\(\left(2x-3\right)\left(x-\dfrac{1}{2}\right)=0\\ \Rightarrow\left[{}\begin{matrix}2x-3=0\\x-\dfrac{1}{2}=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}2x=3\\x=\dfrac{1}{2}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{1}{2}\end{matrix}\right.\) (Thêm KL cuối dòng: Vậy \(x\in\left\{\dfrac{3}{2};\dfrac{1}{2}\right\}\))

(2x-3)x(x-1/2)=0

Đặt từng nhân tử bằng không và giải cho x:

2x - 3 = 0

2x = 3

x = 3/2

x = 0

x - 1/2 = 0

x = 1/2

Bạn xem lại đề, không thấy x

Xin nỗi mik ghi nhầm ko phải tìm x đâu :(
 

\(a,\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{1999\cdot2000}\\ =1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{1999}-\dfrac{1}{2000}\\ =1-\dfrac{1}{2000}\\ =\dfrac{1999}{2000}\\ b,\dfrac{1}{1\cdot4}+\dfrac{1}{4\cdot7}+\dfrac{1}{7\cdot10}+...+\dfrac{1}{100\cdot103}\\ =\dfrac{1}{3}\left(\dfrac{3}{1\cdot4}+\dfrac{3}{4\cdot7}+...+\dfrac{3}{100\cdot103}\right)\\ =\dfrac{1}{3}\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{100}-\dfrac{1}{103}\right)\\ =\dfrac{1}{3}\cdot\left(1-\dfrac{1}{103}\right)\\ =\dfrac{102}{309}\)

\(c,\dfrac{8}{9}-\dfrac{1}{2}-\dfrac{1}{6}-...-\dfrac{1}{72}\\ =\dfrac{8}{9}-\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{8\cdot9}\right)\\ =\dfrac{8}{9}-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{8}-\dfrac{1}{9}\right)\\ =\dfrac{8}{9}-\left(1-\dfrac{1}{9}\right)\\ =\dfrac{8}{9}-\dfrac{8}{9}\\ =0\)

\(\left(5-x\right)\left(7-x\right)< 0\)

\(5-x=0\Rightarrow x=5\)

\(7-x=0\Rightarrow x=7\)

Lập bảng xét dấu

\(\Rightarrow5< x< 7\)

5-x=0,x=5 vì 5-5=0

7-x=0,x=7 vì 7-7=0

⇒5<x<7

chúc bạn học tốt

\(A=\dfrac{2}{4.7}-\dfrac{3}{5.9}+\dfrac{2}{7.10}-\dfrac{3}{9.13}+...+\dfrac{2}{301.304}-\dfrac{3}{401.405}\)

\(A=\dfrac{2}{4.7}+\dfrac{2}{7.10}+\dfrac{2}{301.304}...-\left(\dfrac{3}{5.9}+\dfrac{3}{9.13}+...+\dfrac{3}{401.405}\right)\)

\(A=2\left(\dfrac{1}{4.7}+\dfrac{1}{7.10}+\dfrac{1}{301.304}\right)...-3\left(\dfrac{1}{5.9}+\dfrac{1}{9.13}+...+\dfrac{1}{401.405}\right)\)

\(A=2\left(\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+\dfrac{1}{301}-\dfrac{1}{304}\right)...-3\left(\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+...+\dfrac{1}{401}-\dfrac{1}{405}\right)\)

\(A=2\left(\dfrac{1}{4}-\dfrac{1}{304}\right)-3\left(\dfrac{1}{5}-\dfrac{1}{405}\right)\)

\(A=2\left(\dfrac{76}{304}-\dfrac{1}{304}\right)-3\left(\dfrac{81}{5}-\dfrac{1}{405}\right)\)

\(A=2.\dfrac{75}{304}-3.\dfrac{80}{405}=\dfrac{75}{152}-\dfrac{80}{135}=\dfrac{10125-12160}{152.135}=-\dfrac{2035}{152.135}=-\dfrac{407}{4104}\)

Bạn xem lại đề

\(125.5^2.\dfrac{1}{625}.5^3=5^3.5^2.\dfrac{1}{5^4}.5^3=5^{3+2-4+3}=5^4\\ 8.32.\left(2^4.\dfrac{1}{32}\right)=2^3.2^5.2^4.\dfrac{1}{2^5}=2^{3+5+4-5}=2^7\\ 6^3.5^2.\left(\dfrac{5}{6}\right)^3=6^3.5^2.5^3:6^3=5^{2+3}.6^{3-3}=5^5.6^0=5^5.1=5^5\\ Bài.5A\)

\(Bài.5B\\ a,2401.\left(\dfrac{1}{7}\right)^2.\dfrac{1}{7}.49^2=7^4.\left(\dfrac{1}{7}\right)^3.\left(7^2\right)^2=7^4.\dfrac{1}{7^3}.7^4=7^{4-3+4}=7^5\\ b,9.81:\left(3^5.\dfrac{1}{27}\right)=3^2.3^4:\left(3^5.\dfrac{1}{3^3}\right)=3^{2+4}:\left(3^{5-3}\right)=3^6:3^2=3^{6-2}=3^4\\ c,3^4.7^2.\left(\dfrac{7}{3}\right)^4=3^4.7^2.7^4:3^4=\left(3^4:3^4\right).\left(7^2.7^4\right)=1.7^6=7^6\)

\(x< 100\Leftrightarrow x\le99\Leftrightarrow x-2\le97\)

 Ta thấy \(\left(x-2\right)⋮5,\left(x-2\right)⋮14\) mà \(\left(5,14\right)=1\) nên \(\left(x-2\right)⋮14.5=70\). Mà \(x-2\le97\) nên \(\left(x-2\right)\in\left\{0;70\right\}\) \(\Leftrightarrow x\in\left\{2;72\right\}\) . 

 Vậy \(x=2\) và \(x=72\) là các số thỏa ycbt.