\(\left(x-3,5\right)^2+\left(y-\dfrac{1}{10}\right)^4\le0\)

Vì: \(\left(x-3,5\right)^2\ge0,\left(y-\dfrac{1}{10}\right)^4\ge0\)

\(\Rightarrow\left\{{}\begin{matrix}\left(x-3,5\right)^2=0\\\left(y-\dfrac{1}{10}\right)^4=0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x-3,5=0\\y-\dfrac{1}{10}=0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=3,5\\y=\dfrac{1}{10}\end{matrix}\right.\)

Ta có: \(\left(x-3.5\right)^2\ge0\forall x\)

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

Do đó: \(\left(x-\dfrac{7}{2}\right)^2+\left(y-\dfrac{1}{10}\right)^4\ge0\forall x,y\)

Dấu '=' xảy ra khi \(\left(x,y\right)=\left(\dfrac{7}{2};\dfrac{1}{10}\right)\)

\(\left(x-3,5\right)^2\ge0\forall x\)

\(\left(y-\frac{1}{10}\right)^4\ge0\forall y\)

Do đó: \(\left(x-3,5\right)^2+\left(y-\frac{1}{10}\right)^4\ge0\forall x,y\)

mà \(\left(x-3,5\right)^2+\left(y-\frac{1}{10}\right)^4\le0\)

nên \(\begin{cases}x-3,5=0\\ y-\frac{1}{10}=0\end{cases}\Rightarrow\begin{cases}x=3,5\\ y=0,1\end{cases}\)

a)\(25,5+y-12,5=4.7\)

⇔\(13+y=28\)

⇔\(y=15\)

b)\(76,22-y-25,7=30+5,52\)

⇔\(50,52-y=35,52\)

⇔\(y=15\)

c)\(4,5-y+1,2=3,5\)

⇔\(5,7-y=3,5\)

⇔\(y=2,2\)

\(a,\Rightarrow10+y=28\\ \Rightarrow y=18\\ b,\Rightarrow50,52-y=35,52\\ \Rightarrow y=15\\ c,\Rightarrow5,7-y=3,5\\ \Rightarrow y=2,2\)

y=tất cả các số

(3.36 *3.5) :(8.36 *3.5) -y =1

                               1 -y=1

                                 y=1-1

                                 y=0

các bạn nha!cảm ơn

\(=>y=30:\left(6,5+3,5\right)=30:10=3\)

\(Y x ( 6.5+3.5) = 30\)

\(Y x 10 = 30\)

\(Y = 30:10\)

\(Y = 3\)

y=3,28 nhé

24-(Y:4+Yx6)=3,5

24-(Yx0,25+Yx6)=3,5

24-Yx(0,25+6)=3,5

24-Yx6,25=3.5

Yx6,25=24-3,5

Yx6,25=20,5

Y=20,5:6,25

Y=3,28

`#hungg`

`24-(y:4+y xx 6)=3,5`

`y xx 0,25 + y xx 6 =24-3,5`

`y xx(0,25+6)=20,5`

`y xx 6,25=20,5`

`y=20,5:6,25`

`y=3,28`

đb