a:=>3x=15

=>x=5

b: =>8-11x<52

=>-11x<44

=>x>-4

c: \(VT=\left(\dfrac{x^2-\left(x-6\right)^2}{x\left(x+6\right)\left(x-6\right)}\right)\cdot\dfrac{x\left(x+6\right)}{2x-6}+\dfrac{x}{6-x}\)

\(=\dfrac{12x-36}{2x-6}\cdot\dfrac{1}{x-6}-\dfrac{x}{x-6}=\dfrac{6}{x-6}-\dfrac{x}{x-6}=-1\)

\(\frac{x+4}{5}+\frac{3x+2}{10}< \frac{x-1}{3}\)

\(\Leftrightarrow\frac{6\left(x+4\right)}{30}+\frac{3\left(3x+2\right)}{30}< \frac{10\left(x-1\right)}{30}\)

\(\Leftrightarrow6x+24+9x+6< 10x-10\)

\(\Leftrightarrow5x+40< 0\)

\(\Leftrightarrow x< -8\)

Tự biểu diễn nha bạn

\(\frac{x+4}{5}+\frac{3x+2}{10}< \frac{x-1}{3}\)

\(\Rightarrow\frac{6\left(x+4\right)}{30}+\frac{3\left(3x+2\right)}{30}< \frac{10\left(x-1\right)}{30}\)

\(\Rightarrow6x+24+9x+6< 10x-10\)

\(5x< -40\)

\(\Rightarrow x< -8\)

=>2x^2-3x-4x+6+3x+8<2x^2+4x+2-4x

=>2x^2-4x+14<2x^2+2

=>-4x<-12

=>x>3

A, 3X+6>0

(=)3X>-6

(=)X>-2

VẬY ...

B,10-2X≥-4

(=)-2X≥-4-10

(=)-2X≥-14

(=)X≤7

VẬY....

C,

(=)

(=) -15X+10>-3+3X

(=)-15X-3X>-3-10

(=)-18X>-13

(=)X<

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

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

\(ĐK:x\ne3;-2\)

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

\(\Leftrightarrow\left(x+2\right)\left(x+2\right)+x\left(x-3\right)=x^2+6\)

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

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

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=1\left(tm\right)\\x=-2\left(ktm\right)\end{matrix}\right.\)

Vậy \(S=\left\{1\right\}\)

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

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

Xét \(\left(x+1\right)^2+x-1=x^2+4\)

\(\Leftrightarrow x^2+2x+1+x-1-x^2-4=0\)

\(\Leftrightarrow3x-4=0\)

\(\Leftrightarrow x=\dfrac{4}{3}\)

Xét \(\left(x+1\right)^2+1-x=x^2+4\)

\(\Leftrightarrow x^2+2x+1+1-x-x^2-4=0\)

\(\Leftrightarrow x-2=0\)

\(\Leftrightarrow x=2\)

Vậy \(S=\left\{\dfrac{4}{3};2\right\}\)

2.\(1-\dfrac{x-1}{3}< \dfrac{x+3}{3}-\dfrac{x-2}{2}\)

\(\Leftrightarrow\dfrac{6-2\left(x-1\right)}{6}< \dfrac{2\left(x+3\right)-3\left(x-2\right)}{6}\)

\(\Leftrightarrow6-2\left(x-1\right)< 2\left(x+3\right)-3\left(x-2\right)\)

\(\Leftrightarrow6-2x+2< 2x+6-3x+6\)

\(\Leftrightarrow-x< 4\)

\(\Leftrightarrow x>4\)

Vậy \(S=\left\{x|x>4\right\}\)

0 4