Question

[Phạm Ngọc Gia Huy]

GIÚP E VỚI. KO BỎ BƯỚC Ạ! E CẢM ƠN

Answer 1

[Nguyễn Lê Phước Thịnh]

e: Ta có: \(2x\left(x-5\right)-26=x\left(2x+3\right)\)

\(\Leftrightarrow2x^2-10x-26-2x^2-3x=0\)

\(\Leftrightarrow-13x=26\)

hay x=-2

f: Ta có: \(x^2-9=-2x\left(x-3\right)\)

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

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

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

g: Ta có: \(4x^3-9x=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{3}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=8\\x=-2\end{matrix}\right.\)

Answer 2

[Nhan Thanh]

e. \(2x\left(x-5\right)-26=x\left(3+2x\right)\)

\(\Leftrightarrow2x^2-10x-26=3x+2x^2\)

\(\Leftrightarrow2x^2-10x-3x-2x^2=26\)

\(\Leftrightarrow-13x=26\) \(\Leftrightarrow x=-2\)

g. \(4x^3-9x=0\)

\(\Leftrightarrow x\left(4x^2-9\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\4x^2-9=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=0\\x^2=\dfrac{9}{4}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\pm\dfrac{3}{2}\end{matrix}\right.\)

i. \(x^3-5x=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x^2-5=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=0\\x^2=5\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\pm\sqrt{5}\end{matrix}\right.\)

k. \(x^2=10x-25\)

\(\Leftrightarrow x^2-10x+25=0\)

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

\(\Leftrightarrow x-5=0\)

\(\Leftrightarrow x=5\)

f. \(x^2-9=-2x\left(x-3\right)\)

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

\(\Leftrightarrow x+3=-2x\)

\(\Leftrightarrow3x=-3\)

\(\Leftrightarrow x=-1\)

h. \(x^2-8x+2\left(x-8\right)=0\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x-8=0\\x+2=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=8\\x=-2\end{matrix}\right.\)

j. \(x\left(x-5\right)-x+5=0\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\x-1=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=1\end{matrix}\right.\)

l. \(2x^3-72x=0\)

\(\Leftrightarrow2x\left(x^2-36\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x^2-36=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=0\\x^2=36\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\pm6\end{matrix}\right.\)