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

\(-13x-26=0\)

\(x=-2\)

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

hay x=-2

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

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

<=>2x²-10x-3x-2x²=26

<=>-13x               =26

<=>     x               =-2

Vậy x=-2

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

2x²-10x+3x+2x²=26

4x²-7x=26

x(4x-7)=26

... (tự lập bảng tìm tiếp)

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

<=> 2x² -10x + 3x + 2x² = 26

<=> 13x = 26

<=> 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

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

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

\(\Leftrightarrow-13x=26\)

\(\Leftrightarrow x=-2\)

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

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: \(\left(4-x\right)^2-\left(x-3\right)^2\)

\(=\left(4-x-x+3\right)\left(4-x+x-3\right)\)

\(=\left(-2x+7\right)\cdot1\)

\(=-2x+7\)

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

\(=x^2-25-x^2-10x-25\)

\(=-10x-50\)