A = \(\dfrac{1}{1+2+3}\) + \(\dfrac{1}{1+2+3+4}\) +......+\(\dfrac{1}{1+2+3+4+....+59}\)

A = \(\dfrac{1}{(3+1).3:2}\) + \(\dfrac{1}{(4+1).4:2}\)+......+\(\dfrac{1}{(59+1).59:2}\)

A = \(\dfrac{2}{3.4}\) + \(\dfrac{2}{4.5}\) +.....+ \(\dfrac{2}{59.60}\)

A = 2.(\(\dfrac{1}{3.4}+\dfrac{1}{4.5}+....+\dfrac{1}{59.60}\))

A = 2. ( \(\dfrac{1}{3}\) - \(\dfrac{1}{4}\) + \(\dfrac{1}{4}\) - \(\dfrac{1}{5}\) +....+ \(\dfrac{1}{59}\) - \(\dfrac{1}{60}\))

A = 2. ( \(\dfrac{1}{3}\) - \(\dfrac{1}{60}\))

A = 2. \(\dfrac{19}{60}\)

A = \(\dfrac{19}{30}\)

\(\dfrac{3^{10}\cdot11+9^5\cdot5}{27^3\cdot2^4}\cdot x=-9\)

\(\dfrac{3^{10}\cdot11+3^{10}\cdot5}{3^9\cdot2^4}\cdot x=-9\)

\(\dfrac{3\cdot3^9\cdot16}{3^9\cdot16}\cdot x=-9\)

3x = -9

\(x=-\dfrac{9}{3}=-3\)

\(\dfrac{3\cdot3^9\cdot\left(11+5\right)}{3^9\cdot16}\cdot x=-9\)

Lời giải:

$A=1-3+3^2-3^3+...+3^{2022}-\frac{3^{2023}}{4}$

$3A=3-3^2+3^3-3^4+...+3^{2023}-\frac{3^{2024}}{4}$

$\Rightarrow A+3A=1+3^{2023}-\frac{3^{2023}}{4}-\frac{3^{2024}}{4}$

$\Rightarrow 4A=1$

$\Rightarrow A=\frac{1}{4}$

Lời giải:
Gọi số cây mà 3 lớp trồng được lần lượt là $a,b,c$ (cây)

Theo bài ra ta có: $a+b+c=150$

$\frac{a}{5}=\frac{b}{3}; \frac{b}{2}=\frac{c}{3}$

$\Rightarrow \frac{a}{10}=\frac{b}{6}=\frac{c}{9}$

Áp dụng tính chất dãy tỉ số bằng nhau:

$\frac{a}{10}=\frac{b}{6}=\frac{c}{9}=\frac{a+b+c}{10+6+9}=\frac{150}{25}=6$ 

$\Rightarrow a=10.6=60; b=6.6=36; c=6.9=54$ (cây)

\(4xy+6x-10y=22\)

\(\Leftrightarrow4xy+6x-10y-15=7\)

\(\Leftrightarrow2x\left(2y+3\right)-5\left(2y+3\right)=7\)

\(\Leftrightarrow\left(2x-5\right)\left(2y+3\right)=7\)

Vậy \(\left(x;y\right)=\left(-1;-2\right);\left(2;-5\right);\left(3;2\right);\left(6;-1\right)\)