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a ) \(5\left(x^2\right)+7x+2\)
\(\Leftrightarrow5x^2+7x+2=0\)
\(\Leftrightarrow5x^2+5x+2x+2=0\)
\(\Leftrightarrow\left(5x+2\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{2}{5}\\x=-1\end{matrix}\right.\)
Vậy .............
b ) \(\dfrac{x+1}{17}+\dfrac{x+2}{16}=\dfrac{x+3}{15}+\dfrac{x+4}{14}\)
\(\Leftrightarrow\dfrac{x+1}{17}+1+\dfrac{x+2}{16}+1=\dfrac{x+3}{15}+1+\dfrac{x+4}{14}+1\)
\(\Leftrightarrow\dfrac{x+18}{17}+\dfrac{x+18}{16}=\dfrac{x+18}{15}+\dfrac{x+18}{14}\)
\(\Leftrightarrow\dfrac{x+18}{17}+\dfrac{x+18}{16}-\dfrac{x+18}{15}-\dfrac{x+18}{14}=0\)
\(\Leftrightarrow\left(x+18\right)\left(\dfrac{1}{17}+\dfrac{1}{16}-\dfrac{1}{15}-\dfrac{1}{14}\right)=0\)
Vì \(\left(\dfrac{1}{17}+\dfrac{1}{16}-\dfrac{1}{15}-\dfrac{1}{14}\right)\ne0\)
Ta có : \(x+18=0\Leftrightarrow x=-18\)
Vậy ......
c ) \(\dfrac{x-1}{x-3}=\dfrac{x-4}{x-7}\)
\(\Leftrightarrow\left(x-1\right)\left(x-7\right)=\left(x-3\right)\left(x-4\right)\)
\(\Leftrightarrow x^2-7x-x+7=x^2-4x-3x+12\)
\(\Leftrightarrow-x=5\)
\(\Leftrightarrow x=-5\)
Vậy ..
\(a)\dfrac{1}{3}x+\dfrac{2}{5}\left(x+1\right)=0\)
\(\Leftrightarrow\dfrac{1}{3}x+\dfrac{2}{5}x+\dfrac{2}{5}=0\)
\(\Leftrightarrow x\left(\dfrac{5}{15}+\dfrac{6}{15}\right)=\dfrac{-2}{5}\)
\(\Leftrightarrow x.\dfrac{11}{15}=\dfrac{-2}{5}\)
\(\Leftrightarrow x=\dfrac{-2}{5}.\dfrac{15}{11}\)
\(\Leftrightarrow x=\dfrac{-6}{11}\)
\(\Leftrightarrow x\left(-\dfrac{7}{2}-\dfrac{5}{4}\right)=\dfrac{1}{4}\cdot\dfrac{-4}{3}+\dfrac{1}{6}+\dfrac{3}{4}\)
\(\Leftrightarrow x\cdot\dfrac{-19}{4}=\dfrac{7}{12}\)
hay x=-7/57
Gọi biểu thức trên là A:
Có 1/22=1/2.2<1/1.2; 1/32=1/3.3<1/2.3;.....;1/1002=1/100.100<1/99.100
=>A<1/1.2+1/2.3+1/3.4+....+1/99.100
=>A< 1/1-1/2+1/2-1/3+1/3-1/4+....+1/99-1/100
=>A<1-1/100=99/100
=>A<1 vì 99/100<1 mà A<99/100
Bài 1:
a)
\(\dfrac{x-1}{9}=\dfrac{8}{3}\\ \Leftrightarrow\dfrac{x-1}{9}=\dfrac{24}{9}\\ \Leftrightarrow x-1=24\\ x=24+1\\ x=25\)
b)
\(\left(\dfrac{3x}{7}+1\right):\left(-4\right)=\dfrac{-1}{8}\\ \dfrac{3x}{7}+1=\dfrac{-1}{8}\cdot\left(-4\right)\\ \dfrac{3x}{7}+1=\dfrac{1}{2}\\ \dfrac{3x}{7}=\dfrac{1}{2}-1\\ \dfrac{3x}{7}=\dfrac{-1}{2}\\ 3x=\dfrac{-1}{2}\cdot7\\ 3x=\dfrac{-7}{2}\\ x=\dfrac{-7}{2}:3\\ x=\dfrac{-7}{6}\)
c)
\(x+\dfrac{7}{12}=\dfrac{17}{18}-\dfrac{1}{9}\\ x+\dfrac{7}{12}=\dfrac{5}{6}\\ x=\dfrac{5}{6}-\dfrac{7}{12}\\ x=\dfrac{1}{4}\)
d)
\(0,5x-\dfrac{2}{3}x=\dfrac{7}{12}\\ \dfrac{1}{2}x-\dfrac{2}{3}x=\dfrac{7}{12}\\ x\cdot\left(\dfrac{1}{2}-\dfrac{2}{3}\right)=\dfrac{7}{12}\\ \dfrac{-1}{6}x=\dfrac{7}{12}\\ x=\dfrac{7}{12}:\dfrac{-1}{6}\\ x=\dfrac{-7}{2}\)
e)
\(\dfrac{29}{30}-\left(\dfrac{13}{23}+x\right)=\dfrac{7}{46}\\ \dfrac{29}{30}-\dfrac{13}{23}-x=\dfrac{7}{46}\\ \dfrac{277}{690}-x=\dfrac{7}{46}\\ x=\dfrac{277}{690}-\dfrac{7}{46}\\ x=\dfrac{86}{345}\)
f)
\(\left(x+\dfrac{1}{4}-\dfrac{1}{3}\right):\left(2+\dfrac{1}{6}-\dfrac{1}{4}\right)=\dfrac{7}{46}\\ \left(x-\dfrac{1}{12}\right):\dfrac{23}{12}=\dfrac{7}{46}\\ x-\dfrac{1}{12}=\dfrac{7}{46}\cdot\dfrac{23}{12}\\ x-\dfrac{1}{12}=\dfrac{7}{24}\\ x=\dfrac{7}{24}+\dfrac{1}{12}\\ x=\dfrac{3}{8}\)
g)
\(\dfrac{13}{15}-\left(\dfrac{13}{21}+x\right)\cdot\dfrac{7}{12}=\dfrac{7}{10}\\ \left(\dfrac{13}{21}+x\right)\cdot\dfrac{7}{12}=\dfrac{13}{15}-\dfrac{7}{10}\\ \left(\dfrac{13}{21}+x\right)\cdot\dfrac{7}{12}=\dfrac{1}{6}\\ \dfrac{13}{21}+x=\dfrac{1}{6}:\dfrac{7}{12}\\ \dfrac{13}{21}+x=\dfrac{2}{7}\\ x=\dfrac{2}{7}-\dfrac{13}{21}\\ x=\dfrac{-1}{3}\)
h)
\(2\cdot\left|\dfrac{1}{2}x-\dfrac{1}{3}\right|-\dfrac{3}{2}=\dfrac{1}{4}\\ 2\cdot\left|\dfrac{1}{2}x-\dfrac{1}{3}\right|=\dfrac{1}{4}+\dfrac{3}{2}\\ 2\cdot\left|\dfrac{1}{2}x-\dfrac{1}{3}\right|=\dfrac{7}{4}\\ \left|\dfrac{1}{2}x-\dfrac{1}{3}\right|=\dfrac{7}{4}:2\\ \left|\dfrac{1}{2}x-\dfrac{1}{3}\right|=\dfrac{7}{8}\Rightarrow\left[{}\begin{matrix}\dfrac{1}{2}x-\dfrac{1}{3}=\dfrac{7}{8}\\\dfrac{1}{2}x-\dfrac{1}{3}=\dfrac{-7}{8}\end{matrix}\right.\\ \dfrac{1}{2}x-\dfrac{1}{3}=\dfrac{7}{8}\\ \dfrac{1}{2}x=\dfrac{7}{8}+\dfrac{1}{3}\\ \dfrac{1}{2}x=\dfrac{29}{24}\\ x=\dfrac{29}{24}:\dfrac{1}{2}\\ x=\dfrac{29}{12}\\ \dfrac{1}{2}x-\dfrac{1}{3}=\dfrac{-7}{8}\\ \dfrac{1}{2}x=\dfrac{-7}{8}+\dfrac{1}{3}\\ \dfrac{1}{2}x=\dfrac{-13}{24}\\ x=\dfrac{-13}{24}:\dfrac{1}{2}\\ x=\dfrac{-13}{12}\)
i)
\(3\cdot\left(3x-\dfrac{1}{2}\right)^3+\dfrac{1}{9}=0\\ 3\cdot\left(3x-\dfrac{1}{2}\right)^3=0-\dfrac{1}{9}\\ 3\cdot\left(3x-\dfrac{1}{2}\right)^3=\dfrac{-1}{9}\\ \left(3x-\dfrac{1}{2}\right)^3=\dfrac{-1}{9}:3\\ \left(3x-\dfrac{1}{2}\right)^3=\dfrac{-1}{27}\\ \left(3x-\dfrac{1}{2}\right)^3=\left(\dfrac{-1}{3}\right)^3\\ \Leftrightarrow3x-\dfrac{1}{2}=\dfrac{-1}{3}\\ 3x=\dfrac{-1}{3}+\dfrac{1}{2}\\ 3x=\dfrac{1}{6}\\ x=\dfrac{1}{6}:3\\ x=\dfrac{1}{18}\)
mấy câu này trên mạng và chtt có đầy, bn nên tìm trc khi hỏi để tránh câu hỏi nhiều lần gây nhàm chán
A=(1-\(\dfrac{1}{2} \))(1-\(\dfrac{1}{3}\))(1-\(\dfrac{1}{4}\))....(1-\(\dfrac{1}{2014}\))
=\(\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}.....\dfrac{2013}{2014}\)
=\(\dfrac{1.2.3.4.....2013}{2.3.4....2014}\\ \)
=\(\dfrac{1}{2014}\)
B=12015:2015
=1:2015
=\(\dfrac{1}{2015}\)
So sánh: \(\dfrac{1}{2014}>\dfrac{1}{2015}\)
Câu 1:
a: AC=5-3=2(cm)
b: Trên tia CD, ta có: CA<CD
nên điểm A nằm giữa hai điểm C và D
mà CA=1/2CD
nên A là trung điểm của CD
Ta có: \(3\frac{1}{3}+7\frac{3}{17}\cdot\frac{2}{15}-2\frac{3}{17}\cdot\frac{2}{15}\)
\(=\frac{10}{3}+\frac{122}{17}\cdot\frac{2}{15}-\frac{37}{17}\cdot\frac{2}{15}\)
\(=\frac{10}{3}+\frac{244}{255}-\frac{74}{255}\)
\(=\frac{10}{3}+\frac{2}{3}=\frac{12}{3}=4\)
\(1+\dfrac{1}{2}\left(1+2\right)+\dfrac{1}{3}\left(1+2+3\right)+...+\dfrac{1}{16}\left(1+2+...+16\right)\)
\(=1+\dfrac{2.3}{2.2}+\dfrac{3.4}{3.2}+...+\dfrac{16.17}{16.2}\)
\(=\dfrac{2}{2}+\dfrac{3}{2}+\dfrac{4}{2}+...+\dfrac{17}{2}\)
\(=\dfrac{1}{2}\left(2+3+4+...+17\right)\)
\(=\dfrac{1}{2}.152\)
\(=76\)