\(\frac{2}{x-2}-\frac{3}{x+2}=\frac{x+1}{x^2-4}\left(x\ne\pm2\right)\)

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

\(\Leftrightarrow\frac{2}{x-2}-\frac{3}{x+2}-\frac{x+1}{\left(x-2\right)\left(x+2\right)}=0\)

\(\Leftrightarrow\frac{2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\frac{3\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}-\frac{x+1}{\left(x-2\right)\left(x+2\right)}=0\)

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

\(\Leftrightarrow\frac{-2x-9}{\left(x-2\right)\left(x+2\right)}=0\)

=> -2x-9=0

<=> -2x=9

<=> \(x=\frac{-9}{2}\left(tmđk\right)\)

\(\frac{x}{6}=\frac{1}{2}+\frac{1}{3}.\frac{3}{4}\)

<=> \(\frac{x}{6}=\frac{1}{2}+\frac{3}{12}\)

<=>\(\frac{x}{6}=\frac{1}{2}+\frac{1}{4}\)

<=>\(\frac{x}{6}=\frac{2}{4}+\frac{1}{4}\)

<=>\(\frac{x}{6}=\frac{3}{4}\)

<=>\(x=\frac{3}{4}.6\)

<=>\(x=\frac{9}{2}\)

kl:

k minh di mink giai cho de lam

C=(1x3+3x5+...+99x101)+(2x4+4x6+...+98x100)

đặt S=1x3+3x5+...+99x101

=>6S=6x(1x3+3x5+...+99x101)

=1x3x(5+1)+3x5x(7-1)+...+97x99x(101-95)+99x101x(103-97)

=1x3x5+1x3x1+3x5x7-1x3x5+....+97x99x101-95x97x99+99x101x103-97x99x101

=1x3x1+99x101x103

=>S=(3+99x101x103):6=171650

=>C=171650+(2x4+4x6+...+98x100)

đặt A=2x4+4x6+...+98x100

=>6A=6x(2x4+4x6+...+98x100)

=>6A=2x4x6+4x6x(8-2)+...+96x98x(100-94)+98x100x(102-96)

=2x4x6+4x6x8-2x4x6+...+96x98x100-94x96x98+98x100x102-96x98x100

=98x100x102

=>A=98x100x102:6=166600

=>C=166600+171650

=>C=338250

B=2x2+4x4+6x6+...+100x100

=2x(4-2)+4x(6-2)+6x(8-2)+...+100x(102-2)

=2x4-4+4x6-8+6x8-12+...+100x102-200

=(2x4+4x6+6x8+...+100x102)-(4+8+12+...+200)

đặt A=2x4+4x6+...+98x100+100x102

=>6A=6x(2x4+4x6+...+98x100+100x102)

=>6A=2x4x6+4x6x(8-2)+...+96x98x(100-94)+98x100x(102-96)+100x102x(104-98)

=2x4x6+4x6x8-2x4x6+...+96x98x100-94x96x98+98x100x102-96x98x100+100x102x104-98x100x102

=100x102x104

=>A=100x102x104:6=176800

=>B=176800-(4+8+12+...+200)

đặt S=4+8+12+..+200

Số số hạng của S là:

(200-4):4+1=50 số

S=(200+4)x50:2=5100

=>B=176800-5100

=>B=171700

k mình đi mình trả lời cho

a: Để (d)//y=-x+3m thì m-4=-1

hay m=3

Lời giải:
Để $(d)$ song song với $y=-x+3m$ thì:

\(\left\{\begin{matrix} m-4=-1\\ -m+3\neq 3m\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} m=3\\ m\neq \frac{-3}{2}\end{matrix}\right.\Leftrightarrow m=3\)

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

\(=\frac{x-3}{x-2}+\frac{x-2}{x-4}=\frac{16}{5}\)

\(\Rightarrow5\left(x-3\right)\left(x-4\right)+5\left(x-2\right)\left(x-2\right)=16\left(x-2\right)\left(x-4\right)\)

\(\Leftrightarrow5x^2-35x+60+5x^2-20x+20=16x^2-96x+128\)

\(\Leftrightarrow10x^2-55x+80=16x^2-96x+128\)

\(\Leftrightarrow-6x^2+41x-48=0\)

......

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

\(\Leftrightarrow\frac{x-3}{x-2}+\frac{x-2}{x-4}=\frac{16}{5}\)

\(\Leftrightarrow\frac{5\left(x-3\right)\left(x-4\right)+5\left(x-2\right)^2}{5\left(x-2\right)\left(x-4\right)}=\frac{16.\left(x-2\right)\left(x-4\right)}{5\left(x-2\right)\left(x-4\right)}\)

\(\Rightarrow5x^2-20x-15x+60+5x^2-20x+20=16x^2-64x-32x+128\)

\(\Leftrightarrow10x^2-55x+80=16x^2-96x+128\)

\(\Leftrightarrow6x^2-41x+48=0\)

\(\Leftrightarrow x=\frac{16}{3};x=\frac{3}{2}\)

A.   \(\left(x+1\right)+\left(x+2\right)+......+\left(x+100\right)=5750\)

      \(x+1+x+2+....+x+100=5750\)

      \(100x+\left(1+2+3+.......+100\right)=5750\)

      \(100x+5050=5750\)

\(100x=700\)

\(x=700:100=7\)

B.   x+(1+2+......+100) = 2000

       x + 5050 = 2000

            x = 2000 - 5050

           x= -3050

C.   ( x-1 )+(x-2)+......+( x - 100 ) = 50

 x-1+x-2+.........+x-100 = 50

100x + ( -1-2-........-100  ) = 50

100x + ( - 5050 ) = 50

100x = 50 + 5050

100 x = 5100

x = 5100 : 100

x = 51

A . \(\left(x+1\right)+\left(x+2\right)+\left(x+3\right)+...+\left(x+100\right)=5750\)

\(\left(x+x+x+...+x\right)+\left(1+2+3+...+100\right)=5750\)

\(100x+5050=5750\)

\(100x=5750-5050\)

\(100x=700\)

\(\Rightarrow x=\frac{700}{100}=7\)

B. \(x+\left(1+2+3+4+5+....+100\right)=2000\)

 \(x+\frac{\left(100+1\right).100}{2}=2000\)

\(x+5050=2000\)

\(\Rightarrow x=2000-5050=-3050\)

C. \(\left(x-1\right)+\left(x-2\right)+\left(x-3\right)+....+\left(x-100\right)=50\)

\(\left(x+x+x+...+x\right)-\left(1+2+3+...+100\right)=50\)

\(100x-5050=50\)

\(100x=5100\)

\(\Rightarrow x=\frac{5100}{100}=51\)

Ta có : \(\left(3x-2\right)\left(4x+3\right)=\left(2-3x\right)\left(x-1\right)\)

\(\Leftrightarrow12x^2-8x+9x-6=2x-3x^2-2+3x\)

\(\Leftrightarrow12x^2-8x+9x-6-2x+3x^2+2-3x=0\)

\(\Leftrightarrow15x^2-4x-4=0\)​

\(\Leftrightarrow15x^2-10x+6x-4=0\)

Lỗi :vvvv

\(\Leftrightarrow10x\left(\dfrac{3}{2}x-1\right)+4\left(\dfrac{3}{2}x-1\right)=0\)

\(\Leftrightarrow\left(10x+4\right)\left(\dfrac{3}{2}x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{2}{5}\\x=\dfrac{2}{3}\end{matrix}\right.\)

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