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30 tháng 11 2022

\(S=\dfrac{1}{x\left(x+1\right)}+\dfrac{1}{\left(x+1\right)\left(x+2\right)}+...+\dfrac{1}{\left(x+99\right)\left(x+100\right)}\)

\(=\dfrac{1}{x}-\dfrac{1}{x+1}+\dfrac{1}{x+1}-\dfrac{1}{x+2}+...+\dfrac{1}{x+99}-\dfrac{1}{x+100}\)

\(=\dfrac{x+100-x}{x\left(x+100\right)}=\dfrac{100}{x\left(x+100\right)}\)

6 tháng 12 2018

\(S=\frac{1}{x.\left(x+1\right)}+\frac{1}{\left(x+1\right).\left(x+2\right)}+....+\frac{1}{\left(x+99\right).\left(x+100\right)}\)

\(S=\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+...+\frac{1}{x+99}-\frac{1}{x+100}\)

\(S=\frac{1}{x}-\frac{1}{x+100}=\frac{x+100-x}{x.\left(x+100\right)}=\frac{100}{x^2+100x}\)

30 tháng 11 2022

\(S=\dfrac{1}{x\left(x+1\right)}+\dfrac{1}{\left(x+1\right)\left(x+2\right)}+...+\dfrac{1}{\left(x+99\right)\left(x+100\right)}\)

\(=\dfrac{1}{x}-\dfrac{1}{x+1}+\dfrac{1}{x+1}-\dfrac{1}{x+2}+...+\dfrac{1}{x+99}-\dfrac{1}{x+100}\)

\(=\dfrac{x+100-x}{x\left(x+100\right)}=\dfrac{100}{x\left(x+100\right)}\)

1 tháng 3 2022

gfvfvfvfvfvfvfv555

24 tháng 2 2022

(x-1)(2x^2-8)=0

\(\Leftrightarrow\left(x-1\right)\left(2x^2-8\right)=0\\ \left(2x^3-8x-2x^2+8\right)=0\)

\(\Leftrightarrow2x\left(x-1\right)-8\left(x-1\right)=0\)

\(\Leftrightarrow x=1;x=\dfrac{8}{2}\)

3x^2-8x+5=0

áp dụng công thức bậc 2 ta có:

\(x=\dfrac{-\left(-8\right)\pm\sqrt{\left(-8\right)^2-4.3.5}}{2.3}\)

\(\Rightarrow x=\dfrac{5}{3};x=1\)

24 tháng 2 2022

(7x-1).2x-7x+1=0

\(\Leftrightarrow\left(7x-1\right)\left(2x-1\right)=0\)

\(\Leftrightarrow x=\dfrac{1}{7};x=\dfrac{1}{2}\)

26 tháng 4 2018

heoheo lần sau bạn đánh = kí hiệu đi :(((

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

\(\Leftrightarrow2x+2x-1=3\)

<=> 4x = 4 <=> x = 1

Vậy x = 1

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

\(\Leftrightarrow3\left(3x+1\right)+2\left(x-1\right)=x-9\)

\(\Leftrightarrow9x+3+2x-2=x-9\)

\(\Leftrightarrow10x=-10\Leftrightarrow x=-1\)

Vậy pt có nghiệm x = -1

c/ \(\dfrac{x-1}{x-2}=\dfrac{x+3}{x+2}\) ĐKXĐ: \(x\ne\pm2\)

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

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

\(\Leftrightarrow0x=-4\left(voly\right)\)

Vậy pt vô nghiệm

d/ \(\dfrac{3x-1}{3x+1}+\dfrac{x-3}{x+3}=2\) ĐKXĐ: \(\left\{{}\begin{matrix}x\ne-3\\x\ne-\dfrac{1}{3}\end{matrix}\right.\)

pt <=> \(\dfrac{\left(3x-1\right)\left(x+3\right)}{\left(3x+1\right)\left(x+3\right)}+\dfrac{\left(x-3\right)\left(3x+1\right)}{\left(3x+1\right)\left(x+3\right)}=\dfrac{2\left(3x+1\right)\left(x+3\right)}{\left(3x+1\right)\left(x+3\right)}\)

=> (3x-1)(x+3) + (x-3)(3x+1) = 2(3x+1)(x+3)

\(\Leftrightarrow3x^2+8x-3+3x^2-8x-3=6x^2+20x+6\)

\(\Leftrightarrow-20x=12\Leftrightarrow x=-\dfrac{3}{5}\left(tm\right)\)

Vậy pt có nghiệm x=....

e/ như ý d

26 tháng 4 2018

Mơn bn nhe ^^ tại mjk chưa bt ạk

\(M=\left(\dfrac{x-x^2}{\left(x-1\right)^2}+\dfrac{1}{1+x}-\dfrac{x}{x-1}\right)\cdot\left(\dfrac{3x-1}{x}+\dfrac{1}{x+1}-1\right)\)

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

\(=\dfrac{-2x\left(x+1\right)+x-1}{\left(x-1\right)\left(x+1\right)}\cdot\dfrac{3x^2+2x-1+x-x^2-x}{x\left(x+1\right)}\)

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

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

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

14 tháng 4 2023

\(\left(\dfrac{3x-1}{x+1}-1\right)\)bạn sửa lại đề bào thế này

 

a: =>x-2+2=x^2+2x

=>x^2+2x=x

=>x^2+x=0

=>x(x+1)=0

=>x=0(loại) hoặc x=-1(nhận)

b: =>-9(5x-8)+4(7x-12)=-6(x+18)

=>-45x+72+28x-48=-6x-108

=>-17x+24=-6x-108

=>-11x=-132

=>x=12