Ta có: 2x(x – 5) – x(3 + 2x) = 26

      ⇔ 2 x 2  – 10x – 3x – 2 x 2  =26

      ⇔ - 13x = 26

      ⇔ x = - 2

Ta có: 2x(x – 5) – x(3 + 2x) = 26

      ⇔ 2x2 – 10x – 3x – 2x2 =26

      ⇔ - 13x = 26

      ⇔ x = - 2

\(2x\left(x-5\right)-x\left(3+2x\right)=26\)

\(\Leftrightarrow2x^2-10x-3x-2x^2=26\)

\(\Leftrightarrow-13x=26\)

\(\Leftrightarrow x=-2\)

Bài làm:

Ta có: \(2x\left(x-5\right)-x\left(3+2x\right)=26\)

\(\Leftrightarrow2x^2-10x-3x-2x^2-26=0\)

\(\Leftrightarrow-13x=26\)

\(\Rightarrow x=-2\)

2x( x - 5 ) - x( 3 + 2x ) = 26

<=> 2x² - 10x - 3x - 2x² = 26

<=> -13x = 26

<=> x = -2

3x^2-x = 0

x(3x-1) = 0

TH1: x = 0

TH2: 3x-1 = 0

x= 1/3

b) Ta có: 2x(x – 5) – x(3 + 2x) = 26

      ⇔ 2 – 10x – 3x – 2 =26

      ⇔ - 13x = 26

      ⇔ x = - 2

2x (x-5) - x(3+2x) = 26

2x^2 - 10x - 3x - 2x^2 = 26

-13x = 26

x = -2

a) 2x(x - 5) - x(2x + 3) = 26

⇒ 2x² - 10x - 2x² - 3x = 26

⇒ -13x = 26

⇒ x = -2

chúc bạn học tốt 

\(2x^2-10x-3x-2x^2=26\)

\(2x^2-13x-2x^2=26\)

\(-13x=26\)

\(x=-2\)

\(2x\left(x-5\right)-x\left(3+2x\right)=26\)

\(\Leftrightarrow\left(2x^2-10x\right)-\left(3x+2x^2\right)=26\)

\(\Leftrightarrow-13x=26\)

\(\Leftrightarrow x=-2\)

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 

a) 2x(x - 5) - x(3 + 2x) = 26

2x² - 10x - 3x - 2x² = 26

-13x = 26

=> x = 2

b) 3x³ - 48x = x(3x² - 48) = 0

=> \(\orbr{\begin{cases}x=0\\3x^2-48=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\3x^2=48\Rightarrow x^2=16\end{cases}}}\Rightarrow\orbr{\begin{cases}x=0\\x=4;-4\end{cases}}\)

Vậy x = {0 ; 4 ; -4}

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) \(\Rightarrow\left(x-2012\right)\left(5x-1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=2012\\x=\dfrac{1}{5}\end{matrix}\right.\)

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