Tìm x, biết: 2x(x – 5) – x(3 + 2x) = 26.
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Ta có: 2x(x – 5) – x(3 + 2x) = 26
⇔ 2 x 2 – 10x – 3x – 2 x 2 =26
⇔ - 13x = 26
⇔ x = - 2
![](https://rs.olm.vn/images/avt/0.png?1311)
\(2x\left(x-5\right)-x\left(3+2x\right)=26\)
\(\Leftrightarrow2x^2-10x-3x-2x^2=26\)
\(\Leftrightarrow-13x=26\)
\(\Leftrightarrow x=-2\)
Bài làm:
Ta có: \(2x\left(x-5\right)-x\left(3+2x\right)=26\)
\(\Leftrightarrow2x^2-10x-3x-2x^2-26=0\)
\(\Leftrightarrow-13x=26\)
\(\Rightarrow x=-2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
2x(x-5) - x(3+2x) = 26
<=> 2x^2 - 10x - 3x - 2x^2 = 26
<=> -13x=26
<=> x = -2
2x(x-5) - x (3+2x) = 26
=> 2x2 - 10x - 3x - 2x2 = 26
=> (2x2 - 2x2) + (-10x - 3x) = 26
=> -13x = 26
=> x = 26 : (-13)
=> x = -2
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
b) Ta có: 2x(x – 5) – x(3 + 2x) = 26
⇔ 2 – 10x – 3x – 2 =26
⇔ - 13x = 26
⇔ x = - 2
![](https://rs.olm.vn/images/avt/0.png?1311)
a) 2x(x - 5) - x(2x + 3) = 26
⇒ 2x² - 10x - 2x² - 3x = 26
⇒ -13x = 26
⇒ x = -2
chúc bạn học tốt
![](https://rs.olm.vn/images/avt/0.png?1311)
\(2x\left(x-5\right)-x\left(3+2x\right)=26\)
\(\Leftrightarrow\left(2x^2-10x\right)-\left(3x+2x^2\right)=26\)
\(\Leftrightarrow-13x=26\)
\(\Leftrightarrow x=-2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a, 2x(x-5) - x ( 3 + 2x ) = 26
=> 2x^2 - 10x - 3x - 2x ^ 2 = 26
=> - 13 x = 26
=> x = -2
a, \(2x\left(x-5\right)-x\left(3+2x\right)=26\)
\(\Leftrightarrow2x^2-10x-3x-2x^2=26\)
\(\Leftrightarrow-13x=26\)
\(\Leftrightarrow x=-2\)
Vậy x = -2
b, \(3x^3-48x=0\)
\(\Leftrightarrow3x\left(x^2-16\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}3x=0\\x^2-16=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=4;x=-4\end{cases}}\)
Vậy x = 0 hoặc x = 4 hoặc x = -4
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\Rightarrow x\left(x+3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=-3\end{matrix}\right.\)
b) \(\Rightarrow x\left(x^2-4\right)=0\Rightarrow x\left(x-2\right)\left(x+2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-2\end{matrix}\right.\)
c) \(\Rightarrow\left(x-1\right)\left(5x-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{5}\end{matrix}\right.\)
d) \(\Rightarrow2\left(x+5\right)-x\left(x+5\right)=0\Rightarrow\left(x+5\right)\left(2-x\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=-5\\x=2\end{matrix}\right.\)
e) \(\Rightarrow2x^2-10x-3x-2x^2=26\)
\(\Rightarrow-13x=26\Rightarrow x=-2\)
f) \(\Rightarrow\left(x-2012\right)\left(5x-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=2012\\x=\dfrac{1}{5}\end{matrix}\right.\)
Ta có: 2x(x – 5) – x(3 + 2x) = 26
⇔ 2x2 – 10x – 3x – 2x2 =26
⇔ - 13x = 26
⇔ x = - 2