tính nhanh
\(\frac{1997\chi1996-1}{1995\chi1997+1996}\) \(\frac{254\chi339-145}{254+399\chi253}\)
\(\frac{1997\chi1996-995}{1995\chi1997+1002}\) \(\frac{5392+6001\chi5931}{5392\chi6001-69}\)
\(\frac{1996\chi1995-996}{1000+1996\chi1994}\)
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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)
\(\frac{254.399-145}{254+399.253}\)
\(=\frac{253.399+399-145}{254+399.253}\)
\(=\frac{253.399+253}{254+399.253}\)
\(=1\)
\(\frac{254\times399-145}{254+399\times253}\)
=\(\frac{253\times399+399-145}{254+399\times253}\)
=\(\frac{253\times399+254}{254+399\times253}\)
=1
\(\frac{254.399-145}{254+399.253}=\frac{\left(253+1\right).399-145}{254+399.253}=\frac{253.399+399-145}{254+399.253}=\frac{253.399+254}{254+399.253}=1\)
ghép đầu cuối dăt chung 399 ra ngoài
kq vẫn như cũ chắc chắn 100%
202292.4291
C=\(\frac{1}{2}.\frac{2}{3}.......\frac{2016}{2017}\)
C= CÂU HỎI TƯƠNG TỰ
=> đcpm
\(A=\frac{254\cdot399-145}{254+399\cdot253}\)
\(A=\frac{\left(253+1\right)\cdot399-145}{254+399\cdot253}\)
\(A=\frac{253\cdot399+\left(399-145\right)}{254+399\cdot253}\)
\(A=\frac{253\cdot399+254}{254+399\cdot253}\)
\(A=1\)
\(B=\frac{5932+6001\cdot5931}{5932\cdot6001-69}\)
\(B=\frac{5932+6001\cdot5931}{\left(5931+1\right)\cdot6001-69}\)
\(B=\frac{5932+6001\cdot5931}{5931\cdot6001+\left(6001-69\right)}\)
\(B=\frac{5932+6001\cdot5931}{5931\cdot6001+5932}\)
\(B=1\)
\(C=\left(1-\frac{1}{2}\right)\cdot\left(1-\frac{1}{3}\right)\cdot...\cdot\left(1-\frac{1}{2017}\right)\)
\(C=\frac{1}{2}\cdot\frac{2}{3}\cdot...\cdot\frac{2016}{2017}\)
\(C=\frac{1\cdot2\cdot3\cdot...\cdot2016}{2\cdot3\cdot4\cdot...\cdot2017}\)
\(C=\frac{1}{2017}\)
\(\frac{254+399.253}{399.254-145}\)
\(=\frac{254+399.253}{399.\left(253+1\right)-145}\)
\(=\frac{254+399.253}{399.253+399.1-145}\)
\(=\frac{254+1}{1+399-145}=1\)
Chúc bạn học tốt!!!
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 ) \(\frac{1996\cdot1995-996}{1000+1996\cdot1994}\)
\(=\frac{1996\cdot1995-996}{1996\cdot1994+1000}\)
\(=\frac{1996+\left(1994+1\right)-996}{1996\cdot1994+1000}\)
\(=\frac{1996\cdot1994+1996\cdot1-996}{1996\cdot1994+1000}\)
\(=\frac{1996\cdot1994+1000}{1996\cdot1994+1000}\)
\(=1\)