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

\(\Rightarrow2x^2-10x-3x-2x^2=26\)

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

b) \(3x\left(1-2x\right)+2\left(3x+7\right)=29\)

\(\Rightarrow3x-6x^2+6x+14=29\)

\(\Rightarrow-6x^2+9x-15=0\)

\(\Rightarrow-6\left(x^2-\dfrac{3}{2}x+\dfrac{9}{16}\right)-\dfrac{93}{8}=0\)

\(\Rightarrow-6\left(x-\dfrac{3}{4}\right)^2-\dfrac{93}{8}=0\)(vô lý)

Vậy \(S=\varnothing\)

a. \(2x^2-10x-3x-2x^2=26\Leftrightarrow-13x=26\Leftrightarrow x=-2\)

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

hay x=-2

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

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

\(\Leftrightarrow-13x=26\)

\(\Leftrightarrow x=-2\)

\(b,\left(3x^2-x+1\right)\left(x-1\right)+x^2\left(4-3x\right)=\frac{5}{2}\)

\(\Leftrightarrow3x^3-3x^2-x^2+x+x-1+4x^2-3x^3=\frac{5}{2}\)

\(\Leftrightarrow2x=\frac{7}{2}\)

\(\Leftrightarrow x=\frac{7}{4}\)

Ta có: \(2x\left(x-5\right)-x\left(3+2x\right)=26\\ < =>2x^2-10x-3x-2x^2=26\\ < =>-13x=26\\ =>x=\dfrac{26}{-13}=-2\)

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

\(\Leftrightarrow\) 2x² - 10x - 3x - 2x² = 26

\(\Rightarrow\) -13x = 26

\(\Rightarrow\) x = -2

Vậy x = -2

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

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

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

\(\Leftrightarrow\) \(x\left(-13\right)=26\)

\(\Leftrightarrow\) \(x=\dfrac{-26}{13}\)

\(\Leftrightarrow\) \(x=-2\)

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

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

=> -13x= 26

=> x= -2

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

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

-10x-3x=26

-x(10+3)=26

-x.13=26

-x=2

=>x=-2

2x.x-2x.5-x.3-x.2x=26

3x-10x-3x-3x=26

-13x=26

x=26:(-13)

x=-2

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

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

\(\Leftrightarrow-13x=26\)

\(\Leftrightarrow x=26:-13\)

\(\Leftrightarrow x=-2\)

Vậy \(x=-2\)

a. \(2x\left(x-5\right)-x\left(2x+3\right)=26\Rightarrow2x^2-10x-2x^2-3x=26\)

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

b. \(\left(3y^2-y+1\right)\left(y-1\right)+y^2\left(4-3y\right)=\frac{5}{2}\)

\(\Rightarrow3y^3-3y^2-y^2+y+y-1+4y^2-3y^3=\frac{5}{2}\)\(\Rightarrow2y=\frac{7}{2}\Rightarrow y=\frac{7}{4}\)

c. \(2x^2+3\left(x+1\right)\left(x-1\right)=5x^2+5x\Rightarrow5x^2-3=5x^2+5x\)

\(\Rightarrow x=-\frac{3}{5}\)

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