Mình chưa học lớp 7

Mình mới học lớp 5 thôi

Xin lỗi nha

to cung vay

thờ lịu chỉ mới học lớp 5

\(\left(\frac{x-1}{x+2}\right)^2-4\left(\frac{x^2-1}{x^2-4}\right)^2+3\left(\frac{x+1}{x-2}\right)^2=0\left(1\right)\)

\(ĐKXĐ:x\ne\pm2\)

Đặt \(\frac{x-1}{x+2}=a;\frac{x+1}{x-2}=b\)

=> Phương trình (1) <=> \(a^2-4ab+3b^2=0\)

\(\Leftrightarrow a^2-3ab-ab+3b^2=0\)

\(\Leftrightarrow a\left(a-b\right)-3b\left(a-b\right)=0\)

\(\Leftrightarrow\left(a-3b\right)\left(a-b\right)=0\)

\(\Leftrightarrow\left(a-3b\right)\left(a-b\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}a-3b=0\\a-b=0\end{cases}\Leftrightarrow\orbr{\begin{cases}a=3b\\a=b\end{cases}}}\)

=>  \(b=0;a=0\)

Bạn cùng trường :">

1. Có 1 thừa số là \(1-\frac{5}{5}=0\) nên tích sẽ bằng 0

2.\(B=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.......\frac{49}{50}=\frac{1}{50}\)

\(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)....\left(1-\frac{1}{2009}\right)\)

=\(\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{2008}{2009}\)

=\(\frac{1}{2009}\)

\(H=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)\left(1-\frac{1}{5}\right)\cdot\cdot\cdot\cdot\cdot\left(1-\frac{1}{100}\right)\)

\(\Leftrightarrow H=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot\frac{4}{5}\cdot\cdot\cdot\cdot\cdot\frac{99}{100}\)

\(\Leftrightarrow H=\frac{1.2.3.4.....99}{2.3.4.5.....100}\)

\(\Leftrightarrow H=\frac{1}{100}\)

\(H=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.\frac{4}{5}...\frac{99}{100}\)

\(H=\frac{1.2.3.4...99}{2.3.4.5...100}\)

\(H=\frac{1}{100}\)

Vậy \(H=\frac{1}{100}.\)

 \(H=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.\frac{4}{5}.....\frac{99}{100}\)

\(H=\frac{1.2.3.4.....99}{2.3.4.5.....100}\)

\(H=\frac{1}{100}\)

  (1-1/2)(1-1/3)(1-1/4)...(1-1/2009)

=1/2*2/3*3/4*...*2008/2009

=\(\frac{1\cdot2\cdot3\cdot...\cdot2008}{2\cdot3\cdot4\cdot...\cdot2009}\)

=1/2009

=1/2x2/3x3/4x...../2008/2009

=1/2009

(1-1/3).(1-1/5).(1-1/7).(1-1/9).(1-1/11).(1-1/13).(1-1/2).(1-1/4).(1-1/6).(1-1/8).(1-1/10)

=2/3.4/5.6/7.8/9.10/11.12/13.1/2.3/4.5/6.7/8.9/10

=8/15.48/63.120/143.3/8.35/48.9/10

=384/945.360/1144.315/480

=138240/1081080.315/480

=43545600/518918400=84/1001

khó quá

\(\left(1-\frac{1}{3}\right)\left(1-\frac{1}{5}\right)\left(1-\frac{1}{7}\right)\)    \(\left(1-\frac{1}{11}\right)\)\(\left(1-\frac{1}{13}\right)\left(1-\frac{1}{2}\right)\left(1-\frac{1}{4}\right)\left(1-\frac{1}{6}\right)\)
 \(=1.\left(\frac{1}{3}-\frac{1}{2}\right).\left(\frac{1}{5}-\frac{1}{4}\right).\left(\frac{1}{7}-\frac{1}{6}\right)\)\(.\left(\frac{1}{11}-\frac{1}{10}\right).\frac{1}{13}\)
  \(=1.1.1.1.1.\frac{1}{13}\)
   \(=\frac{1}{13}\)
Cái này là chỉ do mình nghĩ thôi, ko biết đúng ko nữa
Có sai cko mình sorry nha!
Tk và kb hộ mình  với! thanks 

P=(1-1/1+2).(1-1/1+2+3)....(1-1/1+2+3+4+...+2011)

=[1-1/(2+1).2:2].[1-1/(3+1).3:2].....[1-1/(2011+1).2011:2]

=(1-2/2.3).(1-2/3.4)...(1-2/2011.2012)

=4/2.3.10/3.4....4046130/2011.2012

=1.4/2.3 .2.5/3.4 ....2010.2013/2011.2012

=1.2....2010/2.3...2011 .4.5....2013/3.4....2012

=1/2011.2013/3

=671/2011

\(\left[\frac{139}{84}-\frac{577}{924}-\frac{98}{205}\right].\frac{2197}{27}\)

\(\left[\frac{34}{33}-\frac{98}{205}\right]\frac{2197}{27}\)

\(\frac{3736}{6765}.\frac{2197}{27}=44,93713285\)