\(U\left(8\right)=\left\{-1;1;-2;2-4;4;-8;8\right\}\)

\(U\left(11\right)=\left\{-1;1;-11;11\right\}\)

\(U\left(19\right)=\left\{-1;1;-19;19\right\}\)

\(U\left(28\right)=\left\{-1;1;-2;2;-4;4;-7;7;-14;14;-28;28\right\}\)

Ư(8)={−8;−4;−2;−1;1;2;4;8}Ư(11)={−11;−1;1;11}Ư(19)={−19;−1;1;19}Ư(28)={−28;−14;−7;−4;−2;−1;1;2;4;7;14;28}

\(B=\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{2022.2023}\)

\(B=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{2022}-\dfrac{1}{2023}\)

\(B=\dfrac{1}{2}-\dfrac{1}{2023}=\dfrac{2021}{4046}\)

x : 4 = 3

=> x = 3 x 4 = 12

Vậy x =12

\(\dfrac{x}{4}=3\)

\(x=3\times4\)

\(x=12\)

Số học sinh nữ chiếm số phần trăm so với số học sinh cả lớp là

    \(\dfrac{21\times100}{48}\%=43,75\%\) 

                         Đáp số: \(43,75\%\)

21 học sinh nữ chiếm: 21 : 48 = 43,75% ( học sinh của lớp)

Bạn xem lại đề nha .

\(A=\dfrac{1}{2^0.5}+\dfrac{1}{2^1.5}+\dfrac{1}{2^2.5}+...+\dfrac{1}{2^8.5}\)

\(5A=\dfrac{1}{2^0}+\dfrac{1}{2^1}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^8}\)

\(5A=2-1+1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{4}+...++\dfrac{1}{128}+\dfrac{1}{256}\)

\(5A=2-\dfrac{1}{256}=\dfrac{511}{256}\)

\(A=\dfrac{511}{1280}\)

1/5 + 1/5  - 1/10 + 1/10 - 1/20 + 1/20 - 1/40 + ... + 1/640 - 1/1280

= 1/5 + 1/5 - 1/1280 = 511/1280

Diện tích hình thang là: 

\(\dfrac{\left(13+17\right).15}{2}=225\) cm vuông

\(\left\{10^2-\left[3^2.5+2^3.\left(4+2^{12}:2^{11}\right)\right]\right\}.2\)

\(=\left\{100-\left[3^2.5+2^3.\left(4+2\right)\right]\right\}.2\)

\(=\left\{100-\left[9.5+8.6\right]\right\}.2\)

\(=\left\{100-93\right\}.2\)

\(=7.2=14\)

\(\left\{10^2-\left[3^2\times5+2^3\times\left(4+2^{12}:2^{11}\right)\right]\right\}\times2\)

\(=\left\{100-\left[9\times5+8\times\left(4+2\right)\right]\right\}\times2\)

\(=\left\{100-\left[9\times5+8\times6\right]\right\}\times2\)

\(=\left\{100-93\right\}\times2\)

\(=7\times2\)

\(=14\)

\(B=8+8^3+8^5+...+8^{99}\\ \\ 64B=8^3+8^5+8^7+...+8^{101}\\ \\ \Rightarrow64B-B=8^{101}-8\\ \\ 63B=8^{101}-8\\ \\ \Rightarrow B=\dfrac{8^{101}-8}{63}\)

\(B=8+8^3+8^5+8^7+...+8^{99}\)

\(\Rightarrow B=8\left(1+8^2+8^4+8^6+...+8^{98}\right)\)

\(\Rightarrow B+8B=8.\left(8+8^3+8^5+8^7+...+8^{99}\right)+8\left(1+8^2+8^4+8^6+...+8^{98}\right)\)

\(\Rightarrow9B=8\left(1+8^1+8^2+8^3+8^4+...+8^{99}\right)\)

\(\Rightarrow9B=8.\dfrac{8^{99+1}}{8-1}-1\)

\(\Rightarrow B=8.\dfrac{8^{99+1}-1}{7.9}=\dfrac{8^{101}-8}{63}\)

\(\left\{{}\begin{matrix}a+b=221\\UCLN\left(a;b\right)=13\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=13m\\b=13n\\\left(m;n\right)=1\end{matrix}\right.\)

\(\Rightarrow13m+13n=221\)

\(\Rightarrow13\left(m+n\right)=221\)

\(\Rightarrow m+n=17\)

- Với \(\left\{{}\begin{matrix}m=16\\n=1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=208\\b=13\end{matrix}\right.\)

- Với \(\left\{{}\begin{matrix}m=14\\n=3\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=182\\b=39\end{matrix}\right.\)

- Với \(\left\{{}\begin{matrix}m=12\\n=5\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=156\\b=65\end{matrix}\right.\)

- Với \(\left\{{}\begin{matrix}m=10\\n=7\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=130\\b=91\end{matrix}\right.\)

- Với \(\left\{{}\begin{matrix}m=6\\n=11\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=78\\b=143\end{matrix}\right.\)

\(\Rightarrow\left(a;b\right)\in\left\{\left(108;13\right);\left(182;39\right);\left(156:65\right);\left(130;91\right);\left(78;143\right)\right\}\)