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a) \(\left(x+3\right)\left(2x-1\right)-\left(x-3\right)\left(x+1\right)=0\)
\(\Leftrightarrow2x^2+5x-3-x^2+2x+3=0\)
\(\Leftrightarrow x^2+7x=0\Leftrightarrow x\left(x+7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-7\end{matrix}\right.\)
b) \(\left(x+4\right)\left(2x-3\right)-3\left(x-2\right)\left(x+2\right)=0\)
\(\Leftrightarrow2x^2+5x-12-3x^2+12=0\)
\(\Leftrightarrow x^2-5x=0\Leftrightarrow x\left(x-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=5\end{matrix}\right.\)
\(a,\left(x+2\right)^2+\left(x+3\right)^2-2\left(x-2\right)\left(x-3\right)=19\\ \Leftrightarrow x^2+4x+4+x^2+6x+9-2x^2+10x-12=19\\ \Leftrightarrow20x=20\\ \Leftrightarrow x=1\\ b,\left(x+2\right)\left(x^2-2x+4\right)-x\left(x^2-5\right)=15\\ \Leftrightarrow x^3+8-x^3+5x=15\\ \Leftrightarrow5x=7\\ \Leftrightarrow x=\dfrac{7}{5}\\ c,\left(x-1\right)^3+\left(2-x\right)\left(4+2x+x^2\right)+3x\left(x+2\right)=17\\ \Leftrightarrow x^3-3x^2+3x+1+8-x^3+3x^2+6x=17\\ \Leftrightarrow9x=8\\ \Leftrightarrow x=\dfrac{8}{9}\)
a. (x + 2)2 + (x + 3)2 - 2(x - 2)(x - 3) = 19
<=> (x2 + 4x + 4) + (x2 + 6x + 9) - (2x + 4)(x - 3) = 19
<=> x2 + 4x + 4 + x2 + 6x + 9 - 2x2 + 6x - 4x + 12 = 19
<=> x2 + x2 - 2x2 + 4x + 6x + 6x - 4x + 9 + 4 + 12 - 19 = 0
<=> 12x + 6 = 0
<=> 6(2x + 1) = 0
<=> 2x + 1 = 0
<=> 2x = -1
<=> x = \(\dfrac{-1}{2}\)
\(\Rightarrow x^3-3x^2+3x-1+8-x^3=17-3x^2-6x\\ \Rightarrow9x=10\Rightarrow x=\dfrac{10}{9}\)
a. x^3 -2x^2 +4x+2x^2 -4x+8 -x^3 -2x=15
8-2x=15
-2x=7
x=-7/2
a/ (x + 3)(x - 2) + 3x = 4(x + 3/4)
=> x2 + x - 6 + 3x = 4x + 3
=> x2 = 9 => x = 3 hoặc x = -3
Vậy x = 3 , x = -3
b/ (x2 - 5)(x + 2) + 5x = 2x2 + 17
=> x3 + 2x2 - 5x - 10 + 5x - 2x2 - 17 = 0
=> x3 = 27 => x3 = 33 => x = 3
Vậy x = 3
(x-1)3+(2-x)(4+2x+x2)+3x(x+2)=17
<=>x3-3x2+3x-1+8-x3+3x2+6x=17
<=>9x+7=17
<=>9x=10
<=>x=10/9
a) \(x\left(x-4\right)-\left(x^2-8\right)=0\)
\(\Leftrightarrow x^2-4x-x^2+8=0\)
\(\Leftrightarrow-4\left(x-2\right)=0\)
\(\Leftrightarrow x-2=0\)
\(\Leftrightarrow x=2\)
b) \(\left(3x+2\right)\left(x-1\right)-3\left(x+1\right)\left(x-2\right)=4\)
\(\Leftrightarrow3x^2-3x+2x-2-3\left(x^2-2x+x-2\right)=0\)
\(\Leftrightarrow3x^2-3x+2x-2-3x^2+6x-3x+6=0\)
\(\Leftrightarrow2x=-4\)
\(\Leftrightarrow x=-2\)
c) \(x\left(x+5\right)\left(x-5\right)-\left(x+2\right)\left(x^2-2x+4\right)=17\)
\(\Leftrightarrow x\left(x^2-25\right)-\left(x^3+8\right)=17\)
\(\Leftrightarrow x^3-25x-x^3-8=17\)
\(\Leftrightarrow-25x=25\)
\(\Leftrightarrow x=-1\)
a) x(x - 4 ) - ( x^2 - 8 ) = 0
=>x2-4x-x2+8=0
=>8-4x=0
=>4x=8
=>x=2
b) ( 3x + 2 )( x - 1 ) - 3( x + 1 )( x - 2 ) = 4
=>3x2-x-2-3x2+3x+6=4
=>2x+4=4
=>2x=0
=>x=0
c) x( x + 5 )( x - 5 ) - ( x + 2 )( x^2 - 2x + 4 ) = 17
=>x(x2-25)-(x3+8)=17
=>x3-25x-x3-8=17
=>-25x-8=17
=>-25x=25
=>x=-1
12 doc tieng anh
) (x+3)^2-(x+2)(x-2)=4x+17
⇔ X² + 6X +9 - ( X² -4) = 4X +17
⇔ X² + 6X +9 - X² + 4 - 4X -17 =0
⇔ 2X -4 =0
⇔2X =4
⇔ X = 2
c. 3X² +7X =10
⇔ 3X² +7X -10 =0
⇔ (X -1) .(3X +10 ) =0
⇔ X -1 = 0 hoặc 3X + 10 =0
⇔ X =1 hoặc X =-10/3