Tập hợp các phần x tử nguyên dương thỏa mãn

|2x-7|\(\le\)25

a)\(\left|x+1\right|-x+1=2\)

\(\Rightarrow\left|x+1\right|=1+x\)

\(\Rightarrow\left[{}\begin{matrix}x+1=-1-x\\x+1=1+x\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-1\\x=0\end{matrix}\right.\)

b)\(\left|x+1\right|=\left|2x+3\right|\)

\(\Rightarrow\left|x+1\right|-\left|2x+3\right|=0\)

Mà |x+1|\(\ge0\);\(\left|2x-3\right|\ge0\)

\(\Rightarrow\left\{{}\begin{matrix}x+1=0\\2x-3=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-1\\x=\dfrac{3}{2}\end{matrix}\right.\)

\(\dfrac{1}{1.4}+\dfrac{1}{4.7}+\dfrac{1}{7.10}+...+\dfrac{1}{100.103}\)

\(=\dfrac{1}{1}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{100}-\dfrac{1}{103}\)

\(=\dfrac{1}{1}-\dfrac{1}{103}\)

\(=\dfrac{102}{103}\)

Ta có:\(\left|x\left(x+2\right)\right|=x\)

\(\Rightarrow\left[{}\begin{matrix}x\left(x+2\right)=x\\x\left(x+2\right)=-x\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-1\\x=-3\end{matrix}\right.\)

Vậy \(x\in\left\{-3;-1\right\}\)

Thuyết trình về cột sắt không gỉ

a) \(7^6:7^2=7^5\)

b)\(16^2:4^2=4^2\)

c)\(25^2:5^2=5^2\)

d)\(3^2:9^3=3^2:3^6=3^{-4}\)

2

a)\(\dfrac{x-1}{x+5}=\dfrac{6}{7}\)

\(\Rightarrow6x+30=7x-7\)

\(\Rightarrow7+30=7x-6x\)

\(\Rightarrow x=37\)

Vậy x=37

b)\(\left(x-1\right)^{x+2}=\left(x-1\right)^{x+4}\)

\(\Rightarrow\left(x-1\right)^{x+2}-\left(x-1\right)^{x+4}=0\)

\(\Rightarrow\left(x-1\right)^{x+2}\left[1-\left(x-1\right)^2\right]=0\)

\(\Rightarrow\left[{}\begin{matrix}\left(x-1\right)^{x+2}=0\\1-\left(x-1\right)^2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)

Vậy \(x\in\left\{1;2\right\}\)

\(x^{10}=x\)

\(\Rightarrow x^{10}-x=0\)

\(\Rightarrow x\left(x^9-1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x^9-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)

Vậy x \(\in\left\{0;1\right\}\)

18.1+4+7+10+13+16+19+...+.97+100

=18+(100+1)((100-1);3+1):2

=18+1717

=1735