\(y=-2x\)

 Vậy đồ thị hàm số \(y=-2x\)đi qua 2 điểm \(O\left(0;0\right),A\left(1;-2\right)\)

Mình vẽ hơi bị xấu, mong bạn thông cảm nha.

\(f\left(-1\right)=1-8\cdot\left(-1\right)\)

               \(=9\)

  \(f\left(\frac{1}{2}\right)=1-8\cdot\frac{1}{2}\)

               \(=-3\)

f(-1)=1-8(-1)=1+8=9

f(1/2)=1-8(1/2)=1-4=-3

\(2x-\frac{1}{2}=\frac{5}{6}\)

\(=>2x=\frac{5}{6}+\frac{1}{2}\)

\(=>x=\frac{4}{3}:2\)

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

\(2x-\frac{1}{2}=\frac{5}{6}\)

\(2x=\frac{5}{6}+\frac{1}{2}\)

\(2x=\frac{5}{6}+\frac{3}{6}\)

\(2x=\frac{8}{6}=\frac{4}{3}\)

\(x=\frac{4}{3}\div2\)

\(x=\frac{4}{3}\times\frac{1}{2}\)

\(x=\frac{4}{6}=\frac{2}{3}\)

#Học tốt!!!

Ta có: 2³⁰⁰=(2⁶)⁵⁰=64⁵⁰>25⁵⁰

  Vậy 25⁵⁰ <2³⁰⁰

Ta có: \(\frac{a}{b}=\frac{c}{d}\)

=> ad = bc 

=> 3ac + ad = 3ac + bc

=> a(3c + d) = c(3a + b)

=> \(\frac{a}{3a+b}=\frac{c}{3a+d}\) (ĐPCM)

b) Ta có:

\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\)

đặt \(\frac{a}{c}=k\Rightarrow\frac{b}{d}=k\)

=> a = c.k; b = d.k

=> a² = c².k²; b² = d².k²

=> \(\frac{a^2+c^2}{b^2+d^2}=\frac{\left(c^2.k^2\right)+c^2}{\left(d^2.k^2\right)+d^2}\)= \(\frac{c^2\left(k^2+1\right)}{d^2\left(k^2+1\right)}\)=\(\frac{c^2}{d^2}=\frac{a^2}{b^2}=\frac{ac}{bd}\)

=> ĐPCM