Question

Answer 1

Bài 3: 

a) \(6x^2-\left(2x-3\right)\left(3x+2\right)-1=0\)

\(\Leftrightarrow6x^2-\left(6x^2+4x-9x-6\right)-1=0\)

\(\Leftrightarrow6x^2-6x^2+5x+6-1=0\)

\(\Leftrightarrow5x+5=0\)

\(\Leftrightarrow5x=-5\)

\(\Leftrightarrow x=-1\)

b) \(x\left(x+6\right)-7x-42=0\)

\(\Leftrightarrow x\left(x+6\right)-7\left(x+6\right)=0\)

\(\Leftrightarrow\left(x+6\right)\left(x-7\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+6=0\\x-7=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-6\\x=7\end{matrix}\right.\)

Answer 2

Bài 4:

Ta có: 

\(x+y=4\)

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

\(\Rightarrow x^2+y^2+2xy=16\)

\(\Rightarrow10+2xy=16\)

\(\Rightarrow2xy=16-10\)

\(\Rightarrow2xy=6\)

\(\Rightarrow xy=\dfrac{6}{2}\)

\(\Rightarrow xy=3\)

Từ đó ta tính được M:
\(M=x^3+y^3\)

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

\(M=\left(x+y\right)\left[\left(x^2+y^2\right)-xy\right]\)

\(M=4\cdot\left(10-3\right)\)

\(M=4\cdot7\)

\(M=28\)