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a) \(\dfrac{1}{3x-2}-\dfrac{1}{3x+2}-\dfrac{3x-6}{9x^2-4}\)
\(=\dfrac{3x+2-3x+2-3x+6}{\left(3x-2\right)\left(3x+2\right)}\)
\(=\dfrac{-3x+10}{\left(3x-2\right)\left(3x+2\right)}\)
b) \(\dfrac{x+25}{2x^2-50}-\dfrac{x+5}{x^2-5x}-\dfrac{5-x}{2x^2+10x}\)
\(=\dfrac{x+25}{2\left(x-5\right)\left(x+5\right)}-\dfrac{x+5}{x\left(x-5\right)}+\dfrac{x-5}{2x\left(x+5\right)}\)
\(=\dfrac{x^2+25x-2\left(x+5\right)^2+\left(x-5\right)^2}{2x\left(x-5\right)\left(x+5\right)}\)
\(=\dfrac{x^2+25x-2x^2-20x-50+x^2-10x+25}{2x\left(x-5\right)\left(x+5\right)}\)
\(=\dfrac{-5x-25}{2x\left(x-5\right)\left(x+5\right)}\)
\(=\dfrac{-5\left(x+5\right)}{2x\left(x-5\right)\left(x+5\right)}=\dfrac{-5}{2x\left(x-5\right)}\)
c) Ta có: \(\dfrac{1-2x}{2x}-\dfrac{4x}{2x-1}-\dfrac{3}{2x-4x^2}\)
\(=\dfrac{-\left(2x-1\right)^2-8x^2+3}{2x\left(2x-1\right)}\)
\(=\dfrac{-\left(4x^2-4x+1\right)-8x^2+3}{2x\left(2x-1\right)}\)
\(=\dfrac{-4x^2+4x-1-8x^2+3}{2x\left(2x-1\right)}\)
\(=\dfrac{-12x^2+4x+2}{2x\left(2x-1\right)}\)
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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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Bài 2:
a: \(\Leftrightarrow\left(x-2\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)
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\(\text{2x - (x - 3)(5 - x) = (x+4)}^2.\)
\(\Leftrightarrow2x-\left(5x-x^2-15+3x\right)=x^2+8x+16.\)
\(\Leftrightarrow2x-5x+x^2+15-3x-x^2-8x-16=0.\)
\(\Leftrightarrow-14x-1=0.\Leftrightarrow x=\dfrac{-1}{14}.\)
\(\text{(4x + 1)(x - 2) + 25 = (2x+3)}^2-4x.\)
\(\Leftrightarrow4x^2-8x+x-2+25=4x^2+12x+9-4x.\)
\(\Leftrightarrow-15x+14=0.\Leftrightarrow x=\dfrac{14}{15}.\)
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a) \(\left(x+3\right)^2-\left(x-2\right)^3=\left(x+5\right)\left(x^2-5x+25\right)-108\)
\(\Leftrightarrow x^2+6x+9-x^2+4x-4=x^3-5x^2+25x+5x^2-25x+125-108\)
\(\Leftrightarrow x^3-10x+12=0\Leftrightarrow\left(x-2\right)\left(x^2+2x+6\right)=0\)
\(\Leftrightarrow x=2\)( do \(x^2+2x+6=\left(x+1\right)^2+4\ge4>0\))
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+) (5x-1). (2x+3)-3. (3x-1)=0
10x^2+15x-2x-3 - 9x+3=0
10x^2 +8x=0
2x(5x+4)=0
=> x=0 hoặc x= -4/5
+) x^3 (2x-3)-x^2 (4x^2-6x+2)=0
2x^4 -3x^3 -4x^4 + 6x^3 - 2x^2=0
-2x^4 + 3x^3-2x^2=0
x^2(-2x^2+x-2)=0
-2x^2(x-1)^2=0
=> x=0 hoặc x=1
+) x (x-1)-x^2+2x=5
x^2 -x -x^2+2x=5
x=5
+) 8 (x-2)-2 (3x-4)=25
8x - 16-6x+8=25
2x=33
x=33/2
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( 2x + 1 )3 - ( 3x + 2 )2 = ( 2x - 5 )( 4x2 + 10x + 25 ) + 6x( 2x + 1 ) - 9x2
⇔ 8x3 + 12x2 + 6x + 1 - ( 9x2 + 12x + 4 ) = 8x3 - 125 + 12x2 + 6x - 9x2
⇔ 8x3 + 12x2 + 6x + 1 - 9x2 - 12x - 4 = 8x3 + 3x2 + 6x - 125
⇔ 8x3 + 3x2 - 6x - 3 = 8x3 + 3x2 + 6x - 125
⇔ 8x3 + 3x2 - 6x - 3 - 8x3 - 3x2 - 6x + 125 = 0
⇔ -12x + 122 = 0
⇔ -12x = -122
⇔ x = 61/6
\(\frac{3}{2x+5}-\frac{40}{4x^2-25}+\frac{1}{2x-5}\)
\(=\frac{3\cdot\left(2x-5\right)}{4x^2-25}-\frac{40}{4x^2-25}+\frac{2x+5}{4x^2-25}\)
\(=\frac{6x-15-40+2x+5}{\left(2x+5\right)\cdot\left(2x-5\right)}=\frac{8x-50}{\left(2x+5\right)\cdot\left(2x-5\right)}\)