Question

Tìm x:

Answer 1

a,\(< =>x^2-16x+64-x^2-4x=0< =>-20x+64=0\)

\(< =>x=3,2\)

b,\(< =>\left(x+5\right)\left(x-5\right)+\left(x-5\right)\left(x+3\right)=0\)

\(< =>\left(x-5\right)\left(x+5+x+3\right)=0< =>\left(x-5\right)\left(2x+8\right)=0\)

\(< =>\left[{}\begin{matrix}x=5\\x=-4\end{matrix}\right.\)

c,\(< =>x^2-4x+10x+5=0< =>x^2+6x+5=0\)

\(< =>x^2+x+5x+5=0< =>\left(x+1\right)\left(x+5\right)=0\)

\(< =>\)\(\left[{}\begin{matrix}x=-5\\x=-1\end{matrix}\right.\)

Answer 2

a, \(\left(x-8\right)^2-x.\left(x-4\right)=0\)

\(=>x^2-2.x.8+8^2-x^2-4x=0\)

\(=>x^2-16x+64-x^2-4x=0\)

\(=>-20x+64=0\)

\(=>-20x=-64\)

\(=>x=\dfrac{16}{5}\)

Học tốt

Answer 3

a) Ta có: \(\left(x-8\right)^2-x\left(x+4\right)=0\)

\(\Leftrightarrow x^2-16x+64-x^2-4x=0\)

\(\Leftrightarrow-20x=-64\)

hay \(x=\dfrac{16}{5}\)

b) Ta có: \(x^2-25=\left(x+3\right)\left(5-x\right)\)

\(\Leftrightarrow\left(x-5\right)\left(x+5\right)+\left(x+3\right)\left(x-5\right)=0\)

\(\Leftrightarrow\left(x-5\right)\left(2x+8\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-4\end{matrix}\right.\)

Answer 4

c) Ta có: \(x\left(x-4\right)+10x+5=0\)

\(\Leftrightarrow x^2+6x+5=0\)

\(\Leftrightarrow\left(x+1\right)\left(x+5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-5\end{matrix}\right.\)