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5 tháng 10 2021
\(a,=8\left(x^3-125\right)=8\left(x-5\right)\left(x^2+5x+25\right)\\ b,=\left(0,1+4x\right)\left(0,01-0,4x+16x^2\right)\\ c,=\left(x+\dfrac{1}{5}y\right)\left(x^2-\dfrac{1}{5}xy+\dfrac{1}{25}y^2\right)\\ d,=\left(3x-\dfrac{1}{2}y\right)\left(9x^2+\dfrac{3}{2}xy+\dfrac{1}{4}y^2\right)\\ e,=\left(x-1+3\right)\left[\left(x-1\right)^2-3\left(x-1\right)+9\right]\\ =\left(x+2\right)\left(x^2-2x+1-3x+3+9\right)\\ =\left(x+2\right)\left(x^2-5x+13\right)\\ f,=\left(\dfrac{x^2}{2}-y^2\right)\left(\dfrac{x^4}{4}+\dfrac{x^2y^2}{2}+y^4\right)\)
31 tháng 7 2021
1, x3-9x2y+27xy2-27y3=(x-3y)3
2, 27x3-9x2y+xy2-\(\dfrac{1}{27}\)y3=(3x-\(\dfrac{1}{3}\)y)3
3)x6-3x4y+3xy2-y3=(x2-y)3
31 tháng 7 2021
1) \(x^3-9x^2y+27xy^2-27y^3=\left(x-3y\right)^3\)
2) \(27x^3-9x^2y+xy^2-\dfrac{1}{27}y^3=\left(3x-\dfrac{1}{3}y\right)^3\)
3) \(x^6-3x^4y+3xy^2-y^3=\left(x^2-y\right)^3\)
20 tháng 9 2020
a) x3 - 125 = x3 - 53 = ( x - 5 )( x2 + 5x + 25 )
b) a3 + 27 = a3 + 33 = ( a + 3 )( a2 - 3a + 9 )
c) -64 + 1/8x3 = ( 1/2x )3 - 43 = ( 1/2x - 4 )( 1/4x2 + 2x + 16 )
d) 0, 001 - 1000x3 = ( 1/10 )3 - ( 10x )3 = ( 1/10 - 10x )( 1/100 + x + 100x2 )
e) ( x + 1 )3 - ( 2 - x )3 = [ ( x + 1 ) - ( 2 - x ) ][ ( x + 1 )2 + ( x + 1 )( 2 - x ) + ( 2 - x )2 ]
= ( x + 1 - 2 + x )( x2 + 2x + 1 - x2 + x + 2 + 4 - 4x + x2 )
= ( 2x - 1 )( x2 - x + 7 )
f) 1/125 + ( x - 1/5 )3 = ( 1/5 )3 + ( x - 1/5 )3
= [ 1/5 + ( x - 1/5 ) ][ 1/25 - 1/5( x - 1/5 ) + ( x - 1/5 )2 ]
= ( 1/5 + x - 1/5 )( 1/25 - 1/5x + 1/25 + x2 - 2/5x + 1/25 )
= x( x2 - 3/5x + 3/25 )
20 tháng 9 2020
Bài làm :
\(a)x^3-125=x^3-5^3=\left(x-5\right)\left(x^2+5x+25\right)\)
\(b)a^3+27=a^3+3^3=\left(a+3\right)\left(a^2-3a+9\right)\)
T
4 tháng 8 2023
(x-1)3+(2x+3)3=27x3+8
=> (x - 1 + 2x + 3)[(x - 1)2 - (x - 1)(2x + 3) + (2x + 3)2] = (3x)3 + 23
=> (3x + 2)[x2-2x+1-(2x2+x-3)+4x2+12x+9] = (3x + 2)[(3x)2 - 3x.2 + 22]
=> (3x + 2)(3x2 + 9x + 13) = (3x + 2)(9x2 - 6x + 4)
=> (3x + 2)(3x2 + 9x + 13) - (3x + 2)(9x2 - 6x + 4) = 0
=> (3x + 2)(3x2 + 9x + 13 - 9x2 + 6x - 4) = 0
=> (3x + 2)(-6x2 + 15x + 9) = 0
=>\(\left[{}\begin{matrix}3x+2=0\\-6x^2+15x+9=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}3x=-2\\-3\left(2x^2+5x\right)=-9\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{2}{3}\\2x^2+5x=3\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{2}{3}\\2x^2+6x-x-3=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{2}{3}\\2x\left(x+3\right)-\left(x+3\right)=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{2}{3}\\\left(2x-1\right)\left(x+3\right)=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{2}{3}\\x=\dfrac{1}{2}\\x=-3\end{matrix}\right.\)
Vậy phương trình (x-1)3+(2x+3)3=27x3+8 có nghiệm là {-2/3;1/2;-3}
4 tháng 8 2023
=>x^3-3x^2+3x-1+8x^3+36x^2+54x+27=27x^3+8
=>37x^3+51x^2+57x+26-27x^3-8=0
=>10x^3+51x^2+57x+18=0
=>(5x+3)(2x^2+9x+6)=0
=>x=-3/5 hoặc \(x=\dfrac{-9\pm\sqrt{33}}{4}\)
7 tháng 11 2021
Câu 2:
a: \(\Leftrightarrow3x^2+2x-1=0\)
\(\Leftrightarrow\left(x+1\right)\left(3x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{1}{3}\end{matrix}\right.\)
b: \(\Leftrightarrow x^3-4x-x^3-8=4\)
hay x=-3
2 tháng 5 2020
a) Ta có: \(\frac{7}{8}x-5\left(x-9\right)=\frac{20x+1,5}{6}\)
\(\Leftrightarrow\frac{7x}{8}-5x+45-\frac{20x+1,5}{6}=0\)
\(\Leftrightarrow\frac{21x}{24}-\frac{120x}{24}+\frac{1080}{24}-\frac{4\left(20x+1,5\right)}{24}=0\)
\(\Leftrightarrow-99x+1080-4\left(20x+1,5\right)=0\)
\(\Leftrightarrow-99x+1080-80x-6=0\)
\(\Leftrightarrow1074-179x=0\)
\(\Leftrightarrow179x=1074\)
hay x=6
Vậy: x=6
b) Ta có: \(4\left(0,5-1,5x\right)=-\frac{5x-6}{3}\)
\(\Leftrightarrow2-6x=\frac{6-5x}{3}\)
\(\Leftrightarrow\frac{3\left(2-6x\right)}{3}-\frac{6-5x}{3}=0\)
\(\Leftrightarrow6-18x-6+5x=0\)
\(\Leftrightarrow-13x=0\)
mà -13≠0
nên x=0
Vậy: x=0
c) Ta có: \(\frac{x+4}{5}-x+4=\frac{x}{3}-\frac{x-2}{2}\)
\(\Leftrightarrow\frac{6\left(x+4\right)}{30}+\frac{30\left(-x+4\right)}{30}-\frac{10x}{30}+\frac{15\left(x-2\right)}{30}=0\)
\(\Leftrightarrow6\left(x+4\right)+30\left(4-x\right)-10x+15\left(x-2\right)=0\)
\(\Leftrightarrow6x+24+120-30x-10x+15x-30=0\)
\(\Leftrightarrow-19x+114=0\)
\(\Leftrightarrow-19x=-114\)
hay x=6
Vậy: x=6
d) Ta có: \(\frac{4x+3}{5}-\frac{6x-2}{7}=\frac{5x+4}{3}+3\)
\(\Leftrightarrow\frac{21\left(4x+3\right)}{105}-\frac{15\left(6x-2\right)}{105}-\frac{35\left(5x+4\right)}{105}-\frac{315}{105}=0\)
\(\Leftrightarrow84x+63-90x+30-175x-140-315=0\)
\(\Leftrightarrow-181x-362=0\)
\(\Leftrightarrow-181x=362\)
hay x=-2
Vậy: x=-2
e) Ta có: \(\frac{1}{4}\left(x+3\right)=3-\frac{1}{2}\left(x+1\right)-\frac{1}{3}\left(x+2\right)\)
\(\Leftrightarrow\frac{x+3}{4}=3-\frac{x+1}{2}-\frac{x+2}{3}\)
\(\Leftrightarrow\frac{3\left(x+3\right)}{12}-\frac{36}{12}+\frac{6\left(x+1\right)}{12}+\frac{4\left(x+2\right)}{12}=0\)
\(\Leftrightarrow3x+9-36+6x+6+4x+8=0\)
\(\Leftrightarrow13x-13=0\)
\(\Leftrightarrow13x=13\)
hay x=1
Vậy: x=1
17 tháng 10 2021
\(a,=\left(3x+\dfrac{y}{2}\right)\left(9x^2+\dfrac{3}{2}xy+\dfrac{y^2}{4}\right)\\ b,=\left(5x+3y\right)\left(25x^2+15xy+9y^2\right)\)
17 tháng 9 2021
\(a,=\left(3+x\right)\left(9-3x+x^2\right)\\ b,=\left(4x+0,1\right)\left(16x^2-0,4x+0,01\right)\\ c,=\left(2-3x\right)\left(4+6x+9x^2\right)\\ d,=\left(\dfrac{x}{5}-\dfrac{y}{3}\right)\left(\dfrac{x^2}{25}+\dfrac{xy}{15}+\dfrac{y^2}{9}\right)\)
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1) \(a^6-b^6=\left(a^2-b^2\right)\left(a^4+a^2b^2+b^4\right)=\left(a-b\right)\left(a+b\right)\left(a^4+a^2b^2+b^4\right)\)
2) \(27x^3-a^3b^3=\left(3x-ab\right)\left(9x^2+3xab+a^2b^2\right)\)
3) \(\frac{1}{8}-8x^3=\left(\frac{1}{2}-2x\right)\left(\frac{1}{4}+x+4x^2\right)\)
4) \(8+\left(4x-3\right)^3=\left(2+4x-3\right)\left[4-2\left(4x-3\right)+\left(4x-3\right)^2\right]\)
\(=\left(4x-1\right)\left(4-8x+6+16x^2-24x+9\right)\)
\(=\left(4x-1\right)\left(16x^2-32x+19\right)\)