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\(h)\)\(\left(xy+1\right)^2-\left(x+y\right)^2\)
\(=\)\(\left(xy-x-y+1\right)\left(xy+x+y+1\right)\)
\(=\)\(\left[x\left(y-1\right)-\left(y-1\right)\right].\left[x\left(y+1\right)+\left(y+1\right)\right]\)
\(=\)\(\left(x-1\right)\left(y-1\right)\left(x+1\right)\left(y+1\right)\)
\(i)\)\(16b^2c^2-4\left(b^2+c^2-a^2\right)^2\)
\(=\)\(\left(4bc\right)^2-\left(2b^2+2c^2-2a^2\right)^2\)
\(=\)\(\left(4bc-2b^2-2c^2+2a^2\right)\left(4bc+2b^2+2c^2-2a^2\right)\)
\(=\)\(2\left[a^2-\left(b^2-2bc+c^2\right)\right].2\left[\left(b^2+2bc+c^2\right)-a^2\right]\)
\(=\)\(-4\left[a^2-\left(b-c\right)^2\right].\left[a^2-\left(b+c\right)^2\right]\)
\(=\)\(-4\left(a-b+c\right)\left(a+b-c\right)\left(a-b-c\right)\left(a+b+c\right)\)
Chúc bạn học tốt ~
\(a,\left(3+xy^2\right)^2=9+6xy^2+x^2y^4\)
\(b,\left(10-2m^2n\right)^2=100-40m^2n+4m^4n^2\)
\(c,\left(a-b^2\right)\left(a+b^2\right)=a^2-b^4\)
a) ( 3+ xy2 )2
= 32 + 2.3.xy2 + ( xy2 )2
= 9 + 6xy2 + x2y4
b) ( 10 - 2m2n2 )
= 2 ( 5 - m2n2 )
c) ( a - b2 )( a + b2 )
= a2 + ab2 - ab2 - b4
= a2 - b4
Giải:
a) \(\left(3+xy^2\right)^2\)
\(=3^2+2.3.xy^2+\left(xy^2\right)^2\)
\(=9+6xy^2+x^2y^4\)
Vậy ...
b) \(\left(10-2m^2n\right)^2\)
\(=10^2-2.10.2m^2n+\left(2m^2n\right)^2\)
\(=100-40m^2n+4m^4n^2\)
Vậy ...
c) \(\left(a-b^2\right)\left(a+b^2\right)\)
\(=a^2-\left(b^2\right)^2\)
\(=a^2-b^4\)
Vậy ...
a) \(\left(m+n\right)^2-\left(m-n\right)^2+\left(m+n\right)\left(m-n\right)\)
\(=\left(m+n+m-n\right)\left(m+n-m+n\right)+m^2-n^2\)
\(=m^2-n^2+4mn\)
b) \(\left(a+b\right)^3+\left(a-b\right)^3-2a^3\)
\(=\left(a+b-a+b\right)\left[\left(a+b\right)^2-\left(a+b\right)\left(a-b\right)+\left(a-b\right)^2\right]-2a^3\)
\(=2b\left[a^2+2ab+b^2-a^2+b^2+a^2-2ab+b^2\right]-2a^3\)
\(=2b\left(a^2+3b^2\right)-2a^3\)
\(=2a^2b+6b^3-2a^3.\)
Tương tự áp dụng các HĐT.
a) \(\left(m+n\right)^2-\left(m-n\right)^2=\left[\left(m+n\right)-\left(m-n\right)\right]\left[\left(m+n\right)+\left(m-n\right)\right]=\left(2n\right)\left(2m\right)=4mn\)\(\left(m+n\right)\left(m-n\right)=m^2-n^2\)
A=\(4mn+m^2-n^2\) tối giản rồi
b)
\(\left(a+b\right)^3+\left(a-b\right)^3=\left[\left(a+b\right)+\left(a-b\right)\right]^3-3\left(a+b\right)\left(a-b\right)\left[\left(a+b\right)+\left(a-b\right)\right]=8a^3-3.2a.\left(a^2-b^2\right)\)B=\(8a^3-3.2a.\left(a^2-b^2\right)-2a^3=6a\left[a^2-\left(a^2-b^2\right)\right]=6ab^2\)
Có (a+b+c)2 = 3(ab+bc+ac)
\(\Rightarrow a^2+b^2+c^2+2ab+2bc+2ac=3ab+3bc+3ac\)
\(\Rightarrow a^2+b^2+c^2+2ab+2bc+2ac-3ab-3bc-3ac\)\(=0\)
\(\Rightarrow a^2+b^2+c^2-ab-bc-ac\)\(=0\)
\(\Rightarrow a^2+b^2+c^2=ab+bc+ac\)
\(\Rightarrow2a^2+2b^2+2c^2=2ab+2bc+2ac\)
\(\Rightarrow a^2-2ab+b^2+b^2-2bc+c^2+a^2-2ac+c^2\)\(=0\)
\(\Rightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2=0\)
\(\Rightarrow a=b=c\)
\(1.a\left(a+2b\right)^3-b\left(2a+b\right)^3\)
=\(a\left(a^3+6a^2b+12ab^2+8b^3\right)-b\left(8a^3+12a^2b+6ab^2+b^3\right)\)
=\(a^4+6a^3b+12a^2b^2+8ab^3-8a^3b-12a^2b^2-6ab^3-b^4\)
=\(a^4-b^4\)=\(\left(a^2-b^2\right)\left(a^2+b^2\right)\)
\(\left(3+xy\right)^2=9+6xy+xy^2\)
\(\left(10-m^2n\right)^2=100-20m^2n^2+m^4n^2\)
\(\left(a-b^2\right)\left(a+b^2\right)=a^2-b^4\)
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