\(\dfrac{x+2}{x-2}-\dfrac{x-2}{x+2}=\dfrac{16}{x^2-4}\)
\(\Leftrightarrow\dfrac{\left(x+2\right)^2}{\left(x-2\right)\left(x+2\right)}-\dfrac{\left(x-2\right)^2}{\left(x-2\right)\left(x+2\right)}=\dfrac{16}{\left(x-2\right)\left(x+2\right)}\)
\(\Leftrightarrow\left(x+2-x+2\right)\left(x+2+x-2\right)=16\)
\(\Leftrightarrow4\cdot2x=16\)
\(\Rightarrow x=2\)
\(--------\)
\(\dfrac{x+3}{x-2}+\dfrac{x+2}{x-3}=2\)
\(\Leftrightarrow\dfrac{\left(x+3\right)\left(x-3\right)}{\left(x-2\right)\left(x-3\right)}+\dfrac{\left(x+2\right)\left(x-2\right)}{\left(x-2\right)\left(x-3\right)}=\dfrac{2\left(x-2\right)\left(x-3\right)}{\left(x-2\right)\left(x-3\right)}\)
\(\Leftrightarrow x^2-9+x^2-4=2x^2-10x+12\)
\(\Leftrightarrow2x^2-13=2x^2-10x+12\)
\(\Leftrightarrow-10x+12=-13\)
\(\Rightarrow x=2,5\)
#NgTienDat

\(\left(-2x^5+x^4-3x^3\right):2x^3\)
\(=-x^2+\dfrac{1}{2}x-\dfrac{3}{2}\)

Tuổi con năm nay:
\(40\times\dfrac{1}{5}=8\left(tuổi\right)\)
Số tuổi mẹ hơn con:
\(40-8=32\left(tuổi\right)\)
Tuổi của mẹ khi con 14 tuổi:
\(14+32=46\left(tuổi\right)\)
Đáp số: 46 tuổi

\(\dfrac{5}{9}\times\dfrac{8}{17}+\dfrac{4}{9}\times\dfrac{8}{17}\)
\(=\left(\dfrac{5}{9}+\dfrac{4}{9}\right)\times\dfrac{8}{17}\)
\(=1\times\dfrac{8}{17}=\dfrac{8}{17}\)

1)\(A=\dfrac{2}{1.2}+\dfrac{2}{2.3}+\dfrac{2}{3.4}+...+\dfrac{2}{2023.2024}\)
\(=2\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2023.2024}\right)\)
\(=2\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2023}-\dfrac{1}{2024}\right)\)
\(=2\left(1-\dfrac{1}{2024}\right)\)
\(=2\cdot\dfrac{2023}{2024}=\dfrac{2023}{1012}\)
2)
a/\(\dfrac{n}{n+2}+\dfrac{5}{n+2}=\dfrac{n+5}{n+2}=\dfrac{\left(n+2\right)+3}{n+2}=1+\dfrac{3}{n+2}\)
Để \(\dfrac{n}{n+2}+\dfrac{5}{n+2}\) là số nguyên thì \(n+2\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)
Ta có bảng sau:

Vậy để \(\dfrac{n}{n+2}+\dfrac{5}{n+2}\) nguyên thì \(n\in\left\{-5;-3;-1;1\right\}\)
b/\(S=\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{2023.2025}\)
\(=\dfrac{1}{2}\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{2023.2025}\right)\)
\(=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{2023}-\dfrac{1}{2025}\right)\)
\(=\dfrac{1}{2}\left(1-\dfrac{1}{2025}\right)\)
\(=\dfrac{1}{2}\cdot\dfrac{2024}{2025}=\dfrac{1012}{2025}\)
 

Chiều cao tam giác ABC:
\(12\times60:100=7,2\left(cm\right)\)
Diện tích tam giác ABC:
\(\dfrac{12\times7,2}{2}=43,2\left(cm^2\right)\)
Đáp số: 43,2 cm²

Tổng số tiền quỹ lớp 4A:
\(38\times20000=760000\left(đồng\right)\)
Số tiền cả lớp thống nhất dùng để mua sách vở:
\(760000\times\dfrac{5}{8}=475000\left(đồng\right)\)
Số tiền quỹ lớp 4A còn lại:
\(760000-475000=285000\left(đồng\right)\)
Đáp số: 285000 đồng

a/Thay x = -2 vào A, ta có:
\(A=\dfrac{-2}{-2+3}=-2\)
b/\(M=A+B\)
\(=\dfrac{x}{x+3}+\dfrac{2x}{x-3}-\dfrac{3x^2+9}{x^2-9}\)
\(=\dfrac{x}{x+3}+\dfrac{2x}{x-3}-\dfrac{3x^2+9}{\left(x+3\right)\left(x-3\right)}\)
\(=\dfrac{x\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}+\dfrac{2x\left(x+3\right)}{\left(x+3\right)\left(x-3\right)}-\dfrac{3x^2+9}{\left(x+3\right)\left(x-3\right)}\)
\(=\dfrac{x^2-3x+2x^2+6x-3x^2-9}{\left(x+3\right)\left(x-3\right)}\)
\(=\dfrac{3x-9}{\left(x+3\right)\left(x-3\right)}\)
\(=\dfrac{3\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}=\dfrac{3}{x+3}\)

Cân nặng của quả dưa hấu:
\(6:\dfrac{4}{5}=7,5\left(kg\right)\)
\(\rightarrow B\)

\(\dfrac{17}{13}-\dfrac{14}{23}+\dfrac{9}{13}-\dfrac{9}{23}\)
\(=\left(\dfrac{17}{13}+\dfrac{9}{13}\right)-\left(\dfrac{14}{23}+\dfrac{9}{23}\right)\)
\(=2-1\)
\(=1\)