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

⇒ 2x² - 10x - 2x² - 3x = 26

⇒ -13x = 26

⇒ x = -2

chúc bạn học tốt 

bài 11

a) \(x^2-xy+x\\ =x\left(x-y+1\right)\)

b)

\(x^2-2xy-4+y^2\\ =\left(x^2-2xy+y^2\right)-4\\ =\left(x-y\right)^2-4\\ =\left(x-y-2\right)\left(x-y+2\right)\)

c)

\(x^3-x^2-16x+16\\ =x^2\left(x-1\right)-16\left(x-1\right)\\ =\left(x-1\right)\left(x-4\right)\left(x+4\right)\)

bài 12

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

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

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

b)

\(2\left(x+5\right)-x^2-5x=0\\ 2\left(x+5\right)-x\left(x+5\right)=0\\ \left(x+5\right)\left(2-x\right)=0\\ \left[{}\begin{matrix}x+5=0\\2-x=0\end{matrix}\right.\left[{}\begin{matrix}x=-5\\x=2\end{matrix}\right.\)

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

$\iff$ $x\left[2\left(x-5\right)-\left(3+2x\right)\right]=26$

$\iff$ $x\left(2x-10-3-2x\right)=26$

$\iff$ $x\left(-13\right)=26$

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

a) \(\frac{-x}{2}+\frac{2x}{3}+x+\frac{1}{4}+2x+\frac{1}{6}=\frac{3}{8}.\)

\(\frac{-x}{2}+\frac{2x}{3}+3x+\frac{5}{12}=\frac{3}{8}\)

\(x.\left(-\frac{1}{2}+\frac{2}{3}+3\right)+\frac{5}{12}=\frac{3}{8}\)

\(x\cdot\frac{19}{6}=-\frac{1}{24}\)

x = -1/76

b) \(\frac{3}{2x+1}+\frac{10}{4x+2}-\frac{6}{6x+3}=\frac{12}{26}\)

\(\frac{3}{2x+1}+\frac{2.5}{2.\left(2x+1\right)}-\frac{2.3}{3.\left(2x+1\right)}=\frac{6}{13}\)

\(\frac{3}{2x+1}+\frac{5}{2x+1}-\frac{2}{2x+1}=\frac{6}{13}\)

\(\frac{3+5-2}{2x+1}=\frac{6}{13}\)

\(\frac{6}{2x+1}=\frac{6}{13}\)

=> 2x + 1 = 13

2x = 12

x = 6

Giải như sau.

(1)+(2)⇔x2−2x+1+√x2−2x+5=y2+√y2+4⇔(x2−2x+5)+√x2−2x+5=y2+4+√y2+4⇔√y2+4=√x2−2x+5⇒x=3y(1)+(2)⇔x2−2x+1+x2−2x+5=y2+y2+4⇔(x2−2x+5)+x2−2x+5=y2+4+y2+4⇔y2+4=x2−2x+5⇒x=3y

⇔√y2+4=√x2−2x+5⇔y2+4=x2−2x+5, chỗ này do hàm số f(x)=t2+tf(x)=t2+t đồng biến ∀t≥0∀t≥0
Công việc còn lại là của bạn ! 

\(\left(x+6\right)\left(2x+1\right)=0\)

<=>  \(\orbr{\begin{cases}x+6=0\\2x+1=0\end{cases}}\)

<=>  \(\orbr{\begin{cases}x=-6\\x=-\frac{1}{2}\end{cases}}\)

Vậy....

hk tốt

^^

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

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

\(\Leftrightarrow-13x=26\)

\(\Leftrightarrow x=-2\)

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

B3  a) x=4        b) x=-7         c) x=5          d) x=4

B2  a) -3+ -2+ -1+0+1+2+3+4=4

      b) -6+ -5+ -4+ -3+ -2+ -1+0+1+2+3+4=-11

 c) -18+-17+-16+-15+-14+-13+-12+-11+-10+-9+-8+-7+-6+-5+-4+3+-2+-1+0+1+2+3+4+5+6+7+8+9+10+11+12+13+14+15+16+17+18+19=19