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

Ai biết thì trả lời giùm mình với nhe!!!

1/x(x+1)+1/(x+1)(x+2)+1/(x+2)(x+3)-1/x=1/2010

1/x(x+1)+1/(x+1)-1/(x+2)+1/(x+2)-1/(x+3)-1/x=1/2010

1/x(x+1)+1/(x+1)-1/(x+3)-1/x=1/2010

-1/x+1 +(x+3)-(x+1)/(x+1)(x+3)=1/2010

-1/x+3=1/2010

x+3=-2010

x=-2013

(2% x X -1) +2 = 0,2 : 1/10

(0,02 x X -1) + 2 =0.2 :0.1=2

(0.02 x X -1) = 2-2=0

0.02x X = 0+ 1 =1

1 : 0.02 = 50.

Thử lại :(2% x 50 - 1) + 2 =0.2 : 1/10 ( cả 2 biểu thức đều bằng 2)

b)ta coi biểu thức đầu(1 x2 x3 x........x2010) là A. Ta có :

 A x (x -2010)

vì bất cứ số nào nhân với 0 cũng bằng 0 nên biểu thức chứa x phải có kết quả là 0.

x = 0 +2010 =2010

Ta có:\(\frac{x-1}{2013}+\frac{x-2}{2012}=\frac{x-3}{2011}+\frac{x-4}{2010}\Rightarrow\frac{x-1}{2013}-1+\frac{x-2}{2012}-1=\frac{x-3}{2011}-1+\frac{x-4}{2010}-1\)

\(\Rightarrow\frac{x-1-2013}{2013}+\frac{x-2-2012}{2012}=\frac{x-3-2011}{2011}+\frac{x-4-2010}{2010}\)

\(\Rightarrow\frac{x-2014}{2013}+\frac{x-2014}{2012}=\frac{x-2014}{2011}+\frac{x-2014}{2010}\)

\(\Rightarrow\frac{x-2014}{2013}+\frac{x-2014}{2012}-\frac{x-2014}{2011}-\frac{x-2014}{2010}=0\)

\(\Rightarrow\left(x-2014\right)\left(\frac{1}{2013}+\frac{1}{2012}-\frac{1}{2011}-\frac{1}{2010}\right)=0\)

Vì \(\frac{1}{2013}< \frac{1}{2011};\frac{1}{2012}< \frac{1}{2010}\) nên \(\frac{1}{2013}+\frac{1}{2012}-\frac{1}{2011}-\frac{1}{2010}< 0\)

\(\Rightarrow x-2014=0\Rightarrow x=2014\)

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

1.\(\frac{x+1}{2013}\)+\(\frac{x+2}{2012}\)=\(\frac{x+3}{2011}\)+\(\frac{x+4}{2010}\)

⇔\(\frac{x+1}{2013}\)+1+\(\frac{x+2}{2012}\)+1=\(\frac{x+3}{2011}\)+1+\(\frac{x+4}{2010}\)+1

⇔\(\frac{x+2014}{2013}\)+\(\frac{x+2014}{2012}\)=\(\frac{x+2014}{2011}\)+\(\frac{x+2014}{2010}\)

⇔\(\frac{x+2014}{2013}\)+\(\frac{x+2014}{2012}\)-\(\frac{x+2014}{2011}\)-\(\frac{x+2014}{2010}\)=0

⇔(x+2014)(\(\frac{1}{2013}\)+\(\frac{1}{2012}\)-\(\frac{1}{2011}\)-\(\frac{1}{2010}\))=0

Mà \(\frac{1}{2013}\)+\(\frac{1}{2012}\)-\(\frac{1}{2011}\)-\(\frac{1}{2010}\)≠0

⇔x+2014=0

⇔x=-2014

Vậy tập nghiệm của phương trình đã cho là:S={-2014}

2.\(\frac{3x+2}{4}\)+\(\frac{x+3}{2}\)=\(\frac{x-1}{3}\)-\(\frac{-x-1}{12}\)

⇔\(\frac{3\left(3x+2\right)}{12}\)+\(\frac{6\left(x+3\right)}{12}\)=\(\frac{4\left(x-1\right)}{12}\)+\(\frac{x+1}{12}\)

⇒9x+6+6x+18=4x-4+x+1

⇒15x+24=5x-3

⇒15x-5x=-3-24

⇒10x=-27

⇒ x=-\(\frac{27}{10}\)

Vậy tập nghiệm của phương trình đã cho là S={-\(\frac{27}{10}\)}

\(3.\frac{x+1}{x-1}-\frac{x-1}{x+1}+\frac{x^2+3}{1-x^2}=0ĐKXĐ:x\ne\pm1\)

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

\(-3+7x-5x^2+x^3=0\)

\(\left(x-3\right)\left(x-1\right)\left(x-1\right)=0\)

\(\left[{}\begin{matrix}x=3\\x=1\end{matrix}\right.\)Theo ĐKXĐ => x=3

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

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

\(\Leftrightarrow7x+9=2\)

\(\Leftrightarrow7x=2-9\)

\(\Leftrightarrow7x=-7\)

\(\Leftrightarrow x=\dfrac{-7}{7}=-1\)

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

\(\Leftrightarrow4x^2+12x+9-\left(4x^2-3x+4x-3\right)=-1\)

\(\Leftrightarrow4x^2+12x+9-4x^2+3x-4x+3=-1\)

\(\Leftrightarrow11x+12=-1\)

\(\Leftrightarrow11x=-13\)

\(\Leftrightarrow x=\dfrac{-13}{11}\)

Ai giải giúp mình voi