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\(a,27-x^3\)
\(=3^3-x^3\)
\(=\left(3-x\right)\left(9+3x+x^2\right)\)
Các câu còn lại lm tương tự nhé.
hok tốt!
a) \(27-x^3=\left(3-x\right)\left(9+3x+x^2\right)\)
b) \(8x^3+0,001=\left(2x+0,1\right)\left(4x^2-0,2x+0,01\right)\)
c) \(\frac{x^3}{125}-\frac{y^3}{27}=\left(\frac{x}{5}-\frac{y}{3}\right)\left(\frac{x^2}{25}+\frac{xy}{15}+\frac{y^2}{9}\right)\)
p/s: chúc bạn học tốt
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)
\(27-x^3\)
\(=3^3-x^3\)
\(=\left(3-x\right)\left(9+3x+x^2\right)\)
\(8x^3+0,001\)
\(=\left(2x\right)^3+\left(\dfrac{1}{10}\right)^3\)
\(=\left(2x+\dfrac{1}{10}\right)\left(4x^2-2x\dfrac{1}{10}+\left(\dfrac{1}{10}\right)^2\right)\)
\(=2\left(x+\dfrac{1}{5}\right)\left(4x^2-\dfrac{1}{5}x+\dfrac{1}{100}\right)\)
\(\dfrac{x^3}{125}-\dfrac{y^3}{27}\)
\(=\left(\dfrac{x}{5}\right)^3-\left(\dfrac{y}{3}\right)^3\)
\(=\left(\dfrac{x}{5}-\dfrac{y}{3}\right)\left[\left(\dfrac{x}{5}\right)^2+\dfrac{x}{5}.\dfrac{y}{3}+\left(\dfrac{y}{3}\right)^2\right]\)
\(=\left(\dfrac{x}{5}-\dfrac{y}{3}\right)\left(\dfrac{x^2}{25}+\dfrac{xy}{15}+\dfrac{y^2}{9}\right)\)
b )
Dấu = thứ 3 :
Sửa lại : \(2\left(x+\dfrac{1}{20}\right)\)
a: \(x^4+4=x^4+4x^2+4-4x^2=\left(x^2+2-2x\right)\left(x^2+2+2x\right)\)
c: \(x^3-125=\left(x-5\right)\left(x^2+5x+25\right)\)
\(\dfrac{1}{8}x^3-64=\left(\dfrac{1}{2}x-4\right)\left(\dfrac{1}{4}x^2+2x+16\right)\)
d: \(=\left(2x+5y\right)^3\)
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) \(64+16y+y^2=\left(y+8\right)\left(y+8\right)\)
\(a)27{{\rm{x}}^3} + 1 = {\left( {3{\rm{x}}} \right)^3} + 1 = \left( {3{\rm{x}} + 1} \right).\left[ {{{\left( {3{\rm{x}}} \right)}^2} - 3{\rm{x}}.1 + {1^2}} \right] = \left( {3{\rm{x}} + 1} \right)\left( {9{{\rm{x}}^2} - 3{\rm{x}} + 1} \right)\)
\(b)64 - 8{y^3} = {4^3} - {\left( {2y} \right)^3} = \left( {4 - 2y} \right)\left[ {{4^2} + 4.2y + {{\left( {2y} \right)}^2}} \right] = \left( {4 - 2y} \right)\left( {16 + 8y + 4{y^2}} \right)\)
a: \(x^3+27=\left(x+3\right)\left(x^2-3x+9\right)\)
b: \(x^3-\dfrac{1}{8}=\left(x-\dfrac{1}{2}\right)\left(x^2+\dfrac{1}{2}x+\dfrac{1}{4}\right)\)
c: \(8x^3+y^3=\left(2x+y\right)\left(4x^2-2xy+y^2\right)\)
a) \(\left(x+3\right)\cdot\left(x^2-3x+9\right)\)
b) \(\left(x-\dfrac{1}{2}\right)\cdot\left(x^2+\dfrac{1}{2}x+\dfrac{1}{4}\right)\)
c) \(\left(2x+y\right)\cdot\left(4x^2-2xy+y^2\right)\)