các bn làm giúp mình nhà cảm ơn các bn nhiều

a)30 > y-16 > 19+9

Vậy 30 > y-16 > 28

Số vừa nhỏ hơn 30, vừa lớn hơn 28 là số 29

Suy ra: y-16= 29

            y     = 29+16

            y     = 45

Vậy y = 45

b)24+y-9= 45

   24+y   = 45+9

   24+y   = 54

         y   = 54-24

         y   =30

Vậy y= 30

bn thanh sang làm đúng r nha cảm ơn bn nhiều

 58/7 - (31/9 + 30/7)                  
= 58/7 - 31/9 - 30/7
= (58/7 - 30/7) - 31/9
= 28/7- 31/9
= 4 - 31/9 = 36/9 - 31/9
= 5/9

478/19÷(-5/4)-668/19÷(-5/4)
=(478/19-668/19):(-5/4)
=-10 . (-4/5)
=8

\(\dfrac{9}{\sqrt{x-19}}+\dfrac{16}{\sqrt{y-5}}+\dfrac{25}{\sqrt{z-91}}=24-\sqrt{x-19}-\sqrt{y-5}-\sqrt{z-91}\\ \Leftrightarrow\left(\dfrac{9}{\sqrt{x-19}}+\sqrt{x-19}\right)+\left(\dfrac{16}{\sqrt{y-5}}+\sqrt{y-5}\right)+\left(\dfrac{25}{\sqrt{z-91}}+\sqrt{z-91}\right)=24\)

Áp dụng BDT: Cô-si:

\(\Rightarrow\left(\dfrac{9}{\sqrt{x-19}}+\sqrt{x-19}\right)+\left(\dfrac{16}{\sqrt{y-5}}+\sqrt{y-5}\right)+\left(\dfrac{25}{\sqrt{z-91}}+\sqrt{z-91}\right)\ge2\sqrt{\dfrac{9}{\sqrt{x-19}}\cdot\sqrt{x-19}}+2\sqrt{\dfrac{16}{\sqrt{y-5}}\cdot\sqrt{y-5}}+2\sqrt{\dfrac{25}{\sqrt{z-91}}\cdot\sqrt{z-91}}\\ =2\cdot3+2\cdot4+2\cdot5=24\)Dấu "=" xảy ra khi:\(\left\{{}\begin{matrix}\dfrac{9}{\sqrt{x-19}}=\sqrt{x-19}\\\dfrac{16}{\sqrt{y-5}}=\sqrt{y-5}\\\dfrac{25}{\sqrt{z-91}}=\sqrt{z-91}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-19=9\\y-5=16\\z-91=25\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=28\\y=21\\z=116\end{matrix}\right.\)

Vậy các số \(\left\{x;y;z\right\}=\left\{28;21;116\right\}\)

2Y^3-1=15

=> 2.(y^3)=16

y^3=8

y=2

x+16/9=2+30/16

x+16/9=31/8

x=151/72

a) \(15y\left(4y-9\right)-3\left(4y-9\right)=0\)

\(\Leftrightarrow\left(15y-3\right)\left(4y-9\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}15y-3=0\\4y-9=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}y=\frac{1}{5}\\y=\frac{9}{4}\end{matrix}\right.\)

Vây \(y\in\left\{\frac{1}{5};\frac{9}{4}\right\}\)

b) \(8\left(25z+7\right)-27z\left(25z+7\right)=0\)

\(\Leftrightarrow\left(8-27z\right)\left(25z+7\right)=0\) \(\Leftrightarrow\left[{}\begin{matrix}z=\frac{8}{27}\\z=-\frac{7}{25}\end{matrix}\right.\)

Vậy \(z\in\left\{\frac{8}{27};-\frac{7}{25}\right\}\)

c) \(13y\left(y-8\right)-2y+16=0\)

\(\Leftrightarrow13y\left(y-8\right)-2\left(y-8\right)=0\)

\(\Leftrightarrow\left(13y-2\right)\left(y-8\right)=0\) \(\Leftrightarrow\left[{}\begin{matrix}y=\frac{2}{13}\\y=8\end{matrix}\right.\)

Vậy \(y\in\left\{\frac{2}{13};8\right\}\)

d) \(-10y\left(y+2\right)-y-2=0\)

\(\Leftrightarrow\left(-10y-1\right)\left(y+2\right)=0\) \(\Leftrightarrow\left[{}\begin{matrix}y=-2\\y=-\frac{1}{10}\end{matrix}\right.\)

Vậy \(y\in\left\{-2;-\frac{1}{10}\right\}\)

e) \(x\left(x+19\right)^2-\left(x+19\right)^2=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+19\right)^2=0\) \(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-19\end{matrix}\right.\)

Vậy \(x\in\left\{1;-19\right\}\)

\(=\dfrac{19}{8}.\dfrac{16}{9}+\dfrac{19}{8}.\dfrac{2}{9}-\dfrac{19}{8}.1=\dfrac{19}{8}.\left(\dfrac{16}{9}+\dfrac{2}{9}-1\right)\)

\(=\dfrac{19}{8}.1=\dfrac{19}{8}\)

\(\dfrac{19}{8}\times\dfrac{16}{9}+\dfrac{19}{8}\times\dfrac{2}{9}-\dfrac{19}{8}\)

\(=\dfrac{19}{8}\times\left(\dfrac{16}{9}+\dfrac{2}{9}-1\right)\)

\(=\dfrac{19}{8}\times1\)

\(=\dfrac{19}{8}\)

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