Question

A) \(\frac{\left(x-2\right)\left(x+10\right)}{3}-\frac{\left(x+4\right)\left(x+10\right)}{12}=\frac{\left(x-2\right)\left(x+4\right)}{4}\)

B)\(\frac{\left(x+2\right)^2}{8}-2\left(2x+1\right)=25+\frac{\left(x-2\right)^2}{8}\)

Giải các phương trình trên : 

ĐS: a) x=8          b) x= -9

Answer 1

A) Ta có: \(\frac{\left(x-2\right)\left(x+10\right)}{3}-\frac{\left(x+4\right)\left(x+10\right)}{12}=\frac{\left(x-2\right)\left(x+4\right)}{4}\)

\(\Leftrightarrow4\left(x-2\right)\left(x+10\right)-\left(x+4\right)\left(x+10\right)=3\left(x-2\right)\left(x+4\right)\)

\(\Leftrightarrow4\left(x^2+8x-20\right)-\left(x^2+14x+40\right)=3\left(x^2+2x-8\right)\)

\(\Leftrightarrow4x^2+32x-80-x^2-14x-40=3x^2+6x-24\)

\(\Leftrightarrow4x^2-x^2-3x^2+32x-14x-6x=-24+80+40\)

\(\Leftrightarrow12x=96\)

\(\Leftrightarrow x=8\)

Vậy x = 8

B) Ta có: \(\frac{\left(x+2\right)^2}{8}-2\left(2x+1\right)=25+\frac{\left(x-2\right)^2}{8}\)

\(\Leftrightarrow\left(x+2\right)^2-2.8\left(2x+1\right)=25.8+\left(x-2\right)^2\)

\(\Leftrightarrow x^2+4x+4-32x-16=200+x^2-4x+4\)

\(\Leftrightarrow x^2-x^2+4x-32x+4x=200+4-4+16\)

\(\Leftrightarrow-24x=216\)

\(\Leftrightarrow x=-9\)

Vậy x = -9

Answer 2

999+2819=