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\(\left(x^2+x+1\right)\left(x^2+3x+1\right)+x^2\)
\(=x^4+x^3+x^2+3x^3+3x^2+3x+x^2+x+1+x^2\)
\(=x^4+4x^3+6x^2+4x+1\)
\(=\left(x+1\right)^4\)
\(1,\left(x+2y-3\right)^2-4\left(x+2y-3\right)+4=\left(x+2y-3-2\right)^2=\left(x+2y-5\right)^2\)
\(2,\left(x-y\right)^3-1-3\left(x-y\right)\left(x-y-1\right)=\left(x-y-1\right)\text{[}\left(x-y\right)^2+x-y+1\text{]}-3\left(x-y\right)\left(x-y-1\right)=\left(x-y-1\right)\left(x^2+y^2+x-y+1-3x+3y\right)=\left(x-y-1\right)\left(x^2+y^2-2x+2y+1\right)\)
\(3,\left(x^2+y^2-17\right)^2-4\left(xy-4\right)^2=\left(x^2+y^2-17\right)-\left(2xy-8\right)^2=\left(x^2-2xy+y^2-9\right)\left(x^2+y^2+2xy-25\right)=\text{[}\left(x-y\right)^2-3^2\text{]}\text{[}\left(x+y\right)^2-5^2\text{]}=\left(x-y+3\right)\left(x-y-3\right)\left(x+y+5\right)\left(x+y-5\right)\)
1, x2+3xy+2y2= x2+xy+2xy+2y2=x(x+y)+2y(x+y)=(x+2y)(x+y)
2, x(x+2)(x+3)(x+5)+9=x(x+5)(x+2)(x+3)+9=(x2+5x)(x2+5x+6)+9
Đặt x2+5x=t, ta có
t(t+6)+9=t2+6t+9=(t+3)2=(x2+5x+3)2=(x2+8)2
3, x2+2xy+y2+2x+2y-15=(x+y)2+2(x+y)-15=(x+y)2+2(x+y)+1-16=(x+y+1)2-42
= (x+y+1-4)(x+y+1+4)=(x+y-3)(x+y+5)
4, 4x4y4+1=4x4y4+4x2y2+1-4x2y2=(2x2y2+1)2-(2xy)2=(2x2y2+1-2xy)(2x2y2+1+2xy)
a.
\(5x^2\left(x-2y\right)-15x\left(x-2y\right)\)
\(=\left(x-2y\right)\left(5x^2-15x\right)\)
\(=5x\left(x-2y\right)\left(x-3\right)\)
b.
\(3\left(x-y\right)-5x\left(y-x\right)\)
\(=3\left(x-y\right)+5x\left(x-y\right)\)
\(=\left(x-y\right)\left(3+5x\right)\)
Answer:
Câu đầu bạn xem lại.
\(\left(3x+4\right)^2+\left(4x-3\right)^2+\left(2+5x\right).\left(2-5x\right)\)
\(=\left(3x\right)^2+2.2x.4+4^2+\left(4x\right)^2-2.4x.3+3^2+2^2-\left(5x\right)^2\)
\(=9x^2+24x+16+16x^2-24x+9+4-25x^2\)
\(=\left(9x^2+16x^2-25x^2\right)+\left(24x-24x\right)+\left(16+9+4\right)\)
\(=29\)
\(\left(5x+y\right).\left(25x^2-5xy+y^2\right)-\left(5x-y\right).\left(25x^2+5xy+y^2\right)\)
\(=\left(5x+y\right).[\left(5x\right)^2-5x.y+y^2]-\left(5x-y\right).[\left(5x\right)^2+5x.y+y^2]\)
\(=\left(5x\right)^3+y^3-[\left(5x\right)^3-y^3]\)
\(=\left(5x\right)^3+y^3-\left(5x\right)^3+y^3\)
\(=2y^3\)
(x + 2y - 3)2 - 4(x + 2y - 3) + 4
= (x + 2y - 3)2 - 2. 2. (x + 2y - 3) + 22 (hằng đẳng thức số 2, bình phương của một hiệu)
= ( x + 2y - 3 - 2)2
= ( x + 2y - 5)2