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1. -4x( x + 3 )( x - 4 ) - 3x( x2 - x + 1 )
= -4x( x2 - x - 12 ) - 3x( x2 - x + 1 )
= -4x3 + 4x2 + 48x - 3x3 + 3x2 - 3x
= -7x3 + 7x2 + 45x
2. a) 4x( x - 5 ) - ( x - 1 )( 4x - 3 ) = 5
<=> 4x2 - 20x - ( 4x2 - 7x + 3 ) = 5
<=> 4x2 - 20x - 4x2 + 7x - 3 = 5
<=> -13x - 3 = 5
<=> -13x = 8
<=> x = -8/13
b) 6( x - 3 )( x - 4 ) - 6x( x - 2 ) = 4
<=> 6( x2 - 7x + 12 ) - 6x2 + 12x = 4
<=> 6x2 - 42x + 72 - 6x2 + 12x = 4
<=> -30x + 72 = 4
<=> -30x = -68
<=> x = 34/15
Bài 1 :
\(-4x\left(x+3\right)\left(x-4\right)-3x\left(x^2-x+1\right)\)
\(=-7x^3+7x^2+45x\)
Bài 2 :
a, \(4x\left(x-5\right)-\left(x-1\right)\left(4x-3\right)=5\)
\(\Leftrightarrow4x^2-20x-\left[4x^2-7x+3\right]=5\)
\(\Leftrightarrow4x^2-20x-4x^2+7x-3=5\)
\(\Leftrightarrow-13x-8=0\Leftrightarrow x=-\frac{8}{13}\)
b, \(6\left(x-3\right)\left(x-4\right)-6x\left(x-2\right)=4\)
\(\Leftrightarrow6x^2-42x+72-6x^2+12x=4\)
\(\Leftrightarrow-30x+68=0\Leftrightarrow x=\frac{34}{15}\)
\(1,\)
\(2x\left(x-3\right)-\left(3-x\right)=0\)
\(\Leftrightarrow2x\left(x-3\right)+\left(x-3\right)=0\)
\(\Leftrightarrow\left(2x+1\right)\left(x-3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}2x+1=0\\x-3=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{-1}{2}\\x=3\end{cases}}\)
\(2,\)
\(3x\left(x+5\right)-6\left(x+5\right)=0\)
\(\Leftrightarrow\left(3x-6\right)\left(x+5\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}3x-6=0\\x+5=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=2\\x=-5\end{cases}}\)
\(3,\)
\(x^4-x^2=0\)
\(\Leftrightarrow x^2\left(x^2-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x^2=0\\x^2-1=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=\pm1\end{cases}}\)
\(4,\)
\(x^2-2x=0\)
\(\Leftrightarrow x\left(x-2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x-2=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=2\end{cases}}\)
\(5,\)
\(x\left(x+6\right)-10\left(x-6\right)=0\)
\(\Leftrightarrow x^2+6x-10x+60=0\)
\(\Leftrightarrow x^2-4x+60=0\)
\(\Leftrightarrow x^2-4x+4+56=0\)
\(\Leftrightarrow\left(x-2\right)^2=-56\)(Vô lý)
=> Phương trình vô nghiệm
\(a,\Leftrightarrow\left(5x+1\right)\left(x-4\right)-\left(x-4\right)=0\\ \Leftrightarrow\left(x-4\right)\left(5x+1-x\right)=0\\ \Leftrightarrow5x\left(x-4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\\ b,\Leftrightarrow2x^2-10x-2x^2-3x=26\\ \Leftrightarrow-13x=26\\ \Leftrightarrow x=-2\\ c,\Leftrightarrow x^3+1-x^3+3x=15\\ \Leftrightarrow3x=14\\ \Leftrightarrow x=\dfrac{14}{3}\)
\(d,\Leftrightarrow x^3-5x+2x^2-10+5x-2x^2-17=0\\ \Leftrightarrow x^3-27=0\\ \Leftrightarrow x^3=27\\ \Leftrightarrow x=3\)
c, \(x\)(\(x\) - 2022) + 4.(2022 - \(x\)) = 0
(\(x\) - 2022).(\(x\) - 4) = 0
\(\left[{}\begin{matrix}x-2022=0\\x+4=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=2022\\x=4\end{matrix}\right.\)
(3x - 5) (7 - 5x) - (5x + 2) (2 - 3x) = 4
<=> 21x - 15x2 - 35 + 25x - 10x + 15x2 - 4 + 6x = 4
<=> 42x - 39 = 4
<=> 42x = 43
<=> x = 43/42
1: \(\Leftrightarrow x^2-25-x^2-8x-16+\left(4x+1\right)^3=64x^3+8+48x^2-12x\)
\(\Leftrightarrow-8x-41+64x^3+48x^2+12x+1=64x^3+48x^2-12x+8\)
=>4x-40=-12x+8
=>16x=48
hay x=3
2: \(\Leftrightarrow12x^2-48x-x^3+1+x^3-12x^2+48x-64=x^2-2x-3-x^2-10x-25\)
\(\Leftrightarrow-63=-12x-28\)
=>12x+28=63
=>12x=35
hay x=35/12
Answer:
\(\left(x-5\right)^2+\left(x+4\right)\left(3-x\right)=4\)
\(\Rightarrow x^2-10x+25+3x-x^2+12-4x-4=0\)
\(\Rightarrow-11x+33=0\)
\(\Rightarrow x=3\)