Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
1. \(\left(x+1\right)^3-125\)
\(=\left(x+1\right)^3-5^3\)
\(=\left(x+1-5\right).\left[\left(x+1\right)^2+\left(x+1\right).5+5^2\right]\)
2. \(\left(x+4\right)^3-64\)
\(=\left(x+4\right)^3-4^3\)
\(=\left(x+4-4\right).\left[\left(x+4\right)^2+\left(x+4\right).4+4^2\right]\)
3. \(x^3-\left(y-1\right)^3\)
\(=(x^3-y+1).\left[\left(x^2\right)+x.\left(y+1\right)+\left(y+1\right)^2\right]\)
\(\)4. \(\left(a+b\right)^3-c^3\)
\(=\left[\left(a+b\right)-c\right].\left[\left(a+b\right)^2+\left(a+b\right).c+c^2\right]\)
5. \(125-\left(x+2\right)^3\)
\(=5^3-\left(x+2\right)^3\)
\(=\left(5-x-2\right).\left[5^2+5.\left(x+2\right)+\left(x+2\right)^2\right]\)
6. \(\left(x+1\right)^3+\left(x-2\right)^3\)
\(=\left[\left(x+1\right)+\left(x-2\right)\right].\left[\left(x+1\right)^2-\left(x+1\right).\left(x-2\right)+\left(x-2\right)^2\right]\)
a) = a3+b3+c3 +3a2b +3ab2 -3ab(a+b) - 3abc
= (a+b)3+c3-3ab(a+b)-3abc (áp dụng A3+B3 ta có)
=(a+b+c) ( (a+b)2 - (a+b)c +c2) - 3ab(a+b+c)
=(a+b+c) ( (a+b)2 - (a+b)c +c2 - 3ab) (nhân tử chung là a+b+c)
=(a+b+c) ( a2+2ab+b2- ac-bc +c2 -3ab)
=(a+b+c) (a2+b2+c2-ab-ac-bc)
Phần b tương tự
1, \(x^3+3^3=\left(x+3\right)\left(x^2-3x+9\right)\)
2, đề sai
3, \(x^3+8=\left(x+2\right)\left(x^2-2x+4\right)\)
4, \(x^3-64=\left(x-4\right)\left(x^2+4x+16\right)\)
5, \(1000-y^3=\left(10-y\right)=\left(100+10y+y^2\right)\)
tương tự ...
8, \(8x^3+27y^3=\left(2x+3y\right)\left(4x^2-6xy+9y^2\right)\)
Câu 2 đề ko sai nha bạn.
2) x2 - (\(\sqrt{y^3}\))2 ( y>0)
= ( x -\(\sqrt{y^3}\)) ( x +\(\sqrt{y^3}\))
a: \(64x^3-27y^3=\left(4x-3y\right)\left(16x^2+12xy+9y^2\right)\)
c: \(125-\left(x+1\right)^3\)
\(=\left(5-x-1\right)\left(25+5x+5+x^2+2x+1\right)\)
\(=\left(4-x\right)\left(x^2+7x+31\right)\)
a) \(64x^3-27y^3=\left(4x\right)^3-\left(3y\right)^3=\left(4x-3y\right)\left(16x^2+12xy+9y^2\right)\)
\(b)\) \(27x^3+\dfrac{y^3}{8}=\left(3x\right)^3+\left(\dfrac{y}{2}\right)^3\)
\(=\left(3x+\dfrac{y}{2}\right)\left(9x^2-\dfrac{3}{2}xy+\dfrac{1}{4}y^2\right)\)
\(c)\) \(125-\left(x+1\right)^3=5^3-\left(x+1\right)^3=\left(5-x-1\right)\left(25+5\left(x+1\right)+\left(x+1\right)^2\right)\)
\(=\left(4-x\right)\left(x^2+7x+31\right)\)
1) \(\left[\left(a+b\right)-c\right]^2=\left(a+b\right)^2-2c\left(a+b\right)+c^2\)
\(=\left(a^2+2ab+b^2\right)-2ac-2bc+c^2\)
\(=a^2+b^2+c^2+2ab-2ac-2bc\)
2)Phần này tg tự
3)\(\left(x+y+z\right)\left(x+y-z\right)=\left(x+y\right)^2-z^2=x^2+2xy+y^2-z^2\)
a: \(\left(x+y+z\right)^3=x^3+y^3+z^3+3\left(x+y\right)\left(y+z\right)\left(x+z\right)\)
b: \(\left(y-z+1\right)^3=y^3-z^3+1+3\left(y-z\right)\left(y+1\right)\left(-z+1\right)\)
c: \(8x^3-125=\left(2x-5\right)\left(4x^2+10x+25\right)\)
d: \(27y^3+64z^3=\left(3y+4z\right)\left(9y^2-12yz+16z^2\right)\)