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1: \(x^2-2x-24=\left(x-6\right)\left(x+4\right)\)
2: \(x^2-8x+15=\left(x-3\right)\left(x-5\right)\)
3: \(x^2-9x+14=\left(x-2\right)\left(x-7\right)\)
\(=x^2\left(x^2+2x+1\right)+x+1\)
\(=x^2\left(x+1\right)^2+x+1\)
\(=\left(x+1\right)\left[x^2\left(x+1\right)+1\right]\)
\(=\left(x+1\right)\left(x^3+x^2+1\right)\)
\(x^4+2x^3+x^2+x+1\)
\(=x^2\left(x+1\right)^2+\left(x+1\right)\)
\(=\left(x+1\right)\left(x^3+x^2+1\right)\)
Ta có: \(\left(4x+1\right)\left(12x-1\right)\left(3x+2\right)\left(x+1\right)-4\)
\(=\left(12x^2+8x+3x+2\right)\left(12x^2+12x-x-1\right)-4\)
\(=\left(12x^2+11x+2\right)\left(12x^2+11x-1\right)-4\)
\(=\left(12x^2+11x\right)^2+\left(12x^2+11x\right)-6\)
\(=\left(12x^2+11x+3\right)\left(12x^2+11x-2\right)\)
\(5x^2-4\left(x^2-2x+1\right)-5=\left(5x^2-5\right)-4\left(x-1\right)^2=5\left(x^2-1\right)-4\left(x-1\right)^2=5\left(x-1\right)\left(x+1\right)-4\left(x-1\right)^2=\left(x-1\right)\left[5\left(x+1\right)-4\left(x-1\right)\right]=\left(x-1\right)\left(5x+5-4x+4\right)=\left(x-1\right)\left(x+9\right)\)
Phân tích đa thức thành nhân tử:
a) \(xy+y^2-x-y=y\left(x+y\right)-\left(x+y\right)=\left(x+y\right)\left(y-1\right)\)
b) \(25-x^2+4xy-4y^2=25-\left(x^2-4xy+4y^2\right)=25-\left(x-2y\right)^2\)
\(=\left(5-x+2y\right)\left(5+x-2y\right)\)
Rút gọn biểu thức;
\(A=\left(6x+1\right)^2+\left(3x-1\right)^2-2\left(3x-1\right)\left(6x+1\right)\)
\(=\left[\left(6x+1\right)-\left(3x-1\right)\right]^2=\left(6x+1-3x+1\right)=\left(3x+2\right)^2\)
Tìm a để đa thức.. Bạn chia cột dọ thì da
\(xy+y^2-x-y=\left(xy+y^2\right)-\left(x+y\right)=y\left(x+y\right)-\left(x+y\right)=\left(y-1\right)\left(x+y\right)\)b)\(25-\left(x^2-4xy+4y^2\right)=5^2-\left(x-2y\right)^2=\left(x-2y+5\right)\left(5-x+2y\right)\)
a) x2+x-2
= x2-x+2x-2
= x(x-1)+2(x-1)
= (x+2)(x-1)
b) 2x2+5x+3
= 2x2+2x+3x+3
= 2x(x+1)+3(x+1)
= (2x+3)(x+1)
c) 3x2+5x-2
= 3x2+6x-1x-2
= 3x(x+2)-1(x+2)
= (3x-1)(x+2)
\(=2\left(x^2-y^2\right)-6\left(x+y\right)=2\left(x-y\right)\left(x+y\right)-6\left(x+y\right)=\left(x+y\right)\left(2x-2y-6\right)\) Đảm bảo chuẩn ko cần chỉnh (•••
check mk nhá
Giúp mình nha!!!!!
Như nài cx được nè :3
\(A=\left(x^2-3x+1\right)\left(x^2+2x+1\right)-6x^2\)
\(=\left(x^4+2x^3+x^2-3x^3-6x^2-3x+x^2+2x+1\right)-6x^2\)
\(=\left(x^4-x^3-4x^2-x+1\right)-\left(\sqrt{6}x\right)^2\)
\(=\left(x^4-x^3-4x^2-x+1-\sqrt{6}x\right)\left(x^4-x^3-4x^2-x+1+\sqrt{6}x\right)\)