`(3x+2)/3 <= (x-4)/7`

`<=>7(3x+2) <= 3(x-4)`

`<=> 21x+14<=3x-12`

`<=>18x <= -26`

`<=> x <=-13/9`

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

\(\Rightarrow4x-10=3x+6\)

\(\Rightarrow x=16\)

\(\dfrac{2x+5}{3}=\dfrac{x+2}{2}\)

MTC : 6

Quy đồng mẫu thức :

\(\Rightarrow\) \(\dfrac{2\left(2x+5\right)}{6}=\)\(\dfrac{3\left(x+2\right)}{6}\)

 Suy ra : 2(2x + 5) = 3(x + 2)

          \(\Leftrightarrow\) 4x + 10 = 3x + 6

          \(\Leftrightarrow\) 4x + 10 - 3x - 6 = 0

          \(\Leftrightarrow\) x + 4 = 0

          \(\Leftrightarrow\) x = - 4

       Vậy S = \(\left\{-4\right\}\)

 Chúc bạn học tốt

\(=6x^4y^4-x^3y^3+\dfrac{1}{2}x^4y^2\)

ĐK: ` x\ne \pm 3`

`(x+1)/(x-3)+(x-1)/(x+3)=(x+6)/(x^2-9)`

`<=>(x+1)(x+3)+(x-1)(x-3)=x+6`

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

`<=>2x^2+6=x+6`

`<=>2x^2-x=0`

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

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

Vậy `S={0; 1/2}`.

ĐKXĐ: x ≠ -3, x ≠ 3

\(\dfrac{x+1}{x-3}+\dfrac{x-1}{x+3}=\dfrac{x+6}{x^2-9}\)

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

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

\(\Leftrightarrow2x^2-x=0\)

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

\(\Rightarrow\left[{}\begin{matrix}x=0\left(tm\right)\\x=\dfrac{1}{2}\left(tm\right)\end{matrix}\right.\)

Vậy...

\(\dfrac{x+1}{x-3}+\dfrac{x-1}{x+3}=\dfrac{x+6}{x^2-9}\)(a)

ĐKXĐ\(\left\{{}\begin{matrix}x-3\ne0\\x+3\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne3\\x\ne-3\end{matrix}\right.\)

(a)\(\Leftrightarrow\dfrac{x+1}{x-3}+\dfrac{x-1}{x+3}=\dfrac{x+6}{\left(x-3\right)\left(x+3\right)}\)

\(\Leftrightarrow\left(x+1\right).\left(x+3\right)+\left(x-1\right).\left(x-3\right)=x+6\)

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

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

\(\Leftrightarrow2x^2-x=0\)

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(thỏa-mãn-ĐKXĐ\right)\\x=\dfrac{1}{2}\left(thỏa-mãn-ĐKXĐ\right)\end{matrix}\right.\)

Vậy S = \(\left\{0;\dfrac{1}{2}\right\}\)

`(3x+2)/3 <= (x-4)/7`

`<=>7(3x+2) <= 3(x-4)`

`<=>21x+14 <= 3x-12`

`<=> 18x <=-26`

`<=>x <= -13/9`

Vậy `x<=-13/9`.

\(5,\dfrac{4}{x-2}+\dfrac{x}{x+1}-\dfrac{x^2-2}{\left(x-2\right)\left(x+1\right)}=0\left(dkxd:x\ne2;-1\right)\)

\(\Rightarrow4\left(x+1\right)+x\left(x-2\right)-x^2-2=0\)

\(\Rightarrow4x+4+x^2-2x-x^2-2=0\)

\(\Rightarrow2x+2=0\)

\(\Rightarrow x=-1\left(loai\right)\)

Vậy \(S=\varnothing\)

\(4,\dfrac{x}{x-3}-\dfrac{1}{x+2}=0\left(dkxd:x\ne3;-2\right)\)

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

\(\Rightarrow x^2+3x-x+3=0\)

\(\Rightarrow x^2+2x+3=0\)

\(\Rightarrow S=\varnothing\)

giúp em tl những câu tl trên vs

Đáp án:x+2/2x

Ta có:2x+6=2(x+3);x²+3x=x(x+3)

➞MTC:2x(x+3)

Ta co:(x+1/2x+6)+(2x+3/x²+3x)={[(x+1)x]+[(2x+3)2]}/2x(x+3)=x²+5x+6/2x(x+3)=(x+2)(x+3)/2x(x+3)=x+2/2x

Lời giải:

$\frac{x^3+8}{x^2-2x+1}.\frac{x^2+3x+2}{1-x^2}=\frac{(x^3+8)(x^2+3x+2)}{(x^2-2x+1)(1-x^2)}$

$=\frac{(x+2)(x^2-2x+4)(x+1)(x+2)}{(x-1)^2(1-x)(x+1)}$

$=\frac{(x+2)^2(x^2-2x+4)}{-(x-1)^3}$