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a) \(\left(3x+1\right)^2-\left(x+1\right)^2=\left(3x+1-x-1\right)\left(3x+1+x+1\right)=2x\left(4x+2\right)=2x.2\left(2x+1\right)=4x\left(2x+1\right)\)
a) \(\left(3x+1\right)^2-\left(x+1\right)^2=\left(3x+1-x-1\right)\left(3x+1+x+1\right)=2x\left(4x+2\right)=4x\left(2x+1\right)\)
b) \(x^3+y^3+z^3-3xyz=\left(x+y\right)^3-3xy\left(x+y\right)+z^3-3xyz=\left(x+y+z\right)\left(x^2+y^2+2xy+xz+yz+z^2\right)-3xy\left(x+y+z\right)\)
\(=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)=\frac{\left(x+y+z\right)\left[\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\right]}{2}\)
a) Đề bài phải là : \(\left(x+y\right)^2-\left(x-y\right)^2\)thì mới phân tích được.
Nếu đề bài như trên ta có:
\(\left(x+y\right)^2-\left(x-y\right)^2=\)\(\left(x+y-x+y\right)\left(x+y+x-y\right)=2x\cdot2y=4xy\)
b) Ta có: \(\left(3x+1\right)^2-\left(x+1\right)^2=\left(3x+1-x-1\right)\left(3x+1+x+1\right)\)
= \(2x\cdot\left(4x+2\right)=2x\cdot2\cdot\left(2x+1\right)=4x\cdot\left(2x+1\right)\)
c) Ta có : \(x^3+y^3+z^3-3xyz\)
= \(\left(x+y\right)^3+z^3-3x^2y-3xy^2-3xy\)
=\(\left(x+y+z\right)\left(\left(x+y\right)^2-\left(x+y\right)z+z^2\right)-3xy\left(x+y+z\right)\)
=\(\left(x+y+z\right)\left(x^2+2xy+y^2-xz-yz+z^2-3xy\right)\)
=\(\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)\)
\(\left(3x+1\right)^2-\left(3x-1\right)^2\)
\(=\left(3x+1-3x+1\right)\left(3x+1+3x-1\right)\)
\(=2\cdot6x\)
\(=12x\)
_________
\(\left(x+y\right)^2-\left(x-y\right)^2\)
\(=\left(x+y+x-y\right)\left(x+y-x+y\right)\)
\(=2x\cdot2y\)
\(=4xy\)
\(\left(x+y\right)^3+\left(x-y\right)^3\)
\(=\left(x+y+x-y\right)\left[\left(x+y\right)^2-\left(x+y\right)\left(x-y\right)+\left(x-y\right)^2\right]\)
\(=2x\cdot\left(x^2+2xy+y^2-x^2+y^2+x^2-2xy+y^2\right)\)
\(=2x\cdot\left(x^2+3y^2\right)\)
______
\(x^3+y^3+z^3-3xyz\)
\(=\left(x+y\right)^3-3xy\left(x-y\right)+z^3+3xyz\)
\(=\left[\left(x+y\right)^3+z^3\right]-3xy\left(x+y+z\right)\)
\(=\left(x+y+z\right)^3-3z\left(x+y\right)\left(x+y+z\right)-3xy\left(x-y-z\right)\)
\(=\left(x+y+z\right)\left[\left(x+y+z\right)^2-3z\left(x+y\right)-3xy\right]\)
\(=\left(x+y+z\right)\left(x^2+y^2+z^2+2xy+2xz+2yz-3xz-3yz-3xy\right)\)
\(=\left(x+y+z\right)\left(x^2+y^2-xy-xz-yz\right)\)
a/ \(=x^4+x^3+x^2+5x^2+5x+5\)
\(=x^2\left(x^2+x+1\right)+5\left(x^2+x+1\right)=\left(x^2+5\right)\left(x^2+x+1\right)\)
b/ \(=x^3+x^2+2x-x^2-x-2\)
\(=x\left(x^2+x+2\right)-\left(x^2+x+2\right)=\left(x-1\right)\left(x^2+x+2\right)\)
c/ \(=x^3+4x^2+4x-x^2-4x-4\)
\(=x\left(x^2+4x+4\right)-\left(x^2+4x+4\right)=\left(x-1\right)\left(x+2\right)^2\)
câu d khó quá , mk lm k nổi , sr nha ^^
a) x4 + x3 + 6x2 + 5x + 5
= x4 + x3 + x2 + 5x2 + 5x + 5
= x2 ( x2 + x + 1) + 5 (x2 + x + 1)
= (x2 + x + 1) (x2 + 5)
b) x3 + x - 2
= x3 + x2 + 2x - x2 - x - 2
= x (x2 + x + 2) - (x2 + x + 2)
= (x2 + x + 2) (x - 1)
c) x3 + 3x2 - 4
= x3 + 4x2 + 4x - x2 - 4x - 4
= x (x2 + 4x + 4) - (x2 + 4x + 4)
= (x2 + 4x + 4) (x - 1)
= (x + 2)2 (x - 1)
d) xy(x + y) + yz(y + z) + xz(x + z) + 3xyz
= xy(x + y) + xyz + yz(y + z) + xyz + xz(x + z) + xyz
= xy(x + y + z) + yz(x + y + z) + xz(x + y + z)
= (x + y + z) (xy + yz + xz)
a) \(\left(x+y\right)^2-\left(x-y\right)^2\)
\(\Leftrightarrow\left[\left(x+y\right)+\left(x-y\right)\right]\left[\left(x+y\right)-\left(x-y\right)\right]\)
\(\Leftrightarrow\left(x+y+x-y\right)\left(x+y-x+y\right)\)
\(\Leftrightarrow2x.2y=4xy\)
b) \(\left(3x+1\right)^2-\left(x+1\right)^2\)
\(\Leftrightarrow\left[\left(3x+1\right)+\left(x+1\right)\right]\left[\left(3x+1\right)-\left(x+1\right)\right]\)
\(\Leftrightarrow\left(3x+1+x+1\right)\left(3x+1-x-1\right)\)
\(\Leftrightarrow\left(4x+2\right).2x\)
\(\Leftrightarrow8x^2+4x\)
\(\Leftrightarrow x\left(8x+4\right)\)
nếu làm đến đoạn (4x + 2). 2x đó rồi dừng cx đc phải ko
1/ ( x+y)^2 -( x-y)^2 = (x+y-x+y)(x+y+x-y)
=4xy
2/( 3x+1)^2-(x+1)^2= (3x+1-x-1)(3x+1+x+1)
=2x(4x+2)
3/
a) (x+y)2-(x-y)2
=(x+y)(x-y)
b)(3x+1)2-(x+1)2
=[(3x+1)+(x+1)].[(3x+1)-(x+1)]
=(3x+1+x+1)(3x+1-x-1)