Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
ĐK: \(x\in Z\)
a) Giải:
Để \(A\) đạt giá trị lớn nhất
\(\Leftrightarrow\dfrac{2002}{\left|x\right|+2002}\) đạt giá trị lớn nhất
\(\Leftrightarrow\left|x\right|+2002\) phải nhỏ nhất \(\Leftrightarrow\left|x\right|=0\)
\(\Rightarrow A_{Max}=\dfrac{2002}{0+2002}=\dfrac{2002}{2002}=1\)
Vậy giá trị lớn nhất của \(A\) là \(1\)
b) Để \(B\) đạt giá trị lớn nhất
\(\Leftrightarrow\dfrac{\left|x\right|+2002}{-2003}\) phải lớn nhất
Vì \(\left\{{}\begin{matrix}\left|x\right|+2002>0\\-2003< 0\end{matrix}\right.\)\(\Rightarrow\dfrac{\left|x\right|+2002}{-2003}< 0\)
Mà \(\forall-a< 0\) nếu muốn \(-a\) lớn nhất \(\Leftrightarrow a\) nhỏ nhất
\(\Leftrightarrow\left|x\right|+2002\) phải nhỏ nhất \(\Leftrightarrow\left|x\right|=0\)
\(\Rightarrow B_{Max}=\dfrac{0+2002}{-2003}=\dfrac{2002}{-2003}\)
Vậy giá trị lớn nhất của \(B\) là \(\dfrac{2002}{-2003}\)
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) \((5x-1)^6=729=3^6=(-3)^6\)
\(\Rightarrow \left[\begin{matrix} 5x-1=3\\ 5x-1=-3\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=\frac{4}{5}\\ x=\frac{-2}{5}\end{matrix}\right.\)
b)
\(\frac{8}{25}=\frac{2^x}{5^{x-1}}=\frac{2^x}{5^x:5}=5.(\frac{2}{5})^x\)
\(\Rightarrow \frac{8}{125}=(\frac{2}{5})^x\)
\(\Rightarrow (\frac{2}{5})^3=(\frac{2}{5})^x\Rightarrow x=3\)
c)
\((\frac{1}{16})^x=(\frac{1}{2})^{10}\)
\(\Rightarrow (\frac{1}{2^4})^x=(\frac{1}{2})^{10}\)
\(\Rightarrow (\frac{1}{2})^{4x}=(\frac{1}{2})^{10}\Rightarrow 4x=10\Rightarrow x=\frac{5}{2}\)
d)
\(9^{x}:3^x=3\Rightarrow (\frac{9}{3})^x=3\)
\(\Rightarrow 3^x=3^1\Rightarrow x=1\)
a)\(\frac{5}{2}-3\left(\frac{1}{3}-x\right)=\frac{1}{4}-7x\)
\(\Leftrightarrow\frac{5}{2}-1+x=\frac{1}{4}-7x\)
\(\Leftrightarrow8x=-\frac{5}{4}\)
\(\Leftrightarrow x=-\frac{5}{32}\)
c)\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2001}{2003}\)
\(\Leftrightarrow2\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2001}{2003}\)
\(\Leftrightarrow\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}=\frac{2001}{4006}\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2001}{4006}\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2001}{4006}\)
\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{2003}\)
\(\Leftrightarrow x+1=2003\)
\(\Leftrightarrow x=2002\)
a) (1/7.x-2/7).(-1/5.x-2/5)=0
=> 1/7.x-2/7=0hoặc-1/5.x-2/5=0
*1/7.x-2/7=0
1/7.x=0+2/7
1/7.x=2/7
x=2/7:1/7
x=2
b)1/6.x+1/10.x-4/5.x+1=0
(1/6+1/10-4/5).x+1=0
(1/6+1/10-4/5).x=0-1
(1/6+1/10-4/5).x=-1
(-8/15).x=-1
x=-1:(-8/15) =15/8
a)Ta thấy:
\(\dfrac{1}{x}-\dfrac{1}{x+a}=\dfrac{x+a}{x\left(x+a\right)}-\dfrac{x}{x\left(x+a\right)}\)
\(=\dfrac{\left(x+a\right)-x}{x\left(x+a\right)}\)
\(=\dfrac{a}{x\left(x+a\right)}\)
\(\Rightarrowđpcm\)
b)Ta thấy:
\(\dfrac{1}{x\left(x+1\right)}-\dfrac{1}{\left(x+1\right)\left(x+2\right)}\)
\(=\dfrac{\left(x+1\right)\left(x+2\right)}{x\left(x+1\right)^2\left(x+2\right)}-\dfrac{x\left(x+1\right)}{x\left(x+1\right)^2\left(x+2\right)}\)
\(=\dfrac{x+2}{x\left(x+1\right)\left(x+2\right)}-\dfrac{x}{x\left(x+1\right)\left(x+2\right)}\)
\(=\dfrac{\left(x+2\right)-x}{x\left(x+1\right)\left(x+2\right)}=\dfrac{2}{x\left(x+1\right)\left(x+2\right)}\Rightarrowđpcm\)
c)Ta thấy:
\(\dfrac{1}{x\left(x+1\right)\left(x+2\right)}-\dfrac{1}{\left(x+1\right)\left(x+2\right)\left(x+3\right)}\)
\(=\dfrac{\left(x+1\right)\left(x+2\right)\left(x+3\right)}{x\left(x+1\right)^2\left(x+2\right)^2\left(x+3\right)}-\dfrac{x\left(x+1\right)\left(x+2\right)}{x\left(x+1\right)^2\left(x+2\right)^2\left(x+3\right)}=\dfrac{x+3}{x\left(x+1\right)\left(x+2\right)\left(x+3\right)}-\dfrac{x}{x\left(x+1\right)\left(x+2\right)\left(x+3\right)}=\dfrac{x+3-x}{x\left(x+1\right)\left(x+2\right)\left(x+3\right)}=\dfrac{3}{x\left(x+1\right)\left(x+2\right)\left(x+3\right)}\Rightarrowđpcm\)
a/ \(\dfrac{1}{x}-\dfrac{1}{x+a}=\dfrac{a}{x\left(x+a\right)}\)
Ta có: \(\dfrac{1}{x}-\dfrac{1}{x+a}=\dfrac{x+a}{x\left(x+a\right)}-\dfrac{x}{x\left(x+a\right)}\)
\(=\dfrac{\left(x-x\right)+a}{x\left(x+a\right)}\) hay \(\dfrac{a}{x\left(x+a\right)}\)
\(\Rightarrow\dfrac{1}{x}-\dfrac{1}{x+a}=\dfrac{a}{x\left(x+a\right)}\left(đpcm\right)\)
Bài 3: A=2018-|x+2019|. Vì |x+2019|\(\ge\)0 nên -|x+2019|\(\le\)0=>2018-|x+2019|\(\le\) 2. Vậy A có GTLN = 2 khi x+2019=0 hay x=-2019. B=-10-\(\left|2x-\dfrac{1}{1009}\right|\). Vì \(\left|2x-\dfrac{1}{1009}\right|\ge0\Rightarrow-\left|2x-\dfrac{1}{1009}\right|\le0\Rightarrow-10-\left|2x-\dfrac{1}{1009}\right|\le-10\). Vậy B có GTLN = -10 khi 2x-\(\dfrac{1}{1009}=0\) => \(2x=\dfrac{1}{1009}\Rightarrow x=\dfrac{1}{1009}:2=\dfrac{1}{2018}\)
Bài 2: A=\(\left|5x+1\right|-\dfrac{3}{8}\). Vì \(\left|5x+1\right|\ge0\Rightarrow\left|5x+1\right|-\dfrac{3}{8}\ge\dfrac{-3}{8}\). Vậy A có GTNN = \(\dfrac{-3}{8}\) khi 5x+1= 0=> 5x= -1=> x = \(\dfrac{-1}{5}\). B=\(\left|2-\dfrac{1}{6}x\right|+0,25\) , vì \(\left|2-\dfrac{1}{6}x\right|\ge0\Rightarrow\left|2-\dfrac{1}{6}x\right|+0,25\ge0,25\) . Vậy B có GTNN = 0,25 khi \(2-\dfrac{1}{6}x=0\Rightarrow\dfrac{x}{6}=2\Rightarrow x=2.6=12\)
ak đề bài là tìm giá trị nhỏ nhất của các biểu thức
Đề bài là gì ạ?