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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)
a) \(\dfrac{1997x1996-1}{1995x1997+1996}=\dfrac{1997x\left(1995+1\right)-1}{1995x1997+1996}\)
\(=\dfrac{1997x1995+1997-1}{1995x1997+1996}=\dfrac{1997x1995+1996}{1995x1997+1996}=1\)
b) \(\dfrac{1997x1996-995}{1995x1997+1002}=\dfrac{1997x\left(1995+1\right)-995}{1995x1997+1002}\)
\(=\dfrac{1997x1995+1997-995}{1995x1997+1002}=\dfrac{1997x1995+1002}{1995x1997+1002}=1\)
a)37/53x23/48x535353/373737x242424/232323
=37/53x23/48x53/37x24/23
=851/2544x1272/851
=1/2
\(\frac{1996\cdot1995-996}{1000+1996\cdot1994}\)
= \(\frac{1996\cdot1994+1996-996}{1000+1996\cdot1994}\)
= \(\frac{1996\cdot1994+1000}{1000+1996\cdot1994}=1\)
Bài làm ( Lưu ý : dấu . là dấu x )
a ) \(\frac{1997\cdot1996-1}{1995\cdot1997+1996}\) b ) \(\frac{254\cdot399-145}{254+399\cdot253}\)
\(=\frac{1997\cdot\left(1995+1\right)-1}{1995\cdot1997+1996}\) \(=\frac{\left(253+1\right)\cdot399-145}{399\cdot253+254}\)
\(=\frac{1997\cdot1995+1997\cdot1-1}{1995\cdot1997+1996}\) \(=\frac{253\cdot399+399\cdot1-145}{253\cdot399+254}\)
\(=\frac{1997\cdot1995+1996}{1995\cdot1997+1996}\) \(=\frac{253\cdot399+254}{253\cdot399+254}\)
\(=1\) \(=1\)
c ) \(\frac{1997\cdot1996-995}{1995\cdot1997+1002}\) d ) \(\frac{5932+6001\cdot6391}{5392\cdot6001-69}\)
\(=\frac{1997\cdot\left(1995+1\right)-995}{1995\cdot1997+1002}\) \(=\frac{6001\cdot5391+5932}{5391\cdot6001-69}\)
\(=\frac{1997\cdot1995+1997\cdot1-995}{1995\cdot1997+1002}\) \(=\frac{6001\cdot5391+5932}{\left(5931+1\right)\cdot6001-69}\)
\(=\frac{1997\cdot1995+1002}{1995\cdot1997+1002}\) \(=\frac{6001\cdot5931+5932}{5931\cdot6001+6001\cdot1-69}\)
\(=1\) \(=\frac{6001\cdot5391+5932}{5391\cdot6001+5932}\)
\(=1\)
d ) \(\frac{1996\cdot1995-996}{1000+1996\cdot1994}\)
\(=\frac{1996\cdot1995-996}{1996\cdot1994+1000}\)
\(=\frac{1996+\left(1994+1\right)-996}{1996\cdot1994+1000}\)
\(=\frac{1996\cdot1994+1996\cdot1-996}{1996\cdot1994+1000}\)
\(=\frac{1996\cdot1994+1000}{1996\cdot1994+1000}\)
\(=1\)