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bam may tinh di tim nghiem roi them bot hang tu bien thanh nhan tu chung roi chia
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f/ \(3xy\left(x+y\right)-\left(x+y\right)\left(x^2+y^2+2xy\right)+y^3=27\)
\(3x^2y+3xy^2-\left(x+y\right)\left(x+y\right)^2+y^3=27\)
\(3x^2y+3xy^3-\left(x+y\right)^3+y^3=27\)
\(3x^2y+3xy^3-\left(x^3+3x^2y+3xy^2+b^3\right)+y^3=27\)
\(-x^3=27\)
\(x=-3\)
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1.a) \(\Leftrightarrow\) 3x+10-2x =0
\(\Leftrightarrow\text{ 3x-2x=-10}\)
\(\Leftrightarrow x=-10\)
b) coi lại có thiếu ngoặc ko nhé
cứ nhân vào dấu ngoặc rồi làm như thường
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a: \(=\dfrac{4x-8+2x+4-8}{\left(x-2\right)\left(x+2\right)}=\dfrac{6x-12}{\left(x-2\right)\left(x+2\right)}=\dfrac{6}{x+2}\)
b: \(=\dfrac{-x+7x-4}{3x-2}=\dfrac{6x-4}{3x-2}=2\)
c: \(=\dfrac{x}{2x+1}-\dfrac{1}{\left(2x+1\right)\left(2x-1\right)}-\dfrac{\left(x-2\right)}{2x-1}\)
\(=\dfrac{2x^2-x-1-\left(x-2\right)\left(2x+1\right)}{\left(2x+1\right)\left(2x-1\right)}\)
\(=\dfrac{2x^2-x-1-2x^2-x+4x+2}{\left(2x+1\right)\left(2x-1\right)}\)
\(=\dfrac{2x+1}{\left(2x+1\right)\left(2x-1\right)}=\dfrac{1}{2x-1}\)
d: \(=\dfrac{5}{2x-3}+\dfrac{2}{2x+3}+\dfrac{2x-33}{4x^2-99}\)
\(=\dfrac{10x+15+4x-6+2x-33}{\left(2x-3\right)\left(2x+3\right)}=\dfrac{16x-24}{\left(2x-3\right)\left(2x+3\right)}=\dfrac{8}{2x+3}\)
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a) ( 3x - 2 )( 4x + 5 ) - 6x( 2x - 1 )
= 12x2 + 7x - 10 - 12x2 + 6x
= 13x - 10
b) ( 2x - 5 )2 - 4( x + 3 )( x - 3 )
= 4x2 - 20x + 25 - 4( x2 - 9 )
= 4x2 - 20x + 25 - 4x2 + 36
= 61 - 20x
c) 2x3 - 5x2 + 7x - 6
= 2x3 - 3x2 - 2x2 + 3x + 4x - 6
= x2( 2x - 3 ) - x( 2x - 3 ) + 2( 2x - 3 )
= ( 2x - 3 )( x2 - x + 2 )
=> ( 2x3 - 5x2 + 7x - 6 ) : ( 2x - 3 ) = x2 - x + 2
a, (3x - 2 ) (4x + 5) - 6x (2x -1) = ( 7x + 15x -8x - 10 ) - ( 12x2 -6x ) = 7x2 + 15x - 8x -10 -12x2 + 6x = -5x2 + x - 10
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Mấy bài dài dài kia tí mình làm cho :)
( x - 1 )3 - x( x - 2 )2 + 1
= x3 - 3x2 + 3x - 1 - x( x2 - 4x + 4 ) + 1
= x3 - 3x2 + 3x - x3 + 4x2 - 4x
= x2 - x = x( x - 1 )
2x( 3x + 2 ) - 3x( 2x + 3 )
= 6x2 + 4x - 6x2 - 9x
= -5x
( x + 2 )3 + ( x - 3 )2 - x2( x + 5 )
= x3 + 6x2 + 12x + 8 + x2 - 6x + 9 - x3 - 5x2
= 2x2 + 6x + 17
( 2x + 3 )( x - 5 ) + 2x( 3 - x ) + x - 10
= 2x2 - 7x - 15 + 6x - 2x2 + x - 10
= -25
( x + 5 )( x2 - 5x + 25 ) - x( x - 4 )2 + 16x
= x3 + 53 - x( x2 - 8x + 16 ) + 16x
= x3 + 125 - x3 + 8x2 - 16x + 16
= 8x2 + 125
( -x - 2 )3 + ( 2x - 4 )( x2 + 2x + 4 ) - x2( x - 6 )
= -x3 - 6x2 - 12x - 8 + 2x3 - 16 - x3 + 6x2
= -12x - 24 = -12( x + 2 )
Tương tự ...
a, \(\left(x-1\right)^3-x\left(x-2\right)^2+1=x^3-3x^2+3x-1-x^3+4x^2-4x+1=x^2-x\)
b, \(2x\left(3x+2\right)-3x\left(2x+3\right)=6x^2+4x-6x^2-9x=-5x\)
c, \(\left(x+2\right)^3+\left(x-3\right)^2-x^2\left(x+5\right)=x^3+6x^2+12x+8+x^2+6x+9-x^3-5x^2=2x^2+18x+17\)
a) = 8x-6x-20+15x-15-6x+55-22x-30x+75
b)3x3+5x2-6x2+10x+9x-15-2x3+7x2+2x2+7x-2x-7
bạn chỉ cần chép lại đầu bài rồi ghi kết quả này vào là xong nhá bởi vì đây là nhân đa thức với đa thức mà
a) (2x-5)(4-3x)-(3x+11)(5-2x)-15(2x-5)
=(2x-5)(4-3x)+(3x+11)(2x-5)-15(2x-5)
=(2x-5)[(4-3x)+(3x+11)-15]
=(2x-5)(4-3x+3x+11-15)
= 2x-5