Question

1) 5(x-3) (x-7)-(5x+1) (x-2)= -8

2) x(x+1) (x+2)-(x+4) (3x-5)= 84-5x

3) (9x²-5) (x+3)-3x²(3x+9)=(x-5) (x+4)-x(x-11)

4) (x²-4x+16) (x+4)-x(x+1) (x+2)+3x²=0

5) (8x+2) (1-3x)+(6x-1) (4x-10)=-50

6) (x²+2x+4) (2-x)+x(x-3) (x+4)-x²+24=0

7) (\(\dfrac{x}{2}\)+3) (5-6x)+(12x-2) (\(\dfrac{x}{4}\)+3)=0

Answer 1

1) Ta có: \(5\left(x-3\right)\left(x-7\right)-\left(5x+1\right)\left(x-2\right)=-8\)

\(\Leftrightarrow5\left(x^2-10x+21\right)-\left(5x^2-10x+x-2\right)=-8\)

\(\Leftrightarrow5x^2-50x+105-5x^2+9x+2+8=0\)

\(\Leftrightarrow-41x=-115\)

hay \(x=\dfrac{115}{41}\)

2) Ta có: \(x\left(x+1\right)\left(x+2\right)-\left(x+4\right)\left(3x-5\right)=84-5x\)

\(\Leftrightarrow x\left(x^2+3x+2\right)-\left(3x^2+7x-20\right)=84-5x\)

\(\Leftrightarrow x^3+3x^2+2x-3x^2-7x+20-84+5x=0\)

\(\Leftrightarrow x^3=64\)

hay x=4

3) Ta có: \(\left(9x^2-5\right)\left(x+3\right)-3x^2\left(3x+9\right)=\left(x-5\right)\left(x+4\right)-x\left(x-11\right)\)

\(\Leftrightarrow9x^3+27x^2-5x-15-9x^3-27x^2=x^2-x-20-x^2+11x\)

\(\Leftrightarrow-5x-15=10x-20\)

\(\Leftrightarrow-5x-10x=-20+15\)

\(\Leftrightarrow x=\dfrac{-5}{-15}=\dfrac{1}{3}\)