Tính nhanh :

A = \(2016.20152015-2015.20162016\)

= \(2016.2015.10001-2015.2016.1001\)

=0

\(A=2016.20152015-2015.20162016\)

\(=2016.2015.10001-2015.2016.10001\)

\(=0\)

Tính :

\(\dfrac{1993+1993.1994}{1992.1995+1995}\)= \(\dfrac{1993.1+1993.1994}{1992.1995+1995.1}\)=\(\dfrac{1993\left(1+1994\right)}{1995\left(1992+1\right)}\)=\(\dfrac{1993.1995}{1995.1993}=1\)

A= \(\dfrac{1993\left(1994+1\right)}{1995\left(1992+1\right)}\)=1

B=\(\dfrac{399\left(45+55\right)}{1995\left(1996-1991\right)}\)=\(\dfrac{399.5.100}{399.5.5}\)=100

C=\(\dfrac{1996.1994+1996-996}{1000+1996.1994}\)=1

\(\dfrac{x-1}{1992}+\dfrac{x-2}{1993}=\dfrac{x-3}{1994}+\dfrac{x-4}{1995}\)

\(\Rightarrow\left(\dfrac{x-1}{1992}+1\right)+\left(\dfrac{x-2}{1993}+1\right)=\left(\dfrac{x-3}{1994}+1\right)+\left(\dfrac{x-4}{1995}+1\right)\)

\(\Rightarrow\left(\dfrac{x-1+1992}{1992}\right)+\left(\dfrac{x-2+1993}{1993}\right)=\left(\dfrac{x-3+1994}{1994}\right)+\left(\dfrac{x-4+1995}{1995}\right)\)

\(\Rightarrow\dfrac{x+1991}{1992}+\dfrac{x+1991}{1993}=\dfrac{x+1991}{1994}+\dfrac{x+1991}{1995}\)

\(\Rightarrow\dfrac{x+1991}{1992}+\dfrac{x+1991}{1993}-\dfrac{x+1991}{1994}-\dfrac{x+1991}{1995}=0\)

\(\Rightarrow\left(x+1991\right)\left(\dfrac{1}{1992}+\dfrac{1}{1993}-\dfrac{1}{1994}-\dfrac{1}{1995}\right)=0\)

\(\Rightarrow\left(x+1991\right)=0\) ( vì \(\left(\dfrac{1}{1992}+\dfrac{1}{1993}-\dfrac{1}{1994}-\dfrac{1}{1995}\right)\ne0\)

\(\Rightarrow x=-1991\)

Mỗi số hạng của vế trái cộng thêm 1, vế phải = 5. Mỗi số hạng vế trái có mẫu số giống nhau, bạn đặt x+ 2020 làm nhân tử chung, phần còn lại tự làm nhé.

mấy bài còn lại bạn đăng cx làm tương tự

\(\frac{x+24}{1996}+\frac{x+25}{1995}+\frac{x+26}{1994}+\frac{x+27}{1993}+\frac{x+2036}{4}=0\)

\(\Leftrightarrow\left(\frac{x+24}{1996}+1\right)+\left(\frac{x+25}{1995}+1\right)+\left(\frac{x+26}{1994}+1\right)+\left(\frac{x+27}{1993}+1\right)+\left(\frac{x+2036}{4}-4\right)=0\)

\(\Leftrightarrow\frac{x+2020}{1996}+\frac{x+2020}{1995}+\frac{x+2020}{1994}+\frac{x+2020}{1993}+\frac{x+2020}{4}=0\)

\(\Leftrightarrow\left(x+2020\right)\left(\frac{1}{1996}+\frac{1}{1995}+\frac{1}{1994}+\frac{1}{1993}+\frac{1}{4}\right)=0\)

\(\Leftrightarrow x+2020=0\)

\(\Leftrightarrow x=-2020\)

Vậy ....

A = 1995 x 1993 - 18 / 1975 + 1993 x 1994

= (1994 + 1) x 1993 - 18 / 1975 + 1993 x 1994

= 1994 x 1993 + 1993 - 18 / 1975 + 1993 x 1994

= 1994 x 1993 + 1975 / 1975 + 1993 x 1994

= 1

A = 1995 x 1993 - 18 / 1975 + 1993 x 1994= (1994 + 1) x 1993 - 18 / 1975 + 1993 x 1994= 1994 x 1993 + 1993 - 18 / 1975 + 1993 x 1994= 1994 x 1993 + 1975 / 1975 + 1993 x 1994= 1

A = 1995 x 1993 - 18/1975 + 1993 x 1994

   = ( 1994 + 1 ) x 1993 - 18/1975 + 1993 x 1994

  = 1994 x 1993 + 1993 - 18/1975 + 1993 x 1994

  = 1994 x 1993 + 1975/1975 + 1993 x 1994

  = 1

\(\dfrac{x+24}{1996}+\dfrac{x+25}{1995}+\dfrac{x+26}{1994}+\dfrac{x+27}{1993}+\dfrac{x+2036}{4}=0\)

\(\Rightarrow\left(\dfrac{x+24}{1996}+1\right)+\left(\dfrac{x+25}{1995}+1\right)+\left(\dfrac{x+26}{1994}+1\right)+\left(\dfrac{x+27}{1993}+1\right)+\left(\dfrac{x+2036}{4}-4\right)=0\)\(\Rightarrow\dfrac{x+2020}{1996}+\dfrac{x+2020}{1995}+\dfrac{x+2020}{1994}+\dfrac{x+2020}{1993}+\dfrac{x+2020}{4}=0\)\(\Rightarrow\left(x+2020\right)\left(\dfrac{1}{9996}+\dfrac{1}{1995}+\dfrac{1}{1994}+\dfrac{1}{1993}+\dfrac{1}{4}\right)=0\)

\(\Rightarrow x+2020=0\Rightarrow x=-2020\)

\(A=\dfrac{1995\times1994-1}{1993\times1995+1994}\)

\(A=\dfrac{1995\times\left(1993+1\right)-1}{1993\times1995+1994}\)

\(A=\dfrac{1995\times1993+1995-1}{1995\times1993+1994}\)

\(A=\dfrac{1995\times1993+1994}{1995\times1993+1994}\)

\(A=1\)

mik cần gấp

(1995 × 1994 - 1)/(1993 × 1995 + 1994)

= 3978029/3978029

= 1

Lời giải:
PT đã cho tương đương với:

\(\frac{x+24}{1996}+1+\frac{x+25}{1995}+1+\frac{x+26}{1994}+1+\frac{x+27}{1993}+1+\frac{x+2036}{4}-4=0\)

\(\Leftrightarrow \frac{x+2020}{1996}+\frac{x+2020}{1995}+\frac{x+2020}{1994}+\frac{x+2020}{1993}+\frac{x+2020}{4}=0\)

\(\Leftrightarrow (x+2020)\left(\frac{1}{1996}+\frac{1}{1995}+\frac{1}{1994}+\frac{1}{1993}+\frac{1}{4}\right)=0\)

Dễ thấy \(\frac{1}{1996}+\frac{1}{1995}+\frac{1}{1994}+\frac{1}{1993}+\frac{1}{4}\neq 0\) nên \(x+2020=0\Rightarrow x=-2020\) là nghiệm của pt.

Vậy............

bằng 1 nhé bạn