Question

[phuong]

có bạn nào biết làm ko vậy ?? giúp mình với

Answer 1

[ひまわり(In my personal page there....]

b,\(4x^2-20x=0\)

⇔\(4x\left(x-5\right)=0\)

⇔\(\left\{{}\begin{matrix}4x=0\\x-5=0\end{matrix}\right.\)

⇔\(\left\{{}\begin{matrix}x=0\\x=5\end{matrix}\right.\)

c,\(\left(3x-2\right)\left(4x+5\right)=0\)

⇔\(\left\{{}\begin{matrix}3x-2=0\\4x+5=0\end{matrix}\right.\)

⇔\(\left\{{}\begin{matrix}x=\dfrac{2}{3}\\x=-1.25\end{matrix}\right.\)

e,\(\left(x^2+1\right)\left(x-2\right)=0\)

⇔\(\left\{{}\begin{matrix}x^2+1=0\\x-2=0\end{matrix}\right.\)

⇔\(\left\{{}\begin{matrix}x^2=-1\left(loai\right)\\x=2\left(nhan\right)\end{matrix}\right.\)

⇔\(x=2\)

Answer 2

[Nguyễn Lê Phước Thịnh]

a) Ta có: \(\dfrac{x-3}{2011}+\dfrac{x-2}{2012}=\dfrac{x-2012}{2}+\dfrac{x-2011}{3}\)

\(\Leftrightarrow\dfrac{x-3}{2011}+\dfrac{x-2}{2012}-\dfrac{x-2012}{2}-\dfrac{x-2011}{3}=0\)

\(\Leftrightarrow\dfrac{x-3}{2011}-1+\dfrac{x-2}{2012}-1-\dfrac{x-2012}{2}+1-\dfrac{x-2011}{3}+1=0\)

\(\Leftrightarrow\dfrac{x-2014}{2011}+\dfrac{x-2014}{2012}-\dfrac{x-2014}{2}-\dfrac{x-2014}{3}=0\)

\(\Leftrightarrow\left(x-2014\right)\left(\dfrac{1}{2011}+\dfrac{1}{2012}-\dfrac{1}{2}-\dfrac{1}{3}\right)=0\)

mà \(\dfrac{1}{2011}+\dfrac{1}{2012}-\dfrac{1}{2}-\dfrac{1}{3}\ne0\)

nên x-2014=0

hay x=2014

Vậy: S={2014}

b) Ta có: \(4x^2-20x=0\)

\(\Leftrightarrow4x\left(x-5\right)=0\)

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

Vậy: S={0;5}

c) Ta có: \(\left(3x-2\right)\left(4x+5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}3x-2=0\\4x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=2\\4x=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-\dfrac{5}{4}\end{matrix}\right.\)

Vậy: \(S=\left\{\dfrac{2}{3};-\dfrac{5}{4}\right\}\)

d) Ta có: \(\dfrac{x-5}{75}+\dfrac{x-2}{78}+\dfrac{x-6}{74}+\dfrac{x-68}{12}=4\)

\(\Leftrightarrow\dfrac{x-5}{75}-1+\dfrac{x-2}{78}-1+\dfrac{x-6}{74}-1+\dfrac{x-68}{12}-1=0\)

\(\Leftrightarrow\dfrac{x-80}{75}+\dfrac{x-80}{78}+\dfrac{x-80}{74}+\dfrac{x-80}{12}=0\)

\(\Leftrightarrow\left(x-80\right)\left(\dfrac{1}{75}+\dfrac{1}{78}+\dfrac{1}{74}+\dfrac{1}{12}\right)=0\)

mà \(\dfrac{1}{75}+\dfrac{1}{78}+\dfrac{1}{74}+\dfrac{1}{12}>0\)

nên x-80=0

hay x=80

Vậy: S={80}

e) Ta có: \(\left(x^2+1\right)\left(x-2\right)=0\)

mà \(x^2+1>0\forall x\)

nên x-2=0

hay x=2

Vậy: S={2}