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a: Sửa đề: (5-2x)(5+2x)+2x(x+3)=4-2x^2
=>25-4x^2+2x^2+6x=4-2x^2
=>6x+25=4
=>6x=-21
=>x=-7/2
b: (3x-2)(-2x)+5x^2=-x(x-3)
=>-6x^2+4x+5x^2=-x^2+3x
=>4x=3x
=>x=0
c: =>7-(4x^2-9)=x^2+8x+16
=>7-4x^2+9-x^2-8x-16=0
=>-5x^2-8x=0
=>5x^2+8x=0
=>x(5x+8)=0
=>x=0 hoặc x=-8/5
\(a,\dfrac{x-3}{4}+\dfrac{2x-1}{3}=-\dfrac{x}{6}\)
\(\Leftrightarrow\dfrac{3\left(x-3\right)+4\left(2x-1\right)+2x}{12}=0\)
\(\Leftrightarrow3x-9+8x-4+2x=0\)
\(\Leftrightarrow13x-13=0\)
\(\Leftrightarrow13x=13\)
\(\Leftrightarrow x=1\)
\(b,\left(x-3\right)\left(2x-1\right)=\left(2x-1\right)\left(2x+3\right)\)
\(\Leftrightarrow\left(x-3\right)\left(2x-1\right)-\left(2x-1\right)\left(2x+3\right)=0\)
\(\Leftrightarrow\left(2x-1\right)\left(x-3-2x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\-x-6=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-6\end{matrix}\right.\)
a) \(4x\left(a-b\right)+6xy\left(b-a\right)\)
\(=4x\left(a-b\right)-6xy\left(a-b\right)\)
\(=\left(4x-6xy\right)\left(a-b\right)\)
\(=2x\left(2-3y\right)\left(a-b\right)\)
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, \(\left(x+2\right)^3+\left(x-3\right)^2-x^2\left(x+5\right)=x^3+6x^2+12x+8+x^2+6x+9-x^3-5x^2=2x^2+18x+17\)