Ủa lần trước tớ nói với thầy cũng khóa nick tên mạo danh đáng ghét này rồi màToshiro Kiyoshi

\(3^x+3^{x+2}=90\\ \Leftrightarrow3^x+3^x\cdot3^2=90\\\Leftrightarrow 3^x\left(1+3^2\right)=90\\ \Leftrightarrow3^x\cdot10=90\\ \Leftrightarrow3^x=9\\ \Leftrightarrow3^x=3^2\\ \Leftrightarrow x=2\)

Vậy \(x=2\)

Theo bài ra ta có :

\(\dfrac{x-1}{2}=\dfrac{y+3}{4}=\dfrac{z-5}{6}\\ \Rightarrow\dfrac{3\left(x-1\right)}{6}=\dfrac{4\left(y+3\right)}{16}=\dfrac{5\left(z-5\right)}{30}\\ \Rightarrow\dfrac{3x-3}{6}=\dfrac{4y+12}{16}=\dfrac{5z-25}{30}\\ \Rightarrow\dfrac{5z-25}{30}=\dfrac{3x-3}{6}=\dfrac{4y+12}{16}\)

\(5z-3x-4y=50\)

Áp dụng tính chất dãy tỉ số bằng nhau ta được :

\(\dfrac{5z-25}{30}=\dfrac{3x-3}{6}=\dfrac{4y+12}{16}\\ =\dfrac{\left(5z-25\right)-\left(3x-3\right)-\left(4y+12\right)}{30-6-16}\\ =\dfrac{5z-25-3x+3-4y-12}{8}\\ =\dfrac{\left(5z-3x-4y\right)-\left(25-3+12\right)}{8}\\ =\dfrac{50-34}{8}\\ =\dfrac{16}{8}\\ =2\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{3x-3}{6}=2\Rightarrow3x-3=12\Rightarrow3x=15\Rightarrow x=5\\\dfrac{4y+12}{16}=2\Rightarrow4y+12=32\Rightarrow4y=20\Rightarrow y=5\\\dfrac{5z-25}{30}=2\Rightarrow5z-25=60\Rightarrow5z=85\Rightarrow z=17\end{matrix}\right.\)

\(\text{Vậy }x=5\\ y=5\\ z=17\)

Mấy câu trên kia dễ rồi mình chữa mình câu \(c\) bài \(3\) thôi nhé Kazuto Kirikaya

d) \(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{9\cdot10}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{9}-\dfrac{1}{10}\)

\(=1-\dfrac{1}{10}\)

\(=\dfrac{9}{10}\)

Theo bài ra ta có :

\(\dfrac{a}{2}=\dfrac{b}{3}=\dfrac{c}{4}\Rightarrow\dfrac{a}{2}=\dfrac{2b}{6}=\dfrac{3c}{12}\)

\(a+2b-3c=-20\)

Áp dụng tính chất dãy tỉ số bằng nhau ta được :

\(\dfrac{a}{2}=\dfrac{2b}{6}=\dfrac{3c}{12}=\dfrac{a+2b-3c}{2+6-12}=\dfrac{-20}{-4}=5\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{a}{2}=5\Rightarrow a=10\\\dfrac{2b}{6}=5\Rightarrow2b=30\Rightarrow b=15\\\dfrac{3c}{12}=5\Rightarrow3c=60\Rightarrow c=20\end{matrix}\right.\)

\(\text{Vậy }a=10\\ b=15\\ c=20\)