2xy=3yz => x=3/2z

2xy=4zx=> y=2z

xyz=3

thế vào ta có:3/2z.2z.z=3=> z = 1

x = 3/2

y= 2

x,y,z=0

Làm rõ ra bn nhé!

a) 

\(2^{x+3}+5\cdot2^{x+2}=224\)

\(2^x\cdot2^3+5\cdot2^x\cdot2^2=224\)

\(2^x\cdot8+2^x\cdot20=224\)

\(2^x\cdot\left(20+8\right)=224\)

\(2^x\cdot28=224\)

\(2^x=8\)

\(x=3\)

2^x+3+5*2^x+2 = 224

VT=7*2^x+2

pt trở thành 7*2^x+2=224

<=>7*2^x+2=2⁵*7

<=>2^x+2=2⁵

<=>x+2=5

<=>x=3

x=3 nhé

\(x+y+z=1\Rightarrow z=1-x-y\)Thay vào A ta được:

\(A=2xy+3y\left(1-x-y\right)+4\left(1-x-y\right)x\)

\(\Leftrightarrow2xy+3y-3xy-3y^2+4x-4x^2-4xy-A=0\)

\(\Leftrightarrow3y-3y^2+4x-4x^2-5xy-A=0\)

\(\Leftrightarrow-4x^2-\left(5y-4\right)x-3y^2+3y-A=0\)

\(\Leftrightarrow4x^2+\left(5y-4\right)x+3y^2-3y+A=0\)

\(\Delta=\left(5y-4\right)^2-16\left(3y^2-3y+A\right)\)

Để pt có nghiệm \(\Leftrightarrow\Delta\ge0\)

\(\Leftrightarrow\left(5y-4\right)^2-16\left(3y^2-3y+A\right)\ge0\)

\(\Leftrightarrow25y^2-40y+16-48y^2+48y-16A\ge0\)

\(\Leftrightarrow-23y^2+8y+16\ge16A\)

\(\Leftrightarrow16A\le-23\left(y^2-\frac{8}{23}y-\frac{12}{23}\right)=-23\left(y-\frac{4}{23}\right)^2+\frac{384}{23}\le\frac{384}{23}\)

\(\Rightarrow A\le\frac{24}{23}\)

Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}2xy+3y\left(1-x-y\right)+4\left(1-x-y\right)x=\frac{24}{23}\\\left(y-\frac{4}{23}\right)^2=0\\x+y+z=1\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x=\frac{9}{23}\\y=\frac{4}{23}\\z=\frac{10}{23}\end{cases}}\)

Vậy Max A = \(\frac{24}{23}\)\(\Leftrightarrow\hept{\begin{cases}x=\frac{9}{23}\\y=\frac{4}{23}\\z=\frac{10}{23}\end{cases}}\)

\(2=4\sqrt{xy}+2\sqrt{xz}\le2x+2y+x+z=3x+2y+z\)

Ta có:

\(VT=\dfrac{3yz}{x}+\dfrac{4zx}{y}+\dfrac{5xy}{z}=2\left(\dfrac{xy}{z}+\dfrac{zx}{y}+\dfrac{yz}{x}\right)+\left(\dfrac{yz}{x}+\dfrac{xy}{z}\right)+2\left(\dfrac{zx}{y}+\dfrac{xy}{z}\right)\)

\(VT\ge2\left(x+y+z\right)+2y+4x\)

\(VT\ge2\left(3x+2y+z\right)\ge4\)

Dấu "=" xảy ra khi \(x=y=z=\dfrac{1}{3}\)

gọi a là 1 giá trị của biểu thức P, khi đó ta có a = 2xy + 3yz + 4xz

Thay z = 1 - x - y, ta được :

a = 2xy + 3y ( 1 - x - y ) + 4x ( 1 - x - y )

\(\Leftrightarrow4x^2+\left(5y-4\right)x+3y^2-3y+a=0\)

PT có nghiệm \(\Leftrightarrow\Delta\ge0\Leftrightarrow\left(5y-4\right)^2-4.4\left(3y^2-3y+a\right)\ge0\)

\(\Leftrightarrow-23y^2+8y+16\ge16a\)

Vì \(-23y^2+8y+16=-23\left(y-\frac{4}{23}\right)^2+\frac{384}{23}\le\frac{384}{23}\)

\(\Rightarrow16a\le\frac{384}{23}\Rightarrow a\le\frac{24}{23}\Rightarrow P\le\frac{24}{23}\)

Vậy GTLN của P là \(\frac{24}{23}\)

quên còn dấu "="

Dấu "=" xảy ra khi \(\hept{\begin{cases}x+y+z=1\\y=\frac{4}{23}\\x=\frac{4-5y}{8}\end{cases}\Leftrightarrow\hept{\begin{cases}x=\frac{9}{23}\\y=\frac{4}{23}\\z=\frac{10}{23}\end{cases}}}\)