2x . 3 + 2x . 2 = 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
`1,[(-3).3+5]-26=-2x-3`
`=>(-9+5)-26=-2x-3`
`=>-4-26=-2x-3`
`=>-30=-2x-3`
`=>-2x=-27`
`=>x=27/2`
Vậy `x=27/2`
`2)-[(-35)-3]=2x-2`
`=>-(-38)=2x-2`
`=>38=2x-2`
`=>2x=40`
`=>x=20`
Vậy `x=20`
1) \(\left[\left(-3\right)\cdot3+5\right]-26=-2x-3\\ \Rightarrow-9+5-26=-2x-3\\ \Rightarrow-2x=-9+5-26+3\\ \Rightarrow-2x=-27\\ \Rightarrow x=\dfrac{27}{2}\)
Vậy \(x=\dfrac{27}{2}\)
2) \(-\left[\left(-35\right)-3\right]=2x-2\\ \Rightarrow2x-2=-\left(-38\right)\\ \Rightarrow2x=38+2\\ \Rightarrow2x=40\\ \Rightarrow x=20\)
Vậy \(x=20\)
\(\left(2x-1\right)^3+3\left(2x+3\right)^2=26\)
\(\Leftrightarrow\left(2x\right)^3-3\left(2x\right)^2+3.2x-1+3\left(4x^2+12x+9\right)=26\)
\(\Leftrightarrow8x^3-12x^2+6x-1+12x^2+36x+27=26\)
\(\Leftrightarrow8x^3+42x+26=26\)
\(\Leftrightarrow8x^3+42x=0\)
\(\Leftrightarrow4x\left(2x^2+13\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}4x=0\\2x^2+13=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\2x^2=-13\left(L\right)\end{matrix}\right.\)
Vậy \(x=0\)
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=\(\dfrac{2}{3}\)
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
1) (2x + 1)(3x – 2) = (5x – 8)(2x + 1)
⇔ (2x + 1)(3x – 2) – (5x – 8)(2x + 1) = 0
⇔ (2x + 1).[(3x – 2) – (5x – 8)] = 0
⇔ (2x + 1).(3x – 2 – 5x + 8) = 0
⇔ (2x + 1)(6 – 2x) = 0
⇔\(\left[{}\begin{matrix}2x+1=0\\6-2x=0\end{matrix}\right.\) ⇔ \(\left[{}\begin{matrix}x=\dfrac{-1}{2}\\x=3\end{matrix}\right.\)
Vậy.....
2) 4x2 -1 = (2x + 1)(3x - 5)
⇔ (2x-1)(2x+1)-(2x+1)(3x-5)=0
⇔ (2x+1)(2x-1-3x+5)=0
⇔ (2x+1)(4-x)=0
⇔ \(\left[{}\begin{matrix}2x+1=0\\4-x=0\end{matrix}\right.\) ⇔ \(\left[{}\begin{matrix}x=\dfrac{-1}{2}\\x=4\end{matrix}\right.\)
Vậy...
3)
(x + 1)2 = 4(x2 – 2x + 1)
⇔ (x + 1)2 - 4(x2 – 2x + 1) = 0
⇔ x2 + 2x +1- 4x2 + 8x – 4 = 0
⇔ - 3x2 + 10x – 3 = 0
⇔ (- 3x2 + 9x) + (x – 3) = 0
⇔ -3x (x – 3)+ ( x- 3) = 0
⇔ ( x- 3) ( - 3x + 1) = 0
⇔\(\left[{}\begin{matrix}x-3=0\\-3x+1=0\end{matrix}\right.\) ⇔\(\left[{}\begin{matrix}x=3\\x=\dfrac{1}{3}\end{matrix}\right.\)
Vậy......
đề bài cứ sai sai thế nào ấy
2x.(3+2)=26
2x=26/5 =>x ko có giá trị nào