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a)\(\left(a^3-b^3\right)+\left(a-b\right)^2\)
\(=\left(a-b\right)\left(a^2+ab+b^2\right)+\left(a-b\right)^2\)
\(\left(a-b\right)\left(a^2+ab+b^2+a-b\right)\)
b) \(\left(8a^3-27b^3\right)-2a\left(4a^2-9b^2\right)\)
\(=\left(2a-3b\right)\left(4a^2+6ab+9b^2\right)-2a\left(2a-3b\right)\left(2a+3b\right)\)
\(=\left(2a-3b\right)\left(4a^2+6ab+9b^2-4a^2-6ab\right)\)
\(=\left(2a-3b\right)\cdot9b^2\)
\(=\left(a-b\right)\left(a^2+ab+b^2\right)+a^2-2ab+b^2\)
= ...........
a, \(\left(8a^3-27b^3\right)-2a\left(4a^2-9b^2\right)\)
\(=\left(2a-3b\right)\left[\left(2a\right)^2+2a.3b+\left(3b\right)^2\right]-2a\left(2a-3b\right)\left(2a+3b\right)\)
\(=\left(2a-3b\right)\left[4a^2+6ab+9b^2-2a\left(2a+3b\right)\right]\)
\(=\left(2a-3b\right)\left(4a^2+6ab+9b^2-4a^2-6ab\right)\)
\(=\left(2a-3b\right).9b^2\)
b, \(\left(x^3-y^3\right)+\left(x-y\right)^2\)
\(=\left(x-y\right)\left(x^2+xy+y^2\right)+\left(x-y\right)^2\)
\(=\left(x-y\right)\left[\left(x^2+xy+y^2\right)+\left(x-y\right)\right]\)
\(=\left(x-y\right)\left(x^2+xy+y^2+x-y\right)\)
c, \(\left(m^3+n^3\right)+\left(m+n\right)^2\)
\(=\left(m+n\right)\left(m^2-mn+n^2\right)+\left(m+n\right)^2\)
\(=\left(m+n\right)\left(m^2-mn+n^2+m+n\right)\)
Chúc bạn học tốt!!!
\(b,=1^2-\left(x-y\right)^2=\left(1+x-y\right)\left(1-x+y\right)\)
\(c,=\left(x^2+1\right)^2-\left(2x\right)^2=\left(x^2+2x+1\right)\left(x^2-2x+1\right)=\left(x+1\right)^2\left(x-1\right)^2\)
x9 + 1
= (x3)3 + 13
= (x3 + 1)(x6 - x3 + 1)
= (x + 1)(x2 - x + 1)(x6 - x3 +1)
8a3 - 12a2 + 6a - 1
= (2a)3 - 3(2a)21 + 3 . 2a . 12 - 1
= (2a - 1)3
27a3 - 54a2b + 36ab2 - 8b3
= (3a)3 - 3(3a)22b + 3 . 3a . (2b)2 - (2b)3
= (3a - 2b)3
a,8a-8a2+3
=-8(a2-a)+3
=-8[a2-2a\(\dfrac{1}{2}\)+\(\left(\dfrac{1}{2}\right)^2\)-\(\dfrac{1}{4}\)]+3
=-8[(a-\(\dfrac{1}{2}\))2-\(\dfrac{1}{4}\)]+3
=-8(a-\(\dfrac{1}{2}\))2+2+3
=-8(a-\(\dfrac{1}{2}\))2+5
mà (a-\(\dfrac{1}{2}\))2\(\ge\)0
=>-8(a-\(\dfrac{1}{2}\))2\(\le\)0
=>-8(a-\(\dfrac{1}{2}\))2+5\(\le\)5
=> Gía trị lớn nhất biểu thức trên đạt được là 5( khi (a-\(\dfrac{1}{2}\))2=0\(\Leftrightarrow\)a=\(\dfrac{1}{2}\))