1 , 00 = 1 

The price of the pencil is  : 

\(\left(1,10-1\right):2=0,05\) ( $ ) 

...

My answer is : C ) 3 

C . Moles are the slowest and largest 

Change : \(1\dfrac{1}{2}=\dfrac{3}{2}\)

Call that number is a 

We have : 

\(a+\dfrac{3}{2}=\dfrac{3}{2}a\)

\(\Rightarrow\dfrac{3}{2}=\dfrac{1}{2}a\)

\(\Rightarrow a=\dfrac{3}{2}:\dfrac{1}{2}\)

\(\Rightarrow a=3\)

So that number is 3 

Now : Bob is 30 years old and Bill is 20 years old 

\(\left(3x-1\right)\left(x^2+2\right)=\left(3x-1\right)\left(7x-10\right)\)

\(\Leftrightarrow x^2+2=7x-10\)

\(\Leftrightarrow x^2+2-7x+10=0\)

\(\Leftrightarrow x^2-7x+12=0\)

\(\Leftrightarrow x^2-2.x.\dfrac{7}{2}+\left(\dfrac{7}{2}\right)^2-\dfrac{1}{4}=0\)

\(\Leftrightarrow\left(x-\dfrac{7}{2}\right)^2=\dfrac{1}{4}\)

\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{7}{2}=\dfrac{1}{2}\\x-\dfrac{7}{2}=-\dfrac{1}{2}\end{matrix}\right.\)

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

So  \(\left[{}\begin{matrix}x=4\\x=3\end{matrix}\right.\)

We have : 

\(\dfrac{x-91}{37}+\dfrac{x-86}{42}+\dfrac{x-78}{50}+\dfrac{x-49}{79}=4\)

\(\Rightarrow\left(\dfrac{x-91}{37}-1\right)+\left(\dfrac{x-86}{42}-1\right)+\left(\dfrac{x-78}{50}-1\right)+\left(\dfrac{x-49}{79}-1\right)=0\)

\(\Rightarrow\dfrac{x-128}{37}+\dfrac{x-128}{42}+\dfrac{x-128}{50}+\dfrac{x-128}{79}=0\)

\(\Rightarrow\left(x-128\right)\left(\dfrac{1}{37}+\dfrac{1}{42}+\dfrac{1}{50}+\dfrac{1}{79}\right)=0\)

But \(\dfrac{1}{37}+\dfrac{1}{42}+\dfrac{1}{50}+\dfrac{1}{79}\ne0\)

\(\Rightarrow x-128=0\)

\(\Rightarrow x=128\)

So the value of x is : \(128\)

\(\dfrac{2x}{5}-0,5x=\dfrac{1-2x}{4}+0,25\)

\(\Leftrightarrow0,4x-0,5x=\dfrac{1-2x}{4}+\dfrac{1}{4}\)

\(\Leftrightarrow-0,1x=\dfrac{2-2x}{4}\)

\(\Leftrightarrow-0,1x.4=2-2x\)

\(\Leftrightarrow-0,4x=2-2x\)

\(\Leftrightarrow-0,4x+2x=2\)

\(\Leftrightarrow1,6x=2\)

\(\Leftrightarrow x=1,25\)

So the value of x is 1 , 25 

\(\dfrac{x}{3}-\dfrac{2x+1}{2}=\dfrac{x}{6}-x\)

\(\Leftrightarrow\dfrac{2x}{6}-\dfrac{3\left(2x+1\right)}{6}=\dfrac{x}{6}-x\)

\(\Leftrightarrow x=\dfrac{x}{6}-\dfrac{2x}{6}+\dfrac{3\left(2x+1\right)}{6}\)

\(\Leftrightarrow x=\dfrac{x-2x+3\left(2x+1\right)}{6}\)

\(\Leftrightarrow x=\dfrac{-x+6x+3}{6}\)

\(\Leftrightarrow x=\dfrac{5x+3}{6}\)

\(\Leftrightarrow6x=5x+3\)

\(\Leftrightarrow6x-5x=3\)

\(\Leftrightarrow x=3\)

So the value of x is 3

We have : 

\(\left\{{}\begin{matrix}x+2y+3z=6\\2x+3y+z=8\\3x+y+2z=10\end{matrix}\right.\)

\(\Rightarrow x+2y+3z+2x+3y+z+3x+y+2z=6+8+10\)

\(\Rightarrow6x+6y+6z=24\)

\(\Rightarrow6\left(x+y+z\right)=24\)

\(\Rightarrow x+y+z=4\)

So, the value of x + y + z equals : \(4\)

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