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a)
Điều kiện: $x\ne-2,-5,-10,-17.$
$\dfrac3{(x+2)(x+5)}+\dfrac5{(x+5)(x+10)}+\dfrac7{(x+10)(x+17)}=\dfrac{x}{(x+2)(x+17)}$
$\Leftrightarrow\dfrac1{x+2}-\dfrac1{x+5}+\dfrac1{x+5}-\dfrac1{x+10}+\dfrac1{x+10}-\dfrac1{x+17}=\dfrac{x}{(x+2)(x+17)}$
$\Leftrightarrow\dfrac1{x+2}-\dfrac1{x+17}=\dfrac{x}{(x+2)(x+17)}$
$\Leftrightarrow\dfrac{15}{(x+2)(x+17)}=\dfrac{x}{(x+2)(x+17)}$
$\Leftrightarrow x=15.$
b)
Điều kiện: $x\ne1,3,8,20.$
$\dfrac2{(x-1)(x-3)}+\dfrac5{(x-3)(x-8)}+\dfrac{12}{(x-8)(x-20)}-\dfrac1{x-20}=-\dfrac34$
$\Leftrightarrow\left(\dfrac1{x-3}-\dfrac1{x-1}\right)+\left(\dfrac1{x-8}-\dfrac1{x-3}\right)+\left(\dfrac1{x-20}-\dfrac1{x-8}\right)-\dfrac1{x-20}=-\dfrac34$
$\Leftrightarrow-\dfrac1{x-1}=-\dfrac34$
$\Leftrightarrow\dfrac1{x-1}=\dfrac34$
$\Leftrightarrow4=3(x-1)$
$\Leftrightarrow3x=7$
$\Leftrightarrow x=\dfrac73.$
\(\left(\frac{4}{9}+\frac{1}{3}\right)^2=\left(\frac{4}{9}+\frac{3}{9}\right)^2=\left(\frac{7}{9}\right)^2=\frac{49}{81}\)
\(\left(\frac{1}{2}-\frac{3}{5}\right)^3=\left(\frac{5}{10}-\frac{6}{10}\right)^3=\left(\frac{-1}{10}\right)^3=\frac{-1}{1000}\)
\(\left(\frac{-1}{5}\right)^5.\left(\frac{-6}{5}\right)^4=\frac{-5}{3125}.\frac{1296}{625}=\frac{-1296}{390625}\)
\(\left(\frac{3}{4}\right)^3:\left(\frac{3}{4}\right)^2:\left(-\frac{2}{5}\right)^3=\frac{3}{4}:\frac{-8}{125}=\frac{3}{4}.\frac{-125}{8}=\frac{-375}{32}\)
\(=\frac{16}{5}.\frac{15}{16}-\left(\frac{3}{4}+\frac{2}{7}\right):\left(\frac{-29}{28}\right)\)
\(=3-\left(\frac{21}{28}+\frac{8}{28}\right):\left(\frac{-29}{28}\right)\)
\(=3-\left(\frac{29}{28}\right).\left(\frac{-28}{29}\right)\)
\(=3-\left(-1\right)\)
\(=4\)
b) \(=\left(\frac{1}{4}+\frac{25}{2}-\frac{5}{16}\right):\left(12-\frac{7}{12}:\left(\frac{3}{8}-\frac{1}{12}\right)\right)\)
\(=\left(\frac{4}{16}+\frac{200}{16}-\frac{5}{16}\right):\left(12-\frac{7}{12}:\left(\frac{3.3}{2.3.4}-\frac{2}{2.3.4}\right)\right)\)
\(=\left(\frac{199}{16}\right):\left(12-\frac{7}{12}:\left(\frac{9}{24}-\frac{2}{24}\right)\right)\)
\(=\frac{199}{16}:\left(12-\frac{7}{12}.\frac{24}{7}\right)\)
\(=\frac{199}{16}:\left(12-2\right)\)
\(=\frac{199}{16}:10\)
\(=\frac{199}{160}\)
c) \(\left(\frac{-3}{5}+\frac{5}{11}\right):\frac{-3}{7}+\left(\frac{-2}{5}+\frac{6}{5}\right):\frac{-3}{7}\)
\(\left(\frac{-33}{55}+\frac{25}{55}\right):\frac{-3}{7}+\left(\frac{4}{5}\right):\frac{-3}{7}\)
\(\left(\frac{-8}{55}\right).\frac{-7}{3}+\frac{4}{5}.\frac{-7}{3}\)
\(\frac{-7}{3}\left(\frac{-8}{55}+\frac{4}{5}\right)\)
\(\frac{-7}{3}.\frac{36}{55}=\frac{-84}{55}\)
#)Giải :
\(\left(\frac{3}{79}\right)^{20}\cdot\left(\frac{3}{-79}\right)^{19}=\left(\frac{3}{79}\right)^{20}\cdot\left(\frac{3}{79}\right)^{19}=\left(\frac{3}{79}\right)^{39}\)
\(\left(\frac{2}{3}\right)^5:\left(\frac{2}{-3}\right)^3=\left(\frac{2}{3}\right)^5:\left(\frac{2}{3}\right)^3=\left(\frac{2}{3}\right)^2\)






\(\left(\frac{\mathbf{1}}{\mathbf{2}}\right)^{\mathbf{5}}\mathbf{+}\left(\frac{\mathbf{3}}{\mathbf{\pi }}\right)^{\mathbf{5}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{32}}\mathbf{+}\frac{\mathbf{243}}{\mathbf{\pi }^{\mathbf{5}}}\)
Các bước tính toán chi tiết:
1. Tính số hạng thứ nhất:
\(\left(\frac{1}{2}\right)^{5}=\frac{1^{5}}{2^{5}}=\frac{1}{32}\)
2. Tính số hạng thứ hai:
\(\left(\frac{3}{\pi }\right)^{5}=\frac{3^{5}}{\pi ^{5}}=\frac{243}{\pi ^{5}}\)
3. Tổng hợp lại dưới dạng phân số chính xác:
\(\frac{1}{32}+\frac{243}{\pi ^{5}}=\frac{\pi ^{5}+7776}{32\pi ^{5}}\)
Giá trị xấp xỉ dạng số thập phân:
Nếu đề bài yêu cầu làm tròn hoặc tính ra số thập phân (với \(\pi \approx 3,14159\)):
• \(\frac{1}{32} = 0,03125\)
• \(\frac{243}{\pi^5} \approx 0,79407\)
• Kết quả xấp xỉ: \(0,03125 + 0,79407 \approx \mathbf{0,82532}\)
Ai hỏi
(1/2)^5 + (3/π)^5
= 1/32 + 243/π^5
≈ 0,82532
Kết quả: ≈0,8253165653