Với a² + b² + c² = 0 , a² = b² = c² = 0 .

a³ + b³ - c³ = 0 + 0 - 0 = 0

\(a^3+b^3-c^2\le0\)

Ta có: a²\(\ge\)0; b² \(\ge\)0; c² \(\ge\)0

Mà a²+b²+c²=0

=> \(\hept{\begin{cases}a^2=0\\b^2=0\\c^2=0\end{cases}}\)=> \(\hept{\begin{cases}a=0\\b=0\\c=0\end{cases}}\)

Suy ra: a³+b³-c²=0

=> a³+b³-c² \(\le\)0 (điều cần phải chứng minh)

Áp dụng tính chất dãy tỉ số bằng nhau, ta có: 

\(\frac{1+3y}{12}=\frac{1+5y}{5x}=\frac{1+7y}{4x}=\frac{1+3y+1+7y}{12+4x}=\frac{2+10y}{2\left(6+2x\right)}=\frac{2\left(1+5y\right)}{2\left(6+2x\right)}=\frac{1+5y}{6+2x}\)

\(\Rightarrow5x=6+2x\)

\(5x-2x=6\)

\(3x=6\)

\(x=2\)

Với \(x=2\), ta có:

\(\frac{1+3y}{12}=\frac{1+5y}{10}\)

\(\Rightarrow10\left(1+3y\right)=12\left(1+5y\right)\)

\(10+30y=12+60y\)

\(10-12=60y-30y\)

\(30y=-2\)

\(y=-\frac{1}{15}\)

Vậy \(x=2,y=\frac{-1}{15}\).

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

\(\Rightarrow\left(\frac{3x-5}{9}\right)^{2018}+\left(\frac{3y+0,4}{3}\right)^{2020}\ge0\forall x,y\)

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

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

Vậy \(x=\frac{5}{3}\)và \(y=-\frac{2}{15}\)

a) \(x^2-2=0\)

     \(x^2\)     \(=0+2\)

      \(x^2\)     \(=2\)
\(\Rightarrow x=2\)hoặc \(x=-2\)

Vậy \(x\in\left\{2;-2\right\}\)

b) \(2\sqrt{x}+3=11\)

     \(2\sqrt{x}\)     \(=11-3\)

     \(2\sqrt{x}\)     \(=8\)

        \(\sqrt{x}\)     \(=8:2\)

        \(\sqrt{x}\)     \(=4\)

        \(\Rightarrow x\)       \(=2\)

Vậy \(x=2\)

c) \(\left(3\sqrt{x+5}+5\right):2-20=-13\)

    \(\left(3\sqrt{x+5}+5\right):2\)        \(=-13+20\)

    \(\left(3\sqrt{x+5}+5\right):2\)        \(=7\)

        \(3\sqrt{x+5}+5\)              \(=\frac{7}{2}\)

        \(3\sqrt{x+5}\)                        \(=\frac{7}{2}-5\)

        \(3\sqrt{x+5}\)                        \(=\frac{7-10}{2}\)

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

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

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

           \(\sqrt{x+5}\)                          \(=\frac{-3\times1}{2\times3}\)

            \(\sqrt{x+5}\)                          \(=\frac{-1\times1}{2\times1}\)

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

Vì \(\sqrt{x+5}>0\)

\(\Rightarrow\sqrt{x+5}=\frac{-1}{2}\)(vô lí)

\(\Rightarrow\left(x+5\right)\in\varnothing\)

\(\Rightarrow x\in\varnothing\)

Vậy \(x\in\varnothing\)

a) x² - 2 = 0

=> x² = 2

=> \(x=\pm\sqrt{2}\)

Vậy \(x=\pm\sqrt{2}\)

b) \(2\sqrt{x}+3=11\)

=> \(2\sqrt{x}=8\)

=> \(\sqrt{x}=4\)

=> \(\sqrt{x}^2=4^2\) 

=> x = 16

Vậy x = 16

c) \(\left(3\sqrt{x+5}+5\right):2-20=-13\)

=> \(\left(3\sqrt{x+5}+5\right):2=7\)

=> \(3\sqrt{x+5}+5=14\)

=> \(3\sqrt{x+5}=9\)

=> \(\sqrt{x+5}=3\)

=> \(\sqrt{x+5}^2=3^2\)

=> x + 5 = 9

=> x = 4

Vậy x = 4

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

\(\Rightarrow-2.\left(2x-1\right)=3.\left(2-x\right)\)

\(-4x+2=6-3x\)

\(-4x+3x=6-2\)

\(-x=4\Rightarrow x=-4\)

\(\frac{x}{4}=\frac{3}{10}\)                   ;           \(\left|x+\frac{1}{5}\right|-4=-2\)                      ;

=> x . 10 = 4 . 3                          => \(\left|x+\frac{1}{5}\right|=\left(-2\right)+4\)

=> x . 10 = 12                             => \(\left|x+\frac{1}{5}\right|=2\)

=> x = 12 : 10                             => \(\left|x+\frac{1}{5}\right|=\left|2\right|=\left|-2\right|\)

=> x = \(\frac{6}{5}\)                              => \(\orbr{\begin{cases}x+\frac{1}{5}=2\\x+\frac{1}{5}=-2\end{cases}}\)=> \(\orbr{\begin{cases}x=2-\frac{1}{5}=\frac{9}{5}\\x=\left(-2\right)-\frac{1}{5}=\frac{-11}{5}\end{cases}}\)

Vậy x = \(\frac{6}{5}\)                            Vậy \(x\in\left\{\frac{9}{5};\frac{-11}{5}\right\}\)

\(\left(x+\frac{1}{2}\right)^2=\frac{1}{16}\)

=> \(\left(x+\frac{1}{2}\right)^2=\left(\frac{1}{4}\right)^2=\left(\frac{-1}{4}\right)^2\)

=> \(\orbr{\begin{cases}x+\frac{1}{2}=\frac{1}{4}\\x+\frac{1}{2}=\frac{-1}{4}\end{cases}}\)=> \(\orbr{\begin{cases}x=\frac{1}{4}-\frac{1}{2}=\frac{-1}{4}\\x=\frac{-1}{4}-\frac{1}{2}=\frac{-3}{4}\end{cases}}\)

Vậy \(x\in\left\{\frac{-1}{4};\frac{-3}{4}\right\}\)