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![](https://rs.olm.vn/images/avt/0.png?1311)
( 3x - 2 )( 9x2 + 6x + 4 ) - ( 2x - 5 )( 2x + 5 ) = ( 3x - 1 )3 - ( 2x + 3 )2 + 9x( 3x - 1 )
⇔ 27x3 - 8 - ( 4x2 - 25 ) = 27x3 - 27x2 + 9x - 1 - ( 4x2 + 12x + 9 ) + 27x2 - 9x
⇔ 27x3 - 8 - 4x2 + 25 = 27x3 - 1 - 4x2 - 12x - 9
⇔ 27x3 - 4x2 + 17 - 27x3 + 4x2 + 12x + 10 = 0
⇔ 12x + 27 = 0
⇔ 12x = -27
⇔ x = -27/12 = -9/4
![](https://rs.olm.vn/images/avt/0.png?1311)
a: Ta có: \(\left(x+2\right)\left(x+3\right)-\left(x-2\right)\left(x+5\right)=6\)
\(\Leftrightarrow x^2+5x+6-x^2-3x+10=6\)
\(\Leftrightarrow2x=-10\)
hay x=-5
b: Ta có: \(\left(3x+2\right)\left(2x+9\right)-\left(x+2\right)\left(6x+1\right)=\left(x+1\right)-\left(x-6\right)\)
\(\Leftrightarrow6x^2+27x+4x+18-6x^2-x-12x-2=x+1-x+6\)
\(\Leftrightarrow18x+16=7\)
hay \(x=-\dfrac{1}{2}\)
c: Ta có: \(3\left(2x-1\right)\left(3x-1\right)-\left(2x-3\right)\left(9x-1\right)=0\)
\(\Leftrightarrow3\left(6x^2-2x-3x+1\right)-\left(18x^2-2x-27x+3\right)=0\)
\(\Leftrightarrow18x^2-15x+3-18x^2+27x-3=0\)
hay x=0
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(x-1\right)^2-\left(x+1\right)^2=2\left(x+3\right)\)
\(\Leftrightarrow\left(x-1+x+1\right)\left(x-1-x-1\right)=2\left(x+3\right)\)
\(\Leftrightarrow2x\left(-2\right)=2\left(x+3\right)\)
\(\Leftrightarrow-4x=2x+6\)
\(\Leftrightarrow-6x=6\)
\(\Leftrightarrow x=-1\)
2) \(\left(2x-1\right)^2-\left(2x+1\right)^2=4\left(x-3\right)\)
\(\Leftrightarrow\left(2x-1+2x+1\right)\left(2x-1-2x-1\right)-4\left(x-3\right)=0\)
\(\Leftrightarrow4x\left(-2\right)-4x+12=0\)
\(\Leftrightarrow-12x=-12\)
\(\Leftrightarrow x=1\)
3)\(\left(2x+3\right)^2-\left(2x+3\right)\left(2x-4\right)+\left(x-2\right)^2=0\)
\(\Leftrightarrow\left(2x+3\right)\left(2x+3-2x+4\right)+\left(x^2-4x+4\right)=0\)
\(\Leftrightarrow7\left(2x+3\right)+x^2-4x+4=0\)
\(\Leftrightarrow x^2+10x+25=0\)
\(\Leftrightarrow\left(x+5\right)^2=0\)
\(\Leftrightarrow x=-5\)
4) \(8x^3-\left(x+1\right)^3=3x-3\)
\(\Leftrightarrow8x^3-\left(x^3+3x+3x^2+1\right)-3x+3=0\)
\(\Leftrightarrow7x^3-3x^2-6x+2=0\)
\(\Leftrightarrow\left(x-1\right)\left(7x^2+4x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\frac{-2+3\sqrt{2}}{7}\\x=\frac{-2-3\sqrt{2}}{7}\end{matrix}\right.\)
5)\(\left(3x-2\right)\left(9x^2+6x+4\right)-\left(3x-1\right)\left(9x^2-3x+1\right)=x-4\)
\(\Leftrightarrow\left(3x\right)^3-2^3-\left(\left(3x\right)^3-1^3\right)=x-4\)
\(\Leftrightarrow27x^3-8-\left(27x^3-1\right)=x-4\)
\(\Leftrightarrow-7=x-4\)
\(\Leftrightarrow x=-3\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Ta có: \(6x\left(x-5\right)+3x\left(7-2x\right)=18\)
\(\Leftrightarrow6x^2-30x+21x-6x^2=18\)
\(\Leftrightarrow-9x=18\)
hay x=-2
Vậy: S={-2}
b) Ta có: \(2x\left(3x+1\right)+\left(4-2x\right)\cdot3x=7\)
\(\Leftrightarrow6x^2+2x+12x-6x^2=7\)
\(\Leftrightarrow14x=7\)
hay \(x=\dfrac{1}{2}\)
Vậy: \(S=\left\{\dfrac{1}{2}\right\}\)
c) Ta có: \(0.5x\left(0.4-4x\right)+\left(2x+5\right)\cdot x=-6.5\)
\(\Leftrightarrow0.2x-2x^2+2x^2+5x=-6.5\)
\(\Leftrightarrow5.2x=-6.5\)
hay \(x=-\dfrac{5}{4}\)
Vậy: \(S=\left\{-\dfrac{5}{4}\right\}\)
d) Ta có: \(\left(x+3\right)\left(x+2\right)-\left(x-2\right)\left(x+5\right)=6\)
\(\Leftrightarrow x^2+5x+6-\left(x^2+3x-10\right)=6\)
\(\Leftrightarrow x^2+5x+6-x^2-3x+10=6\)
\(\Leftrightarrow2x+16=6\)
\(\Leftrightarrow2x=-10\)
hay x=-5
Vậy: S={-5}
e) Ta có: \(3\left(2x-1\right)\left(3x-1\right)-\left(2x-3\right)\left(9x-1\right)=0\)
\(\Leftrightarrow3\left(6x^2-5x+1\right)-\left(18x^2-29x+3\right)=0\)
\(\Leftrightarrow18x^2-15x+3-18x^2+29x-3=0\)
\(\Leftrightarrow14x=0\)
hay x=0
Vậy: S={0}
![](https://rs.olm.vn/images/avt/0.png?1311)
TÌM X
a) (3x+2)(2x+9)-(6x+1)(x+2)=7
=> 6x2 + 31x +18 - 6x2 - 13x - 2 - 7 = 0
=> 18x + 9 = 0 => 9(2x + 1) = 0 => 2x + 1 = 0 => x = -1/2
b) (x-2)(x+5)-(x+3)(x+2)=-6
=> x2 + 3x - 10 - x2 - 5x -6 + 6 = 0 => -2x -10 = 0 => -2(x + 5) = 0
=> x + 5 = 0 => x = -5
c) 3(2x-1)(3x-1)-(2x-3)(9x-1)=0
=> 18x2 - 15x +3 - 18x2 + 29x -3 = 0 => 14x = 0 => x = 0
a) \(\left(3x+2\right)\left(2x+9\right)-\left(6x+1\right)\left(x+2\right)=7\\\Rightarrow 6x^2+31x+18-6x^2-16x-2-7=0\\ \Rightarrow18x+9=0\Rightarrow9\left(2x+1\right)=0\Rightarrow2x+1=0\Rightarrow x=-\frac{1}{2}\)
b) \(\left(x-2\right)\left(x+5\right)-\left(x+3\right)\left(x+2\right)=-6\\ \Rightarrow x^2+3x-10-x^2-5x-6+6=0\\ \Rightarrow-2x-10=0\\ \Rightarrow-2\left(x+5\right)=0\\ \Rightarrow x+5=0\\ \Rightarrow x=-5\)
c) \(3\left(2x-1\right)\left(3x-1\right)-\left(2x-3\right)\left(9x-1\right)=0\\ \Rightarrow18x^2-15x+3-18x^2+29x-3=0\\ \Rightarrow14x=0\\ \Rightarrow x=0\)
![](https://rs.olm.vn/images/avt/0.png?1311)
A = ( 3x )3 + 23 - 27x3 + 6 = 27x3 + 8 - 27x3 + 6 = 14 ( đpcm )
B = x3 + 3x2 + 3x + 1 - ( x3 - 1 ) - 3x2 - 3x = x3 + 1 - x3 + 1 = 2 ( đpcm )
C = 6( x + 2 )( x2 - 2x )( x2 - 2x + 4 ) - 6x3 - 2 ( bạn xem lại đề bài nhé ._. )
D = 2[ ( 3x )3 + 13 ] - 54x3 = 2( 27x3 + 1 ) - 54x3 = 54x3 + 2 - 54x3 = 2 ( đpcm )
\(\left(2x+3\right)^2+2\left(2x+3\right)\left(x-2\right)+\left(2-x\right)^2=4\)
\(\left(2x+3\right)^2+2\left(2x+3\right)\left(x-2\right)+\left(x-2\right)^2=4\)
\(\left(2x+3+x-2\right)^2=\left(\pm2\right)^2\)
\(\left(3x+1\right)^2=\left(\pm2\right)^2\)
\(\left[\begin{array}{nghiempt}3x+1=2\\3x+1=-2\end{array}\right.\)
\(\left[\begin{array}{nghiempt}3x=2-1\\3x=-2-1\end{array}\right.\)
\(\left[\begin{array}{nghiempt}3x=1\\3x=-3\end{array}\right.\)
\(\left[\begin{array}{nghiempt}x=\frac{1}{3}\\x=-1\end{array}\right.\)
***
\(\left(x+3\right)\left(3-x\right)=5\)
\(3^2-x^2=5\)
\(x^2=9-5\)
\(x^2=4\)
\(x^2=\left(\pm2\right)^2\)
\(x=\pm2\)
***
\(\left(3x+1\right)\left(9x^2-3x+1\right)=2\)
\(27x^3+3=2\)
\(27x^3=2-3\)
\(\left(3x\right)^3=-1\)
\(3x=-1\)
\(x=-\frac{1}{3}\)
Đâu có y đâu bạn