Vì số mới là khi viết thêm chữ số 7 vào bên trái số cần tìm có ba chữ số nên số mới hơn số cũ là 7000 đơn vị.

Hiệu số phần bằng nhau là:

\(36-1=35\) (phần)

Số cần tìm là:

\(7000:35\times1=200\)

Quãng đường người đó đi được dài:

\(16\times3+11\times2=70\left(km\right)\)

Trung bình mỗi giờ người đó đi được:

\(70:\left(3+2\right)=14\left(km\right)\)

\(\left|x-2016\right|-2015=2\\ \Rightarrow\left|x-2016\right|=2+2015\\ \Rightarrow\left|x-2016\right|=2017\\ \Rightarrow\left[{}\begin{matrix}x-2016=2017\\x-2016=-2017\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=4033\\x=-1\end{matrix}\right.\)

\(\dfrac{5.4^{15}.9^9-4.3^{20}.8^9}{5.2^{10}.6^{12}-7.2^{29}.27^6}\\ =\dfrac{5.\left(2^2\right)^{15}.\left(3^2\right)^9-2^2.3^{20}.\left(2^3\right)^9}{5.2^{10}.\left(2.3\right)^{12}-7.2^{29}.\left(3^3\right)^6}\\ =\dfrac{5.2^{30}.3^{18}-2^{29}.3^{20}}{5.2^{22}.3^{12}-7.2^{29}.3^{18}}\\ =\dfrac{2^{29}.3^{18}.\left(5.2-3^2\right)}{2^{22}.3^{12}.\left(5-7.2^7.3^6\right)}\\ =\dfrac{2^7.3^6}{5-7.2^7.3^6}\)

a)

\(A=1152-\left(374+1152\right)+\left(-65+374\right)\\ =\left(1152-1152\right)+\left(-374+374\right)-65\\ =-65\)

b)

\(B=\dfrac{1}{1.6}+\dfrac{1}{6.11}+\dfrac{1}{11.16}+...+\dfrac{1}{96.101}\\ =\dfrac{1}{5}\left(\dfrac{5}{1.6}+\dfrac{5}{6.11}+\dfrac{5}{11.16}+...+\dfrac{5}{96.101}\right)\\ =\dfrac{1}{5}\left(1-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{16}+...+\dfrac{1}{96}-\dfrac{1}{101}\right)\\ =\dfrac{1}{5}\left(1-\dfrac{1}{101}\right)\\ =\dfrac{1}{5}.\dfrac{100}{101}=\dfrac{20}{101}\)

\(3^{x+2}+3^{x+1}-3^x=33\\ 3^x.3^2+3^x.3-3^x=33\\ 3^x.\left(9+3-1\right)=33\\ 3^x.11=33\\ 3^x=33:11\\ 3^x=3\\ \Rightarrow x=1\)

Với \(a+b+c=0\Rightarrow\left\{{}\begin{matrix}a+b=-c\\b+c=-a\\c+a=-b\end{matrix}\right.\). Khi đó:

+,

\(\dfrac{a^2}{a^2-b^2-c^2}+\dfrac{b^2}{b^2-a^2-c^2}+\dfrac{c^2}{c^2-a^2-b^2}\\ =\dfrac{a^2}{a^2-\left(b^2+c^2\right)}+\dfrac{b^2}{b^2-\left(a^2+c^2\right)}+\dfrac{c^2}{c^2-\left(a^2+b^2\right)}\\ =\dfrac{a^2}{a^2-\left(b+c\right)^2+2bc}+\dfrac{b^2}{b^2-\left(a+c\right)^2+2ac}+\dfrac{c^2}{c^2-\left(a+b\right)^2+2ab}\\ =\dfrac{a^2}{a^2-\left(-a\right)^2+2bc}+\dfrac{b^2}{b^2-\left(-b\right)^2+2ac}+\dfrac{c^2}{c^2-\left(-c\right)^2+2ab}\\ =\dfrac{a^3}{2abc}+\dfrac{b^3}{2abc}+\dfrac{c^3}{2abc}\\ =\dfrac{\left(a+b+c\right)^3-3\left(a+b\right)\left(b+c\right)\left(c+a\right)}{2abc}\\ =\dfrac{0-3.\left(-c\right).\left(-a\right).\left(-b\right)}{2abc}=\dfrac{3}{2}\)

+,

\(\dfrac{1}{b^2+c^2-a^2}+\dfrac{1}{c^2+a^2-b^2}+\dfrac{1}{a^2+b^2-c^2}\\ =\dfrac{1}{\left(b+c\right)^2-a^2-2bc}+\dfrac{1}{\left(c+a\right)^2-b^2-2ac}+\dfrac{1}{\left(a+b\right)^2-c^2-2ab}\\ =\dfrac{1}{\left(-a\right)^2-a^2-2bc}+\dfrac{1}{\left(-b\right)^2-b^2-2ac}+\dfrac{1}{\left(-c\right)^2-c^2-2ab}\\ =\dfrac{1}{-2bc}+\dfrac{1}{-2ac}+\dfrac{1}{-2ab}\\ =-\dfrac{a+b+c}{2abc}=0\)

#$\mathtt{Toru}$