Đưa về dạng tích:
a) 216x3 + (x + y)3
b) (2x + 1)3 + 8x3
c) (5x - 2)3 - 27x3
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a, <=> x = -4
b, <=> 6x + 2 = -2x + 5 <=> 8x = 3 <=> x = 3/8
c, <=> 5x + 2x - 2 = 4x + 7 <=> 2x = 9 <=> x = 9 /2
d, <=> 10x^2 - 10x^2 - 15x = 15 <=> x = -1
a, <=> x = -4
b, <=> 6x + 2 = -2x + 5 <=> 8x = 3 <=> x = 3/8
c, <=> 5x + 2x - 2 = 4x + 7 <=> 2x = 9 <=> x = 9 /2
d <=> 10x^2 - 10x^2 - 15x = 15 <=> x = -1
a: =>-x+2x=3-7
=>x=-4
b: =>6x+2+2x-5=0
=>8x-3=0
hay x=3/8
c: =>5x+2x-2-4x-7=0
=>3x-9=0
hay x=3
d: =>10x2-10x2-15x=15
=>-15x=15
hay x=-1
`a,(x+3)(x^2+2021)=0`
`x^2+2021>=2021>0`
`=>x+3=0`
`=>x=-3`
`2,x(x-3)+3(x-3)=0`
`=>(x-3)(x+3)=0`
`=>x=+-3`
`b,x^2-9+(x+3)(3-2x)=0`
`=>(x-3)(x+3)+(x+3)(3-2x)=0`
`=>(x+3)(-x)=0`
`=>` $\left[ \begin{array}{l}x=0\\x=-3\end{array} \right.$
`d,3x^2+3x=0`
`=>3x(x+1)=0`
`=>` $\left[ \begin{array}{l}x=0\\x=-1\end{array} \right.$
`e,x^2-4x+4=4`
`=>x^2-4x=0`
`=>x(x-4)=0`
`=>` $\left[ \begin{array}{l}x=0\\x=4\end{array} \right.$
1) a) \(\left(x+3\right).\left(x^2+2021\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x+3=0\\x^2+2021=0\end{matrix}\right.\\\left[{}\begin{matrix}x=-3\left(nhận\right)\\x^2=-2021\left(loại\right)\end{matrix}\right. \)
=> S={-3}
a) \(3x-2=2x-3\)
\(\Leftrightarrow x+1=0\)
\(\Leftrightarrow x=-1\)
b) \(3-4y+24+6y=y+27+3y\)
\(\Leftrightarrow-2y=0\Leftrightarrow y=0\)
c) \(7-2x=22-3x\)
\(\Leftrightarrow x-15=0\)
\(\Leftrightarrow x=15\)
d) \(8x-3=5x+12\)
\(\Leftrightarrow3x-15=0\Leftrightarrow x=5\)
a) 3x-2=2x-3
3x=2x-1
Bớt mỗi vế 2x
x=-1
b)3-4y+24+6y=y+27+3y
3-4y+6y=y+3+3y
3-4y+3y=y+3
<=> y=0
c.7-2x=22-3x
2x=15-3x
15=x
d.8x-3=5x+12
3x-3=12
3x=15
x=5
câu e hình như bạn thiếu đề
f)x+2x+3x-19=3x+5
6x-19=3x+5
3x-19=5
3x=24
<=>x=8
g)11=8x-3=5x-3+x
11=8x-3
11=6x-3
<=> x không tồn tại
h)4-2x+15=9x+4x-2x
4-2x+15=11x
<=> nghiệm trên có số thập phân vô hạn tuần hoàn nhé
T
Ngập mặt ~
Mình làm 1;2 câu thôi. Các câu sau bạn làm tương tự nhé.
a/ 3x - 2 = 2x - 3
<=> 3x - 2 - 2x + 3 = 0
<=> x + 1 = 0
<=> x = -1
b/ 3 - 4y + 24 + 6y = y + 27 + 3y
<=> 3 - 4y + 24 + 6y - y - 27 - 3y = 0
<=> -2y = 0
<=> y = 0
\(a,\Leftrightarrow\left(4-5x\right)\left(4+5x\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{4}{5}\\x=-\dfrac{4}{5}\end{matrix}\right.\\ b,\Leftrightarrow\left(x+1-2\right)\left(x+1+2\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-3\end{matrix}\right.\\ c,\Leftrightarrow\left(3x+1-2x\right)\left(3x+1+2x\right)=0\\ \Leftrightarrow\left(x+1\right)\left(5x+1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-\dfrac{1}{5}\end{matrix}\right.\\ d,Sửa:\left(4x+1\right)^2-\left(x-2\right)^2=0\\ \Leftrightarrow\left(4x+1-x+2\right)\left(4x+1+x-2\right)=0\\ \Leftrightarrow\left(3x+3\right)\left(5x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{1}{5}\end{matrix}\right.\\ e,\Leftrightarrow\left(2x+1-x-3\right)\left(2x+1+x+3\right)=0\\ \Leftrightarrow\left(x-2\right)\left(3x+4\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-\dfrac{4}{3}\end{matrix}\right.\)
a) \(\left(x-1\right)^3\)
\(=x^3-3x^2+3x-1\)
b) \(\left(2x-3y\right)^3\)
\(=\left(2x\right)^3-3\left(2x\right)^23y+3.2x\left(3y\right)^3+\left(3y\right)^3\)
\(=8x^3-36x^2y+54xy^2-27y^3\)
Bài 3:
a: Ta có: \(\left(x-2\right)^3-x^2\left(x-6\right)=5\)
\(\Leftrightarrow x^3-6x^2+12x-8-x^3+6x^2=5\)
\(\Leftrightarrow12x=13\)
hay \(x=\dfrac{13}{12}\)
b: Ta có: \(\left(x-1\right)\left(x^2+x+1\right)-x\left(x+2\right)\left(x-2\right)=4\)
\(\Leftrightarrow x^3-1-x^3+4x=4\)
\(\Leftrightarrow4x=5\)
hay \(x=\dfrac{5}{4}\)
a) 216x3 + ( x + y )3 = ( 6x )3 + ( x + y )3
= [ 6x + ( x + y ) ][ ( 6x )2 - 6x( x + y ) + ( x + y )2 ]
= ( 6x + x + y )( 36x2 - 6x2 - 6xy + x2 + 2xy + y2 )
= ( 7x + y )( 31x2 - 4xy + y2 )
b) ( 2x + 1 )3 + 8x3 = ( 2x + 1 )2 + ( 2x )3
= [ ( 2x + 1 ) + 2x ][ ( 2x + 1 )2 - ( 2x + 1 )2x + ( 2x )2
= ( 2x + 1 + 2x )( 4x2 + 4x + 1 - 4x2 - 2x + 4x2 )
= ( 4x + 1 )( 4x2 + 2x + 1 )
c) ( 5x - 2 )3 - 27x3 = ( 5x - 2 ) - ( 3x )3
= [ ( 5x - 2 ) - 3x ][ ( 5x - 2 )2 + ( 5x - 2 )3x + ( 3x )2
= ( 5x - 2 - 3x )( 25x2 - 20x + 4 + 15x2 - 6x + 9x2 )
= ( 2x - 2 )( 49x2 - 26x + 4 )
= 2( x - 1 )( 49x2 - 26x + 4 )
a) \(216x^3+\left(x+y\right)^3=\left(6x\right)^3+\left(x+y\right)^3\)
\(=\left(6x+x+y\right)\left[\left(6x\right)^2-6x\left(x+y\right)+\left(x+y\right)^2\right]\)
\(=\left(7x+y\right)\left(36x^2-6x^2-6xy+x^2+2xy+y^2\right)\)
\(=\left(7x+y\right)\left(31x^2-4xy+y^2\right)\)
b) \(\left(2x+1\right)^3+8x^3=\left(2x+1\right)^3+\left(2x\right)^3\)
\(=\left(2x+1+2x\right)\left[\left(2x+1\right)^2-2x\left(2x+1\right)+\left(2x\right)^2\right]\)
\(=\left(4x+1\right)\left(4x^2+4x+1-\left(4x^2-2x\right)+4x^2\right)\)
\(=\left(4x+1\right)\left(4x^2+1+2x\right)\)
c) \(\left(5x-2\right)^3-27x^3=\left(5x-2\right)^3-\left(3x\right)^3\)
\(=\left(5x-2-3x\right)\left[\left(5x-2\right)^2+3x\left(5x-2\right)+\left(3x\right)^2\right]\)
\(=\left(2x-2\right)\left(25x^2-20x+4+15x^2-6x+9x^2\right)\)
\(=\left(2x-2\right)\left(49x^2-26x+4\right)\)