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1 tháng 12 2019

\(\frac{2}{2x+3}+\frac{5}{2x-3}-\frac{2x-33}{9-4x^2}\)

\(\frac{2}{2x+3}+\frac{5}{2x-3}+\frac{2x-33}{4x^2-9}\)

\(\frac{2\left(2x-3\right)}{\left(2x+3\right)\left(2x-3\right)}+\frac{5\left(2x+3\right)}{\left(2x-3\right)\left(2x+3\right)}+\frac{2x-33}{\left(2x-3\right)\left(2x+3\right)}\)

\(\frac{4x-6+10x-15+2x-33}{\left(2x-3\right)\left(2x+3\right)}\)

\(\frac{16x-54}{\left(2x-3\right)\left(2x+3\right)}\)

1 tháng 12 2019

\(\frac{2}{2x+3}+\frac{5}{2x-3}-\frac{2x-33}{9-4x^2}\)\(=\frac{2}{2x+3}+\frac{5}{2x-3}+\frac{2x-33}{4x^2-9}\)

\(=\frac{2\left(2x-3\right)}{\left(2x+3\right)\left(2x-3\right)}+\frac{5\left(2x+3\right)}{\left(2x+3\right)\left(2x-3\right)}+\frac{2x-33}{\left(2x+3\right)\left(2x-3\right)}\)

\(=\frac{4x-6+10x+15+2x-33}{\left(2x+3\right)\left(2x-3\right)}=\frac{16x-24}{\left(2x+3\right)\left(2x-3\right)}=\frac{8\left(2x-3\right)}{\left(2x+3\right)\left(2x-3\right)}=\frac{8}{2x+3}\)

29 tháng 11 2019

Làm ngắn gọn thôi nhé :v

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

\(A=\frac{x^5-3x^4-3x^3+11x^2-6x}{x^5-8x^2+22x^2-24x+9}\)

\(A=\frac{x^4-3x^3-3x^2+11x-6}{x^4-8x^3+22x^2-24x+9}\)

\(A=\frac{\left(x-1\right)\left(x-1\right)\left(x+2\right)\left(x-3\right)}{\left(x-1\right)\left(x-1\right)\left(x-3\right)\left(x-3\right)}\)

\(A=\frac{x+2}{x-3}\)

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

\(B=\frac{-x^4-4x^3+16x+16}{-x^4+8x^2-16}\)

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

\(B=\frac{x+2}{x-2}\)

\(C=\frac{1+x}{3-x}-\frac{1-2x}{3+x}-\frac{x\left(1-x\right)}{9-x^2}\)

\(C=\frac{1+x}{3-x}-\left(\frac{1-2x}{3+x}\right)-\frac{x\left(1-x\right)}{9-x^2}\)

\(C=\frac{10x}{-x^2+9}\)

\(D=\frac{5}{2x^2+6x}-\frac{4-3x^2}{x^2-9}-3\)

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

\(D=\frac{51x^2+138x-45}{2x^4+6x^2-18x^2-54x}\)

\(D=\frac{3\left(17x-5\right)\left(x+3\right)}{2x\left(x+3\right)\left(x+3\right)\left(x-2\right)}\)

\(D=\frac{51x-15}{2x^3-18x}\)

\(E=\frac{3x+2}{x^2-2x+1}-\frac{6}{x^2-1}-\frac{3x-2}{x^2+2x+1}\)

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

\(E=\frac{10x^4-10}{x^6-3x^4+3x^2-1}\)

\(E=\frac{10\left(x^2+1\right)\left(x+1\right)\left(x-1\right)}{\left(x+1\right)\left(x+1\right)\left(x+1\right)\left(x-1\right)\left(x-1\right)\left(x-1\right)}\)

\(E=\frac{10x^2+10}{x^4-2x+1}\)

14 tháng 12 2018

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

b, \(\frac{1}{xy-x^2}-\frac{1}{y^2-xy}=\frac{y^2-xy-xy+x^2}{\left(xy-x^2\right)\left(y^2-xy\right)}=\frac{x^2+y^2}{xy^3-xyxy-xyxy+x^3y}\)Tu rut gon tiep

c, tt

d, cx r

14 tháng 12 2018

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

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

b) \(\frac{1}{xy-x^2}-\frac{1}{y^2-xy}=\frac{1}{x\left(y-x\right)}-\frac{1}{y\left(y-x\right)}\)

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

c) \(\frac{9x-3}{4x-1}-\frac{3x}{1-4x}=\frac{9x-3}{4x-1}+\frac{3x}{4x-1}\)

\(=\frac{9x-3+3x}{4x-1}=\frac{6x-3}{4x-1}\)

26 tháng 2 2019

a,\(A=\left(\frac{2x-x^2}{2\left(x^2+4\right)}-\frac{2x^2}{\left(x^2+4\right)\left(x-2\right)}\right)\left(\frac{2x+x^2\left(1-x\right)}{x^3}\right)\left(ĐKXĐ:x\ne2;x\ne0\right)\)

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

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

\(=\frac{-x\left(x^2+4\right)}{2\left(x^2+4\right)\left(x-2\right)}.\frac{\left(x-2\right)\left(x+1\right)}{-x^2}=\frac{x+1}{2x}\)

b, \(A=x\Leftrightarrow\frac{x+1}{2x}=x\Rightarrow2x^2=x+1\Leftrightarrow2x^2-x-1=0\)

\(\Leftrightarrow\left(2x+1\right)\left(x-1\right)=0\Leftrightarrow\orbr{\begin{cases}x=-\frac{1}{2}\\x=1\end{cases}}\)(thỏa mãn điều kiện)

c, \(A\in Z\Leftrightarrow\frac{x+1}{2x}\in Z\Leftrightarrow x+1⋮\left(2x\right)\)

\(\Leftrightarrow2x+2⋮2x\Leftrightarrow2⋮2x\Leftrightarrow1⋮x\Leftrightarrow x=\pm1\) (thỏa mãn ĐKXĐ)

1 tháng 11 2019

a) \(P=\frac{2}{2x+3}+\frac{3}{2x+1}-\frac{6x+5}{\left(2x+3\right)\left(2x-3\right)}\)

\(=\frac{2\left(2x+1\right)\left(2x-3\right)}{\left(2x+3\right)\left(2x-3\right)\left(2x+1\right)}+\frac{3\left(2x+3\right)\left(2x-3\right)}{\left(2x+1\right)\left(2x+3\right)\left(2x-3\right)}-\frac{\left(6x+5\right)\left(2x+1\right)}{\left(2x+3\right)\left(2x-3\right)\left(2x+1\right)}\)

\(=\frac{\left(4x+2\right)\left(2x-3\right)+3\left(4x^2-9\right)-12x^2-16x-5}{\left(2x+3\right)\left(2x-3\right)\left(2x+1\right)}\)

\(=\frac{8x^2-8x-6+12x^2-27-12x^2-16x-5}{\left(2x+3\right)\left(2x-3\right)\left(2x+1\right)}\)

\(=\frac{8x^2-24x-38}{\left(2x+3\right)\left(2x-3\right)\left(2x+1\right)}\)

Check hộ mình xem nghi nghi sai sai

1 tháng 11 2019

b) \(Q=\left(\frac{x+1}{2x-1}+\frac{3}{x^2-1}-\frac{x+3}{2x+2}\right).\frac{4x^2-4}{5}\)

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

\(=\left(\frac{2\left(x+1\right)\left(x-1\right)\left(x+1\right)}{2\left(2x-1\right)\left(x-1\right)\left(x+1\right)}+\frac{2.3\left(2x-1\right)}{2\left(x-1\right)\left(x+1\right)\left(2x-1\right)}-\frac{\left(x+3\right)\left(2x-1\right)\left(x-1\right)}{2\left(x+1\right)\left(2x-1\right)\left(x-1\right)}\right).\frac{4x^2-4}{5}\)

\(=\frac{2\left(x+1\right)\left(x^2-1\right)+12x-6-\left(2x^2+5x-3\right)\left(x-1\right)}{2\left(2x-1\right)\left(x+1\right)\left(x-1\right)}.\frac{4x^2-4}{5}\)

\(=\frac{2\left(x^3+x^2-x-1\right)+12x-6-2x^3-5x^2+3x+2x^2+5x-3}{2\left(2x-1\right)\left(x+1\right)\left(x-1\right)}.\frac{4x^2-4}{5}\)

\(=\frac{2x^3+2x^2-2x-2+20x-2x^3-3x^2-9}{2\left(2x-1\right)\left(x+1\right)\left(x-1\right)}.\frac{4x^2-4}{5}\)

\(=\frac{-x^2+18x-11}{2\left(2x-1\right)\left(x+1\right)\left(x-1\right)}.\frac{4\left(x-1\right)\left(x+1\right)}{5}\)

\(=\frac{-x^2+18x-11}{\left(2x-1\right)}.\frac{2}{5}\)

\(=\frac{-2x^2+36x-22}{5\left(2x-1\right)}\)

2 tháng 1 2020

\(\frac{2x-9}{x^2-5x+6}-\frac{x+3}{x-2}-\frac{2x+1}{3-x}\)

\(=\frac{2x-9}{x^2-2x-3x+6}-\frac{\left(x+3\right)\left(x-3\right)}{\left(x-2\right)\left(x-3\right)}+\frac{\left(2x+1\right)\left(x-2\right)}{\left(x-3\right)\left(x-2\right)}\)

\(=\frac{2x-9}{x\left(x-2\right)-3\left(x-2\right)}-\frac{x^2-9}{\left(x-2\right)\left(x-3\right)}+\frac{2x^2-3x-2}{\left(x-3\right)\left(x-2\right)}\)

\(=\frac{2x-9-x^2+9+2x^2-3x-2}{\left(x-2\right)\left(x-3\right)}\)

\(=\frac{x^2-x-2}{\left(x-2\right)\left(x-3\right)}\)

\(=\frac{x^2-2x+x-2}{\left(x-2\right)\left(x-3\right)}\)

\(=\frac{x\left(x-2\right)+\left(x-2\right)}{\left(x-2\right)\left(x-3\right)}\)

\(=\frac{\left(x-2\right)\left(x+1\right)}{\left(x-2\right)\left(x-3\right)}=\frac{x+1}{x-3}\)

2 tháng 1 2020

x^2-5x+6= (x-2)(x-3)

Áp dụng vào biểu thức ta đc

\(=(2x+9)/((x-2)(x-3))-((x+3)(x-3)/((x-2)x-3))+((2x+1)(x-2))/((x-3)(x-2)\)

=(2x+9-x^2+9+2x^2-3x-2)/(x-2)(x-3)

=(16-x+x^2)/(x-2)(x-3)

con lại bạn tự khai triển ra nhé!

2 tháng 8 2017

Ta có

4x-8=9x-3-2x+1

<=>-6=-3x(chuyển vế đổi dấu)

<=>x=2

b)

Ta có

Căn cả 2 vế ta đcx-5/ cawn3 =3 

<=>x=10.2

3 tháng 8 2020

\(5X\left(X-2020\right)+X=2020\)

\(\Leftrightarrow5X^2-10100X+X=2020\)

\(\Leftrightarrow5X^2-10099X=2020\)

\(\Leftrightarrow5X^2-10099X-2020=0\)

\(\Leftrightarrow5X^2-10100X+x-2020=0\)

\(\Leftrightarrow5X\left(X-2020\right)+X-2020=0\)

\(\Leftrightarrow\left(X-2020\right)\left(5X+1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=2020\\x=-\frac{1}{5}\end{cases}}\)

3 tháng 8 2020

\(4\left(x-5\right)^2-\left(2x+1\right)^2=0\)

\(\Leftrightarrow\left[2\left(x-5\right)\right]^2-\left(2x+1\right)^2=0\)

\(\Leftrightarrow\left[2\left(x-5\right)-2x-1\right]\left[2\left(x-5\right)+2x+1\right]=0\)

\(\Leftrightarrow\left(2x-10-2x-1\right)\left(2x-10+2x+1\right)=0\)

\(\Leftrightarrow-11\left(4x-9\right)=0\)

\(\Leftrightarrow x=\frac{9}{4}\)