Rút gọn
a) A=(3x+1)2−2(3x+1)(3x+5)+(5x+5)2A=\left(3x+1\right)^2-2\left(3x+1\right)\left(3x+5\right)+\left(5x+5\right)^2A=(3x+1)2−2(3x+1)(3x+5)+(5x+5)2
b) B=(3+1)(32+1)(34+1)(38+1)(318+1)(332+1)B=\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{18}+1\right)\left(3^{32}+1\right)B=(3+1)(32+1)(34+1)(38+1)(318+1)(332+1)
c) C=(a+b−c)2+(a−b+c)2−2(b−c)2C=\left(a+b-c\right)^2+\left(a-b+c\right)^2-2\left(b-c\right)^2C=(a+b−c)2+(a−b+c)2−2(b−c)2
d) D=(a+b+c)2+(a−b−c)2+(b−c−a)2+(c−b−a)2D=\left(a+b+c\right)^2+\left(a-b-c\right)^2+\left(b-c-a\right)^2+\left(c-b-a\right)^2D=(a+b+c)2+(a−b−c)2+(b−c−a)2+(c−b−a)2
e)E=(a+b+c+d)2+(a+b−c−d)2+(a+c−b−d)2+(a+d−b−c)2E=\left(a+b+c+d\right)^2+\left(a+b-c-d\right)^2+\left(a+c-b-d\right)^2+\left(a+d-b-c\right)^2E=(a+b+c+d)2+(a+b−c−d)2+(a+c−b−d)2+(a+d−b−c)2
A = (3x + 1)2 - 2(3x + 1)(5x + 5) + (5x + 5)2
= [(3x + 1)-(5x + 5)]2
= (3x + 1 - 5x - 5)2
= [(-2x) - 4]2
B = (3 + 1)(32 + 1)(34 + 1)(38 + 1)(316 +1)(332 + 1)
=> (3 - 1)B = (3 - 1)(3 + 1)(32 + 1)(34 + 1)(38 + 1)(316 +1)(332 + 1)
=>2B = (32 - 1)(32 + 1)(34 + 1)(38 + 1)(316 +1)(332 + 1)
= (34 - 1)(34 + 1)(38 + 1)(316 +1)(332 + 1)
= (38 - 1)(38 + 1)(316 +1)(332 + 1)
= (316 - 1)316 +1)(332 + 1)
= (332 - 1)(332 + 1)
= 364 - 1
vì 2B = 364 - 1
=> B = 364−12\dfrac{3^{64}-1}{2}2364−1
C = a2 + b2 + c2 + 2ab - 2ac - 2bc + a2 + b2 + c2 - 2ab + 2ac - 2bc - 2( b2 - 2bc + c2)
= 2a2 + 2b2 + 2c2 - 4bc - 2b2 + 4bc - 2c2
= 2a2
Tính giá trị của
Q=x2−y2−z2−2yz−20xQ=x^2-y^2-z^2-2yz-20xQ=x2−y2−z2−2yz−20xbiết x+y+z=10.
Giúp với nha
B1 : Chứng minh:
x2+5y2+2x−4xy−10y+2014>0∀xyx^2+5y^2+2x-4xy-10y+2014>0\forall xyx2+5y2+2x−4xy−10y+2014>0∀xy
B2:Tính nhanh:
A=3(22+1)(24+1)−.(264+1)+1A=3\left(2^2+1\right)\left(2^4+1\right)-.\left(2^{64}+1\right)+1A=3(22+1)(24+1)−.(264+1)+1
cho x + y = 1; x - y = 3. Tính x2^22 - y2^{^{ }2}2
Khai triển : (x+1)3 , (x-2)3 , (x+3)3 , (2x-1)3 , (3x-2y)3
Tìm x
A)(x+3)3−x(3x+1)2+(2x+1)(4x2−2x+1)=28\left(x+3\right)^3-x\left(3x+1\right)^2+\left(2x+1\right)\left(4x^2-2x+1\right)=28(x+3)3−x(3x+1)2+(2x+1)(4x2−2x+1)=28
B)(x2−1)3−(x4+x2+1)(x2−1)=0\left(x^2-1\right)^3-\left(x^4+x^2+1\right)\left(x^2-1\right)=0(x2−1)3−(x4+x2+1)(x2−1)=0
Tính nhanh:
E=7802−22021252+150.125+752E=\dfrac{780^2-220^2}{125^2+150.125+75^2}E=1252+150.125+7527802−2202
A= a3 + 11 + 3a + 3a2 với a=9
B= x3 + 3x + 3x3 + 31 với x=19
C= x2 - y2 với x=87, y=13
D= 4x2 - 28x + 45 với x=4
M= 25x2 - 30x +15 với x=2
Phân tích đa thức thành nhân tử bằng nhiều cách :
a, x2+x−12x^2+x-12x2+x−12
b, 4x2−5x+14x^2-5x+14x2−5x+1
c, 2x2−5x+22x^2-5x+22x2−5x+2
d, 7x−3x2−27x-3x^2-27x−3x2−2
e, x3+x2−x+2x^3+x^2-x+2x3+x2−x+2
C/m a=b=c a, (a+b+c)2=3(a2+b2+c2)\left(a+b+c\right)^2=3\left(a^2+b^2+c^2\right)(a+b+c)2=3(a2+b2+c2) b, (a+b+c)2=3(ab+bc+ca)\left(a+b+c\right)^2=3\left(ab+bc+ca\right)(a+b+c)2=3(ab+bc+ca)
Chứng minh rằng các biểu thức sau có giá trị dương với mọi giá trị của x : 9x2-12x+5
*Chứng minh: a, ( a+b)4 = a4+4a3b+6a2b2+4ab2+b2
b, (a+b)5 = a5+5a4b+10a2b3+5ab4+b5
* Áp dụng (a-b)4; (a-b)5; (a+1)4; (a-5)5 Giải hộ tui vs
