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Gợi ý:
a) Đặt \(t=x^2+x+1\)
b) Đặt \(t=x^2+8x+11\)
c) \(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24\)
\(=\left[\left(x+2\right)\left(x+5\right)\right].\left[\left(x+3\right)\left(x+4\right)\right]-24\)
\(=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\)
Đặt: \(t=x^2+7x+11\)
x4 - 3x2 +1
= x4 - 2x2 + 1 - x2
= ( x2 - 1 )2 - x2
= ( x2- 1 +x ) ( x2 - 1 - x )
B1:
a) \(5\left(x^2+y^2\right)-20x^2y^2\)
\(=5\left(x^2-4x^2y^2+y^2\right)\)
b) \(=2\left(x^8-16\right)=2\left(x^4-4\right)\left(x^4+4\right)=2\left(x^2-2\right)\left(x^2+2\right)\left(x^4+4\right)\)
B2:
a) Đặt \(x^2-3x+1=y\)
=> \(y^2-12y+27\)
\(=\left(y^2-12y+36\right)-9\)
\(=\left(y-6\right)^2-3^2\)
\(=\left(y-9\right)\left(y-3\right)\)
\(=\left(x^2-3x-10\right)\left(x^2-3x-4\right)\)
\(=\left(x+1\right)\left(x-4\right)\left(x^2-3x-10\right)\)
b) Đặt \(x^2+7x+11=t\)
Ta có: \(\left[\left(x+2\right)\left(x+5\right)\right]\cdot\left[\left(x+3\right)\left(x+4\right)\right]-24\)
\(=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\)
\(=\left(t-1\right)\left(t+1\right)-24\)
\(=t^2-25\)
\(=\left(t-5\right)\left(t+5\right)\)
\(=\left(x^2+7x+6\right)\left(x^2+7x+16\right)\)
\(=\left(x+1\right)\left(x+6\right)\left(x^2+7x+16\right)\)
F=x2+2xy+y2-x-y-12
= (x + y)^2 - (x + y) - 12
= (x + y)(x + y - 1) - 12
đặt x + y = t
F = t(t - 1) - 12
= t^2 - t - 12
= (t - 4)(t + 3)
G=(x2-3x-1)2-12(x2-3x-1)+27
đăth x^2 - 3x - 1 = t
G = t^2 - 12t + 27
= (t - 3)(t - 9)
có t = x^2 - 3x - 1
thay vào
Câu F ( kiểm tra lại đề )
Câu G . Đặt x^2 -3x-1=t
t^2 -12t+27 ( thực hiện pp tách)
a) x.(1-x)+(x-1)2
=x.1-x2+x2-2.x2.1+12
=x-x2+x2-2.x2+1
=(-x2+x2-2x2)+1
=-2x2+1
b)(x+1)2-3.(x+1)
=x2+2.x2.1+12-3.x+3
=x2+2.x2+1-3x+3
=(x2+2x2)+(1+3)-3x
=3x2-3x+4
c)3x.(x-1)2-(1-x)3
=3x.x2-2,x2.1+12-13-3.12.x+3.x.12=3x.x2-2x2+1-1-3x+3x=(3x-3x+3x)(x2-2x2)(1-1)=3x.(-x2)
\(G=2x^2-3x+1=2x^2-2x-x+1\)
\(=2x\left(x-1\right)-\left(x-1\right)=\left(2x-1\right)\left(x-1\right)\)
\(H=-x^2+5x-4=-x^2+4x+x-4\)
\(=-x\left(x-4\right)+\left(x-4\right)=\left(1-x\right)\left(x-4\right)\)
\(I=x^2+4x+3=x^2+3x+x+3\)
\(=x\left(x+3\right)+\left(x+3\right)=\left(x+1\right)\left(x+3\right)\)
\(K=2x^2+7x+5=2x^2+2x+5x+5\)
\(=2x\left(x+1\right)+5\left(x+1\right)=\left(2x+5\right)\left(x+1\right)\)
\(L=-3x^2-5x-2=-3x^2-3x-2x-2\)
\(=-3x\left(x+1\right)-2\left(x+1\right)=\left(-3x-2\right)\left(x+1\right)\)
G = 2x2 - 3x +1 = 2x2 -2x -x +1 =(x-1).(2x-1)
H = -x2 + 5x - 4 = -x2 + 4x +x-4 = (x-4).(1-x)
I = x2 + 4x + 3 = x2 + 3x + x + 3 =(x+3).(x+1)
K = 2x2 + 7x + 5 = 2x2 + 2x + 5x + 5 = (x+1).(2x+5)
L = -3x2 -5x -2 = -3x2 - 3x - 2x - 2 = -3.x(x+1) - 2.(x+1) = (x+1).(-3x-2)
a) 2x + 2y - x2 - xy
= 2(x + y) + x(x + y)
= (x + y) (x + 2)
mk ko bít phân tích đúng ko đúng thì t i c k nhé!! 245433463463564564574675687687856856846865855476457
a)\(2x+2y-x^2-xy=2\left(x+y\right)-x\left(x+y\right)=\left(2-x\right)\left(x+y\right)\)
b)\(\left(x+3\right)^2-\left(2x-5\right)\left(x+3\right)\)
\(=\left(x+3\right)\left[\left(x+3\right)-\left(2x-5\right)\right]\)
\(=\left(x+3\right)\left(8-x\right)\)
c)\(\left(3x+2\right)^2+\left(3x-2\right)^2-2\left(9x^2-4\right)\)
\(=\left(3x+2\right)^2+\left(3x-2\right)^2-2\left(3x-2\right)^2\)
\(=\left(3x+2\right)\left[\left(3x+2\right)-\left(3x-2\right)\right]+\left(3x-2\right)\left[\left(3x-2\right)-\left(3x+2\right)\right]\)
\(=4\left(3x+2\right)-4\left(3x-2\right)\)
\(=4\left(3x+2-3x+2\right)\)
=4.4=16
Đặt \(x^2+1=t\)
Ta có: \(\left(x^2+1\right)^2+3x\left(x^2+1\right)+2x^2\)
\(=t^2+3xt+2t^2\)
\(=t^2+xt+2xt+2t^2\)
\(=t\left(t+x\right)+2x\left(t+x\right)\)
\(=\left(t+x\right)\left(t+2x\right)\)
\(=\left(x^2+1+x\right)\left(x^2+1+2x\right)\)
\(=\left(x^2+x+1\right)\left(x+1\right)^2\)
Chúc bạn học tốt
xin lỗi bạn, mình viết nhầm 2x2 thành 2t2