( 3x-5 /9 )^2002 > 0 ; ( 3y+0,4/3 )^2004 > 0

=> (3x-5/9 )^2002 = 0 và ( 3y + 0,4 / 3 )^2004 = 0

=> 3x - 5 = 0

3x = 5

x = 5/3

=> 3y + 0,4 = 0

3y = -0,4

y= -2/15

post từng câu một thôi bn nhìn mệt quá

We have:

\(A=\Sigma_{cyc}\frac{1}{3xy+3zx+x+y+z}\le\frac{1}{3xy+3zx+3\sqrt[3]{xyz}}=\Sigma_{cyc}\frac{1}{3xy+3zx+3}=\Sigma_{cyc}\frac{1}{3\left(xy+zx+1\right)}\)

Dat \(\left(\frac{1}{x};\frac{1}{y};\frac{1}{z}\right)=\left(a;b;c\right)\Rightarrow abc=1\)

\(\Rightarrow A\le\Sigma_{cyc}\frac{1}{3\left(\frac{1}{ab}+\frac{1}{ca}+1\right)}=\Sigma_{cyc}\frac{a}{3\left(a+b+c\right)}=\frac{1}{3}\)

Dau '=' xay ra khi \(x=y=z=1\)

a/ \(M=\left(x^2+\frac{1}{y^2}\right)\left(y^2+\frac{1}{x^2}\right)=x^2y^2+\frac{1}{x^2y^2}+2=\left(xy-\frac{1}{xy}\right)^2+4\ge4\)

Suy ra Min M = 4 . Dấu "=" xảy ra khi x=y=1/2

b/ Đề đúng phải là \(\frac{1}{3x+3y+2z}+\frac{1}{3x+2y+3z}+\frac{1}{2x+3y+3z}\ge\frac{3}{2}\)

Ta có \(6=\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}\ge\frac{9}{2\left(x+y+z\right)}\Rightarrow x+y+z\ge\frac{3}{4}\)

Lại có \(\frac{1}{3x+3y+2z}+\frac{1}{3x+2y+3z}+\frac{1}{2x+3y+3z}\ge\frac{9}{8\left(x+y+z\right)}\ge\frac{9}{8.\frac{3}{4}}=\frac{3}{2}\)

\(\left(\frac{3x-5}{9}\right)^{2018}+\left(\frac{3y+0,4}{3}\right)^{2020}=0\)

Ta có : \(\hept{\begin{cases}\left(\frac{3x-5}{9}\right)^{2018}\ge0\forall x\\\left(\frac{3y+0,4}{3}\right)^{2020}\ge0\forall y\end{cases}}\Rightarrow\left(\frac{3x-5}{9}\right)^{2018}+\left(\frac{3y+0,4}{3}\right)^{2020}\ge0\forall x,y\)

Dấu "=" xảy ra <=> \(\hept{\begin{cases}\frac{3x-5}{9}=0\\\frac{3y+0,4}{3}=0\end{cases}}\Leftrightarrow\hept{\begin{cases}3x-5=0\\3y+0,4=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\frac{5}{3}\\y=-\frac{2}{15}\end{cases}}\)

\(3xy-1=x+y\ge2\sqrt{xy}\)

\(\Leftrightarrow\left(\sqrt{xy}-1\right)\left(3\sqrt{xy}+1\right)\ge0\)

\(\Leftrightarrow\sqrt{xy}\ge1\Leftrightarrow xy\ge1\)

Và \(xy+x+y+1=4xy\)

\(\Leftrightarrow\left(x+1\right)\left(y+1\right)=4xy\)

Ta có: \(\frac{3x}{y\left(x+1\right)}-\frac{1}{y^2}=\frac{3xy-x-1}{y^2\left(x+1\right)}=\frac{y}{y^2\left(x+1\right)}=\frac{1}{y\left(x+1\right)}\)

\(M=\frac{1}{y\left(x+1\right)}+\frac{1}{x\left(y+1\right)}=\frac{2xy+x+y}{4x^2y^2}=5xy-1\)

Xét hàm số \(f\left(t\right)=\frac{20t^2-8t\left(5t-1\right)}{16t^4}=\frac{8t-20t^2}{16t^4}\le0\) 

Nên hàm số nghịch biến với \(t\ge1\)

\(\Rightarrow f\left(t\right)_{Max}=f\left(1\right)=1\Leftrightarrow M_{Max}=1\)

Đặt \(\frac{1}{x}=a,\frac{1}{y}=b\Rightarrow a+b+ab=3\)

Ta có:\(3=a+b+ab\ge3\sqrt[3]{a^2b^2}\Rightarrow ab\le1\)

Suy ra

\(M=\frac{ab}{a+1}+\frac{ab}{b+1}=ab\left(\frac{a+1+b+1}{ab+a+b+1}\right)=\frac{ab.\left(5-ab\right)}{4}=\frac{-\left[\left(ab\right)^2-2ab+1\right]+3ab+1}{4}=\frac{-\left(ab-1\right)^2+3ab+1}{4}\le1\)Dấu bằng xảy ra khi a=b=1

Áp dụng bất đẳng thức Cô-si, ta có: \(\left(3x+1\right)\left(y+z\right)+x=3xy+3xz+\left(x+y+z\right)\ge3xy+3xz+3\sqrt[3]{xyz}\)\(=3xy+3xz+3\Rightarrow\frac{1}{\left(3x+1\right)\left(y+z\right)+x}\le\frac{1}{3\left(xy+xz+1\right)}\)

Tiếp tục áp dụng bất đẳng thức dạng \(u^3+v^3\ge uv\left(u+v\right)\), ta được: \(\frac{1}{3\left(xy+xz+1\right)}=\frac{1}{3\left[x\left(\left(\sqrt[3]{y}\right)^3+\left(\sqrt[3]{z}\right)^3\right)+1\right]}\le\frac{1}{3\left[x\sqrt[3]{yz}\left(\sqrt[3]{y}+\sqrt[3]{z}\right)+1\right]}\)\(=\frac{\sqrt[3]{xyz}}{3\left[\sqrt[3]{x^2}\left(\sqrt[3]{y}+\sqrt[3]{z}\right)+\sqrt[3]{xyz}\right]}=\frac{\sqrt[3]{yz}}{3\left(\sqrt[3]{xy}+\sqrt[3]{yz}+\sqrt[3]{zx}\right)}\)

Tương tự rồi cộng lại theo vế, ta được: \(P\le\frac{1}{3}\)

Đẳng thức xảy ra khi x = y = z = 1

Ta có : \(9^{x-1}=\frac{1}{9}\)

=> \(9^{x-1}=9^{-1}\)

=> x - 1 = -1

=> x = 0 

ko biết bạn học mũ âm chưa nêu chưa thì mk xin lỗi 

=> 

Cảm ơn bạn nha. Còn mấy phần kia bạn biết làm không?

2) Ta có : \(\sqrt{\left(x-\sqrt{2}\right)^2}+\sqrt{\left(y+\sqrt{2}\right)^2}+\left|x+y+z\right|=0\)

\(\Leftrightarrow\sqrt{x^2-\sqrt{2}^2}+\sqrt{y^2+\sqrt{2}^2}+\left|x+y+z\right|=0\)

\(\Leftrightarrow\sqrt{x^2-2}+\sqrt{y^2+2}+\left|x+y+z\right|=0\)

Mà : \(\sqrt{x^2-1}\ge0\); \(\sqrt{y^2+2}\ge0\); \(\left|x+y+z\right|\ge0\)

Nên : \(\sqrt{x^2-1}=0;\sqrt{y^2+2};\left|x+y+z\right|=0\)

\(\Rightarrow\hept{\begin{cases}x^2-1=0\\y^2+2=0\\x+y+z=0\end{cases}\Rightarrow\hept{\begin{cases}x^2=1\\y^2=-2\\z=0-x-y\end{cases}\Rightarrow}\hept{\begin{cases}x=1\\y=-\sqrt{2}\\z=0-1+\sqrt{2}\end{cases}\Rightarrow}\hept{\begin{cases}x=1\\y=-\sqrt{2}\\z=-1+\sqrt{2}\end{cases}}}\)