Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Ta có: \(x^2>=0\) với mọi x
suy ra: \(x^2+6x+10>=10\)với mọi x
Dấu "=" xảy ra khi và chỉ khi:
x^2=0 và 6x=0
Suy ra: x=0
\(A=x^2+6x+10\)
\(\Leftrightarrow A=x^2+6x+9+1\Leftrightarrow\left(x+3\right)^2+1\ge1\)với mọi x
dấu "=" xảy ra khi x=-3
vậy x=-3
\(a,=\left[\left(x+2\right)-\left(x-3\right)\right]^2=\left(x+2-x+3\right)^2=5^2=25\)
\(b=x^2-5\)
\(c=\left(x+y-x+y\right)\left(x+y+x-y\right)=2y.2x=4xy\)
\(a,\left(\dfrac{1}{x-1}-\dfrac{x}{x-1^2}.\dfrac{x^2+1+x}{x+1}\right):\dfrac{1}{x^2-1}\\ =\left(\dfrac{1}{x-1}-\dfrac{x\left(x^2+1+x\right)}{\left(x-1\right)\left(x+1\right)}\right):\dfrac{1}{x^2-1}\\ =\left(\dfrac{1\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}-\dfrac{x^3+x+x^2}{\left(x-1\right)\left(x+1\right)}\right):\dfrac{1}{x^2-1}\)
\(\dfrac{x+1-x^3-x-x^2}{\left(x-1\right)\left(x+1\right)}:\dfrac{1}{\left(x-1\right)\left(x+1\right)}\\ =\dfrac{\left(x+1-x^3-x-x^2\right)\left(x-1\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}=1-x^3-x^2\)
b,
thay x=\(\dfrac{1}{2}\) vào bt M ta được:
\(1-\left(\dfrac{1}{2}\right)^3-\left(\dfrac{1}{2}\right)^2=\dfrac{5}{8}\)
a) \(\left(x-2\right)\left(x^2-5x+1\right)-x\left(x^2+11\right)\)
\(=x\left(x^2-5x+1\right)-2\left(x^2-5x+1\right)-x\left(x^2+11\right)\)
\(=x^3-5x^2+x-2x^2+10x-2-x^3-11x\)
\(=-7x^2-2\)
b) \(\left(x-1\right)\left(x^2+x+1\right)+x^3-2\)
\(=x\left(x^2+x+1\right)-1\left(x^2+x+1\right)+x^3-2\)
\(=x^3+x^2+x-x^2-x-1+x^3-2\)
\(=2x^3-3\)
c) \(\left(x-y\right)\left(x+y\right)-2x\left(x-y\right)\)
\(=x\left(x+y\right)-y\left(x+y\right)-2x\left(x-y\right)\)
\(=x^2+xy-yx-y^2-2x^2+2xy\)
\(=-x^2-y^2+2xy\)
a, \(\left(x-2\right)\left(x^2-5x+1\right)-x\left(x^2+11\right)\)
\(=x^3-7x^2+11x-2-x^3-11x=-7x^2-2\)
b, \(\left(x-1\right)\left(x^2+x+1\right)+\left(x^3-2\right)\)
\(=x^3-1+x^3-2=2x^3-3\)
c, \(\left(x-y\right)\left(x+y\right)-2x\left(x-y\right)\)
\(=x^2-y^2-2x^2+2xy=-x^2-y^2+2xy\)
a) Ta có: \(\left(x-2\right).\left(x^2+2x+4\right)+\left(x-2\right)^3-\left(x-2\right).\left(x+2\right)\)
\(=\left(x^3-8\right)+\left(x-2\right)^3-\left(x^2-4\right)\)
\(=x^3-8+x^3-6x^2+12x-8-x^2+4\)
\(=2x^3-7x^2+12x-12\)
b) Ta có: \(\left(3-2x\right)^2-\left(x+3\right)^2-\left(2x+1\right)\left(2x-1\right)\)
\(=9-12x+4x^2-x^2-6x-9-4x^2+1\)
\(=3x^2-18x+1\)
a) \(A=\left(x-1\right).\left(x+1\right)+\left(x+2\right).\left(x^2+2x+4\right)-x.\left(x^2+x+2\right)\)
\(=x^2-1+x^3+2x^2+4x+2x^2+4x+8-x^3-x^2-2x\)
\(=\left(x^3-x^3\right)+\left(x^2+2x^2+2x^2-x^2\right)+\left(4x+4x-2x\right)+\left(-1+8\right)\)
\(=4x^2+6x+7\)
b) Thay vào ta được
\(A=4.\left(\frac{1}{2}\right)^2+6.\frac{1}{2}+7=1+3+7=11\)
Hướng dẫn thôi :
a) x ( x + 2 ) ( x^2 - 6x + 4 )
b) ( x + 1 ) ( x + 2 ) ( x - 2 )
(a+b)2-(a-b)2=[(a+b)+(a-b)][(a+b)-(a-b)]=(a+b+a-b)(a+b-a+b)=2a.2b=4ab
Cảm ơn bạn nhìu