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

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

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

=>x=7 hoặc x=-1/2

20 tháng 2 2022

a) \(4x^2-1=\left(2x+1\right)\left(3x-8\right)\)

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}2x+1=0\\7-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-1}{2}\\x=7\end{matrix}\right.\)

Vậy phương trình có nghiệm \(x=\dfrac{-1}{2};7\)

19 tháng 2 2022

ĐKXĐ:\(x\ne-1\)

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

19 tháng 2 2022

đk : x khác -1 

\(\dfrac{1-x-3-2x}{x+1}=0\Leftrightarrow\dfrac{-3x-2}{x+1}=0\Rightarrow-3x-2=0\Leftrightarrow x=-\dfrac{2}{3}\)(tm) 

4: =>1-x+3x+3=2x+3

=>2x+4=2x+3

=>0x+1=0(vô lý)

5: =>1+4x-8=2x-3

=>4x-7-2x+3=0

=>2x-4=0

hay x=2(loại)

6: =>7+2x-3=6-2x

=>2x+4+2x-6=0

=>4x-2=0

hay x=1/2(nhận)

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

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

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

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

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

\(\Leftrightarrow4\left(x+2\right)=0\)

\(\Leftrightarrow x+2=0\)

\(\Leftrightarrow x=-2\)

tìm a,b,c,d thỏa mãn

a2-2a+b2+4b+4c2-4c+6=0

13: \(\Leftrightarrow x^2+4x+4-x^2-5x-4=x^2\)

\(\Leftrightarrow x^2+x=0\)

=>x(x+1)=0

=>x=0(loại) hoặc x=-1(nhận)

14: \(\Leftrightarrow x-2-5x-5=-15\)

=>-4x-7=-15

=>-4x=-8

hay x=2(loại)

15: \(\Leftrightarrow\left(1-6x\right)\left(x+2\right)+\left(9x+4\right)\left(x-2\right)=3x^2-2x+1\)

\(\Leftrightarrow x+2-6x^2-12x+9x^2-18x+4x-8=3x^2-2x+1\)

\(\Leftrightarrow3x^2-25x-6-3x^2+2x-1=0\)

=>-23x-7=0

hay x=-7/23(nhận)

31 tháng 5 2019

\(\frac{1000}{x}-\frac{1000}{x+10}=5\)

\(1000\left(\frac{1}{x}-\frac{1}{x+10}\right)=5\)

\(\frac{x+10-x}{x\left(x+10\right)}=\frac{5}{1000}\)

\(\frac{10}{x^2+10x}=\frac{1}{200}\)

\(x^2+10x-200=0\)

\(x^2-10x+20x-200=0\)

\(x\left(x-10\right)+20\left(x-10\right)=0\)

\(\left(x+20\right)\left(x-10\right)=0\)

=>x=-20 hoặc x=10

18 tháng 3 2016

\(\frac{1}{^{^{2^2}}}+\frac{1}{3^2}+\frac{1}{4^2}+........+\frac{1}{100^2}<\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.........+\frac{1}{99.100}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.....+\frac{1}{99}-\frac{1}{100}=1-\frac{99}{100}<1\)

20 tháng 2 2022

\(\dfrac{x+3}{x-3}-\dfrac{x-3}{x+3}=\dfrac{9}{x^2-9}\left(ĐKXĐ:x\ne\pm3\right)\)

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

\(\Rightarrow x^2+6x+9-x^2+6x-9=9\)

\(\Leftrightarrow12x=9\)

\(\Leftrightarrow x=\dfrac{3}{4}\) (nhận).

Vậy \(S=\left\{\dfrac{3}{4}\right\}\)

 

=>(50x+50x+250+65x+11050)*1,1=216500

=>165x+11300=196818,1818

=>165x=185518,1818

=>\(x\simeq124.353\)

9 tháng 4 2022

c,\(\dfrac{5-x}{2}-\dfrac{3x+4}{3}=\dfrac{1}{4}\)

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

\(\dfrac{6\left(5-x\right)}{12}+\dfrac{4\left(-3x-4\right)}{12}=\dfrac{3}{12}\)

⇔6(5-x)+4(-3x-4)=3

⇔   30-6x-12x-16=3

⇔            30-16-3=12x+6x

⇔                     11=18x

⇔                       x=\(\dfrac{11}{18}\)

Vậy S=\(\left\{\dfrac{11}{18}\right\}\)

d)x2-5x=9(x-5)

⇔x(x-5)=9(x-5)

⇔x(x-5)-9(x-5)=0

⇔(x-9)(x-5)=0

\(\left\{{}\begin{matrix}x-9=0\Leftrightarrow x=9\\x-5=0\Leftrightarrow x=5\end{matrix}\right.\)

Vậy S=\(\left\{5;9\right\}\)