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10 tháng 9 2017

Gọi 1997 là A a có

1996=A-1

1998=A+1

\(A^2-\left(A-1\right)\left(A+1\right)\)

\(=A^2-\left(A^2+A-A+1\right)\)

\(=A^2-\left(A^2+1\right)\)

\(=A^2-A^2-1\)

\(=-1\)

10 tháng 9 2017

1997*1997-1996*1998

(1996+1)*1997-1996*1998

1996*1997+1997-1996*1998

1996*(1997-1998)+1997

1996*(-1)+1997

-1996+1997=1

=>(x+1/1998+1)+(x+2/1997+1)=(x+3/1996+1)+(x+4/1995+1)

=>x+1999=0

=>x=-1999

14 tháng 12 2017

Vào trang cá nhân của t mà xem.T vừa làm r.Lười gõ lại lắm T^T

14 tháng 12 2017

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

\(=\dfrac{x+1}{1998}+\dfrac{x+2}{1997}-\dfrac{x+3}{1996}-\dfrac{x+4}{1995}=0\)

\(=\dfrac{x+1}{1998}+1+\dfrac{x+2}{1997}+1-\dfrac{x+3}{1996}-1-\dfrac{x+4}{1995}-1=0\)

\(=\dfrac{x+1999}{1998}+\dfrac{x+1999}{1998}-\left(\dfrac{x+3}{1996}+1\right)-\left(\dfrac{x+4}{1995}+1\right)=0\)

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

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

\(x+1999=0\)

Vậy \(x=-1999\)

9 tháng 2 2019

Ta có :

\(\dfrac{1997^2-1996^2}{1997^2+1996^2}=\dfrac{1.\left(1997+1996\right)}{1997^2+1996^2}=\dfrac{3993}{1997^2+1996^2}\)

Lại có : \(\dfrac{1}{3993}=\dfrac{3993}{3993^2}\)

Do \(3993^2=\left(1997+1996\right)^2>1997^2+1996^2\)

\(\Rightarrow\dfrac{3993}{3993^2}< \dfrac{3993}{1997^2+1996^2}\)

\(\Rightarrow\dfrac{1}{3993}< \dfrac{1997^2-1996^2}{1997^2+1996^2}\)

12 tháng 12 2017

a)

\(x^4+1996x^2+1995x+1996\)

\(=\left(x^4-x\right)+\left(1996x^2+1996x+1996\right)\)

\(=x\left(x^3-1\right)+1996\left(x^2+x+1\right)\)

\(=x\left(x-1\right)\left(x^2+x+1\right)+1996\left(x^2+x+1\right)\)

\(=\left(x^2+x+1\right)\left[x\left(x-1\right)+1996\right]\)

\(=\left(x^2+x+1\right)\left(x^2-x+1996\right)\)

b)

\(x^4+1997x^2+1996x+1997\)

\(=\left(x^4-x\right)+\left(1997x^2+1997x+1997\right)\)

\(=x\left(x^3-1\right)+1997\left(x^2+x+1\right)\)

\(=x\left(x-1\right)\left(x^2+x+1\right)+1997\left(x^2+x+1\right)\)

\(=\left(x^2+x+1\right)\left[x\left(x-1\right)+1997\right]\)

\(=\left(x^2+x+1\right)\left(x^2-x+1997\right)\)

12 tháng 12 2017

x4+1996x2+1995x+1996

=(x4_x)+(1996x2+1996x+1996)

=x(x3-1)+1996(x2+x+1)

=x(x-1)(x2+x+1)+1996(x2+x+1)

=(x2+x+1)((x2-1)+1996)

=(x2+x+1)((x+1)(x-1)+1996)

Câu 2 tương tự bạn nhé!

12 tháng 3 2020

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

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

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

12 tháng 3 2020

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

14 tháng 7 2015

1998^2-1997(1998+1)

=1998^2-(1998-1)(1998+1)

=1998^2-(1998^2-1)

=1998^2-1998^2+1

=1