Question

[Lương Công Nghĩa]

giai pt giup minh voi a

Answer 1

[Đỗ Tuệ Lâm]

d) . đkxđ : x khác 0 , x khác -5 

\(\Leftrightarrow\left(2x+5\right)\left(x+5\right)=2x.x\)

<=> \(2x^2+10x+5x+25=2x^2\)

<=> \(2x^2+15x+25-2x^2=0\)

<=> \(15x+25=0\)

<=> \(15x=-25\Rightarrow x=\dfrac{-25}{15}=-\dfrac{5}{3}\left(nhận\right)\)

Vậy.....

e).

\(\left|x+2\right|=3x+5\Leftrightarrow\left\{{}\begin{matrix}x+2=3x+5\left(khi\right)x+2\ge0\Leftrightarrow x\ge-2\left(1\right)\\-\left(x+2\right)=3x+5\left(khi\right)x+2< 0\Leftrightarrow x< -2\left(2\right)\end{matrix}\right.\)

Giải pt ( 1) khi  \(x\ge-2\) :

\(x+2=3x+5\\ \Leftrightarrow x+2-3x-5=0\\ \Leftrightarrow-2x-3=0\)

<=> \(-2x=3\)

\(\Rightarrow x=\dfrac{-3}{2}\left(nhận\right)\)

Giải pt (2) khi \(x< -2\) :

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

\(\Leftrightarrow-x-2-3x-5=0\)

<=> \(-4x-7=0\)

<=> \(-4x=7\)

<=> \(x=\dfrac{-7}{4}\left(loại\right)\)

Vậy \(S=\left\{-\dfrac{3}{2}\right\}\)

f).

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

\(\Leftrightarrow\left\{{}\begin{matrix}2x+1=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{1}{2}\\x=3\end{matrix}\right.\)

Vậy .. 

Answer 2

[YangSu]

\(\dfrac{2x+5}{2x}=\dfrac{x}{x+5}\)

\(\Leftrightarrow\dfrac{\left(2x+5\right)\left(x+5\right)-2x^2}{2x\left(x+5\right)}=0\)

\(\Leftrightarrow2x^2+10x+5x+25-2x^2=0\)

\(\Leftrightarrow15x=-25\)

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

\(\left|x+2\right|=3x+5\)

\(\Leftrightarrow\left[{}\begin{matrix}x+2=3x+5\\x+2=-3x-5\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x+2-3x-5=0\\x+2+3x+5=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}-2x-3=0\\4x+7=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=-\dfrac{7}{4}\end{matrix}\right.\)

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

\(\Leftrightarrow\left[{}\begin{matrix}2x+1=0\\x-3=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=3\end{matrix}\right.\)