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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)
\(\frac{1996\cdot1995-996}{1000+1996\cdot1994}\)
= \(\frac{1996\cdot1994+1996-996}{1000+1996\cdot1994}\)
= \(\frac{1996\cdot1994+1000}{1000+1996\cdot1994}=1\)
\(\frac{2}{7}\)+ \(\frac{5}{14}\)+\(\frac{1}{7}\)+ \(\frac{3}{14}\)=\(\frac{4}{14}\)+\(\frac{5}{14}\)+\(\frac{2}{14}\)+\(\frac{3}{14}\)=\(\frac{14}{14}\)=1
469x281+489x719=469x281+(469+20)x719=469x281+469x719+20x719=469x(281+719)+1438=469x1000+1438=469000+1438=470438
a\(\frac{2}{5}\)+\(\frac{5}{14}\)+\(\frac{1}{7}\)+\(\frac{3}{14}\)=\(\frac{53}{70}\)+\(\frac{1}{7}\)=\(\frac{9}{10}\)+\(\frac{3}{14}\)=\(\frac{39}{35}\)
b\(\frac{1995.1997-1}{1996.1995+1994}\)=3984008001
c 469x281+489x719
=(489-469)x(281+719)
=20x1000
=20000
\(B=\)\(\frac{3+33+333+3333+33333}{4+44+444+4444+44444}\)
\(B=\frac{3.1+3.11+3.111+3.1111+3.11111}{4.1+4.11+4.111+4.1111+4.11111}\)
\(B=\frac{3.\left(1+11+111+1111+11111\right)}{4.\left(1+11+111+1111+11111\right)}\)
\(B=\frac{3}{4}\)
\(A=\frac{1}{3}+\frac{1}{6}+\frac{1}{12}+\frac{1}{24}+\frac{1}{48}+\frac{1}{96}+\frac{1}{192}\)
\(A.2=\left(\frac{1}{3}+\frac{1}{6}+\frac{1}{12}+\frac{1}{24}+\frac{1}{48}+\frac{1}{96}+\frac{1}{192}\right).2\)
\(A.2=\frac{2}{3}+\frac{1}{3}+\frac{1}{6}+\frac{1}{12}+\frac{1}{24}+\frac{1}{48}+\frac{1}{96}\)
=>\(A.2-A=\left(\frac{2}{3}+\frac{1}{3}+\frac{1}{6}+\frac{1}{12}+\frac{1}{24}+\frac{1}{48}+\frac{1}{96}\right)-\left(\frac{1}{3}+\frac{1}{6}+\frac{1}{12}+\frac{1}{24}+\frac{1}{48}+\frac{1}{96}+\frac{1}{192}\right)\)
\(A=\frac{2}{3}-\frac{1}{192}\)
\(A=\frac{127}{192}\)
\(\frac{1995}{1997}.\frac{1990}{1993}.\frac{1997}{1994}.\frac{1993}{1995}.\frac{997}{995}\)
Đặt \(C=\frac{1995}{1997}.\frac{1990}{1993}.\frac{1997}{1994}.\frac{1993}{1995}.\frac{997}{995}\)
\(C=\frac{1995.1990.1997.1993.997}{1997.1993.1994.1995.995}\)
\(C=\frac{1990.997}{1994.995}\)
\(C=\frac{995.2+997}{997.2+995}=1\)
\(B=\frac{3+33+333+3333+ 33333}{4+44+444+4444+44444}\)
\(\Rightarrow B=\frac{3\left(1+11+111+1111+11111\right)}{4\left(1+11+111+1111+11111\right)}=\frac{3}{4}\)
#)Giải :
A, \(\frac{254x399-145}{254+399x253}\)
\(=\frac{253x399+399-145}{254+399x253}\)
\(=\frac{253x399+254}{254+399x253}\)
\(=1\)
B, \(\frac{5932+6001x5931}{5931x6001-69}\)
\(=\frac{5932+6001x5931}{\left(5931+1\right)x6001-69}\)
\(=\frac{5932+6001x5931}{5931x6001+6001-69}\)
\(=\frac{5932+6001x5932}{5932x6001+5932}\)
\(=1\)
#~Will~be~Pens~#
\(\frac{254x399-145}{254+399x253}=\frac{\left(253+1\right)x399-145}{254+399x253}=\frac{253x399+1x399-145}{254+399x253}=\frac{253x399+254}{254+399x253}\)
\(=1\)
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 ) \(\frac{1996\cdot1995-996}{1000+1996\cdot1994}\)
\(=\frac{1996\cdot1995-996}{1996\cdot1994+1000}\)
\(=\frac{1996+\left(1994+1\right)-996}{1996\cdot1994+1000}\)
\(=\frac{1996\cdot1994+1996\cdot1-996}{1996\cdot1994+1000}\)
\(=\frac{1996\cdot1994+1000}{1996\cdot1994+1000}\)
\(=1\)