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Bài 1 : Phân tích các đa thức sau thành nhân tử :
a) 8x3 - 64
=(2x)3 + 43
=(2x+4)(4x2 - 8x + 16)
c) 125x3 + 1
=5x3 + 13
=(5x+1)(25x2 +5x+1)
d) 8x3 - 27
=(2x)3 - 33
=(2x - 3)(2x2 + 6x + 9)
e) 1 + 8x6y3
=1 + (2x2y)3
=(1 + 2x2y)(4x4y2 -2x2y + 1)
f) 125x3 + 27y3
=(5x)3 + (3y3)
=(5x + 3y)(25x2 - 15xy + 9y2)
Bài 1
a) \(8x^3-64\)
\(=\left(2x\right)^3-4^3\)
\(=\left(2x-4\right)\left(4x^2+8x+16\right)\)
c) \(125x^3+1\)
\(=\left(5x\right)^3+1^3\)
\(=\left(5x+1\right)\left(25x^2-5x+1\right)\)
d) \(8x^3-27\)
\(=\left(2x\right)^3-3^3\)
\(=\left(2x-3\right)\left(4x^2+6x+9\right)\)
e) \(1+8x^6x^3\)
\(=1^3+\left(2x^2y\right)^3\)
\(=\left(1+2x^2y\right)\left(1-2x^2y+4x^4y^2\right)\)
f) \(125x^3+27y^3\)
\(=\left(5x\right)^3+\left(3y\right)^3\)
\(=\left(5x+3y\right)\left(25x^2-15xy+9x^2\right)\)
Làm bài 1 thôi !! Mấy bài kia tương tự . Tìm nhân tử chung ra .
a) \(m^2-n^2=\left(m-n\right)\left(m+n\right)\)
b) \(\left(x^2+x-1\right)^2-\left(x^2+2x+3\right)^2=\left(x^2+x-1+x^2+2x+3\right)\left(x^2+x-1-x^2-2x-3\right)\)
\(=\left(2x^2+3x+2\right)\left(-x-4\right)\)
c) \(-16+\left(x-3\right)^2=\left(x-3+4\right)\left(x-3-4\right)=x\left(x-7\right)\)
d) \(64+16y+y^2=\left(y+8\right)\left(y+8\right)\)
\(1,a,\left(12x-5\right)^2=12^2x^2-2.12.5x+5^2\)
\(b,\left(4x^2-y\right)^3=\left(4x^2\right)^3-3.\left(4x^2\right)^2y+3.4x^2.y^2-y^3=4x^6-3.16x^4y+12x^2y^2\)
\(c,\left(7x+8\right)^3=\left(7x\right)^3+3.\left(7x\right)^28+3.7x.8+8^3\)
\(a,x^3-16x=x\left(x^2-16\right)=x\left(x^2-4^2\right)=x\left(x-4\right)\left(x+4\right)\)
\(b,x^2-12x+36=x^2-2.x.6+6^2=\left(x-6\right)^2\)
\(1-8x^3=1^3-\left(2x\right)^3=\left(1-2x\right)\left(1^2+1.2x+\left(2x\right)^2\right)=\left(1-2x\right)\left(1+2x+4x^2\right)\)
\(d,\dfrac{1}{25}x^2-\dfrac{1}{64}y^2=\left(\dfrac{1}{5}x\right)^2-\left(\dfrac{1}{8}x\right)^2=\left(\dfrac{1}{5}x-\dfrac{1}{8}x\right)\left(\dfrac{1}{5}x+\dfrac{1}{8}x\right)=x\left(\dfrac{1}{5}-\dfrac{1}{8}\right)\left(\dfrac{1}{5}+\dfrac{1}{8}\right)\)
\(A=x^2-8x+13=\left(x^2-8x+16\right)-3\ge-3\)Vậy \(Min_A=-3\) khi \(x+4=0\Leftrightarrow x=-4\)
\(B=2x^2+10x+5=2\left(x^2+5x+\dfrac{25}{4}\right)-\dfrac{5}{4}=2\left(x+\dfrac{5}{2}\right)^2-\dfrac{5}{4}\ge\dfrac{-5}{4}\)Vậy \(Min_B=-\dfrac{5}{4}\) khi \(x+\dfrac{5}{2}=0\Rightarrow=\dfrac{-5}{2}\)
\(C=4x-x^2=4-\left(4-4x+x^2\right)=4-\left(2-x\right)^2\le4\)Vậy \(Max_C=4\) khi \(2-x=0\Rightarrow x=2\)
Bài 1:
a, \(A=x^2-8x+13\)
\(A=x^2-4x-4x+16-3\)
\(A=\left(x-4\right)^2-3\)
Với mọi giá trị của \(x\in R\) ta có:
\(\left(x-4\right)^2\ge0\Rightarrow\left(x-4\right)^2-3\ge-3\)
Hay \(A\ge-3\) với mọi giá trị của \(x\in R\).
Để \(A=-3\) thì \(\left(x-4\right)^2-3=-3\Rightarrow x=4\)
Vậy......
Câu b tương tự
c, \(4x-x^2\)
\(C=-\left(x^2-4x\right)=-\left(x^2-2x-2x+4-4\right)\)
\(=-\left[\left(x-2\right)^2-4\right]\)
Với mọi giá trị của \(x\in R\) ta có:
\(\left(x-2\right)^2\ge0\Rightarrow\left(x-2\right)^2-4\ge-4\)
\(\Rightarrow-\left[\left(x-2\right)^2-4\right]\le4\)
Hay \(A\le4\) với mọi giá trị của \(x\in R\).
Để \(A=4\) thì \(-\left[\left(x-2\right)^2-4\right]=4\Rightarrow x=2\)
Vậy......
Chúc bạn học tốt!!!
a) 1-8x^3=1^3-(2x)^3=(1-2x)(1+2x+4x^2)
b) 125x^3+8=(5x)^3+2^3=(5x+2)(25x^2-10x+4)
c) x^6-1=(x^3)^2-1^2=(x^3-1)(x^3+1)
d) x^9-1=(x^3)^3-1^3=(x^3-1)(x^6+x^3+1)
e) x^6+1=(x^2)^3+1^3=(x^2+1)(x^4-x^2+1)
f) x^9+1=(x^3)^3+1^3=(x^3+1)(x^6-x^3+1)
a) Ta có: \(x^2+2x+1\)
\(=x^2+2\cdot x\cdot1+1^2\)
\(=\left(x+1\right)^2\)
b) Ta có: \(1-2y+y^2\)
\(=y^2-2\cdot y\cdot1+1^2\)
\(=\left(y-1\right)^2\)
c) Ta có: \(x^3-3x^2+3x-1\)
\(=x^3-x^2-2x^2+2x+x-1\)
\(=x^2\left(x-1\right)-2x\left(x-1\right)+\left(x-1\right)\)
\(=\left(x-1\right)\left(x^2-2x+1\right)\)
\(=\left(x-1\right)^3\)
d) Ta có: \(27+27x+9x^2+x^3\)
\(=x^3+3x^2+6x^2+18x+9x+27\)
\(=x^2\left(x+3\right)+6x\left(x+3\right)+9\left(x+3\right)\)
\(=\left(x+3\right)\left(x^2+6x+9\right)\)
\(=\left(x+3\right)^3\)
e) Ta có: \(8-125x^3\)
\(=2^3-\left(5x\right)^3\)
\(=\left(2-5x\right)\left(4+10x+25x^2\right)\)
f) Ta có: \(64x^3+\frac{1}{8}\)
\(=\left(4x\right)^3+\left(\frac{1}{2}\right)^3\)
\(=\left(4x+\frac{1}{2}\right)\left(16x^2-2x+\frac{1}{4}\right)\)
g) Ta có: \(1-x^2y^4\)
\(=1^2-\left(xy^2\right)^2\)
\(=\left(1-xy^2\right)\left(1+xy^2\right)\)
a) \(x^2+2x+1=x^2+2x.1+1^2=\left(x+1\right)^2\)
b) \(1-2y+y^2=1^2-2y.1+y^2=\left(1-y\right)^2\)
c) \(x^3-3x^2+3x-1=\left(x-1\right)^3\)
d) \(27+27x+9x^2+x^3=3^3+3.3^2x+3.3x^2+x^3=\left(3+x\right)^3\)
e) \(8-125x^3=2^3-\left(5x\right)^3=\left(2-5x\right)\left[2^2+2.5x+\left(5x\right)^2\right]=\left(2-5x\right)\left(4+10x+25x^2\right)\)
f) \(64x^3+\frac{1}{8}=\left(4x\right)^3+\left(\frac{1}{2}\right)^3=\left(4x+\frac{1}{2}\right)\left[\left(4x\right)^2-4x.\frac{1}{2}+\left(\frac{1}{2}\right)^2\right]=\left(4x+\frac{1}{2}\right)\left(16x^2-2x+\frac{1}{4}\right)\)
Ko chắc ạ!
BÀi 1 : xem lại đề
bài 2
a) 27 - x^3
= ( 3 -x )( 9 + 3x + x^2)
b) 8x^3 + 0,001
= (2x + 0,1) ( 4x^2 - 0,2x + 0,01)
\(\frac{x^3}{64}-\frac{y^3}{125}=\left(\frac{x}{4}-\frac{y}{5}\right)\left(\frac{x^2}{16}-\frac{xy}{20}+\frac{y^2}{25}\right)\)
a+b=7=>(a+b)2=49
=>a2+2ab+b2=49
Do ab=3
=>2ab=6
=>b2+a2=43
Ta có:a3+b3=(a+b)(a2-ab+b2)
Thay a2+b2=43 ab=3 a+b=7
=> a3+b3=7.(43-3)=7.40=280
a)27-x3=(3-x)(9+3x+x2)
b)8x3+0,001=(2x+0,1)(4x2-0,2x+0,01)
c)x3/64-y3/125=(x/4-y/5)(x2/16+xy/20+y2/25)
a) \(M=x^2-100=\left(x-10\right)\left(x+10\right)\)
b) \(N=x^3+1000=\left(x+10\right)\left(x^2-10x+100\right)\)
c) \(125x^3-1=\left(5x-1\right)\left(25x^2+5x+1\right)\)
d) \(8x^3-27=\left(2x-3\right)\left(4x^2+6x+9\right)\)
a, \(M=x^2-100=\left(x-10\right)\left(x+10\right)\)
b, \(N=x^3+1000=\left(x+10\right)\left(x^2-10x+100\right)\)
c, \(P=125x^3-1=\left(5x-1\right)\left(25x^2+5x+1\right)\)
d, \(Q=8x^3-27=\left(2x-3\right)\left(2x+3\right)\)