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`2x^2 + 10x + 3x - 2x^2 = 26`
`<=> 13x = 26`
`<=> 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: \(2x\left(x-5\right)-x\left(3+2x\right)=26\)
\(\Leftrightarrow2x^2-10x-3x-2x^2-26=0\)
\(\Leftrightarrow-13x=26\)
\(\Rightarrow x=-2\)
\(2x^2-10x-3x-2x^2-26=0\)
\(-13x-26=0\)
\(x=-2\)
2x2−10x−3x−2x2−26=02x2−10x−3x−2x2−26=0
13x−26=0−13x−26=0
x=−2