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(3x-1)^2 - 16 = (3x-1)^2 - 4^2
= (3x-1-4)(3x-1+4)
= (3x-5)(3x+3)
\(\left(3x-1\right)^2-16\)
\(=\left(3x-1\right)^2-4^2\)
\(=\left(3x-1-4\right)\left(3x-1+4\right)\)
\(=\left(3x-5\right)\left(3x+3\right)\)
\(=3\left(x+1\right)\left(3x-5\right)\)
Bài 2:
1) \(x^2-4x+4=\left(x-2\right)^2\)
2) \(x^2-9=x^2-3^2=\left(x-3\right)\left(x+3\right)\)
3) \(1-8x^3=\left(1-2x\right)\left(1+2x+4x^2\right)\)
4) \(\left(x-y\right)^2-9x^2=\left(x-y\right)^2-\left(3x\right)^2=\left(x-y-3x\right)\left(x-y+3x\right)=\left(-2x-y\right)\left(4x-y\right)\)
5) \(\dfrac{1}{25}x^2-64y^2=\left(\dfrac{1}{5}x-8y\right)\left(\dfrac{1}{5}x+8y\right)\)
6) \(8x^3-\dfrac{1}{8}=\left(2x-\dfrac{1}{2}\right)\left(4x^2+x+\dfrac{1}{4}\right)\)
a) Đặt a + b = x ; a - b = y. Khi đó:
\(\left(a+b\right)^3-\left(a-b\right)^3\)
\(\Leftrightarrow x^3-y^3\)
\(\Leftrightarrow\left[x-y\right]\left[x^2+xy+y^2\right]\)
Thế lại vào ta có:
\(\Leftrightarrow\left[\left(a+b\right)-\left(a-b\right)\right]\left[\left(a+b\right)^2+\left(a+b\right)\left(a-b\right)+\left(a-b\right)^2\right]\)
\(\Leftrightarrow\left[\left(a-a\right)+\left(b+b\right)\right]\left[\left(a^2+b^2+2ab\right)+\left(a^2-b^2\right)+\left(a^2+b^2-2ab\right)\right]\)
\(\Leftrightarrow2b\left[\left(a^2+a^2+a^2\right)+\left(b^2-b^2+b^2\right)+\left(2ab-2ab\right)\right]\)
\(\Leftrightarrow2b\left[3a^2+b^2\right]\)
Mik làm tuỳ theo mình piết thôi nhé
a) ( a + b )3- ( a - b )3= a3 + b3 - a3 - b3 = a3 - a3 + b3 - b3 = 0
b) tương tự như ở trên!!! Hơi khác một tí!!!
c) ( 6x - 1 )2 - ( 3x + 2 ) = ..........
\(\left(a+b\right)^3-\left(a-b\right)^3\)
\(=\left(a+b-a+b\right)\left(\left(a+b\right)^2+\left(a+b\right)\left(a-b\right)+\left(a-b\right)^2\right)\)
\(=2b\left(\left(a+b\right)^2+\left(a^2-b^2\right)+\left(a-b\right)^2\right)\)
\(\left(a+b\right)^3+\left(a-b\right)^3\)
\(=\left(a+b+a-b\right)\left(\left(a+b\right)^2-\left(a+b\right)\left(a-b\right)+\left(a-b\right)^2\right)\)
\(=2a\left(\left(a+b\right)^2-\left(a^2-b^2\right)+\left(a-b\right)^2\right)\)
a) (a+b)3 -(a-b)3 = a3 + 3a2b + 3ab2 +b3 - a3 + 3a2b - 3ab2 +b3
= 2a3 + 6a2b + 2b3
(a+b)3-(a-b)3=a3+3a2b+3ab2+b3-(a3-3a2b+3ab2-b3)
=a3+3a2b+3ab2+b3-a3+3a2b-3ab2+b3
=6a2b+2b3
Áp dụng hđt a3-b3=(a-b)(a2+ab+b2) ấy
\(\left(a+b\right)^3-\left(a-b\right)^3=\left[\left(a+b\right)-\left(a-b\right)\right]\left[\left(a+b\right)^2+\left(a+b\right)\left(a-b\right)+\left(a-b\right)^2\right]\)
\(=\left(a+b-a+b\right)\left(a^2+2ab+b^2+a^2-b^2+a^2-2ab+b^2\right)\)
\(=2b\left(3a^2+b^2\right)\)
\(=\left[\left(x+y\right)^3-1\right]-3xy\left(x+y-1\right)\)
\(=\left(x+y-1\right)\left[\left(x+y\right)^2+1+2\left(x+y\right)\right]-3xy\left(x+y-1\right)\)
\(=\left(x+y-1\right)\left(x^2+y^2+2xy+1+2x+2y-3xy\right)\)
\(=\left(x+y+1\right)\left(x^2+y^2-xy+1+2x+2y\right)\)
\(=\left(x+y-1\right)\left[\left(x^2+1+2x\right)\left(y^2-xy+2y\right)\right]\)
\(=\left(x+y-1\right)\left(x+1\right)^2\left(y-x+2\right)y\)
bạn ơi hình như sai đề thì phải a bạn mình nghĩ phải là \(\left(x^2-x+2\right)^2\)
\(\left(x^2-x+2\right)+\left(x-2\right)^2=\left(x^2-x+2\right)+x^2-2^2\)
\(=x^2-x+2+x^2-2^2\)\(=\left(x^2+x^2\right)+\left(2-2^2\right)-x\)
\(=2x^2-\left(2-4\right)-x=2x^2-\left(-2\right)-x\)
\(=2x^2+2-x=2x^2+2.1-x=2\left(x^2+1\right)-x\)
1/(x+2)2 -(3x-1)2=(x+2+3x-1)(x+2-3x+1)=4x(-2x+3)=-8x2+12x
2/(x4+x2)(-2x3-2x)=x2(x2+1)-2x(x2+1)=(x2+1)(x2-2x)
Ta có: x6 -y6= (x3)2 -(y3)2 = (x3 - y3)(x3 + y3)
\(x^6-y^6\)
\(=\left(x^3-y^3\right)\left(x^3+y^3\right)\)
\(=\left(x-y\right)\left(x^2+xy+y^2\right)\left(x+y\right)\left(x^2-xy+y^2\right)\)
hk
tốt