Giải:

a) \(x+\left(-\dfrac{31}{12}\right)^2=\left(\dfrac{49}{12}\right)^2-x=y\)

\(\Leftrightarrow x+\left(-\dfrac{31}{12}\right)^2=\left(\dfrac{49}{12}\right)^2-x\)

\(\Leftrightarrow x+\left(-\dfrac{31}{12}\right)^2-\left(\dfrac{49}{12}\right)^2+x=0\)

\(\Leftrightarrow2x+\left(-\dfrac{31}{12}\right)^2-\left(\dfrac{49}{12}\right)^2=0\)

\(\Leftrightarrow2x+\dfrac{\left(-31\right)^2}{12^2}-\dfrac{49^2}{12^2}=0\)

\(\Leftrightarrow2x+\dfrac{\left(-31\right)^2-49^2}{144}=0\)

\(\Leftrightarrow2x+\dfrac{961-2401}{144}=0\)

\(\Leftrightarrow2x+\dfrac{-1440}{144}=0\)

\(\Leftrightarrow2x+\left(-10\right)=0\)

\(\Leftrightarrow2x=10\)

\(\Leftrightarrow x=5\)

Mà \(x+\left(-\dfrac{31}{12}\right)^2=y^2\)

\(\Leftrightarrow5+\dfrac{961}{144}=y^2\)

\(\Leftrightarrow y^2=\dfrac{1681}{144}\)

\(\Leftrightarrow y=\pm\dfrac{41}{12}\)

Vậy ...

b) \(\left(\dfrac{1}{2}x-5\right)^{20}+\left(y^2-\dfrac{1}{4}\right)^{10}\le0\)

Vì \(\left(\dfrac{1}{2}x-5\right)^{20}\ge0;\forall x\)

và \(\left(y^2-\dfrac{1}{4}\right)^{10}\ge0;\forall y\)

\(\Rightarrow\left(\dfrac{1}{2}x-5\right)^{20}+\left(y^2-\dfrac{1}{4}\right)^{10}\le0\)

\(\Leftrightarrow\left(\dfrac{1}{2}x-5\right)^{20}+\left(y^2-\dfrac{1}{4}\right)^{10}=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{2}x-5=0\\y^2-\dfrac{1}{4}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{2}x=5\\y^2=\dfrac{1}{4}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=10\\y=\pm\dfrac{1}{2}\end{matrix}\right.\)

Vậy ...

Chúc bạn học tốt!

X+(-31/12)^2 = (49/12)^2 -x=y

(-31/12)^2 - (49/12)^2 = -x-x = y

961/144 - 2410/144 = -2x

-10=-2x

10=2x

10:2=x

5=x

X+961/144=y^2

5+961/144=y^2

1681/144=y^2

=>y=41/144

Dấu phân số mình ký hiệu là / đó nha

Chứng minh gì bạn?

\(x+\left(\frac{-31}{12}\right)^2\left(\frac{49}{12}\right)^2-x=y\)

\(x+\frac{961}{144}.\frac{2401}{144}-x=y\)

\(x+\frac{2307361}{20736}-x=y\)

\(y=\frac{2307361}{20736}\)

         Thay vào \(x+\frac{961}{144}.\frac{2401}{144}-x=y\) ta được

\(x+\frac{2307361}{20736}-x=y\)

\(x-x+\frac{2307361}{20736}=\frac{2307361}{20736}\)

Vậy x thuộc N;\(y=\frac{2307361}{20736}\)

\(\Leftrightarrow\left(x-x\right)+\frac{961}{144}\cdot\frac{2401}{144}=y\Leftrightarrow y=\frac{2307361}{20736}\)

\(\Rightarrow x\in R\)

( hình như đề sai :v )

đề saI

​

Vì 

 \(x+\left(-\frac{31}{12}\right)^2=\left(\frac{49}{12}\right)^2-x\Rightarrow2x=\left(\frac{49}{12}\right)^2-\left(-\frac{31}{12}\right)^2=10\)

=> 2x = 10 => x = 5

Thay x = 5 vào ta có :

     \(5+\left(-\frac{31}{12}\right)^2=y^2\Leftrightarrow\frac{1681}{144}=y^2=\left(\frac{41}{12}\right)^2=\left(-\frac{41}{12}\right)^2\)

=> y = 41/12 hoặc y = -41/12 

Tìm x,y

x*(x-y)=3/10 và y*(x-y)=–3/20

\(x+\left(\frac{-31}{12}\right)^2=\left(\frac{49}{12}\right)^2-x=y^2\)

Xét \(x+\left(\frac{-31}{12}\right)^2=\left(\frac{49}{12}\right)^2-x\)

\(\Rightarrow2x=\left(\frac{49}{12}\right)^2-\left(\frac{-31}{12}\right)^2=\frac{2401}{144}+\frac{961}{144}\)

\(\Rightarrow2x=\frac{3362}{144}\)

\(\Rightarrow x=\frac{3362}{144}.\frac{1}{2}=\frac{1681}{144}\)

Ta lai xét :

\(x+\left(\frac{-31}{12}\right)^2=y^2\)

\(\Rightarrow\frac{1681}{144}+\frac{-961}{144}=y^2\)

\(\Rightarrow\frac{720}{144}=y^2\)

\(\Rightarrow y^2=5\)

\(\Rightarrow y=2,236067977\)