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1 tháng 3 2017

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

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

tự tính nhé bạn

1 tháng 3 2017

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

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

\(\Leftrightarrow\frac{x^2+x+2x+2+x^2-x+2x-2+x^2-x}{\left[\left(x-1\right)\left(x+2\right)\right]\left[x\left(x+1\right)\right]}=\frac{3}{4}\)

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

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

=> x2 + x - 2 = 4

=> x2 + x - 6 = 0

=> \(\left(x+\frac{1}{2}\right)^2=\frac{25}{4}\)

Pt có nghiệm nhỏ nhất khi \(x+\frac{1}{2}=-\frac{5}{2}\)\(\Leftrightarrow x=-3\)

1 tháng 3 2017

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

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

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

ĐKXĐ: \(x\notin\left\{-1;-2;-3;-4\right\}\)

Ta có: \(\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}+\dfrac{1}{\left(x+3\right)\left(x+4\right)}=\dfrac{1}{6}\)

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

\(\Leftrightarrow\dfrac{1}{x+1}-\dfrac{1}{x+4}=\dfrac{1}{6}\)

\(\Leftrightarrow\dfrac{x+4}{\left(x+1\right)\left(x+4\right)}-\dfrac{x+1}{\left(x+1\right)\left(x+4\right)}=\dfrac{1}{6}\)

\(\Leftrightarrow\dfrac{x+4-x-1}{\left(x+1\right)\left(x+4\right)}=\dfrac{x^2+5x+4}{6\left(x+1\right)\left(x+4\right)}\)

\(\Leftrightarrow\dfrac{18}{6\left(x+1\right)\left(x+4\right)}=\dfrac{x^2+5x+4}{6\left(x+1\right)\left(x+4\right)}\)

Suy ra: \(x^2+5x+4=18\)

\(\Leftrightarrow x^2+5x-14=0\)

\(\Leftrightarrow x^2+7x-2x-14=0\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x+7=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-7\left(nhận\right)\\x=2\left(nhận\right)\end{matrix}\right.\)

Vậy: S={-7;2}

22 tháng 3 2021

thank

1: \(\Leftrightarrow x^2+6x+9-6x+3>x^2-4x\)

=>-4x<12

hay x>-3

2: \(\Leftrightarrow6+2x+2>2x-1-12\)

=>8>-13(đúng)

4: \(\dfrac{2x+1}{x-3}\le2\)

\(\Leftrightarrow\dfrac{2x+1-2x+6}{x-3}< =0\)

=>x-3<0

hay x<3

6: =>(x+4)(x-1)<=0

=>-4<=x<=1

1 tháng 3 2017

Ta có: \(\dfrac{1}{\left(x-1\right)\left(x-3\right)}+\dfrac{1}{\left(x-3\right)\left(x-5\right)}+\dfrac{1}{\left(x-5\right)\left(x-7\right)}=\dfrac{3}{16}\) \(\Rightarrow\dfrac{1}{2}\left(\dfrac{1}{x-1}-\dfrac{1}{x-3}+\dfrac{1}{x-3}-\dfrac{1}{x-5}+\dfrac{1}{x+5}-\dfrac{1}{x-7}\right)=\dfrac{3}{16}\)

\(\Rightarrow\) \(\dfrac{1}{2}\left(\dfrac{1}{x-1}-\dfrac{1}{x-7}\right)=\dfrac{3}{16}\)

\(\Rightarrow\) \(\dfrac{6}{x^2-8x+7}=\dfrac{3}{8}\)

\(\Rightarrow\) \(x^2-8x+7=16\)

\(\Rightarrow\) \(x^2-8x-9=0\)

\(\Rightarrow\) \(\left(x-9\right)\left(x+1\right)=0\)

\(\Rightarrow\) \(\left\{{}\begin{matrix}x=9\\x=-1\end{matrix}\right.\)

Vậy: Nghiệm lớn nhất của phương trình là: \(x=9\).

1 tháng 3 2017

Cho mình hỏi là sao lại đặt 1/2 ra ngoài vậy bạn

b)

ĐKXĐ: \(x\notin\left\{2;3;\dfrac{1}{2}\right\}\)

Ta có: \(\dfrac{x+4}{2x^2-5x+2}+\dfrac{x+1}{2x^2-7x+3}=\dfrac{2x+5}{2x^2-7x+3}\)

\(\Leftrightarrow\dfrac{x+4}{\left(x-2\right)\left(2x-1\right)}+\dfrac{x+1}{\left(x-3\right)\left(2x-1\right)}=\dfrac{2x+5}{\left(2x-1\right)\left(x-3\right)}\)

\(\Leftrightarrow\dfrac{\left(x+4\right)\left(x-3\right)}{\left(x-2\right)\left(2x-1\right)\left(x-3\right)}+\dfrac{\left(x+1\right)\left(x-2\right)}{\left(x-2\right)\left(x-3\right)\left(2x-1\right)}=\dfrac{\left(2x+5\right)\left(x-2\right)}{\left(2x-1\right)\left(x-3\right)\left(x-2\right)}\)

Suy ra: \(x^2-3x+4x-12+x^2-2x+x-2=2x^2-4x+5x-10\)

\(\Leftrightarrow2x^2-14=2x^2+x-10\)

\(\Leftrightarrow2x^2-14-2x^2-x+10=0\)

\(\Leftrightarrow-x-4=0\)

\(\Leftrightarrow-x=4\)

hay x=-4(nhận)

Vậy: S={-4}

8 tháng 2 2021

Không hiểu câu hỏi lắm :vvv

26 tháng 2 2022

đkxđ: x khác 0

\(\Leftrightarrow8.\left(x+\dfrac{1}{x}\right)\left(x+\dfrac{1}{x}\right)-4\left(x^2+\dfrac{1}{x^2}\right)\left(x+\dfrac{1}{x}\right)+4\left(x^2+\dfrac{1}{x^2}\right)^2=x^2+8x+16\)

\(\Leftrightarrow\left(x+\dfrac{1}{x}\right)\left[\left(8.x+\dfrac{1}{x}\right)-4\left(x^2+\dfrac{1}{x^2}\right)\right]+4\left(x^4+2+\dfrac{1}{x^2}\right)-x^2-8x-16=0\)

\(\Leftrightarrow\left(x+\dfrac{1}{x}\right)\left[\left(\dfrac{8x^2+1}{x}-4x^2-\dfrac{4}{x^2}\right)\right]+4x^4+8+\dfrac{4}{x^2}-x^2-8x-16=0\)

\(\Leftrightarrow\left(x+\dfrac{1}{x}\right)\left(\dfrac{x\left(8x^2+1\right)}{x^2}-\dfrac{4x^2.x^2}{x^2}-\dfrac{4}{x^2}\right)+......=0\)

\(\Leftrightarrow\left(x+\dfrac{1}{x}\right)\left(\dfrac{8x^3+x-4x^4-4}{x^2}\right)+...=0\)

\(\Leftrightarrow\dfrac{x^2}{x}.-\dfrac{4x^4+8x^3+x-4}{x^2}+.....=0\)

\(\Leftrightarrow-\dfrac{4x^6+8x^5+x^3-4x^2}{x^3}+\dfrac{4x^4+8+4x^2}{1}-\dfrac{x^2-8x-16}{1}=0\)

\(\Leftrightarrow......+\dfrac{x^3.\left(4x^4+8+4x^2\right)}{x^3}-\dfrac{x^3\left(x^2-8x-16\right)}{x^3}=0\)

\(\Leftrightarrow-4x^6+8x^5+x^3-4x^2+4x^7+8x^3+4x^5-x^5+8x^4+16x^3=0\)

\(\Leftrightarrow4x^7-4x^6+12x^5+8x^4+25x^3-4x^2=0\)

=> x=0 ( loại , ko tm)

Vậy pt vô nghiệm

26 tháng 2 2022

oho