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Bài 1:
a) \(\dfrac{19}{12}+\left|\dfrac{-5}{2}\right|+\left(\dfrac{3}{2}\right)^2=\dfrac{19}{12}+\dfrac{5}{2}+\dfrac{9}{4}\)
\(=\dfrac{19+5.6+9.3}{12}=\dfrac{76}{12}=\dfrac{19}{3}\)
b) \(\dfrac{2}{11}.\dfrac{16}{9}-\dfrac{2}{11}.\dfrac{7}{9}=\dfrac{2}{11}\left(\dfrac{16}{9}-\dfrac{7}{9}\right)=\dfrac{2}{11}.1=\dfrac{2}{11}\)
Bài 2:
Áp dụng t/c dtsbn:
\(\dfrac{a}{8}=\dfrac{b}{3}=\dfrac{a-b}{8-3}=\dfrac{55}{5}=11\)
\(\Rightarrow\left\{{}\begin{matrix}x=11.8=88\\b=11.3=33\end{matrix}\right.\)
\(\frac{1}{5}A=\frac{1}{5}.\left(\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{20}}\right)\)
\(\Rightarrow\frac{1}{5}A=\frac{1}{5^2}+\frac{1}{5^3}+...+\frac{1}{5^{20}}\)
\(\Rightarrow\frac{1}{5}A-A=\left(\frac{1}{5^2}+...+\frac{1}{5^{21}}\right)-\left(\frac{1}{5}+...+\frac{1}{5^{20}}\right)\)
\(-\frac{4}{5}A=\frac{1}{5^{21}}-\frac{1}{5}\)
\(\Rightarrow A=\left(\frac{1}{5^{21}}-\frac{1}{5}\right):\left(-\frac{4}{5}\right)\)
các câu còn lại tương tự thôi
B1 c2
dùng xích ma \(\text{∑}^{20}_1\left(\frac{1}{5^x}\right)=0,25=\frac{1}{4}\)
chỗ phía dưới là 1 nha nó bị che
\(\frac{2^{19}\cdot27^3}{4^9\cdot3}=\frac{2^{19}\cdot\left(3^3\right)^3}{\left(2^2\right)^9\cdot3}=\frac{2^{19}\cdot3^9}{2^{18}\cdot3}=\frac{2\cdot3^8}{1\cdot1}=2\cdot3^8\)
\(C=1+3+3^2+3^3+....+3^{99}\)
\(\Rightarrow3C=3+3^2+3^3+3^4+...+3^{100}\)
\(\Rightarrow3C-C=\left(3+3^2+3^3+3^4+...+3^{100}\right)-\left(1+3+3^2+3^3+....+3^{99}\right)\)
\(\Rightarrow2C=3^{100}-1\)
\(C=\frac{3^{100}-1}{2}\)
1/
a, xem lại đề
b, \(\sqrt{6}+\sqrt{12}+\sqrt{30}+\sqrt{56}< \sqrt{6,25}+\sqrt{12,25}+\sqrt{30,25}+\sqrt{56,26}=2,5+3,5+5,5+7,5=19\)
2/
a, \(\sqrt{26}+\sqrt{17}>\sqrt{25}+\sqrt{16}=5+4=9\)
b, xem lại
4/
\(B=\frac{\sqrt{x}+1}{\sqrt{x}-2}=\frac{\sqrt{x}-2+3}{\sqrt{x}-2}=1+\frac{3}{\sqrt{x}-2}\)
Để \(B\in Z\Leftrightarrow\sqrt{x}-2\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)
Ta có bảng
\(\sqrt{x}-2\) | 1 | -1 | 3 | -3 |
\(\sqrt{x}\) | 3 | 1 | 5 | -1 |
x | loại | 1 | loại | loai |
Vậy...
bài 19 :
a) \(\left(x+1\right)\left(x-2\right)< 0\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x+1< 0\\x-2>0\end{matrix}\right.\\\left[{}\begin{matrix}x+1>0\\x-2< 0\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x< -1\\x>2\end{matrix}\right.\\\left[{}\begin{matrix}x>-1\\x< 2\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\in\varnothing\\-1< x< 2\end{matrix}\right.\) vậy \(-1< x< 2\)
b) \(\left(x-2\right)\left(x+\dfrac{2}{3}\right)>0\) \(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x-2>0\\x+\dfrac{2}{3}>0\end{matrix}\right.\\\left[{}\begin{matrix}x-2< 0\\x+\dfrac{2}{3}< 0\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x>2\\x>\dfrac{-2}{3}\end{matrix}\right.\\\left[{}\begin{matrix}x< 2\\x< \dfrac{-2}{3}\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x>2\\x< \dfrac{-2}{3}\end{matrix}\right.\) vậy \(x>2\) hoặc \(x< \dfrac{-2}{3}\)