightarrow6x-9-2x+4=-3\)

\(\Leftrightarrow4x=2\)

hay \(x=\dfrac{1}{2}\)

b: Ta có: \(x\left(5-2x\right)+2x\left(x-1\right)=13\)

\(\Leftrightarrow5x-2x^2+2x^2-2x=13\)

\(\Leftrightarrow3x=13\)

hay \(x=\dfrac{13}{3}\)

c: Ta có: \(5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\)

\(\Leftrightarrow5x^2-5x-5x^2+7x-10x+14=6\)

\(\Leftrightarrow-8x=-8\)

hay x=1

a: Ta có: \(4\left(2-x\right)+x\left(x+6\right)=x^2\)

\(\Leftrightarrow8-4x+x^2+6x-x^2=0\)

\(\Leftrightarrow2x=-8\)

hay x=-4

b: Ta có: \(x\left(x-7\right)-\left(x-2\right)\left(x+5\right)=0\)

\(\Leftrightarrow x^2-7x-x^2-3x+10=0\)

\(\Leftrightarrow-10x=-10\)

hay x=1

c: Ta có: \(\left(2x+3\right)\left(3-2x\right)+\left(2x-1\right)^2=2\)

\(\Leftrightarrow9-4x^2+4x^2-4x+1=2\)

\(\Leftrightarrow-4x=-8\)

hay x=2

a: Ta có: \(3\left(2x-3\right)+2\left(2-x\right)=-3\)

\(\Leftrightarrow6x-9+4-2x=-3\)

\(\Leftrightarrow4x=2\)

hay \(x=\dfrac{1}{2}\)

giải phần còn lại giúp mình được ko?

1) \(\dfrac{3x}{4x-8}\)

\(ĐKXĐ:4x-8\ne0\Leftrightarrow x\ne2\)

2) \(\dfrac{2x}{x^2-9}\)

\(ĐKXĐ:x^2-9\ne0\Leftrightarrow\)\(\left\{{}\begin{matrix}x\ne3\\x\ne-3\end{matrix}\right.\)

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

\(ĐKXĐ:\)\(x+1\ne0\Leftrightarrow x\ne-1\)

(do \(x^2-x+1=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}>0\))

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

\(ĐKXĐ:x-1\ne0\Leftrightarrow x\ne1\)

5) \(\dfrac{x-2}{x^2+3}\)

Do \(x^2+3>0\forall x\in R\)

Vậy biểu thức trên xác định với mọi x

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

\(ĐKXĐ:\)\(\left\{{}\begin{matrix}x+1\ne0\\x+2\ne0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\ne-1\\x\ne-2\end{matrix}\right.\)