Ta có: \(4\ge2\left(x^2+y^2\right)\ge\left(x+y\right)^2\)

    \(\Rightarrow x+y\le2\)

Ta có: \(P=\sqrt{x\left(14x+10y\right)}+\sqrt{y\left(14y+10x\right)}\)

              \(=\sqrt{\dfrac{24x\left(14x+10y\right)}{24}}+\sqrt{\dfrac{24y\left(14y+10x\right)}{24}}\le\dfrac{\dfrac{24x+14x+10y}{2}}{\sqrt{24}}+\dfrac{\dfrac{24y+14y+10x}{2}}{\sqrt{24}}\)

\(\Leftrightarrow P\le\dfrac{24\left(x+y\right)}{2\sqrt{6}}\le\dfrac{24.2}{2\sqrt{6}}=4\sqrt{6}\)

Dấu "=" xảy ra ⇔ x = y = 1

Đây nè bạn.

a) Ta có: \(M=x^2-2xy+y^2-10x+10y\)

\(=\left(x-y\right)^2-10\left(x-y\right)\)

\(=9^2-10\cdot9=-9\)

\(-8x=-10y\Leftrightarrow\frac{-8x}{40}=\frac{-10y}{40}\Leftrightarrow\frac{-x}{5}=\frac{-y}{4}\Leftrightarrow\frac{-5x}{25}=\frac{-2y}{8}\)

Áp dụng tính chất của dãy tỉ số bằng nhau

\(\Rightarrow\frac{-5x}{25}=\frac{-2y}{8}=\frac{-5x-2y}{25+8}=\frac{396}{33}=12\)

\(\Leftrightarrow\frac{-x}{5}=12\Leftrightarrow-x=60\Leftrightarrow x=-60\)

\(\Leftrightarrow\frac{-y}{4}=12\Leftrightarrow-y=48\Leftrightarrow y=-48\)

Vậy \(x=-60;y=-48\)

15x = -10y = 6z

<=> \(\frac{15x}{30}=\frac{-10y}{30}=\frac{6z}{30}\) 

<=> \(\frac{x}{2}=\frac{y}{-3}=\frac{z}{5}\)

Đặt \(\frac{x}{2}=\frac{y}{-3}=\frac{z}{5}=k\Rightarrow\hept{\begin{cases}x=2k\\y=-3k\\z=5k\end{cases}}\)

Ta có: xyz = -30000

=> 2k.(-3k).5k = -30000

=> -30k³ = -30000

=> k³ = 1000

=> k = 10

=> x = 20, y = -30, z = 50

Vì  15x = -10y = 6z => \(\frac{15x}{30}=\frac{-10y}{30}=\frac{6z}{30}\)  => \(\frac{x}{2}=\frac{-y}{3}=\frac{z}{5}\)

Đặt : \(\frac{x}{2}=\frac{-y}{3}=\frac{z}{5}=k\), ta có : x = 2k ; y = (-3).k ; x = 5k

=> x.y.z = 2   .k. ( -3 ). k.5.k = -30.k³ = -30000

=> k³ = 1000 => k = 10 => x = 10. 2 = 20

                                     => y = 10. ( - 3 ) = -30

                                      => z = 10.5 = 50

a)6x=10y=>x/10=y/6=>x/50=y/30

10y=15z=>y/15=z/10=>y/30=z/20

x+y+z/50+30+10=90/90=1

x=50;y=30;z=10

b)đề 2 có sai ko z pn

\(15x=10y=6z\)

\(\Rightarrow\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{5}\)

Áp dụng t/c dtsbn:

\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{5}=\dfrac{x+y+z}{2+3+5}=\dfrac{20}{10}=2\)

\(\Rightarrow\left\{{}\begin{matrix}x=2.2=4\\y=2.3=6\\z=2.5=10\end{matrix}\right.\)

Ta có : 15x = 6z

=> x = 6/15z

-10y = 6z

=>  y= -3/5z

=> xyz = -30000

<=> (6/15z) . (-3/5z) . z = -30000

<=> z^3 .( -6/25) = -30000

<=> z^3       = 125000

<=> z   = 50

=> y = -30

=> x = 20

20 k mk nhe !