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A) (a-b-c)^2=a^2-b^2-c^2-2ab+2bc-2bc
B) (a-b)^3=a^3-3a^2b+3ab^2-b^3
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\(a,=10x-8y+3\\ b,=\left(x+y\right)^2:\left(x+y\right)=x+y\\ c,=\left(a+b\right)\left(3a-2c+d\right):\left(a+b\right)=3a-2c+d\\ d,=\left(x-1\right)^3:\left(x-1\right)^2=x-1\)
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a)\(x^4-6x^2+2x+28\)
\(=\left(x^4-x^3\right)+\left(x^3-x^2\right)-\left(5x^2-5x\right)-\left(3x-3\right)+25\)
\(=\left(x-1\right)\left(x^3+x^2-5x-3\right)+25\)
=> số dư là 25
b) Cách làm tương tự câu a nhé
![](https://rs.olm.vn/images/avt/0.png?1311)
a,\(A=\dfrac{x+4}{x^2-2x}+\dfrac{2}{x}\)
\(=\dfrac{x+4}{x\left(x-2\right)}+\dfrac{2}{x}\)
\(=\dfrac{x+4}{x\left(x-2\right)}+\dfrac{2\left(x-2\right)}{x\left(x-2\right)}\)
\(=\dfrac{x+4+2x-4}{x\left(x-2\right)}\)
\(=\dfrac{3x}{x\left(x-2\right)}=\dfrac{3}{x-2}\)
b, Để A có giá trị bằng - 3
\(\Leftrightarrow\dfrac{3}{x-2}=-3\)
\(\Leftrightarrow x-2=-1\)
\(\Leftrightarrow x=1\) ( t/m )
Vậy x = 1 thì A có giá trị bằng -3
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1a) -3x2(2x3 - 2x + 1/3) = -6x5 + 6x3 - x2
b) (x4 + 2x3 - 2/3).(-3x4) = -3x8 - 6x7 + 2x4
c) (x + 3)(x - 4) = x2 - 4x + 3x - 12 = x2 - x - 12
d)(x - 4)(x2 + 4x + 16) = (x - 4)(x2 + 4x + 42) = x3 - 64
e) 4(x - 1/2)(x + 1/2)(4x2 + 1) =4(x2 - 1/4)(4x2 + 1) = 4(4x4 + x2 - x2 - 1/4) = 4(4x4 - 1/4) = 16x4 - 1
B2. a) (2 - x)(x2 + 2x + 4) + x(x - 3)(x + 4) - x2 + 24 = 0
=> 8 - x3 + x(x2 + 4x - 3x - 12) - x2 + 24 = 0
=> 8 - x3 + x3 + x2 - 12x - x2 + 24 = 0
=> -12x + 32 = 0
=> -12x = -32
=> x = -32 : (-12) = 8/3
b) (x/2 + 3)(5 - 6x) + (12x - 2)(x/4 + 3) = 0
=> 5x/2 - 3x2 + 15 - 18x + 3x2 + 36x - x/2 - 6 = 0
=> 20x + 9 = 0
=> 20x = -9
=> x = -9/20
\(\left(a+b\right)^2\)
\(=\left(a+b\right)\left(a+b\right)\)
\(=a^2+ab+ba+b^2\)
\(=a^2+2ab+b^2\)
\(\left(a-b\right)^2\)
\(=\left(a-b\right)\left(a-b\right)\)
\(=a^2-ab-ba+b^2\)
\(=a^2-2ab+b^2\)
Thanks