C1:Cho $\frac{a}{b}$= $\frac{c}{d}$ CM:$\frac{a+b}{a}$= $\frac{c+d}{c}$ C2:So sánh $2^{450}$ với $3^{300}$ C3:tìm x biết: $\frac{x+2}{11}$+ $\fr

Bài giải: 

Có: $\dfrac{a}{b}=\dfrac{c}{d}$

$\to \dfrac{b}{a}=\dfrac{d}{c}$

$\to 1+\dfrac{b}{a}=1+\dfrac{d}{c}$

$\to \dfrac{a+b}{a}=\dfrac{c+d}{c}(đpcm)$

Câu 2: 

Có: $2^{450}=$ $ $($2^{3})^{150}=$ $8^{150}$

      $3^{300}=$ $(3^{2})^{150}=$ $9^{150}$ 

Nhận thấy: $8^{150}<$ $9^{150}⇒$ $2^{450}<$ $3^{300}$ 

Vậy $2^{450}<$ $3^{300}$ 

Câu 3:

$\dfrac{x+2}{11}+$ $\dfrac{x+2}{12}+$ $\dfrac{x+2}{13}=$ $\dfrac{x+2}{14}+$ $\dfrac{x+2}{15}$

$⇒$ $\dfrac{x+2}{11}+$ $\dfrac{x+2}{12}+$ $\dfrac{x+2}{13}-$$\dfrac{x+2}{14}-$ $\dfrac{x+2}{15}=0$

$⇒$ $(x+2)(\dfrac{1}{11}+$ $\dfrac{1}{12}$+ $\dfrac{1}{13}$ – $\dfrac{1}{14}$ -$\dfrac{1}{15})=0$ 

$⇒$ $x + 2 = 0$ ( vì $\dfrac{1}{11}+$ $\dfrac{1}{12}$+ $\dfrac{1}{13}$ – $\dfrac{1}{14}$ -$\dfrac{1}{15}$ # 0)

$⇒$ $x = -2$

Vậy $x = -2$