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Câu 1:
a) 2x(3x+2) - 3x(2x+3) = 6x^2+4x - 6x^2-9x = -5x
b) \(\left(x+2\right)^3+\left(x-3\right)^2-x^2\left(x+5\right)\)
\(=x^3+6x^2+12x+8+x^2-6x+9-x^3-5x^2\)
\(=2x^2+6x+17\)
c) \(\left(3x^3-4x^2+6x\right)\div\left(3x\right)=x^2-\dfrac{4}{3}x+2\)
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a/ \(=x^3-2x^5\)
b/\(=5x^2+5-x^3-x\)
c/ \(=x^3+3x^2-4x-2x^2-6x+8=x^3=x^2-10x+8\)
d/ \(=x^2-x^3+4x-2x+2x^2-8=3x^2-x^3+2x-8\)
e/ \(=x^4-x^2+2x^3-2x\)
f/ \(=\left(6x^2+x-2\right)\left(3-x\right)=17x^2+5x-6-6x^3\)
Bài 1:
a) (3x - 2)(4x + 5) = 0
<=> 3x - 2 = 0 hoặc 4x + 5 = 0
<=> 3x = 2 hoặc 4x = -5
<=> x = 2/3 hoặc x = -5/4
b) (2,3x - 6,9)(0,1x + 2) = 0
<=> 2,3x - 6,9 = 0 hoặc 0,1x + 2 = 0
<=> 2,3x = 6,9 hoặc 0,1x = -2
<=> x = 3 hoặc x = -20
c) (4x + 2)(x^2 + 1) = 0
<=> 4x + 2 = 0 hoặc x^2 + 1 # 0
<=> 4x = -2
<=> x = -2/4 = -1/2
d) (2x + 7)(x - 5)(5x + 1) = 0
<=> 2x + 7 = 0 hoặc x - 5 = 0 hoặc 5x + 1 = 0
<=> 2x = -7 hoặc x = 5 hoặc 5x = -1
<=> x = -7/2 hoặc x = 5 hoặc x = -1/5
Bài 1:
a: \(=6x^3-10x^2+6x\)
b: \(=-2x^3-10x^2-6x\)
Bài 4:
a: =>3x+10-2x=0
=>x=-10
c: =>3x2-3x2+6x=36
=>6x=36
hay x=6
Bài 1:
\(a,=6x^3-10x^2+6x\\ b,=-2x^3-10x^2-6x\)
Bài 4:
\(a,\Leftrightarrow3x+10-2x=0\Leftrightarrow x=-10\\ b,\Leftrightarrow x\left(2x^2+9x-5\right)-\left(2x^3+9x^2+x+4,5\right)=3,5\\ \Leftrightarrow2x^3+9x^2-5x-2x^3-9x^2-x-4,5=3,5\\ \Leftrightarrow-6x=8\Leftrightarrow x=-\dfrac{4}{3}\\ c,\Leftrightarrow3x^2-3x^2+6x=36\Leftrightarrow x=6\)
Bài 1:
\(a,=7xy\left(2x-3y+4xy\right)\\ b,=x\left(x+y\right)-5\left(x+y\right)=\left(x-5\right)\left(x+y\right)\\ c,=\left(x-y\right)\left(10x+8\right)=2\left(5x+4\right)\left(x-y\right)\\ d,=\left(3x+1-x-1\right)\left(3x+1+x+1\right)\\ =2x\left(4x+2\right)=4x\left(2x+1\right)\\ e,=5\left[\left(x-y\right)^2-4z^2\right]=5\left(x-y-2z\right)\left(x-y+2z\right)\\ f,=x^2+8x-x-8=\left(x+8\right)\left(x-1\right)\\ g,\left(x+y\right)^3-\left(x+y\right)=\left(x+y\right)\left[\left(x+y\right)^2-1\right]\\ =\left(x+y\right)\left(x+y-1\right)\left(x+y+1\right)\\ h,=x^2+3x+x+3=\left(x+3\right)\left(x+1\right)\)
Bài 12:
1) A = x2 - 6x + 11
= (x2 - 6x + 9) + 2
= (x - 3)2 + 2
Ta có: (x - 3)2 ≥ 0 ∀ x
Dấu ''='' xảy ra khi x - 3 = 0 ⇔ x = 3
Do đó: (x - 3)2 + 2 ≥ 2
Hay A ≥ 2
Dấu ''='' xảy ra khi x = 3
Vậy Min A = 2 tại x = 3
2) B = x2 - 20x + 101
= (x2 - 20x + 100) + 1
= (x - 10)2 + 1
Ta có: (x - 10)2 ≥ 0 ∀ x
Dấu ''='' xảy ra khi x - 10 = 0 ⇔ x = 10
Do đó: (x - 10)2 + 1 ≥ 1
Hay B ≥ 1
Dấu ''='' xảy ra khi x = 10
Vậy Min B = 1 tại x = 10
2:
a: \(9x^2-1=\left(3x\right)^2-1=\left(3x-1\right)\left(3x+1\right)\)
b: \(2\left(x-1\right)+x^2-x\)
\(=2\left(x-1\right)+x\left(x-1\right)\)
\(=\left(x-1\right)\left(x+2\right)\)
c: \(3x^2+14x-5\)
\(=3x^2+15x-x-5\)
\(=3x\left(x+5\right)-\left(x+5\right)=\left(x+5\right)\left(3x-1\right)\)
3:
a: \(2x\left(x-1\right)-2x^2=4\)
=>\(2x^2-2x-2x^2=4\)
=>-2x=4
=>x=-2
b: \(x\left(x-3\right)-\left(x+2\right)\left(x-1\right)=5\)
=>\(x^2-3x-\left(x^2+x-2\right)=5\)
=>\(x^2-3x-x^2-x+2=5\)
=>-4x=3
=>x=-3/4
c: \(4x^2-25+\left(2x+5\right)^2=0\)
=>\(\left(2x-5\right)\left(2x+5\right)+\left(2x+5\right)^2=0\)
=>\(\left(2x+5\right)\left(2x-5+2x+5\right)=0\)
=>4x(2x+5)=0
=>\(\left[{}\begin{matrix}x=0\\x=-\dfrac{5}{2}\end{matrix}\right.\)
a: \(\left(2x-1\right)^2-2\left(2x-3\right)^2+4\)
\(=4x^2-4x+1+4-2\left(4x^2-12x+9\right)\)
\(=4x^2-4x+5-8x^2+24x-18\)
\(=-4x^2+20x-13\)
e: \(\left(2x+3y\right)\left(4x^2-6xy+9y^2\right)=8x^3+27y^3\)
a: Ta có: \(\left(2x-1\right)^2-2\left(2x-3\right)^2+4\)
\(=4x^2-4x+1-2\left(4x^2-12x+9\right)+4\)
\(=4x^2-4x+5-8x^2+24x-18\)
\(=-4x^2+20x-13\)
b: \(\left(3x+2\right)^2+2\left(3x+2\right)\left(1-2y\right)+\left(1-2y\right)^2\)
\(=\left(3x+2+1-2y\right)^2\)
\(=\left(3x-2y+3\right)^2\)
Bài 1:
a) \(x^2+9y^2-y^4-6xy\)
\(=\left(x^2-6xy+9y^2\right)-y^4\)
\(=\left[x^2-2.x.3y+\left(3y\right)^2\right]-\left(y^2\right)^2\)
\(=\left(x-3y\right)^2-\left(y^2\right)^2\)
\(=\left(x-3y-y^2\right)\left(x-3y+y^2\right)\)
b) \(2x^2-x-28\)
\(=2x^2-8x+7x-28\)
\(=2x\left(x-4\right)+7\left(x-4\right)\)
\(=\left(x-4\right)\left(2x+7\right)\)
Bài 2:
a) \(2x\left(x^2-2x+3\right)-2x^3\)
\(=2x\left(x^2-2x+3-x^2\right)\)
\(=2x\left(3-2x\right)\)
b) \(2x\left(x-3\right)-\left(x+5\right)\left(2x-1\right)\)
\(=\left(2x^2-6x\right)-\left(2x^2+9x-5\right)\)
\(=2x^2-6x-2x^2-9x+5\)
\(=-15x+5\)
\(=-5\left(3x-1\right)\)
c) \(\left(5-x\right)^2+\left(x+5\right)^2-\left(2x+10\right)\left(x-5\right)\)
\(=\left(x-5\right)^2-2\left(x+5\right)\left(x-5\right)+\left(x+5\right)^2\)
\(=\left[\left(x-5\right)-\left(x+5\right)\right]^2\)
\(=\left(x-5-x-5\right)^2\)
\(=\left(-10\right)^2=100\)
Bài 3:
a) \(x-2=\left(x-2\right)^2\)
\(\Rightarrow\left(x-2\right)-\left(x-2\right)^2=0\)
\(\Rightarrow\left(x-2\right)\left(1-x+2\right)=0\)
\(\Rightarrow\left(x-2\right)\left(3-x\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-2=0\\3-x=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=2\\x=3\end{matrix}\right.\)
b) \(\left(-3x+9\right)x^2-7x+21=0\)
\(\Rightarrow-3\left(x-3\right)x^2-7\left(x-3\right)=0\)
\(\Rightarrow\left(x-3\right)\left(-3x^2-7\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-3=0\\-3x^2-7=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=3\\x^2=-\dfrac{7}{3}\end{matrix}\right.\)
Mà x2 > 0 hoặc x2 = 0 với mọi x
=> x2 = -7/3 không thỏa mãn
=> x= 3
Phân tích đa thức
a, x^2+9y^2-y^4-6xy
=(x^2-6xy+9y^2)-y^4
=(x-3y)^2-y^4
=(x-3y-y^2)(x-3y+y^2)
b, 2x^2-x-28
=(2x^2-8x)+(7x-28)
=2x(x-4)+7(x-4)
=(x-4)(2x+7)
Rút gọn
a,2x(x^2-2x+3)-2x^3
=2x(x^2-2x+3-x^2)
=2x(-2x+3)
b,2x(x-3)-(x+5)(2x-1)
=2x^2-6x-2x^2-9x+5
=-15x+5
=-5(3x-1)
c,(5-x)^2+(x+5)^2-(2x+10)(x-5)
Ta có:(5-x)^2=(x-5)^2
=(x-5)^2-2(x+5)(x-5)+(x+5)^2
=(x-5-x-5)^2
=100
Tìm x
a,x-2=(x-2)^2=0
=>x-2=0=>x=2
b,(-3x+9)x^2-7x+21=0
=>-3(x-3)x^2-7(x-3)=0
=>(x-3)(-3x^2-7)=0
=>\(\left[{}\begin{matrix}x-3=0=>x=3\\-3x^2-7=0=>x=\sqrt{\dfrac{-7}{3}}\end{matrix}\right.\)