Rút gọn biểu thức:
A=2x(x−2)−x(2x−3)2x\left(x-2\right)-x\left(2x-3\right)2x(x−2)−x(2x−3)
B=(x−1)(2x+1)−(x2−2x−1)\left(x-1\right)\left(2x+1\right)-\left(x^2-2x-1\right)(x−1)(2x+1)−(x2−2x−1)
C=(x+y)(x2−xy+y2)−x3\left(x+y\right)\left(x^2-xy+y^2\right)-x^3(x+y)(x2−xy+y2)−x3
D=(12x−3)(x+4)−x(2x+7)\left(12x-3\right)\left(x+4\right)-x\left(2x+7\right)(12x−3)(x+4)−x(2x+7)
E=(2x+y)(4x2−2xy+y2)\left(2x+y\right)\left(4x^2-2xy+y^2\right)(2x+y)(4x2−2xy+y2)
A=2x(x−2)−x(2x−3)=2x2−4x−2x2+3x=−xB=(x−1)(2x+1)−(x2−2x−1)=x(2x+1)−(2x+1)−x2+2x+1=2x2+x−2x−1−x2+2x+1=x2+xC=(x+y)(x2−xy+y2)−x3=x(x2−xy+y2)+y(x2−xy+y2)−x3=x3−x2y+xy2+x2y−xy2+y3−x3=y3A=2x\left(x-2\right)-x\left(2x-3\right)\\ =2x^2-4x-2x^2+3x\\ =-x\\ B=\left(x-1\right)\left(2x+1\right)-\left(x^2-2x-1\right)\\ =x\left(2x+1\right)-\left(2x+1\right)-x^2+2x+1\\ =2x^2+x-2x-1-x^2+2x+1\\ =x^2+x\\ C=\left(x+y\right)\left(x^2-xy+y^2\right)-x^3\\ =x\left(x^2-xy+y^2\right)+y\left(x^2-xy+y^2\right)-x^3\\ =x^3-x^2y+xy^2+x^2y-xy^2+y^3-x^3\\ =y^3A=2x(x−2)−x(2x−3)=2x2−4x−2x2+3x=−xB=(x−1)(2x+1)−(x2−2x−1)=x(2x+1)−(2x+1)−x2+2x+1=2x2+x−2x−1−x2+2x+1=x2+xC=(x+y)(x2−xy+y2)−x3=x(x2−xy+y2)+y(x2−xy+y2)−x3=x3−x2y+xy2+x2y−xy2+y3−x3=y3
D=(12x−3)(x+4)−x(2x+7)=x(12x−3)+4(12x−3)−2x2−7x=12x2−3x+48x−12−2x2−7x=10x2+38x−12E=(2x+y)(4x2−2xy+y2)=2x(4x2−2xy+y2)+y(4x2−2xy+y2)=8x3−4x2y+2xy2+4x2y−2xy2+y3=8x3+y3D=\left(12x-3\right)\left(x+4\right)-x\left(2x+7\right)\\ =x\left(12x-3\right)+4\left(12x-3\right)-2x^2-7x\\ =12x^2-3x+48x-12-2x^2-7x\\ =10x^2+38x-12\\ E=\left(2x+y\right)\left(4x^2-2xy+y^2\right)\\ =2x\left(4x^2-2xy+y^2\right)+y\left(4x^2-2xy+y^2\right)\\ =8x^3-4x^2y+2xy^2+4x^2y-2xy^2+y^3\\ =8x^3+y^3D=(12x−3)(x+4)−x(2x+7)=x(12x−3)+4(12x−3)−2x2−7x=12x2−3x+48x−12−2x2−7x=10x2+38x−12E=(2x+y)(4x2−2xy+y2)=2x(4x2−2xy+y2)+y(4x2−2xy+y2)=8x3−4x2y+2xy2+4x2y−2xy2+y3=8x3+y3
Chứng minh rằng:
a) (a+b)(a+c)+(c+a)(c+b)=2(b+a)(b+c)\left(a+b\right)\left(a+c\right)+\left(c+a\right)\left(c+b\right)=2\left(b+a\right)\left(b+c\right)(a+b)(a+c)+(c+a)(c+b)=2(b+a)(b+c)biết a2+c2=2b2a^2+c^2=2b^2a2+c2=2b2.
b) a(a+2)−a(a−7)(a−5)⋮7a\left(a+2\right)-a\left(a-7\right)\left(a-5\right)⋮7a(a+2)−a(a−7)(a−5)⋮7với mọi giá trị nguyên của a.
chứng minh rằng giá trị ko phụ thuộc vào biến y
y^4 -( y^2 - 1)( y^2 +1)
BT3/6. Tìm x, biết
a/ 2x(12x-5)-8x(3x-1)=30
b/3x(3-2x)+6x(x-1)=15
Cho a3 + b3 + c3 = 0
CMR a3b3 + 2b3c3 + 3a3c3 ≤\le≤0
Cho 2aa + bb = 3cc
Tính: 2015a-b + 2016b-c + 2017c-a
a) (x+2).(x-5) > 0
b) (x-3).(x+4) < 0
a,5(x-2)+3x (x-2)=0
b,2/3+5/2:x=2/4
c,2x(x-1/7)=0
d,11/12-(2/5+x)=2/3
e,3/4+1/4:x=2/5
g,2/3x+5/7=3/10
Tìm x:
a) 23\dfrac{2}{3}32 - 13\dfrac{1}{3}31 . ( x - 32\dfrac{3}{2}23 ) - 12\dfrac{1}{2}21 . ( 2x + 1 ) = 5
b) ( x + 12\dfrac{1}{2}21 ) . ( 34\dfrac{3}{4}43 - x ) = 0
c) 2x−1−3+2\dfrac{2x-1}{-3+2}−3+22x−1 = 0
Tìm x, biết
a) x(2x-7)-4x+14=0
b) (2x+5)(3x-1)-(2x+5)=0
c) (2x−3)2−6x+9=0\left(2x-3\right)^2-6x+9=0(2x−3)2−6x+9=0
A=(1−x31−x−x):1−x21−x−x2+x3\left(\dfrac{1-x^3}{1-x}-x\right):\dfrac{1-x^2}{1-x-x^2+x^3}(1−x1−x3−x):1−x−x2+x31−x2
Loga.vn - Cộng Đồng Luyện Thi Trực Tuyến