a^2x(x-5)-x(3+2x)=26
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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
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
a, 3x.(x+2) - 5.(x+2) = 0
=> * (x+2) = 0 => x = 2
Và * (3x -5) = 0 => x = \(\frac{5}{3}\)
Vậy x = 2 ; x = \(\frac{5}{3}\)
b, 2x2 + 3x +10x -2x2 =26
13x = 26
=> x = 26 : 13
=> x = 2
Vậy: x=2
a) ( x + 2 ) ( x + 3 ) - ( x - 2 ) ( x + 5 ) = 0
x2 + 3x + 2x + 6 - x2 + 5x - 2x - 10 = 0
8x - 4 = 0
8x = 0 + 4
8x = 4
x = 4 : 8
x = 1/2
b) 2x ( x -5 ) - x ( 3 + 2x ) = 26
2x2 - 10x - 3x - 2x2 = 26
-13x = 26
x = 26 : (-13)
x = -2
2x2−10x−3x−2x2−26=02x2−10x−3x−2x2−26=0
13x−26=0−13x−26=0
x=−2
`2x^2 + 10x + 3x - 2x^2 = 26`
`<=> 13x = 26`
`<=> x = 2`
Ta có: 2x(x – 5) – x(3 + 2x) = 26
⇔ 2 x 2 – 10x – 3x – 2 x 2 =26
⇔ - 13x = 26
⇔ x = - 2
a)\(5\left(2x-1\right)-4\left(8-3x\right)=7\)
\(\Leftrightarrow10x-5+12x-32=7\)
\(\Leftrightarrow22x-37=7\)
\(\Leftrightarrow22x=44\Rightarrow x=2\)
b)\(5x\left(x-5\right)-x\left(2x+3\right)=26\)
\(\Leftrightarrow5x^2-25x-2x^2-3x=26\)
\(\Leftrightarrow3x^2-28x-26=0\)
\(\Leftrightarrow3\left(x-\dfrac{14}{3}\right)^2-\dfrac{274}{3}=0\)
\(\Rightarrow x=\dfrac{14}{3}\pm\dfrac{\sqrt{274}}{3}\)