Bài 2: Làm tính chia
a. (x3 – 3x2 + x – 3) : (x – 3) d. (2x4 – 5x2 + x3 – 3 – 3x) : (x2 – 3)
b. (x – y – z)5 : (x – y – z)3 e. (x2 + 2x + x2 – 4) : (x + 2)
c. (2x3 + 5x2 – 2x + 3) : (2x2 – x + 1) f. (2x3 – 5x2 + 6x – 15) : (2x – 5)
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a) Đa thức thương x 2 – 6x + 9.
b) Đa thức thương 2 x 2 – 5.
c) Đa thức thương x 2 + 4x + 3 và đa thức dư -12.
d) Đa thức x + 5 và đa thức dư x – 4.
Bài 1:
a) \(3x^2\left(2x^3-x+5\right)-6x^5-3x^3+10x^2\)
\(=6x^5-3x^3+10x^2-6x^5-3x^3+10x^2\)
\(=10x^2+10x^2\)
\(=20x^2\)
b) \(-2x\left(x^3-3x^2-x+11\right)-2x^4+3x^3+2x^2-22x\)
\(=-2x^4+6x^3+2x^2-22x-2x^4+3x^3+2x^2-22x\)
\(=-4x^4+9x^3+4x^2-44x\)
a) \(\left(2x^3-x^2+5x\right):x\)
\(=\dfrac{2x^3-x^2+5x}{x}\)
\(=\dfrac{x\left(2x^2-x+5\right)}{x}\)
\(=2x^2-x+5\)
b) \(\left(3x^4-2x^3+x^2\right):\left(-2x\right)\)
\(=\dfrac{3x^4-2x^3+x^2}{-2x}\)
\(=\dfrac{2x\left(\dfrac{3}{2}x^3-x^2+\dfrac{1}{2}x\right)}{-2x}\)
\(=-\left(\dfrac{3}{2}x^3-x^2+\dfrac{1}{2}x\right)\)
\(=-\dfrac{3}{2}x^3+x^2-\dfrac{1}{2}x\)
c) \(\left(-2x^5+3x^2-4x^3\right):2x^2\)
\(=\dfrac{-2x^5+3x^2-4x^3}{2x^2}\)
\(=\dfrac{2x^2\left(-x^3+\dfrac{3}{2}-2x\right)}{2x^2}\)
\(=-x^3-2x+\dfrac{3}{2}\)
d) \(\left(x^3-2x^2y+3xy^2\right):\left(-\dfrac{1}{2}x\right)\)
\(=\dfrac{x^3-2x^2y+3xy^2}{-\dfrac{1}{2}x}\)
\(=\dfrac{\dfrac{1}{2}x\left(2x^2-4xy+6y^2\right)}{-\dfrac{1}{2}x}\)
\(=-\left(2x^2-4xy+6y^2\right)\)
\(=-2x^2+4xy-6y^2\)
e) \(\left[3\left(x-y\right)^5-2\left(x-y\right)^4+3\left(x-y\right)^2\right]:5\left(x-y\right)^2\)
\(=\dfrac{3\left(x-y\right)^5-2\left(x-y\right)^4+3\left(x-y\right)^2}{5\left(x-y\right)^2}\)
\(=\dfrac{5\left(x-y\right)^2\left[\dfrac{3}{5}\left(x-y\right)^3-\dfrac{2}{5}\left(x-y\right)^2+\dfrac{3}{5}\right]}{5\left(x-y\right)^2}\)
\(=\dfrac{3}{5}\left(x-y\right)^3-\dfrac{2}{5}\left(x-y\right)^2+\dfrac{3}{5}\)
f) \(\left(3x^5y^2+4x^3y^3-5x^2y^4\right):2x^2y^2\)
\(=\dfrac{3x^5y^2+4x^3y^3-5x^2y^4}{2x^2y^2}\)
\(=\dfrac{2x^2y^2\left(\dfrac{3}{2}x^3+2xy-\dfrac{5}{2}y^2\right)}{2x^2y^2}\)
\(=\dfrac{3}{2}x^3+2xy-\dfrac{5}{2}y^2\)
\(\left(1-x\right)\left(5x+3\right)=\left(3x-7\right)\left(x-1\right)\)
\(< =>\left(1-x\right)\left(5x+3+3x-7\right)=0\)
\(< =>\left(1-x\right)\left(8x-4\right)=0\)
\(< =>\orbr{\begin{cases}1-x=0\\8x-4=0\end{cases}< =>\orbr{\begin{cases}x=1\\x=\frac{1}{2}\end{cases}}}\)
\(\left(x-2\right)\left(x+1\right)=x^2-4\)
\(< =>\left(x-2\right)\left(x+1\right)=\left(x-2\right)\left(x+2\right)\)
\(< =>\left(x-2\right)\left(x+1-x-2\right)=0\)
\(< =>-1\left(x-2\right)=0\)
\(< =>2-x=0< =>x=2\)
Bài 5:
a. 1 - 2y + y2
= (1 - y)2
b. (x + 1)2 - 25
= (x + 1)2 - 52
= (x + 1 - 5)(x + 1 + 5)
= (x - 4)(x + 6)
c. 1 - 4x2
= 12 - (2x)2
= (1 - 2x)(1 + 2x)
d. 8 - 27x3
= 23 - (3x)3
= (2 - 3x)(4 + 6x + 9x2)
e. (đề hơi khó hiểu ''x3'' !?)
g. x3 + 8y3
= (x + 2y)(x2 - 2xy + y2)
Bài 13:
1: \(A=-x^2+4x+3\)
\(=-\left(x^2-4x-3\right)=-\left(x^2-4x+4-7\right)\)
\(=-\left(x-2\right)^2+7\le7\)
Dấu '=' xảy ra khi x=2
2: \(B=-\left(x^2-6x+11\right)\)
\(=-\left(x-3\right)^2-2\le-2\)
Dấu '=' xảy ra khi x=3
2 b, (x-y-z)^5-3 = (x-y-z)^2
3 c, x^2-1
4 d, 2x^4 + x^2 - 6x^2 + x^3 - 3 - 3x / x^2 - 3
= x^2(2x^2 + x + 1) - 3(2x^2 + x + 1) / x^2 - 3
= (2x^2 + x + 1)(x^2 - 3) / x^2 - 3
= 2x^2 + x + 1
5 e, 2.(x-1)
6 f, (2x3 – 5x2 + 6x – 15) : (2x – 5)
=(2x3−5x2)+(6x−15)=(2x3−5x2)+(6x−15)
=x2(2x−5)+3(2x−5)=x2(2x−5)+3(2x−5)
=(x2+3)(2x−5)=(x2+3)(2x−5)
=(2x3−5x2+6x−15):(2x−5)=x2+3
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