a) \(\frac{7}{13}x\frac{7}{15}-\frac{5}{12}x\frac{21}{39}+\frac{49}{91}x\frac{8}{15}=\frac{7}{13}x\frac{7}{15}-\frac{5}{12}x\frac{7}{13}+\frac{7}{13}x\frac{8}{15}\)

\(=\frac{7}{13}x\left(\frac{7}{15}-\frac{5}{12}+\frac{8}{15}\right)=\frac{7}{13}x\left(1-\frac{5}{12}\right)=\frac{7}{13}x\frac{7}{12}=\frac{49}{156}\)

b) \(\left(\frac{12}{199}+\frac{23}{200}-\frac{34}{201}\right)x\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{6}\right)=\left(\frac{12}{199}+\frac{23}{200}-\frac{34}{201}\right)x0=0\)

Bài 2:

a) \(0,5+\left(x-\frac{15}{2}\right):\frac{1}{2}=\frac{9}{2}\)

\(\left(x-\frac{15}{2}\right):\frac{1}{2}=4\)

\(x-\frac{15}{2}=2\) 

x = 19/2

b) \(2012\times x-2010\times x=2014\)

\(x\times\left(2012-2010\right)=2014\)

\(x\times2=2014\)

x = 1007

c) ( x + 1) + (x+2) + (x+3)+...+(x+100) = 5750

\(x\times100+\left(1+2+3+...+100\right)=5750\)

\(x\times100+5050=5750\)

\(x\times100=700\)

x = 7

a) 199+36+201+184+37

= 199+201+36+184+37

=400+220+37

=620+37=657

b)1+3+5+...+103+105 (sử dụng công thức)

= (105+1)x105:2

=106x105:2

=11130:2=5565

Bài 2 

a) (x-3):4-15=2

    (x-3):4    =2+15

    (x-3):4    =17

    (x-3)       =17x4

     x-3        =68

     x           =68+3

     x           =71

b) (x-5) x ( 2 x X - 6) =0

có 2 TH

TH1: x-5=0

        x  =0+5

        x  =5

TH2: 2.x-6= 0

        2.x  =0+6

        2.x  =6

           x  =6:2

            x  =3

1. a) =657
    b) =5565
2. a) x=71
    b) x=5 hoặc x=3

a) -12x + 60 + 21 - 7x =15

-12x - 7x + 81 = 15

-19x = 15 - 81

-19x = -66

x= -66 : (-19)

x = \(\frac{66}{19}\)

Vậy x = \(\frac{66}{19}\)

b) 30x + 60 - 6x + 30 - 24x= 100

30x - 6x - 24x + 60 +30= 100

x ( 30 - 6- 24 ) + 90 = 100

x. 0 + 90 =100

x. 0 =10 ( vô lý )

\(\Rightarrow\) x \(\in\phi\)

k đi mình lamf cho

\(\dfrac{x+1}{199}+\dfrac{x+2}{198}+\dfrac{x+3}{197}+\dfrac{x+4}{196}+\dfrac{x+220}{5}=0\)

\(\Leftrightarrow\left(\dfrac{x+1}{199}+1\right)+\left(\dfrac{x+2}{198}+1\right)+\left(\dfrac{x+3}{197}+1\right)+\left(\dfrac{x+4}{196}+1\right)+\dfrac{x+200}{5}+\dfrac{20}{5}-4=0\)

\(\Leftrightarrow\dfrac{x+200}{199}+\dfrac{x+200}{198}+\dfrac{x+200}{197}+\dfrac{x+200}{196}+\dfrac{x+200}{5}=0\)

\(\Leftrightarrow\left(x+200\right)\left(\dfrac{1}{199}+\dfrac{1}{198}+\dfrac{1}{197}+\dfrac{1}{196}+\dfrac{1}{5}\right)=0\)

\(\Leftrightarrow x=-200\)( do \(\dfrac{1}{199}+\dfrac{1}{198}+\dfrac{1}{197}+\dfrac{1}{196}+\dfrac{1}{5}>0\))

\(\dfrac{x+1}{199}+\dfrac{x+2}{198}+\dfrac{x+3}{197}+\dfrac{x+4}{196}+\dfrac{x+220}{5}=0\\ \Leftrightarrow\left(\dfrac{x+1}{199}+1\right)+\left(\dfrac{x+2}{198}+1\right)+\left(\dfrac{x+3}{197}+1\right)+\left(\dfrac{x+4}{196}+1\right)+\left(\dfrac{x+220}{5}-4\right)=0\\ \Leftrightarrow\dfrac{x+200}{199}+\dfrac{x+200}{198}+\dfrac{x+200}{197}+\dfrac{x+200}{196}+\dfrac{x+200}{5}=0\\ \Leftrightarrow\left(x+200\right)\left(\dfrac{1}{199}+\dfrac{1}{198}+\dfrac{1}{197}+\dfrac{1}{196}+\dfrac{1}{5}\right)=0\\ \Leftrightarrow x=-200\)

a, \(12-2\left(1-x\right)^2=\left(3x-2\right)\left(2x-3\right)\)

\(< =>12-2\left(1-2x+x^2\right)=6x^2-9x-4x+6\)

\(< =>12-2+4x-2x^2=6x^2-13x+6\)

\(< =>10+4x-2x^2-6x^2+13x-6=0\)

\(< =>-8x^2+17x+4=0< =>\orbr{\begin{cases}x=\frac{17-\sqrt{417}}{16}\\x=\frac{17+\sqrt{417}}{16}\end{cases}}\)

b, \(10x+3-5x=4x+12< =>5x+3-4x-12=0\)

\(< =>x-9=0< =>x=9\)

c, \(11x+42-2x=100-9x-22< =>9x+42-100+9x+22=0\)

\(< =>18x+64-100=0< =>18x-36=0< =>x=\frac{36}{18}=2\)

d, \(2x-\left(3-5x\right)=4\left(x+3\right)< =>2x-3+5x=4x+12\)

\(< =>7x-3-4x-12=0< =>3x-15=0< =>x=\frac{15}{3}=5\)

e, \(2\left(x-3\right)+5x\left(x-1\right)=5x^2< =>2x-6+5x^2-5=5x^2\)

\(< =>2x-11+5x^2-5x^2=0< =>2x-11=0< =>x=\frac{11}{2}\)

f, \(-6\left(1,5-2x\right)=3\left(-15+2x\right)< =>-6\left(\frac{3}{2}-2x\right)=3\left(2x-15\right)\)

\(< =>-9+12x-6x+45=0< =>6x+36=0< =>x=-6\)

g, \(14x-\left(2x+7\right)=3x+12x-13< =>14x-2x-7=15x-13\)

\(< =>12x-7-15x+13=0< =>-3x+6=0< =>x=-2\)

h, \(\left(x-4\right)\left(x+4\right)-2\left(3x-2\right)=\left(x-4\right)^2\)

\(< =>x^2-16-6x+4=x^2-8x+16\)

\(< =>x^2-6x-12-x^2+8x-16=0\)

\(< =>2x-28=0< =>x=\frac{28}{2}=14\)

q, \(4\left(x-2\right)-\left(x-3\right)\left(2x-5\right)=?\)thiếu đề

1/x+x+1+x+2+x+3+...+x+2006+2007=2007

------------------------------------------=2007-2007

------------------------------------------=0

x+x+x+...+x+1+2+3+...+2006=0

2007.x+(1+2+...+2006)=0

2007.x+(2006+1).[(2006-1)+1]:2=0

2007.x+2013021=0

2007.x=0-2013021

x=-2013021:2007

x=-1003

2/x+x+1+x+2+...+x+198=401-201-200-199

199.x+(1+2+...+198)=-199

199.x+(1+198).[(198-1)+1]:2=-199

199.x+19701=-199

199.x=-199-19701

x=-19900:199

x=-100

3/x+x+1+x+2+...+x+2008=2010-2010-2009

2009.x+(2008+1).[(2008-1)+1]:2=-2009

2009.x+2017036=-2009

2009.x=-2009-2017036

x=-2019045:2009

x=-1005

vế phải đâu?

ko có vế pk

Ủa ko có vế phải thì mình làm bằng niềm tin à? :D