\(\dfrac{\left(x+y\right)2}{x2+xy}+\dfrac{\left(x-y\right)2}{x2-xy}=-\left(\dfrac{\left(x-y\right)2}{x2-xy}\right)+\dfrac{\left(x-y\right)2}{x2-xy}=0\)

b: \(\dfrac{x^2-4x}{xy-4x-3y+12}+\dfrac{x-2}{y-4}\)

\(=\dfrac{x\left(x-4\right)}{\left(y-4\right)\left(x-3\right)}+\dfrac{x-2}{y-4}\)

\(=\dfrac{x^2-4x+x^2-5x+6}{\left(y-4\right)\left(x-3\right)}=\dfrac{2x^2-9x+6}{\left(y-4\right)\left(x-3\right)}\)

c: \(=\dfrac{y^2}{\left(y-5\right)\left(x+1\right)}+\dfrac{2}{x+1}\)

\(=\dfrac{y^2+2y-10}{\left(y-5\right)\left(x+1\right)}\)

a/ \(\left(x^2-5\right)\left(x+2\right)+5x=2x^2+17\)

\(\Leftrightarrow x^3+2x^2-5x-10+5x-2x^2=17\)

\(\Leftrightarrow x^3=17+10=27\Leftrightarrow x=3\)

Vậy x = 3

b/ \(\left(x^2-x+1\right)\left(x+1\right)-x^3+3x=15\)

\(\Leftrightarrow x^3+1-x^3+3x=15\)

\(\Leftrightarrow3x=14\Leftrightarrow x=\dfrac{14}{3}\)

Vậy.............................

a )

\(\left(2x-5\right)\left(x+2\right)+5x=2x^2+17\)

\(\Leftrightarrow2x^2-x-10+5x-2x^2-17=0\)

\(\Leftrightarrow4x-27=0\)

\(\Leftrightarrow x=\dfrac{27}{4}\)

b )

\(\left(x^2-x+1\right)\left(x+1\right)-x^3+3x=15\)

\(\Leftrightarrow x^3-1-x^3+3x=15\)

\(\Leftrightarrow3x=16\)

\(\Leftrightarrow x=\dfrac{16}{3}\)

\(\frac{5x+1}{x^2+5}+\frac{5x+2}{x^2+4}+\frac{5x+3}{x^2+3}+\frac{5x+4}{x^2+2}=-4\)

\(\Leftrightarrow\frac{5x+1}{x^2+5}+1+\frac{5x+2}{x^2+4}+1+\frac{5x+3}{x^2+3}+1+\frac{5x+4}{x^2+2}+1=0\)

\(\Leftrightarrow\frac{x^2+5x+6}{x^2+5}+\frac{x^2+5x+6}{x^2+4}+\frac{x^2+5x+6}{x^2+3}+\frac{x^2+5x+6}{x^2+2}=0\)

\(\Leftrightarrow\left(x^2+5x+6\right)\left(\frac{1}{x^2+5}+\frac{1}{x^2+4}+\frac{1}{x^2+3}+\frac{1}{x^2+2}\right)=0\)

\(\Leftrightarrow x^2+5x+6=0\)\(\left(\text{Vì }\frac{1}{x^2+5}+\frac{1}{x^2+4}+\frac{1}{x^2+3}+\frac{1}{x^2+2}\ne0\forall x\right)\)

\(\Leftrightarrow\left(x+3\right)\left(x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-2\end{matrix}\right.\)

Vậy phương trình có tập nghiệm \(S=\left\{-3;-2\right\}.\)

thankiu bạn nha!

\(a\text{) }7-\left(2x+4\right)=-\left(x+4\right)\)

\(\Leftrightarrow7-2x-4=-x-4\)

\(\Leftrightarrow x=7\)

\(b\text{) }\frac{3x-1}{3}=\frac{2-x}{2}\)

\(\Leftrightarrow2\left(3x-1\right)=3\left(2-x\right)\)

\(\Leftrightarrow6x-2=6-3x\)

\(\Leftrightarrow9x=8\Leftrightarrow x=\frac{8}{9}\)

\(c\text{) }\frac{2\left(3x+5\right)}{3}-\frac{x}{2}=5-\frac{3\left(x+1\right)}{4}\)

\(\Leftrightarrow8\left(3x+5\right)-6x=60-9\left(x+1\right)\)

\(\Leftrightarrow24x+40-6x=60-9x-9\)

\(\Leftrightarrow27x=11\Leftrightarrow x=\frac{11}{27}\)

\(d\text{) }x^2-4x+4=9\)

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

\(\Leftrightarrow x-2=3\Leftrightarrow x=5\)

\(e\text{) }\frac{x-1}{x+2}-\frac{x}{x-2}=\frac{5x-8}{x^2-4}\)

\(\Leftrightarrow\left(x-2\right)\left(x-1\right)-x\left(x+2\right)=5x-8\)

\(\Leftrightarrow x^2-x-2x+3-x^2-2x=5x-8\)

\(\Leftrightarrow11-10x=0\Leftrightarrow x=\frac{11}{10}\)

Bổ sung: e) ĐKXĐ: x ≠ \(\pm\) 2

đợi tí

\(=\frac{x-1}{\left(x-1\right)\left(x-4\right)}-\frac{4}{x\left(x-4\right)}=\frac{1}{x-4}-\frac{4}{x\left(x-4\right)}\)

\(=\frac{x-4}{x\left(x-4\right)}=\frac{1}{x}\)

a.\(=\frac{x}{x-y}+\frac{y}{x-y}+1=\frac{x+y+x-y}{x-y}=\frac{2x}{x-y}\)

b. \(=\frac{4}{x+2}+\frac{3}{x-2}-\frac{5x+2}{\left(x-2\right).\left(x+2\right)}\)

\(=\frac{4x-8+3x+6-5x-2}{\left(x-2\right).\left(x+2\right)}\)

\(=\frac{2x-4}{\left(x-2\right).\left(x+2\right)}=\frac{2.\left(x-2\right)}{\left(x-2\right).\left(x+2\right)}=\frac{2}{x+2}\)

k mik nhé. tks bạn nhiều

a)\(\frac{x}{x-y}-\frac{y}{y-x}+1=\frac{x}{x-y}-\frac{y}{-\left(x-y\right)}+1=\frac{x+y}{x-y}+1=\frac{2x}{x+y}\)

b)\(\frac{4}{x+2}-\frac{3}{2-x}+\frac{5x+2}{4-x^2}=\frac{7}{x+2}-\frac{5x+2}{\left(x-2\right)\left(x+2\right)}=\frac{2x-2}{\left(x-2\right)\left(x+2\right)}\)

*\(\dfrac{x-1}{x+2}\)-\(\dfrac{x}{x+2}\)=\(\dfrac{5x-2}{4-x^2}\).ĐKXĐ: x\(\ne\pm2\)

<=>\(\dfrac{\left(x-1\right)\left(2-x\right)}{4-x^2}\)-\(\dfrac{x\left(2-x\right)}{4-x^2}\)=\(\dfrac{5x-2}{4-x^2}\)

=>2x-\(x^2\)-2+x-2x+\(x^2\)=5x-2

<=>x-2=5x-2

<=>x-5x=2-2

<=>-4x=0

<=> x = 0(TM)

Vậy phương trình có tập nghiệm là S={0}

*(x+4)(5x+9)-x-4=0

<=>(x+4)(5x+9)-(x+4)=0

<=>(x+4)(5x+9-1)=0

<=>(x+4)(5x+8)=0

<=>x+4= 0 hoặc 5x+8=0

(+) x+4=0 (+)5x+8=0

<=>x=-4 <=>5x=-8

<=>x=\(\dfrac{-8}{5}\)

Vậy phương trình có tập nghiệm là S={\(-4;\dfrac{-8}{5}\)}