mik trả lời cho bạn rồi đó

a) Ta có:

f(0) = -2.0³ + 3.0² - 0 + 5 = 0 + 0 - 0 + 5 = 5

g(-1) = 2.(-1)³ - 2.(-1)² + (-1) - 9 = -2 - 2 - 1 - 9 = -14

b) f(x) + g(x) = (-2x³ + 3x² - x + 5) + (2x³ - 2x² + x - 9)

                   = -2x³ + 3x² - x + 5 + 2x³ - 2x² + x - 9

                  = (-2x³ + 2x³) + (3x² - 2x²) - (x - x) + (5 - 9)

                 = x² - 4

f(x) - g(x) = (-2x³ + 3x² - x + 5) - (2x³ - 2x² + x - 9)

               = -2x³ + 3x² - x + 5 - 2x³ + 2x² - x + 9

              = -(2x³ + 2x³) + (3x² + 2x²) - (x + x) + (5 + 9)

             = -4x³ + 5x² - 2x + 14

\(a.\)\(10x+2^2.5=10\)

\(\Leftrightarrow10x+4.5=10\)

\(\Leftrightarrow10x+20=10\)

\(\Leftrightarrow10x=10-20\)

\(\Leftrightarrow10x=-10\)

\(\Leftrightarrow x=-10:10\)

\(\Leftrightarrow x=-1\)

\(b.\)\(125-5\left(4+x\right)=15\)

\(\Leftrightarrow5\left(4+x\right)=125-15\)

\(\Leftrightarrow5\left(4+x\right)=110\)

\(\Leftrightarrow4+x=110:5\)

\(\Leftrightarrow4+x=22\)

\(\Leftrightarrow x=22-4\)

\(\Leftrightarrow x=18\)

\(c.\)\(2^6+\left(218-x\right)=73\)

\(\Leftrightarrow64+\left(218-x\right)=73\)

\(\Leftrightarrow218-x=73-64\)

\(\Leftrightarrow218-x=9\)

\(\Leftrightarrow x=218-9\)

\(\Leftrightarrow x=209\)

a)10x+2².5=10

=>10x+4.5=10

=>10x+20 =10

=>10x       =10-20

=>10x       =-10

=>x           =-10:10

=>x           =-1

b)125-5(4+x)=15

=>5(4+x)       =125-15

=>5(4+x)       =110

=>x+4           =110:5

=>x+4           =22

=>x               =22-4

=>x               =18

c)2⁶+(218-x)=73

=>218-x       =73-64

=>218-x       =9

=>x              =218-9

=>x              =209

a) Ta có: \(\hept{\begin{cases}\left|x+\frac{1}{2}\right|\ge0\\\left|2y-1\right|\ge0\end{cases}}\)
\(\Rightarrow\left|x+\frac{1}{2}\right|+\left|2y-1\right|+11\ge11\)
\(\Rightarrow A\ge11\)
\(\Rightarrow\)GTNN của A là 11 \(\Leftrightarrow\hept{\begin{cases}\left|x+\frac{1}{2}\right|=0\\\left|2y-1\right|=0\end{cases}}\)
                                        \(\Leftrightarrow\hept{\begin{cases}x=-\frac{1}{2}\\y=\frac{1}{2}\end{cases}}\)
Vậy ... 
b) Ta có: \(\hept{\begin{cases}\left|x-1,2\right|\ge0\\\left|y+1\right|\ge0\end{cases}}\)
\(\Rightarrow\left|x-1,2\right|+\left|y+1\right|+1\ge1\)
\(\Rightarrow\frac{1}{\left|x-1,2\right|+\left|y+1\right|+1}\le1\)
\(\Rightarrow\frac{7}{\left|x-1,2\right|+\left|y+1\right|+1}\le7\)
\(\Rightarrow B\le7\)
\(\Rightarrow\)GTNN của B là 7 \(\Leftrightarrow\hept{\begin{cases}\left|x-1,2\right|=0\\\left|y+1\right|=0\end{cases}}\)
                                      \(\Leftrightarrow\hept{\begin{cases}x=1,2\\y=-1\end{cases}}\)
Vậy ... 

Mí bạn giúp mik vs chiều nay mình học rồi :(((

Bài 1:

\(a)f\left(x\right)=10x\)

\(\Leftrightarrow f\left(0\right)=10.0=0\)

\(\Leftrightarrow f\left(-1\right)=10\left(-1\right)=-10\)

\(\Leftrightarrow f\left(\frac{1}{2}\right)=\frac{10}{2}=5\)

\(b)\)Vì \(f\left(x\right)=10x\)

Nên: \(f\left(a+b\right)=10\left(a+b\right)\)

Và: \(f\left(a\right)+f\left(b\right)=10a+10b=10\left(a+b\right)\)

Do đó:

\(f\left(a+b\right)=f\left(a\right)+f\left(b\right)\left(đpcm\right)\)

\(c)\)Vì \(\hept{\begin{cases}f\left(x\right)=10x\\f\left(x\right)=x^2\end{cases}\Leftrightarrow x^2=10x}\)

\(\Leftrightarrow x^2-10x=0\)

\(\Leftrightarrow x\left(x-10\right)=0\)

\(\Leftrightarrow\hept{\begin{cases}x=0\\x-10=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=0\\x=10\end{cases}}}\)

Vậy với \(\hept{\begin{cases}x=0\\x=10\end{cases}}\)thì \(f\left(x\right)=x^2\)

Mí bạn giúp mình với mình đang cần làm gấp X ((

MSC:60

Ta có:\(\frac{1}{4}=\frac{1.5.3}{4.5.3}=\frac{15}{60}\)

          \(\frac{4}{5}=\frac{4.4.3}{5.4.3}=\frac{48}{60}\)

           \(\frac{2}{3}=\frac{2.4.5}{3.4.5}=\frac{40}{60}\)

1/4=15/60,

4/5=48/60

2/3=20/60

Ta có \(x^4+x^2+1=x^4+2x^2+1-x^2\)

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

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

\(=a\cdot\frac{x}{x^2+x+1}=\frac{ax}{x^2-x+1+2x}=\frac{ax}{a+2x}\)

(ko biết có đúng ko nữa..)