Không cần tính cũng biết là 1

\(\frac{1995\cdot1996-1000}{1995\cdot1995+995}=\frac{1995.1995+1995-1000}{1995.1995+995}=\frac{1995.1995+995}{1995.1995+995}=1\)

\(\frac{1995\cdot1996-1000}{1995\cdot1995+995}=\frac{1995\cdot1995+1995-1000}{1995\cdot1995+995}=\frac{1995^2+995}{1995^2+995}=1\)

\(\dfrac{1995\cdot\left(1995+1\right)-1000}{1995\cdot1995+995}\) =   \(\dfrac{1995\cdot1995+1995-1000}{1995\cdot1995+995}\)

= \(\dfrac{1995\cdot1995+995}{1995\cdot1995+995}\) = \(1\)

a) > 1

b) = 1

\(a,\frac{1995.1994-1}{1993.1995+1994}>1\)

\(b,\frac{2002.2004-58}{2003.2003-59}=1\)

a) x⁴+x³+2x²+x+1=(x⁴+x³+x²)+(x²+x+1)=x²(x²+x+1)+(x²+x+1)=(x²+x+1)(x²+1)

b)a³+b³+c³-3abc=a³+3ab(a+b)+b³+c³ -(3ab(a+b)+3abc)=(a+b)³+c³-3ab(a+b+c)

=(a+b+c)((a+b)²-(a+b)c+c²)-3ab(a+b+c)=(a+b+c)(a²+2ab+b²-ac-ab+c²-3ab)=(a+b+c)(a²+b²+c²-ab-ac-bc)

c)Đặt x-y=a;y-z=b;z-x=c

a+b+c=x-y-z+z-x=o

đưa về như bài b

d)nhóm 2 hạng tử đầu lại và 2hangj tử sau lại để 2 hạng tử sau ở trong ngoặc sau đó áp dụng hằng đẳng thức dề tính sau đó dặt nhân tử chung

e)x²(y-z)+y²(z-x)+z²(x-y)=x²(y-z)-y²((y-z)+(x-y))+z²(x-y)

=x²(y-z)-y²(y-z)-y²(x-y)+z²(x-y)=(y-z)(x²-y²)-(x-y)(y²-z²)=(y-z)(x²-2y²+xy+xz+yz)

bằng 1

a/   A = B

vì  \(\frac{10^{1993}+10}{10^{1993}+1}=1\)và \(\frac{10^{1994}+10}{10^{1994}+1}=1\)

Học tốt

cảm ơn bạn 

gắng học tốt nhé

A = B

vì \(\frac{10^{1993}+10}{10^{1993}+1}=10\) và \(\frac{10^{1994}+10}{10^{1994}+1}=10\)

học tốt

\(A=\frac{10^{1993}+10}{10^{1993}+1}\)

\(=\frac{10^{1993}+1+9}{10^{1993}+1}\)

\(=\frac{10^{1993}+1}{10^{1993}+1}+\frac{9}{10^{1993}+1}\)

\(=1+\frac{9}{10^{1993}+1}\)( 1 )

\(B=\frac{10^{1994}+10}{10^{1994}+1}\)

\(=\frac{10^{1994}+1+9}{10^{1994}+1}\)

\(=\frac{10^{1994}+1}{10^{1994}+1}+\frac{9}{10^{1994}+1}\)

\(=1+\frac{9}{10^{1994}+1}\)( 2 )

Vì \(\frac{9}{10^{1993}+1}>\frac{9}{10^{1994}+1}\)( 3 )

Từ ( 1 )( 2 )( 3 )\(\Rightarrow1+\frac{9}{10^{1993}+1}>1+\frac{9}{10^{1994}+1}\)

\(\Rightarrow A>B\)

a) 8,40690928

b) 25/98832

c) 1/1000

Ta có : \(\frac{x-1991}{9}+\frac{x-1993}{7}+\frac{x-1995}{5}+\frac{x-1997}{3}+\frac{x-1999}{1}\)\(=\frac{x-9}{1991}+\frac{x-7}{1993}+\frac{x-5}{1995}+\frac{x-3}{1997}+\frac{x-1}{1999}\)

\(\Rightarrow\left(\frac{x-1991}{9}-1\right)+\left(\frac{x-1993}{7}-1\right)+\left(\frac{x-1995}{5}-1\right)+\left(\frac{x-1997}{3}-1\right)+\left(\frac{x-1999}{1}-1\right)\)

\(=\left(\frac{x-9}{1991}-1\right)+\left(\frac{x-7}{1993}-1\right)+\left(\frac{x-5}{1995}-1\right)+\left(\frac{x-3}{1997}-1\right)+\left(\frac{x-1}{1999}\right)\)

\(\Rightarrow\frac{x-2000}{9}+\frac{x-2000}{7}+\frac{x-2000}{5}+\frac{x-2000}{3}\)

\(=\frac{x-2000}{1991}+\frac{x-2000}{1993}+\frac{x-2000}{1995}+\frac{x-2000}{1997}+\frac{x-2000}{1999}\)

\(\Rightarrow\left(x-2000\right)\left(\frac{1}{9}+\frac{1}{7}+\frac{1}{5}+\frac{1}{3}\right)=\left(x-2000\right)\left(\frac{1}{1991}+\frac{1}{1993}+\frac{1}{1995}+\frac{1}{1997}+\frac{1}{1999}\right)\)

\(\Rightarrow\left(x-2000\right)\left(\frac{1}{9}+\frac{1}{7}+\frac{1}{5}+\frac{1}{3}\right)-\left(x-2000\right)\left(\frac{1}{1991}+\frac{1}{1993}+\frac{1}{1995}+\frac{1}{1997}+\frac{1}{1999}\right)=0\)

\(\Rightarrow\left(x-2000\right)\left[\left(\frac{1}{9}+\frac{1}{7}+\frac{1}{5}+\frac{1}{3}\right)-\left(\frac{1}{1991}+\frac{1}{1993}+\frac{1}{1995}+\frac{1}{1997}+\frac{1}{1999}\right)\right]=0\)

Vì \(\left(\frac{1}{9}+\frac{1}{7}+\frac{1}{5}+\frac{1}{3}\right)-\left(\frac{1}{1991}+\frac{1}{1993}+\frac{1}{1995}+\frac{1}{1997}+\frac{1}{1999}\right)\ne0\)

=> x - 2000 = 0 

=> x = 2000