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15 tháng 11 2016

cm cái jz ?????

 

30 tháng 11 2015

\(a.\) Với  \(a+b+c=0\)  thì  \(\frac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{abc}=\frac{\left(-c\right).\left(-a\right).\left(-b\right)}{abc}=\frac{-abc}{abc}=-1\)

\(b.\)   Công thức tổng quát:  \(\frac{1}{n\left(n+1\right)}=\frac{1}{n}-\frac{1}{n+1}\)

Ta có:

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

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

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

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

\(\frac{1}{\left(x+4\right)\left(x+5\right)}=\frac{1}{x+4}-\frac{1}{x+5}\)

Do đó, suy ra được:  \(A=\frac{1}{x}-\frac{1}{x+5}=\frac{x+5-x}{x\left(x+5\right)}=\frac{5}{x\left(x+5\right)}\)

 

 

28 tháng 2 2020

Bài 1:

\(a, \dfrac{1}{2}x(2-x)=x-\dfrac{1}{2}x^2\)

\(b, \dfrac{x-5}{5-x}\)\(=-\dfrac{x-5}{x-5}\)\(=-1\)

Bài 2:

\(a, x+y-x^2+y^2=(x+y)-(x^2-y^2)=(x+y)-(x-y)(x+y)\)

\(=(x+y)(1-x+y)\)

\(b, x(x-3)+3x-1=0 \)

\(⇔x^2-3x+3x-1=0 \)

\(⇔x^2-1=0 \)

\(⇔(x-1)(x+1)=0 \)

\(⇔\left[\begin{array}{} x-1=0\\ x+1=0 \end{array}\right.\)

\(⇔\left[\begin{array}{} x=1\\ x=-1 \end{array}\right.\)

Bài 3:

\(a,A=\dfrac{x(x+2)-x(x-2)+8}{x^2-4}:\dfrac{4}{x-2}\)

\(A=\dfrac{4x+8}{(x-2)(x+2)}.\dfrac{x-2}{4}\)

\(A=\dfrac{4(x+2)}{(x-2)(x+2)}.\dfrac{x-2}{4}\)

\(A=1\)

\(b, B=(1-\dfrac{a+b}{a-b})(1-\dfrac{2b}{a+b})\)

\(B=\dfrac{-2b}{a-b}.\dfrac{a-b}{a+b}\)

\(B=\dfrac{-2b}{a+b}\)

Bài 4:

\(C=(2+1)(2^2+1)(2^4+1)(2^8+1)(2^{16}+1)(2^{32}+1)\)

\(C=(2^2-1)(2^2+1)(2^4+1)(2^8+1)(2^{16}+1)(2^{32}+1)\)

\(C=(2^4-1)(2^4+1)(2^8+1)(2^{16}+1)(2^{32}+1)\)

\(C=(2^8-1)(2^8+1)(2^{16}+1)(2^{32}+1)\)

\(C=(2^{16}-1)(2^{16}+1)(2^{32}+1)\)

\(C=(2^{32}-1)(2^{32}+1)=2^{64}-1\)

28 tháng 2 2020

Thanks bạn nha!!!!ok

8 tháng 2 2017

b)\(\frac{x+1}{2008}+\frac{x+2}{2007}+\frac{x+3}{2006}=\frac{x+4}{2005}+\frac{x+5}{2004}+\frac{x+6}{2003}\)

<=>\(\left(\frac{x+1}{2008}+1\right)+\left(\frac{x+2}{2007}+1\right)+\left(\frac{x+3}{2006}+1\right)=\left(\frac{x+4}{2005}+1\right)+\left(\frac{x+5}{2004}+1\right)+\left(\frac{x+6}{2003}+1\right)\)

<=>\(\frac{x+2009}{2008}+\frac{x+2009}{2007}+\frac{x+2009}{2006}=\frac{x+2009}{2005}+\frac{x+2009}{2004}+\frac{x+2009}{2003}\)<=>\(\frac{x+2009}{2008}+\frac{x+2009}{2007}+\frac{x+2009}{2006}-\frac{x+2009}{2005}-\frac{x+2009}{2004}-\frac{x+2009}{2003}=0\)\( \left(x+2009\right)\left(\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}-\frac{1}{2005}-\frac{1}{2004}-\frac{1}{2003}\right)=0\)

\(\left(\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}-\frac{1}{2005}-\frac{1}{2004}-\frac{1}{2003}\right)\ne0\)

nên phương trình đó xảy ra khi và chỉ khi x+2009=0

<=>x=-2009

Vậy phương trình có no là x=-2009

8 tháng 2 2017

b) \(\frac{x+1}{2008}+\frac{x+2}{2007}+\frac{x+3}{2006}=\frac{x+4}{2005}+\frac{x+5}{2004}+\frac{x+6}{2003}\)

\(\Leftrightarrow\)\(\left(\frac{x+1}{2008}+1\right)+\left(\frac{x+2}{2007}+1\right)+\left(\frac{x+3}{2006}+1\right)\)=

\(\left(\frac{x+4}{2005}+1\right)+\left(\frac{x+5}{2004}+1\right)+\left(\frac{x+6}{2003}+1\right)\)

\(\Leftrightarrow\) \(\frac{x+2009}{2008}+\frac{x+2009}{2007}+\frac{x+2009}{2006}-\)\(\frac{x+2009}{2005}-\frac{x+2009}{2004}-\frac{x+2009}{2003}=0\)

\(\Leftrightarrow\) \(\left(x+2009\right)\)\(\left(\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}-\frac{1}{2005}-\frac{1}{2004}-\frac{1}{2003}\right)\)= 0

\(\Leftrightarrow\)\(x+2009=0\)

( vì \(\left(\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}-\frac{1}{2005}-\frac{1}{2004}-\frac{1}{2003}\right)\) \(\ne0\))

\(\Leftrightarrow\) \(x=-2009\)

Vậy x = -2009