\(\left|\frac{3}{2}x+\frac{1}{9}\right|+\left|\frac{1}{5}y-\frac{1}{2}\right|=0\)

vì \(\left|\frac{3}{2}x+\frac{1}{9}\right|\ge0;\left|\frac{1}{5}y-\frac{1}{2}\right|\ge0=>\left|\frac{3}{2}x+\frac{1}{9}\right|+\left|\frac{1}{5}y-\frac{1}{2}\right|\ge0\) (với mọi x,y)

Mà \(\left|\frac{3}{2}x+\frac{1}{9}\right|+\left|\frac{1}{5}y-\frac{1}{2}\right|=0\) (theo đề)

Nên \(\left|\frac{3}{2}x+\frac{1}{9}\right|=0=>\frac{3}{2}x=-\frac{1}{9}=>x=-\frac{2}{27}\)

      \(\left|\frac{1}{5}y-\frac{1}{2}\right|=0=>\frac{1}{5}y=\frac{1}{2}=>y=\frac{5}{2}\)

Vậy...........

Ta có:

3x-1/2 = 0 

3x= 1/2

x= 1/6

và 1/2y + 3/5 =0

     1/2y = -3/5

         y= -6/5

Vậy x= 1/6 và y = -6/5

\(\left(3x-\frac{1}{2}\right)+\left(\frac{1}{2}y+\frac{3}{5}\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}3x-\frac{1}{2}=0=\frac{1}{6}\\\frac{1}{2}y+\frac{3}{5}=0=\frac{6}{5}\end{cases}}\)

Vậy ......

a.

\(-\frac{2}{3}-\frac{1}{3}\times\left(2x-5\right)=\frac{3}{2}\)

\(-\frac{2}{3}-\frac{2}{3}x+\frac{5}{3}=\frac{3}{2}\)

\(\left(-\frac{2}{3}+\frac{5}{3}\right)-\frac{2}{3}x=\frac{3}{2}\)

\(\frac{3}{3}-\frac{2}{3}x=\frac{3}{2}\)

\(1-\frac{2}{3}x=\frac{3}{2}\)

\(\frac{2}{3}x=1-\frac{3}{2}\)

\(\frac{2}{3}x=\frac{2}{2}-\frac{3}{2}\)

\(\frac{2}{3}x=-\frac{1}{2}\)

\(x=-\frac{1}{2}\div\frac{2}{3}\)

\(x=-\frac{1}{2}\times\frac{3}{2}\)

\(x=-\frac{3}{4}\)

b.

\(\frac{1}{3}x+\frac{2}{5}\times\left(x-1\right)=0\)

\(\frac{1}{3}x+\frac{2}{5}x-\frac{2}{5}=0\)

\(x\times\left(\frac{1}{3}+\frac{2}{5}\right)=\frac{2}{5}\)

\(x\times\left(\frac{5}{15}+\frac{6}{15}\right)=\frac{2}{5}\)

\(x\times\frac{11}{15}=\frac{2}{5}\)

\(x=\frac{2}{5}\div\frac{11}{15}\)

\(x=\frac{2}{5}\times\frac{15}{11}\)

\(x=\frac{6}{11}\)

Chúc bạn học tốt

a ) \(-\frac{2}{3}-\frac{1}{3}\left(2x-5\right)=\frac{3}{2}\)

                  \(\frac{1}{3}\left(2x-5\right)=-\frac{2}{3}-\frac{3}{2}\)

                   \(\frac{1}{3}\left(2x-5\right)=\frac{-13}{6}\)

                        \(\left(2x-5\right)=-\frac{13}{6}:\frac{1}{3}\)

                         \(\left(2x-5\right)=-\frac{13}{6}.\frac{3}{1}\)

                         \(\left(2x-5\right)=-\frac{13}{2}\)

                           \(2x=-\frac{13}{2}+5\)

                            \(2x=-\frac{3}{2}\)

                              \(\Rightarrow x=-\frac{3}{2}:2\)

                              \(\Rightarrow x=-\frac{3}{2}.\frac{1}{2}\)

                              \(\Rightarrow x=-\frac{3}{4}\)

Ta có : \(\left(3x-\frac{y}{5}\right)^2\ge0;\left(2y+\frac{3}{7}\right)^2\ge0\)

\(=>\left(3x-\frac{y}{5}\right)^2+\left(2y+\frac{3}{7}\right)^2\ge0\)

Mà \(\left(3x-\frac{y}{5}\right)^2+\left(2y+\frac{3}{7}\right)^2=0\)nên dấu "=" xảy ra 

\(< =>\hept{\begin{cases}3x-\frac{y}{5}=0\\2y+\frac{3}{7}=0\end{cases}}< =>\hept{\begin{cases}3x-\frac{y}{5}=0\\y=-\frac{3}{14}\end{cases}}\)

\(< =>\hept{\begin{cases}x=-\frac{1}{70}\\y=-\frac{3}{14}\end{cases}}\)

Ta có : \(\left(x+y-\frac{1}{4}\right)^2\ge0;\left(x-y+\frac{1}{5}\right)^2\ge0\)

Cộng theo vế ta được : \(\left(x+y-\frac{1}{4}\right)^2+\left(x-y+\frac{1}{5}\right)^2\ge0\)

Mà \(\left(x+y-\frac{1}{4}\right)^2+\left(x-y+\frac{1}{5}\right)^2=0\)nên dấu "=" xảy ra 

\(< =>\hept{\begin{cases}y+x=\frac{1}{4}\\y-x=\frac{1}{5}\end{cases}}< =>\hept{\begin{cases}y=\frac{9}{40}\\x=\frac{1}{40}\end{cases}}\)

\(\left|x^2-3x\right|+\left|\left(x+1\right)\left(x-3\right)\right|=0\)

\(\Leftrightarrow\hept{\begin{cases}\left|x^2-3x\right|=0\\\left|\left(x+1\right)\left(x-3\right)\right|=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x^2-3x=0\\\left(x+1\right)\left(x-3\right)=0\end{cases}}\)

Xét \(x^2-3x=0\)

\(\Rightarrow x\left(x-3\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=3\end{cases}}\)

Xét \(\left(x+1\right)\left(x-3\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+1=0\\x-3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-1\\x=3\end{cases}}\)

Vì xét 2 trị biểu thức , một cái có 2 giá trị (0 or 3) , một cái (-1 or 3)

Nên ta lấy cái chung 

=> x = 3

Theo BĐT AM - GM cho 3 số dương, ta có: \(\left(3x+1\right)\left(y+z\right)+x=3xy+3zx+x+y+z\)

\(\ge3xy+3zx+3\sqrt[3]{xyz}=3zx+3xy+3=3\left(zx+xy+1\right)\)(Do xyz = 1)

\(\Rightarrow\frac{1}{\left(3x+1\right)\left(y+z\right)+x}\le\frac{1}{3\left(zx+xy+1\right)}\)(1)

Tương tự ta có: \(\frac{1}{\left(3y+1\right)\left(z+x\right)+y}\le\frac{1}{3\left(xy+yz+1\right)}\)(2); \(\frac{1}{\left(3z+1\right)\left(x+y\right)+z}\le\frac{1}{3\left(yz+zx+1\right)}\)(3)

Cộng theo từng vế của 3 BĐT (1), (2), (3), ta được:  \(P\le\frac{1}{3}\left(\frac{1}{xy+yz+1}+\frac{1}{yz+zx+1}+\frac{1}{zx+xy+1}\right)\)

Ta có BĐT: \(a^3+b^3\ge ab\left(a+b\right)\)

Thật vậy, với a, b dương thì (*)\(\Leftrightarrow\left(a+b\right)\left(a^2-ab+b^2\right)\ge ab\left(a+b\right)\Leftrightarrow a^2-ab+b^2\ge ab\)

\(\Leftrightarrow\left(a-b\right)^2\ge0\)(đúng)

Áp dụng BĐT trên và sử dụng giả thiết xyz = 1, ta được: \(\frac{1}{xy+yz+1}=\frac{\sqrt[3]{xyz}}{y\left(z+x\right)+\sqrt[3]{xyz}}\)

\(=\frac{\sqrt[3]{xyz}}{y\left[\left(\sqrt[3]{z}\right)^3+\left(\sqrt[3]{x}\right)^3\right]+\sqrt[3]{xyz}}\le\frac{\sqrt[3]{xyz}}{y\sqrt[3]{zx}\left(\sqrt[3]{z}+\sqrt[3]{x}\right)+\sqrt[3]{xyz}}\)

\(=\frac{\sqrt[3]{xyz}}{\sqrt[3]{y^3zx}\left(\sqrt[3]{z}+\sqrt[3]{x}\right)+\sqrt[3]{xyz}}=\frac{\sqrt[3]{xyz}}{\sqrt[3]{y^2}\left(\sqrt[3]{z}+\sqrt[3]{x}\right)+\sqrt[3]{xyz}}\)

\(=\frac{\sqrt[3]{zx}}{\sqrt[3]{y}\left(\sqrt[3]{z}+\sqrt[3]{x}\right)+\sqrt[3]{zx}}=\frac{\sqrt[3]{zx}}{\sqrt[3]{xy}+\sqrt[3]{yz}+\sqrt[3]{zx}}\)(*)

Tương tự: \(\frac{1}{yz+zx+1}\le\frac{\sqrt[3]{xy}}{\sqrt[3]{xy}+\sqrt[3]{yz}+\sqrt[3]{zx}}\)(**); \(\frac{1}{zx+xy+1}\le\frac{\sqrt[3]{yz}}{\sqrt[3]{xy}+\sqrt[3]{yz}+\sqrt[3]{zx}}\)(***)

Cộng theo từng vế của 3 BĐT (*), (**), (***), ta được: \(\frac{1}{xy+yz+1}+\frac{1}{yz+zx+1}+\frac{1}{zx+xy+1}\le\frac{\sqrt[3]{xy}+\sqrt[3]{yz}+\sqrt[3]{zx}}{\sqrt[3]{xy}+\sqrt[3]{yz}+\sqrt[3]{zx}}=1\)

\(\Rightarrow P\le\frac{1}{3}\left(\frac{1}{xy+yz+1}+\frac{1}{yz+zx+1}+\frac{1}{zx+xy+1}\right)\le\frac{1}{3}\)

Đẳng thức xảy ra khi x = y = z = 1

https://h.vn//hoi-dap/question/873191.html

a) \(\left|3x-\frac{1}{2}\right|+\left|\frac{1}{2}y+\frac{3}{5}\right|=0\)

=>\(3x-\frac{1}{2}=0;\frac{1}{2}y+\frac{3}{5}=0\left(\left|3x-\frac{1}{2}\right|;\left|\frac{1}{2}y+\frac{3}{5}\right|\ge0\right)\)

=>\(x=\frac{1}{6};y=\frac{-6}{5}\)

b)\(\left|\frac{3}{2}x+\frac{1}{9}\right|+\left|\frac{1}{5}y-\frac{1}{2}\right|\le0\)

Ta lại có:

\(\left|\frac{3}{2}x+\frac{1}{9}\right|+\left|\frac{1}{5}y-\frac{1}{2}\right|\ge0\)

=>\(\frac{3}{2}x+\frac{1}{9}=0;\frac{1}{5}y-\frac{1}{2}=0\Rightarrow x=-\frac{2}{27};y=\frac{5}{2}\)

a) \(\left|\frac{1}{2}+x\right|+\left|x+y+z\right|+\left|\frac{1}{3}+y\right|=0\)

=> \(\left|\frac{1}{2}+x\right|=\left|x+y+z\right|=\left|\frac{1}{3}+y\right|=0\)

1/2 + x = 0 => x = -1/2

1/3 + y = 0 => y = -1/3

-1/2 + -1/3 + z = 0 

=> z = 5/6