A=\(\frac{1993\cdot1995+28}{1993\cdot1996-1965}\)
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\(\frac{1995\cdot1996-1000}{1995\cdot1995+995}=\frac{1995.1995+1995-1000}{1995.1995+995}=\frac{1995.1995+995}{1995.1995+995}=1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{1995\cdot\left(1995+1\right)-1000}{1995\cdot1995+995}\) = \(\dfrac{1995\cdot1995+1995-1000}{1995\cdot1995+995}\)
= \(\dfrac{1995\cdot1995+995}{1995\cdot1995+995}\) = \(1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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)
![](https://rs.olm.vn/images/avt/0.png?1311)
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a/ A = B
vì \(\frac{10^{1993}+10}{10^{1993}+1}=1\)và \(\frac{10^{1994}+10}{10^{1994}+1}=1\)
Học tốt
![](https://rs.olm.vn/images/avt/0.png?1311)
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 )\(\Rightarrow1+\frac{9}{10^{1993}+1}>1+\frac{9}{10^{1994}+1}\)
\(\Rightarrow A>B\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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
Ta có : \(A=\frac{1993.1995+28}{1993.\left(1995+1\right)-1965}=\frac{1993.1995+28}{1993.1995+1993-1965}=\frac{1993.1995+28}{1993.1995+28}=1\)
Bài giải
\(A=\frac{1993\cdot1995+28}{1993\cdot1996-1965}=\frac{1993\cdot1995+28}{1993\cdot1995+1993-1965}=\frac{1993\cdot1995+28}{1993\cdot1995+28}=1\)
Vậy A = 1