ta có: 2(x-3) - 3(1-2x)=4+4(1-x)

=>    2x-6 - 3+6x = 4+4 - 4x

=>    2x+6x+4x= 4 + 4 +6+3

=>    12x         = 17

          x= 17/12

chúc bạn học tốt nhé!

\(3\left(2x-6\right)-4\left(1+2x\right)-2\left(x-4\right)=4-3\left(1+2x\right)-5\left(1-2x\right).\)

\(\Leftrightarrow6x-18-4-8x-2x+8=4-3-6x-5+10x\)

\(\Leftrightarrow-4x-14=4x-4\)

\(\Leftrightarrow-4x-4x=-4+14\)

\(\Leftrightarrow-8x=10\)

\(\Leftrightarrow x=-\frac{5}{4}\)

\(\Leftrightarrow x^2-2x+1-2x^2-4x-2=8+6x^2+12x+x^3-4+2x\)

\(\Leftrightarrow-x^2-6x-1=4+6x^2+14x+x^3\)

\(\Leftrightarrow0=5+7x^2+20x+x^3\)

tự giải nốt nha

<=>(x-4)(x+1)(x-4)<0

<=> (x-4)^2(x+1)<0 mà (x-4)^2>=0

<=> x+1<0<=> x<-1

sr bn mình viết sai đề phải là\(\left(x-2\right)^2\left(x+1\right)\left(x-4\right)< 0\)

(x-2)^2(x+1)(x-4) <0 mà (x-2)^2>=0

<=> (x+1)(x-4)<0 mà x+1 >x-4 

<=> x+1 >0 và x-4<0

<=> x>-1 và x<4

d) \(\left|x-1\right|+\left|x-5\right|+\left|2x+5\right|\)

\(=\left|1-x\right|+\left|5-x\right|+\left|2x+5\right|\)

\(\ge\left|1-x+5-x\right|+\left|2x+5\right|\)

\(\ge\left|6-2x+2x+5\right|=11\)

Dấu \(=\)khi \(\hept{\begin{cases}\left(1-x\right)\left(5-x\right)\ge0\\\left(6-2x\right)\left(2x+5\right)\ge0\end{cases}}\Leftrightarrow-\frac{5}{2}\le x\le1\).

e) \(\left|x+2\right|+\left|x-1\right|+\left|x-4\right|+\left|x+5\right|=12\)

\(\Leftrightarrow\left|x+2\right|+\left|1-x\right|+\left|4-x\right|+\left|x+5\right|=12\)

Có \(\left|x+2\right|+\left|1-x\right|+\left|4-x\right|+\left|x+5\right|\ge\left|x+2+1-x\right|+\left|4-x+x+5\right|=3+9=12\)

Dấu \(=\)khi \(\hept{\begin{cases}\left(x+2\right)\left(1-x\right)\ge0\\\left(4-x\right)\left(x+5\right)\ge0\end{cases}}\Leftrightarrow-2\le x\le1\).

f) \(\left|x-1\right|+\left|x-2\right|+\left|x-3\right|+\left|3x-10\right|\)

\(\ge\left|x-1+x-2\right|+\left|3-x+3x-10\right|\)

\(=\left|2x-3\right|+\left|2x-7\right|\)

\(\ge\left|2x-3+7-2x\right|=4\)

Dấu \(=\)khi \(\hept{\begin{cases}\left(x-1\right)\left(x-2\right)\ge0\\\left(3-x\right)\left(3x-10\right)\ge0\\\left(2x-3\right)\left(7-2x\right)\ge0\end{cases}}\Leftrightarrow3\le x\le\frac{10}{3}\).

a,3x-10=2x+13

\(\Rightarrow\)3x-2x=10+13

\(\Rightarrow\)x=23

b,x+12=-5-x

\(\Rightarrow\)x+x=-12-5

\(\Rightarrow\)2x=-17

\(\Rightarrow\)x=-8,5

c,x+5=10-x

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

\(\Rightarrow\)2x=5

\(\Rightarrow\)x=2,5

e,12-x=x+1

\(\Rightarrow\)-x-x=-12+1

\(\Rightarrow\)-2x=-11

\(\Rightarrow\)x=5,5

f,14+4x=3x+20

\(\Rightarrow\)4x-3x=-14+20

\(\Rightarrow\)x=6

g,2.(x-1)+3(x-2)=x-4

\(\Rightarrow\)2x-2+3x-6=x-4

\(\Rightarrow\)2x-2+3x-6-x+4=0

\(\Rightarrow\)4x-4=0

\(\Rightarrow\)4x=4

\(\Rightarrow\)x=1

h,3(4-x)-2(x-1)=x+20

\(\Rightarrow\)12-3x-2x+2-x-20=0

\(\Rightarrow\)-6x-6=0

\(\Rightarrow\)-6x=6

\(\Rightarrow\)x=-1

i,4(2x+7)-3(3x-2)=24

\(\Rightarrow\)8x+28-9x+6=24

\(\Rightarrow\)-x+34=24

\(\Rightarrow\)-x=-10

\(\Rightarrow\)x=10

k,3(x-2)+2x=10

\(\Rightarrow\)3x-6+2x=10

\(\Rightarrow\)5x-6=10

\(\Rightarrow\)5x=16

\(\Rightarrow\)x=3,2

Phần d, tớ không biết làm!!!!

Nhưng bn cũng đã giúp tớ rùi . Thank kiu bạn

Ai làm phần d giúp mk vs ??? Mk sắp xong rùi !!!

Ta có bất đẳng thức giá trị tuyệt đối: 

\(\left|A\right|+\left|B\right|\ge\left|A+B\right|\)

Dấu \(=\)khi \(AB\ge0\).

d) \(\left|x+1\right|+\left|x+2\right|+\left|2x-3\right|\)

\(\ge\left|x+1+x+2\right|+\left|2x-3\right|\)

\(=\left|2x+3\right|+\left|3-2x\right|\)

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

Dấu \(=\)khi \(\hept{\begin{cases}\left(x+1\right)\left(x+2\right)\ge0\\\left(2x+3\right)\left(3-2x\right)\ge0\end{cases}}\Leftrightarrow-1\le x\le\frac{3}{2}\).

e) \(\left|x+1\right|+\left|x+2\right|+\left|x-3\right|+\left|x-5\right|\)

\(=\left(\left|x+1\right|+\left|3-x\right|\right)+\left(\left|x+2\right|+\left|5-x\right|\right)\)

\(\ge\left|x+1+3-x\right|+\left|x+2+5-x\right|\)

\(=4+7=11\)

Dấu \(=\)khi \(\hept{\begin{cases}\left(x+1\right)\left(3-x\right)\ge0\\\left(x+2\right)\left(5-x\right)\ge0\end{cases}}\Leftrightarrow-1\le x\le3\).

Do đó phương trình đã cho vô nghiệm.