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\(8-36x+54x^2-27x^3=0\)
\(=2^3-3.2^2.3x+3.9x^2.2-3x^3\)
\(=2-3x^3=0\)
\(\Rightarrow2-3x=0\Rightarrow x=\frac{2}{3}\)
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a, A = 27x3 + 54x2 + 36x + 4
=> A = (3x)3 + 2 . 3 . 9x2 + 3 . 3x . 4 + 8 - 4
=> A = [(3x)3 + 2 . 3 . (3x)2 + 3 . 3x . 22 + 23] - 4
=> A = (3x + 2)3 - 4
Thay x = -2002 vào A ta có: A = (3x + 2)3 - 4
=> A = [3 x (-2002) + 2]3 - 4 = [6006 + 2]3 - 4 = 60083 - 4
b, B = 2x2 + 4x + xy + 2y
=> B = 2x(x + 2) + y(x + 2)
=> B = (x + 2)(2x + y)
Thay x = 88; y = -76 vào B
=> B = (88 + 2)[2 . 88 + (-76)]
=> B = 90 . [176 + (-76)] = 90 . 100 = 9000
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\(a,M=27x^3+108x^2+144x+64=\left(3x+4\right)^2\)Tại \(x=32\Rightarrow M=\left(3.32+4\right)^3=100^3=1000000\)
\(b,N=27x^3-54x^2+36x-8=\left(3x-2\right)^3\)
Tại \(x=14\Rightarrow N=\left(3.14-2\right)^3=40^3=64000\)
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a) \(49x^2-56x+16\)
\(=\left(7x-4\right)^2\)
\(=\left(7.2-4\right)^2=100\)
b) mk chỉnh lại đề
\(27x^3+54x^2+36x+8\)
\(=\left(3x+2\right)^3\)
\(=\left[3.\left(-2\right)+2\right]^3=-64\)
c) \(\left(x-1\right)^3-4x\left(x+1\right)\left(x-1\right)+3\left(x+1\right)\left(x^2+x+1\right)+3\left(x-1\right)^2\)
\(=6x^2+7x+5\)
\(=6.\left(-\frac{2}{5}\right)^2+7.\left(-\frac{2}{5}\right)+5\)
\(=\frac{79}{25}\)
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Bài 2:
a: \(A=a^2+b^2+c^2+2ab-2ac-2bc+a^2+b^2+c^2-2ab-2bc+2ac\)
\(=2a^2+2b^2+2c^2-4bc\)
\(=2+2\cdot9+2\cdot1-4\cdot3\cdot\left(-1\right)=22+12=34\)
b: \(B=\left(a+b-a+b\right)\left(a+b+a-b\right)=4ab=4\cdot2\cdot5=40\)
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(x - 5)² = (3 + 2x)²
(x - 5)² - (3 + 2x)² = 0
[(x - 5) - (3 + 2x)][(x - 5) + (3 + 2x)] = 0
(x - 5 - 3 - 2x)(x - 5 + 3 + 2x) = 0
(-x - 8)(3x - 2) = 0
-x - 8 = 0 hoặc 3x - 2 = 0
*) -x - 8 = 0
-x = 8
x = -8
*) 3x - 2 = 0
3x = 2
x = 2/3
Vậy x = -8; x = 2/3
--------------------
27x³ - 54x² + 36x = 9
27x³ - 54x² + 36x - 9 = 0
27x³ - 27x² - 27x² + 27x + 9x - 9 = 0
(27x³ - 27x²) - (27x² - 27x) + (9x - 9) = 0
27x²(x - 1) - 27x(x - 1) + 9(x - 1) = 0
(x - 1)(27x² - 27x + 9) = 0
x - 1 = 0 hoặc 27x² - 27x + 9 = 0
*) x - 1 = 0
x = 1
*) 27x² - 27x + 9 = 0
Ta có:
27x² - 27x + 9
= 27(x² - x + 1/3)
= 27(x² - 2.x.1/2 + 1/4 + 1/12)
= 27[(x - 1/2)² + 1/12] > 0 với mọi x ∈ R
⇒ 27x² - 27x + 9 = 0 (vô lí)
Vậy x = 1
A = x² + y²
= x² - 2xy + y² + 2xy
= (x - y)² + 2xy
= 4² + 2.1
= 16 + 2
= 18
B = x³ - y³
= (x - y)(x² + xy + y²)
= (x - y)(x² - 2xy + y² + xy + 2xy)
= (x - y)[(x - y)² + 3xy]
= 4.(4² + 3.1)
= 4.(16 + 3)
= 4.19
= 76
C = x⁴ + y⁴
= (x²)² + (y²)²
= (x²)² + 2x²y² + (y²)² - 2x²y²
= (x² + y²)² - 2x²y²
= (x² - 2x²y² + y² + 2x²y²)² - 2x²y²
= [(x - y)² + 2x²y²]² - 2x²y²
= (4² + 2.1²)² - 2.1²
= (16 + 2)² - 2
= 18² - 2
= 324 - 2
= 322
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\(x^2+x+\dfrac{1}{4}=\left(x+\dfrac{1}{4}\right)^2\)
\(8x^3+27=\left(2x+3\right)\left(4x^2-6x+9\right)\)
\(-x^3+3x^2-3x+1=\left(-x+1\right)^3\)
\(A=27x^3-54x^2+36x-8+15\)
\(=\left(3x-2\right)^3+15\)
Thay x = 12 vào A ta đc:
\(A=\left(3.12-2\right)^3+15\)
\(=34^3+15\)
\(=39319\)
Vậy A =39319 tại x =12
=.= hok tốt!!