mai nah

1) \(\left(x+1\right)\left(x+2\right)-3x\left(x-4\right)=x^2+3x+2-3x^2+12x=-2x^2+15x+2\)

2) \(\left(3x-4\right)\left(x-2\right)=3x\left(x-9\right)\)

\(\Leftrightarrow3x^2-10x+8=3x^2-27x\)

\(\Leftrightarrow17x=-8\Leftrightarrow x=-\dfrac{8}{17}\)

3) \(-3\left(x-4\right)\left(x-2\right)-x^2\left(-3x+18\right)+24x-25\)

\(=-3x^3+6x^2+12x^2-24x+3x^3-18x^2+24x-25=-25\)

a) = (x+3).(x-3)^2-(x-3)(x+3)^2

=(x^2-9)(x-3)-(x^2-9)(x+3)

=(x^2-9)(x-3-x-3)

=-6(x^2-9)

các câu còn lại tương tự

\(a,\left(x+3\right)\left(x^2-3x+9\right)-\left(x-3\right)\left(x^2+3x+9\right)\)

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

\(=x^3+3-x^3+3\)

\(=6\)

\(b,\left(x-5\right)\left(x^2+5x+25\right)-\left(x+5\right)\left(x^2-5x+25\right)\)

\(=x^3-5^3-x^3-5^3\)

\(=-125-125\)

\(=-250\)

9) Ta có: \(\dfrac{2x+5}{x+3}+1=\dfrac{4}{x^2+2x-3}-\dfrac{3x-1}{1-x}\)

\(\Leftrightarrow\left(2x+5\right)\left(x-1\right)+x^2+2x-3=4+\left(3x-1\right)\left(x+3\right)\)

\(\Leftrightarrow2x^2-2x+5x-5+x^2+2x-3-4-3x^2-10x+x+3=0\)

\(\Leftrightarrow-4x=9\)

hay \(x=-\dfrac{9}{4}\)

10) Ta có: \(\dfrac{x-1}{x+3}-\dfrac{x}{x-3}=\dfrac{7x-3}{9-x^2}\)

\(\Leftrightarrow\dfrac{\left(x-1\right)\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}-\dfrac{x\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{3-7x}{\left(x-3\right)\left(x+3\right)}\)
Suy ra: \(x^2-4x+3-x^2-3x-3+7x=0\)

\(\Leftrightarrow0x=0\)(luôn đúng)

Vậy: S={x|\(x\notin\left\{3;-3\right\}\)}

11) Ta có: \(\dfrac{5+9x}{x^2-16}=\dfrac{2x-1}{x+4}+\dfrac{3x-1}{x-4}\)

\(\Leftrightarrow\dfrac{\left(2x-1\right)\left(x-4\right)}{\left(x-4\right)\left(x+4\right)}+\dfrac{\left(3x-1\right)\left(x+4\right)}{\left(x-4\right)\left(x+4\right)}=\dfrac{9x+5}{\left(x-4\right)\left(x+5\right)}\)

Suy ra: \(2x^2-9x+4+3x^2+12x-x-4-9x-5=0\)

\(\Leftrightarrow5x^2-7x=0\)

\(\Leftrightarrow x\left(5x-7\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{7}{5}\end{matrix}\right.\)

12) Ta có: \(\dfrac{2x}{2x-1}+\dfrac{x}{2x+1}=1+\dfrac{4}{\left(2x-1\right)\left(2x+1\right)}\)

\(\Leftrightarrow\dfrac{2x\left(2x+1\right)}{\left(2x-1\right)\left(2x+1\right)}+\dfrac{x\left(2x-1\right)}{\left(2x+1\right)\left(2x-1\right)}=\dfrac{4x^2-1+4}{\left(2x-1\right)\left(2x+1\right)}\)

Suy ra: \(4x^2+2x+2x^2-x-4x^2-3=0\)

\(\Leftrightarrow2x^2+x-3=0\)

\(\Leftrightarrow2x^2+3x-2x-3=0\)

\(\Leftrightarrow\left(2x+3\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=1\end{matrix}\right.\)

13) Ta có: \(\dfrac{x+2}{x-2}-\dfrac{1}{x}=\dfrac{2}{x^2-2x}\)

\(\Leftrightarrow\dfrac{x\left(x+2\right)}{x\left(x-2\right)}-\dfrac{x-2}{x\left(x-2\right)}=\dfrac{2}{x\left(x-2\right)}\)

Suy ra: \(x^2+2x-x+2-2=0\)

\(\Leftrightarrow x^2+x=0\)

\(\Leftrightarrow x\left(x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(loại\right)\\x=-1\left(nhận\right)\end{matrix}\right.\)

a) \(\dfrac{1}{4}+\dfrac{\dfrac{3}{4}}{3x}=-5\)

\(\dfrac{\dfrac{3}{4}}{3x}=-5-\dfrac{1}{4}\)

\(\dfrac{\dfrac{3}{4}}{3x}=-\dfrac{21}{4}\)

\(3x=\dfrac{\dfrac{3}{4}}{-\dfrac{21}{4}}\)

\(3x=-\dfrac{1}{7}\)

\(x=\dfrac{-\dfrac{1}{7}}{3}\)

\(x=-\dfrac{1}{21}\)

a) 1/4 + 3/4 : 3x = -5

3/4 : 3x = -5 - 1/4

3/4 : 3x = -21/4

3x = 3/4: -21/4

3x = 1/-7

x = 1/-7 : 3

x = 1/-21

b) x - 25%x = 0,5

1.x - 1/4 . x = 1/2

(1 - 1/4) . x = 1/2

3/4 . x = 1/2

x = 1/2 : 3/4

x = 1/2 . 4/3

x = 2/3

b) \(x-25\%x=0,5\)

\(x-\dfrac{25}{100}x=\dfrac{1}{2}\)

\(x-\dfrac{1}{4}x=\dfrac{1}{2}\)

\(x\left(1-\dfrac{1}{4}\right)=\dfrac{1}{2}\)

\(\dfrac{3}{4}x=\dfrac{1}{2}\)

\(x=\dfrac{\dfrac{1}{2}}{\dfrac{3}{4}}\)

\(x=\dfrac{2}{3}\)

Ta có: \(C=-3x\left(x-4\right)\left(x-2\right)+x\left(3x-18\right)-25\)

\(=-3x\left(x^2-6x+8\right)+3x^2-18x-25\)

\(=-3x^3+18x^2-24x+3x^2-18x-25\)

\(=-3x^3+21x^2-42x-25\)