a/Viết đề mà cx sai đc nữa: \(\left(\frac{x+2}{98}+1\right)\left(\frac{x+3}{97}+1\right)=\left(\frac{x+4}{96}+1\right)\left(\frac{x+5}{95}+1\right)\)

\(\Leftrightarrow\frac{x+100}{98}.\frac{x+100}{97}-\frac{x+100}{96}.\frac{x+100}{95}=0\)

\(\Leftrightarrow\left(x+100\right)^2\left(\frac{1}{98.97}-\frac{1}{96.95}\right)=0\)

\(\Rightarrow x=-100\)

b/\(\Leftrightarrow\left(\frac{x+1}{1998}+1\right)+\left(\frac{x+2}{1997}+1\right)=\left(\frac{x+3}{1996}+1\right)+\left(\frac{x+4}{1995}+1\right)\)

\(\Leftrightarrow\frac{x+1999}{1998}+\frac{x+1999}{1997}-\frac{x+1999}{1996}-\frac{x+1999}{1995}=0\)

\(\Leftrightarrow\left(x+1999\right)\left(...\right)=0\Rightarrow x=-1999\)

b,\(\frac{x+1}{1998}+\frac{x+2}{1997}=\frac{x+3}{1996}+\frac{x+4}{1995}\)

=>\(\frac{x+1}{1998}+1\frac{x+2}{1997}+1=\frac{x+3}{1996}+1+\frac{x+4}{1995}+1\)

\(\Leftrightarrow\)\(\frac{x+1999}{1998}+\frac{x+1999}{1997}=\frac{x+1999}{1996}+\frac{x+1999}{1995}\)

\(\Leftrightarrow\)\(\frac{x+1999}{1998}+\frac{x+1999}{1997}-\frac{x+1999}{1996}-\frac{x+1999}{1995}\)=0

\(\Leftrightarrow\)\(\left(x+1999\right)\left(\frac{1}{1998}+\frac{1}{1997}-\frac{1}{1996}-\frac{1}{1995}\right)\)=0

\(\Leftrightarrow\)x+1999=0(Vì \(\frac{1}{1998}+\frac{1}{1997}-\frac{1}{1996}-\frac{1}{1995}\ne0\))

\(\Leftrightarrow\)x=-1999

Vậy x=-1999

\(\Leftrightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}\)

\(\Leftrightarrow\frac{ab+bc+ac}{abc}=\frac{1}{a+b+c}\Leftrightarrow\left(a+b+c\right)\left(ab+bc+ac\right)-abc=0\)

\(\Leftrightarrow a^2b+a^2c+b^2a+b^2c+abc+abc+bc^2+ac^2=0\)

\(\Leftrightarrow\left(a+b\right)\left(b+c\right)\left(a+c\right)=0\Leftrightarrow...\)

\(P=0\)

\(\frac{-1997.1996+1}{\left(-1995\right).\left(-1997\right)+1996}=\frac{-1997\left(1995+1\right)+1}{1995.1997+1996}\)

\(=\frac{-1997.1995+\left(-1997\right)+1}{1995.1997+1996}=\frac{-1997.1995+\left(-1996\right)}{1995.1997+1996}=-1\)

\(\dfrac{\left(-1997\right).1996+1}{\left(-1995\right).\left(-1997\right)+1996}=\dfrac{-1997\left(1995+1\right)+1}{1995.1997+1996}=\dfrac{-1997.1995+\left(-1997\right)+1}{1995.1997+1996}=\dfrac{-1997.1995+\left(-1996\right)}{1995.1997+1996}=-1\)

\(\frac{5.18-10.27+15.36}{10.36-20.54+30.72}\)

\(=\frac{5.18-10.27+15.36}{5.2.18.2-10.2.27.2+15.2.36.2}\)

\(=\frac{5.18-10.27+15.36}{5.8.2.2-10.27.2.2+15.36.2.2}\)

\(=\frac{1}{2.2-2.2+2.2}\)

\(=\frac{1}{2.2}=\frac{1}{4}\)

Giúp mik với

trước 5h nha

a) \(\frac{120-\left(-0,5\right).\left(-40\right).\left(-5\right).\left(-0,2\right).20.0,25}{5+10+15+...+1995}\)

\(=\frac{120-\left[\left(-0,5\right).\left(-0,2\right)\right].\left[\left(-40\right).0,25\right].\left[\left(-5\right).\left(20\right)\right]}{\left(1995+5\right).\left[\left(1995-5\right)\div5+1\right]\div2}\)

\(=\frac{120-0,1.\left(-10\right).\left(-100\right)}{2000.399\div2}\)

\(=\frac{120-100}{1000.399}\)

\(=\frac{1}{19950}\)

b) \(\frac{5.18-10.27+15.36}{10.36-20.54+30.72}\)

\(=\frac{5.18-2.5.27+3.5.36}{10.2.18-20.2.27+5.2.3.2.36}\)

\(=\frac{5.18-2.5.27+3.5.36}{20.18-20.2.27+20.3.36}\)

\(=\frac{5.\left(18-2.27+3.36\right)}{20.\left(18-2.27+3.36\right)}\)

\(=\frac{1}{4}\)

c) \(\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right)\left(\frac{1}{4}-1\right)...\left(\frac{1}{1999}-1\right)\)

\(=\left(\frac{-1}{2}\right).\left(\frac{-2}{3}\right).\left(\frac{-3}{4}\right)...\left(\frac{-1998}{1999}\right)\)

\(=\frac{\left(-1\right).\left(-2\right).\left(-3\right)...\left(-1998\right)}{2.3.4...1999}\)

\(=\frac{\left(-1\right).\left(-1\right).\left(-1\right)...\left(-1\right)}{1.1.1...1999}\)

Ta có : 1998 số (-1) mà 1998 là số chẵn

Vậy tích của 1998 số (-1) = 1

\(\Rightarrow\frac{\left(-1\right).\left(-1\right).\left(-1\right)...\left(-1\right)}{1.1.1...1999}\)

\(=\frac{1}{1999}\)

rut gon con 1/-1995 vi rut gon 1996 ca tu va mau

rut gon -1997 ca tu va mau cung deu co vi vay sau khi rut gon cac so giong nh au thi phan so tren con 1/-1995