(x-2).(2x+3)=26
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\(4\left(x+1\right)\left(-x+2\right)+\left(2x-1\right)\left(2x+3\right)=-11\)
\(\text{⇔}-4x^2+4x+8+4x^2+4x-3=-11\)
\(\text{⇔}8x+5=-11\)
\(\text{⇔}8x=-16\)
\(\text{⇔}x=-2\)
Vậy: \(x=-2\)
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\(\left(2x+4\right)\left(3x+1\right)\left(x-2\right)-\left(-3x^2+1\right)\left(-2x+\dfrac{2}{3}\right)=-\dfrac{26}{3}\)
\(\text{⇔}6x^3+2x^2-24x-8-6x^3-2x^2-2x+\dfrac{2}{3}=-\dfrac{26}{3}\)
\(\text{⇔}-26x-\dfrac{22}{3}=-\dfrac{26}{3}\)
\(\text{⇔}-26x=-\dfrac{4}{3}\)
\(\text{⇔}x=\dfrac{2}{39}\)
x(2x - 3) - 2(3 - 2x) = 0
x(2x - 3) + 2(2x - 3) = 0
(2x - 3)(x + 2) = 0
\(\left[\begin{array}{nghiempt}2x-3=0\\x+2=0\end{array}\right.\)
\(\left[\begin{array}{nghiempt}x=\frac{3}{2}\\x=-2\end{array}\right.\)
2x(x - 5) - x(3 + 2x) = 26
2x2 - 10x - 3x - 2x2 = 26
- 13x = 26
x = - 26 : 13
x = - 2
a) (x - 1)3 + 3(x + 1)2 = (x2 - 2x + 4)(x + 2)
x3 - 3x2 + 3x - 1 + 3(x2 + 2x+ 1) = x3 + 8
\(\Rightarrow\)x3 - 3x2 + 3x - 1 + 3x2 + 6x - x3 - 8 = 0
\(\Rightarrow\) 9x - 9 = 0
\(\Rightarrow\) 9x = 9
\(\Rightarrow\) x = 1
b) (2x - 1)(x + 3) - x(3 + 2x) = 26
2x2 + 6x - x - 3 - 3x - 2x2 = 26
2x - 3 = 26
\(\Rightarrow\) 2x = 29
\(\Rightarrow\) x = 14.5
a) ( x - 1)3 + 3(x+1)2 = (x2 - 2x + 4 )( x+ 2)
=>x3-3x2+3x-1+3(x2+2x+1)=x3+8
=>x3-3x2+3x-1+3x2+6x+3-x3-8=0
=>(x3-x3)+(-3x2+3x2)+(3x+6x)+(-1+3-8)=0
=>9x-6=0
=>9x=6
=>x=\(\dfrac{2}{3}\)
\(\Leftrightarrow\left(x+2\right)\left(x^2-2x+2^2\right)-x\left(x^2-3^2\right)=26\\ \Leftrightarrow x^3+2^3-x^3+9x=26\\ \Leftrightarrow9x+8=26\\ \Rightarrow x=\dfrac{26-8}{9}=2\)
Lời giải:
$(x+2)(x^2-2x+4)-x(x+3)(x-3)=26$
$\Leftrightarrow x^3+8-x(x^2-9)=26$
$\Leftrightarrow 9x+8=26$
$\Leftrightarrow 9x=18$
$\Leftrightarrow x=2$
1) Ta có: \(\left(x+2\right)^2+\left(x-3\right)^2\)
\(=x^2+4x+4+x^2-6x+9\)
\(=2x^2-2x+13\)
2) Ta có: \(\left(4-x\right)^2-\left(x-3\right)^2\)
\(=\left(4-x-x+3\right)\left(4-x+x-3\right)\)
\(=-2x+7\)
3) Ta có: \(\left(x-5\right)\left(x+5\right)-\left(x+5\right)^2\)
\(=x^2-25-x^2-10x-25\)
=-10x-50
4) Ta có: \(\left(x-3\right)^2-\left(x-4\right)\left(x+4\right)\)
\(=x^2-6x+9-x^2+16\)
=-6x+25
5) Ta có: \(\left(y^2-6y+9\right)-\left(y-3\right)^2\)
\(=y^2-6y+9-y^2+6y-9\)
=0
6) Ta có: \(\left(2x+3\right)^2-\left(2x-3\right)\left(2x+3\right)\)
\(=4x^2+12x+9-4x^2+9\)
=12x+18
\(x^3+8-x\left(x^2-9\right)=26\)
\(\Leftrightarrow x^3+8-x^3+9x=26\)
\(\Leftrightarrow9x+8=26\Leftrightarrow9x=18\)
\(\Leftrightarrow x=2\)
Đúng thì TICK nka !
2x(x-5)-x(3+2x)= 26
2x^2-10x-3x-2x^2=26
-13x=26 x=-2
Vậy x=-2.
b) giống