Rút gọn:
\(A=\dfrac{6!}{\left(m-2\right)\left(m-3\right)}.\left[\dfrac{1}{\left(m+1\right)\left(m-4\right)}.\dfrac{\left(m+1\right)!}{\left(m-5\right)!5!}-\dfrac{m\left(m-1\right)!}{12.\left(m-4\right)!3!}\right]\) với \(m\ge5\)
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6 tháng 12 2022
\(=\dfrac{1}{x}-\dfrac{1}{x+1}+\dfrac{1}{x+1}-\dfrac{1}{x+2}+...+\dfrac{1}{x+5}-\dfrac{1}{x+6}\)
=1/x-1/x+6
\(=\dfrac{x+6-x}{x\left(x+6\right)}=\dfrac{6}{x\left(x+6\right)}\)
1 tháng 1 2019
\(A=\dfrac{\left(1+\dfrac{1}{4}\right)\left(3^4+\dfrac{1}{4}\right)........\left(51^4+\dfrac{1}{4}\right)}{\left(2^4+\dfrac{1}{4}\right)\left(4^4+\dfrac{1}{4}\right).......\left(52^4+\dfrac{1}{4}\right)}\)
\(=\dfrac{\left(1+1+\dfrac{1}{2}\right)\left(1-1+\dfrac{1}{2}\right)........\left(11^2-11+\dfrac{1}{2}\right)}{\left(2^2+2+\dfrac{1}{2}\right)\left(2^2-2+\dfrac{1}{2}\right)........\left(12^2-12+\dfrac{1}{2}\right)}\)
\(=\dfrac{\dfrac{1}{2}\left(1.2+\dfrac{1}{2}\right)\left(2.3+\dfrac{1}{2}\right)........\left(11.12+\dfrac{1}{2}\right)}{\left(2.3+\dfrac{1}{2}\right)\left(3.4+\dfrac{1}{2}\right)......\left(12.13+\dfrac{1}{2}\right)}\)
\(=\dfrac{\dfrac{1}{2}}{12.13+\dfrac{1}{2}}\)
\(=\dfrac{1}{313}\)
24 tháng 9 2017
a,\(\dfrac{2^{19}.27^3+15.4^9.9^4}{6^9.2^{10}+12^{10}}\)=\(\dfrac{2^{19}.\left(3^3\right)^3+3.5.\left(2^2\right)^9.\left(3^2\right)^4}{6^9.2^{10}+6^{10}.2^{10}}\)
=\(\dfrac{2^{19}.3^9+2^{18}.3^9.5}{6^9.2^{10}.\left(1+6\right)}\)=\(\dfrac{2^{18}.3^9.\left(2+5\right)}{6^9.2^{10}.7}\)=\(\dfrac{2^{18}.3^9}{6^9.2^{10}}=\dfrac{2^{10}.2^8.3^9}{2^9.3^9.2^{10}}=\dfrac{2^8}{2^8.2}=\dfrac{1}{2}\)
b, \(\dfrac{\left(\dfrac{-1}{2}\right)^3-\left(\dfrac{3}{4}\right)^3.\left(-2\right)^2}{2.\left(-1\right)^5+\left(\dfrac{3}{4}\right)^2-\dfrac{3}{8}}=\dfrac{\dfrac{-1}{8}-\dfrac{27}{64}.4}{-2+\dfrac{9}{16}-\dfrac{3}{8}}=\dfrac{\dfrac{-1}{8}-\dfrac{27}{16}}{\dfrac{-23}{16}-\dfrac{3}{8}}=\dfrac{\dfrac{-29}{16}}{\dfrac{-29}{16}}=1\)
AH
Akai Haruma
Giáo viên
29 tháng 8 2018
Bài 1:
a)
\(|x+\frac{4}{15}|-|-3,75|=-|-2,15|\)
\(\Leftrightarrow |x+\frac{4}{15}|-3,75=-2,15\)
\(\Leftrightarrow |x+\frac{4}{15}|=-2,15+3,75=\frac{8}{5}\)
\(\Rightarrow \left[\begin{matrix} x+\frac{4}{15}=\frac{8}{5}\\ x+\frac{4}{15}=-\frac{8}{5}\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=\frac{4}{3}\\ x=\frac{-28}{15}\end{matrix}\right.\)
b )
\(|\frac{5}{3}x|=|-\frac{1}{6}|=\frac{1}{6}\)
\(\Rightarrow \left[\begin{matrix} \frac{5}{3}x=\frac{1}{6}\\ \frac{5}{3}x=-\frac{1}{6}\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=\frac{1}{10}\\ x=-\frac{1}{10}\end{matrix}\right.\)
c)
\(|\frac{3}{4}x-\frac{3}{4}|-\frac{3}{4}=|-\frac{3}{4}|=\frac{3}{4}\)
\(\Leftrightarrow |\frac{3}{4}x-\frac{3}{4}|=\frac{3}{2}\)
\(\Rightarrow \left[\begin{matrix} \frac{3}{4}x-\frac{3}{4}=\frac{3}{2}\\ \frac{3}{4}x-\frac{3}{4}=-\frac{3}{2}\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=3\\ x=-1\end{matrix}\right.\)
AH
Akai Haruma
Giáo viên
29 tháng 8 2018
Bài 3:
a) Ta thấy:
\(|x+\frac{15}{19}|\geq 0, \forall x\Rightarrow A\ge 0-1=-1\)
Vậy GTNN của $A$ là $-1$ khi \(x+\frac{15}{19}=0\Leftrightarrow x=-\frac{15}{19}\)
b)Vì \(|x-\frac{4}{7}|\geq 0, \forall x\Rightarrow B\geq \frac{1}{2}+0=\frac{1}{2}\)
Vậy GTNN của $B$ là $\frac{1}{2}$ khi \(x-\frac{4}{7}=0\Leftrightarrow x=\frac{4}{7}\)
29 tháng 7 2022
Bài 1:
a: \(\Leftrightarrow\left|x+\dfrac{4}{15}\right|=-2.15+3.75=\dfrac{8}{5}\)
=>x+4/15=8/5 hoặc x+4/15=-8/5
=>x=4/3 hoặc x=-28/15
b: \(\Leftrightarrow\left[{}\begin{matrix}\dfrac{5}{3}x=-\dfrac{1}{6}\\\dfrac{5}{3}x=\dfrac{1}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-1}{6}:\dfrac{5}{3}=\dfrac{-3}{30}=\dfrac{-1}{10}\\x=\dfrac{1}{10}\end{matrix}\right.\)
c: \(\Leftrightarrow\left|x-1\right|-1=1\)
=>|x-1|=2
=>x-1=2 hoặc x-1=-2
=>x=3 hoặc x=-1
Bài 2:
b: \(\Leftrightarrow\left\{{}\begin{matrix}x-y=0\\y+\dfrac{9}{25}=0\end{matrix}\right.\Leftrightarrow x=y=-\dfrac{9}{25}\)
Bài 3:
a: \(A=\left|x+\dfrac{15}{19}\right|-1>=-1\)
Dấu '=' xảy ra khi x=-15/19
b: \(\left|x-\dfrac{4}{7}\right|+\dfrac{1}{2}>=\dfrac{1}{2}\)
Dấu '=' xảy ra khi x=4/7
TV
20 tháng 7 2018
bai 1
a) \(\left|x+\dfrac{4}{15}\right|-\left|-3,75\right|=-\left|2,15\right|\)
\(\left|x+\dfrac{4}{15}\right|-3,75=-2,,15\)
\(\left|x+\dfrac{4}{15}\right|=-2,15+3,75=1,6\)
\(\Rightarrow\left[{}\begin{matrix}x+\dfrac{4}{15}=1,6\\x+\dfrac{4}{15}=-1,6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{4}{3}\\x=-\dfrac{28}{15}\end{matrix}\right.\)
Vậy ....
b) \(\left|\dfrac{5}{3}x\right|=\left|-\dfrac{1}{6}\right|\)
\(\left|\dfrac{5}{3}x\right|=\dfrac{1}{6}\)
\(\Rightarrow\left[{}\begin{matrix}\dfrac{5}{3}x=-\dfrac{1}{6}\\\dfrac{5}{3}x=\dfrac{1}{6}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{10}\\x=\dfrac{1}{10}\end{matrix}\right.\)
c) \(\left|\dfrac{3}{4}x-\dfrac{3}{4}\right|-\dfrac{3}{4}=\left|-\dfrac{3}{4}\right|\)
\(\left|\dfrac{3}{4}x-\dfrac{3}{4}\right|-\dfrac{3}{4}=\dfrac{3}{4}\)
\(\left|\dfrac{3}{4}x-\dfrac{3}{4}\right|=\dfrac{3}{2}\)
\(\Rightarrow\left[{}\begin{matrix}\dfrac{3}{4}x-\dfrac{3}{4}=\dfrac{3}{2}\\\dfrac{3}{4}x-\dfrac{3}{4}=-\dfrac{3}{2}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\\-1\end{matrix}\right.\)
TV
20 tháng 7 2018
bai 2
a) \(\left|\dfrac{1}{2}-\dfrac{1}{3}+x\right|=\dfrac{1}{4}-\left|y\right|\)
\(\left|\dfrac{1}{6}+x\right|=\dfrac{1}{4}-\left|y\right|\) (*)
với mọi x ta luôn có \(\left|\dfrac{1}{6}+x\right|\ge0\)
\(\Rightarrow\dfrac{1}{4}-\left|y\right|\ge0\)
\(\Rightarrow\left|y\right|\le\dfrac{1}{4}\) \(\Rightarrow\dfrac{1}{4}-\left|y\right|=\left|\dfrac{1}{4}-y\right|\)
Nên từ * \(\Rightarrow\left|\dfrac{1}{6}+x\right|=\left|\dfrac{1}{4}-y\right|\)
\(\Rightarrow\left|\dfrac{1}{6}+x\right|-\left|\dfrac{1}{4}-y\right|=0\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{6}+x=0\\\dfrac{1}{4}-y=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{1}{6}\\y=\dfrac{1}{4}\end{matrix}\right.\)
b) \(\left|x-y\right|+\left|y+25\right|=0\)
với mọi x, y tao luôn có \(\left\{{}\begin{matrix}\left|x-y\right|\ge0\\\left|y+25\right|\ge0\end{matrix}\right.\)
mà \(\left|x-y\right|+\left|y+25\right|=0\)
\(\Rightarrow\left\{{}\begin{matrix}\left|x-y\right|=0\\\left|y+25\right|=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=y\\y=-25\end{matrix}\right.\Rightarrow}\left\{{}\begin{matrix}x=-25\\y=-25\end{matrix}\right.\)
29 tháng 7 2022
Bài 1:
a: \(\Leftrightarrow\left|x+\dfrac{4}{15}\right|=-2.15+3.75=\dfrac{8}{5}\)
=>x+4/15=8/5 hoặc x+4/15=-8/5
=>x=4/3 hoặc x=-28/15
b: \(\Leftrightarrow\left[{}\begin{matrix}\dfrac{5}{3}x=-\dfrac{1}{6}\\\dfrac{5}{3}x=\dfrac{1}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-1}{6}:\dfrac{5}{3}=\dfrac{-3}{30}=\dfrac{-1}{10}\\x=\dfrac{1}{10}\end{matrix}\right.\)
c: \(\Leftrightarrow\left|x-1\right|-1=1\)
=>|x-1|=2
=>x-1=2 hoặc x-1=-2
=>x=3 hoặc x=-1
Bài 2:
b: \(\Leftrightarrow\left\{{}\begin{matrix}x-y=0\\y+\dfrac{9}{25}=0\end{matrix}\right.\Leftrightarrow x=y=-\dfrac{9}{25}\)
Bài 3:
a: \(A=\left|x+\dfrac{15}{19}\right|-1>=-1\)
Dấu '=' xảy ra khi x=-15/19
b: \(\left|x-\dfrac{4}{7}\right|+\dfrac{1}{2}>=\dfrac{1}{2}\)
Dấu '=' xảy ra khi x=4/7
Lời giải:
\(A=\frac{6!}{(m-2)(m-3)}\left[\frac{m!}{(m-4)!.5!}-\frac{m!}{(m-4)!3.4!}\right]\)
\(=\frac{6!}{(m-2)(m-3)}.\frac{m!}{(m-4)!}(\frac{1}{5!}-\frac{1}{3.4!})=\frac{-4}{(m-2)(m-3)}.\frac{m!}{(m-4)!}\)
\(=\frac{-4}{(m-2)(m-3)}.(m-3)(m-2)(m-1)m=-4m(m-1)\)