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a) 4(x+2) - 7(2x - 1) + 9(3x - 4)=30
⇔4x+8 - 14x + 7 + 27x - 36 = 30
⇔ 17x = 51
⇔ x = 3
b) 2(5x - 8) - 3(4x - 5) = 4(3x - 4) + 11
⇔ 10x - 16 - 12x + 15 = 12x - 16 + 11
⇔ -14x = -4
⇔ x= \(\frac{2}{7}\)
c) 5x(1 - 2x) - 3x(x + 18) = 0
⇔ 5x - 10x\(^2\) - 3x\(^2\) -54x =0
⇔ -13x\(^2\) -49 x = 0
⇔ -x ( 13x + 49 ) =0
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\13x+49=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\frac{-49}{13}\end{matrix}\right.\)
d) 5x - 3{4x - 2[4x - 3(5x - 2)]} = 182
⇔ 5x - 3[ 4x - 2( 4x - 15x + 6 ) ]= 182
⇔5x - 3 ( 4x - 8x + 30x - 12 ) = 182
⇔ 5x - 3 ( 26x - 12 ) = 182
⇔ 5x - 78x + 36 = 182
⇔ - 73x = 146
⇔ x = -2
a)4(18 - 5x) - 12(3x - 7) = 15(2x - 16) - 6(x + 14)
<=>72 - 20x - 36x +84 = 30x - 240 - 6x 84
<=> -80x = -480
<=> x = 6
b) 5(3x+5)-4(2x-3) =5x+3(2x+12)+1
<=> 15x + 25 - 8x + 12 = 5x + 6x + 36 + 1
<=> 15x + 25 - 8x + 12 - 5x - 6x - 36 - 1 = 0
<=> -4x = 0
<=> x = 0
c) 2(5x-8)-3(4x-5)=4(3x-4)+11
= 10x - 16 - 12x + 15 = 12x - 16 + 11
= -14x = -4
= x =\(\frac{2}{7}\)
d) 5x-3{4x-2[4x-3(5x-2)]}=182
= 5x - 3 . [4x - 2(4x - 15x + 6)]
= 5x - 3 . (4x - 8x + 30x - 12)
= 5x - 12x + 24x - 90x + 36
= -73x + 36 = 182
=> -73x = 182 - 36 = 146
=> x = 146 : (-73) = -2
~Hok tốt~
2:
a: =>x-1=0 hoặc 3x+1=0
=>x=1 hoặc x=-1/3
b: =>x-5=0 hoặc 7-x=0
=>x=5 hoặc x=7
c: =>\(\left[{}\begin{matrix}x-1=0\\x+5=0\\3x-8=0\end{matrix}\right.\Leftrightarrow x\in\left\{1;-5;\dfrac{8}{3}\right\}\)
d: =>x=0 hoặc x^2-1=0
=>\(x\in\left\{0;1;-1\right\}\)
Bài 1.
1) ( 2x + 1 )3 - ( 2x + 1 )( 4x2 - 2x + 1 ) - 3( 2x - 1 ) = 15
<=> 8x3 + 12x2 + 6x + 1 - [ ( 2x )3 - 13 ] - 6x + 3 = 15
<=> 8x3 + 12x2 + 4 - 8x3 + 1 = 15
<=> 12x2 + 15 = 15
<=> 12x2 = 0
<=> x = 0
2) x( x - 4 )( x + 4 ) - ( x - 5 )( x2 + 5x + 25 ) = 13
<=> x( x2 - 16 ) - ( x3 - 53 ) = 13
<=> x3 - 16x - x3 + 125 = 13
<=> 125 - 16x = 13
<=> 16x = 112
<=> x = 7
Bài 2.
A = ( x + 5 )( x2 - 5x + 25 ) - ( 2x + 1 )3 - 28x3 + 3x( -11x + 5 )
= x3 + 53 - ( 8x3 + 12x2 + 6x + 1 ) - 28x3 - 33x2 + 15x
= -27x3 + 125 - 8x3 - 12x2 - 6x - 1 - 33x2 + 15x
= -33x3 - 45x2 + 9x + 124 ( có phụ thuộc vào biến )
B = ( 3x + 2 )3 - 18x( 3x + 2 ) + ( x - 1 )3 - 28x3 + 3x( x - 1 )
= 27x3 + 54x2 + 36x + 8 - 54x2 - 36x + x3 - 3x2 + 3x - 1 - 28x3 + 3x2 - 3x
= 7 ( đpcm )
C = ( 4x - 1 )( 16x2 + 4x + 1 ) - ( 4x + 1 )3 + 12( 4x + 1 )3 + 12( 4x + 1 ) - 15
= ( 4x )3 - 13 - [ ( 4x + 1 )3 - 12( 4x + 1 )3 - 12( 4x + 1 ) ] - 15
= 64x3 - 1 - ( 4x + 1 )[ ( 4x + 1 )2 - 12( 4x + 1 )2 - 12 ] - 15
= 64x3 - 16 - ( 4x + 1 )[ 16x2 + 8x + 1 - 12( 16x2 + 8x + 1 ) - 12 ]
= 64x3 - 16 - ( 4x + 1 )( 16x2 + 8x - 11 - 192x2 - 96x - 12 )
= 64x3 - 16 - ( 4x + 1 )( -176x2 - 88x - 23 )
= 64x3 - 16 - ( -704x3 - 528x2 - 180x - 23 )
= 64x3 - 16 + 704x3 + 528x2 + 180x + 23
= 768x3 + 528x2 + 180x + 7 ( có phụ thuộc vào biến )
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a) x3 -2x2 +5x-4
=x3-x2-x2+x+4x-4
=x2(x-1)-x(x-1)+4(x-1)
=(x2-x+4)(x-1)
b) x3-x2+x+3
=x3+x2-2x2-2x+3x+3
=x2(x+1) -2x(x+1)+3(x+1)
=(x2-2x+3)(x+1)
c) 6x3+x2+x+1
=6x3+ 3x2-2x2-x+2x+1
=6x2(x+\(\frac{1}{2}\)) - 2x(x+\(\frac{1}{2}\)) +2(x+\(\frac{1}{2}\))
=(6x2-2x+2) (x+\(\frac{1}{2}\))
=2( 3x2-x+1) (x+\(\frac{1}{2}\))
d) 4x3 + 6x2+4x+1
= 4x3+2x2+4x2+2x+2x+1
= 4x2(x+\(\frac{1}{2}\))+ 4x(x+\(\frac{1}{2}\))+2(x+\(\frac{1}{2}\))
= 2(2x2 +2x+1)( x+\(\frac{1}{2}\))
e) x6 -9x3+8
Ta có : 6x2 - 11x + 3
= 6x2 - 2x - 9x + 3
= (6x2 - 2x) - (9x - 3)
= 2x(3x - 1) - 3(3x - 1)
= (2x - 3)(3x - 1)
bạn viết rõ đề ra nhé
b, \(\left|4x-8\right|=1-x\)ĐK : \(x\le1\)
TH1 : \(4x-8=1-x\Leftrightarrow5x=9\Leftrightarrow x=\dfrac{9}{5}\)( ktm )
TH2 : \(4x-8=x-1\Leftrightarrow3x=7\Leftrightarrow x=\dfrac{7}{3}\)( ktm )
b) Ta có: \(\left|4x-8\right|=1-x\)
\(\Leftrightarrow\left[{}\begin{matrix}4x-8=1-x\left(x\ge2\right)\\4x-8=x-1\left(x< 2\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}4x+x=1+8\\4x-x=-1+8\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}5x=9\\3x=7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{9}{5}\left(loại\right)\\x=\dfrac{7}{3}\left(loại\right)\end{matrix}\right.\)