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Câu 1:
Ta có: \(\left[\dfrac{1}{2.5}+\dfrac{1}{5.8}+...+\dfrac{1}{65.68}\right]x-\dfrac{7}{34}=\dfrac{19}{68}\)
\(\Rightarrow\left[\dfrac{1}{3}\left(\dfrac{3}{2.5}+\dfrac{3}{5.8}+...+\dfrac{3}{65.68}\right)\right]x=\dfrac{33}{68}\)
\(\Rightarrow\left[\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+...+\dfrac{1}{65}-\dfrac{1}{68}\right)\right]x=\dfrac{33}{68}\)
\(\Rightarrow\left[\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{68}\right)\right]x=\dfrac{33}{68}\)
\(\Rightarrow\dfrac{11}{68}x=\dfrac{33}{68}\)
\(\Rightarrow x=3\)
Vậy \(x=3.\)
a/ \(A=\dfrac{2012}{\left|x\right|+2013}\)
vì: \(\left|x\right|\ge0\Rightarrow\left|x\right|+2013\ge2013\)
=> \(\dfrac{2012}{\left|x\right|+2013}\le\dfrac{2012}{2013}\)
Dấu ''='' xảy ra khi x = 0
Vậy MAXA = 2012/2013 khi x = 0
b/ \(B=\dfrac{\left|x\right|+2012}{-2013}\)
Vì: \(\left|x\right|\ge0\Rightarrow\left|x\right|+2012\ge2012\)
=> \(\Rightarrow\dfrac{\left|x\right|+2012}{-2013}\le-\dfrac{2012}{2013}\)
Dấu ''='' xảy ra khi x = 0
Vậy.........
Bài 2: Ăn cơm xoq lm cho
Bài 2:
a, Để C nhỏ nhất thì /x/+2012 phải nhỏ nhất
Mà /x/ luôn lớn hơn hoặc bằng 0 => /x/+2012 nhỏ nhất khi /x/ =0
=> x+0, GTNN của C=\(\dfrac{0+2012}{2013}=\dfrac{2012}{2013}\)khi x=0
a) Ta có:
\(\frac{x+11}{12}+\frac{x+11}{13}+\frac{x+11}{14}=\frac{x+11}{15}+\frac{x+11}{16}\)
\(\Rightarrow\left(x+11\right)\left(\frac{1}{12}+\frac{1}{13}+\frac{1}{14}\right)=\left(x+11\right)\left(\frac{1}{15}+\frac{1}{16}\right)\)
Mà ta có:
\(\frac{1}{12}+\frac{1}{13}+\frac{1}{14}\ne\frac{1}{15}+\frac{1}{16}\)
\(\Rightarrow x+11=0\Rightarrow x=-11\)
Ta có:
\(A=1+x+x^2+x^3+...+x^{100}\)
Đặt \(B=x+x^2+x^3+...+x^{100}\)
\(\Rightarrow B=\left(-11\right)+\left(-11\right)^2+\left(-11\right)^3+...+\left(-11\right)^{100}\)
\(\Rightarrow-11B=\left(-11\right)^2+\left(-11\right)^3+\left(-11\right)^4+...+\left(-11\right)^{101}\)
\(\Rightarrow-11B-B=\left(-11\right)^{101}-\left(-11\right)\)
\(\Rightarrow-12B=\left(-11\right)^{101}+11\Rightarrow B=\frac{\left(-11\right)^{101}+11}{-12}\)
\(\Rightarrow A=1+B=\frac{\left(-11\right)^{101}+11}{-12}+1\)
a) 9.33.\(\dfrac{1}{81}\) .32 = 32. 33.\(\dfrac{1}{3^4}\) . 32 = 33
b) 4. 25: \(\) (23.\(\dfrac{1}{16}\))= 22. 25: 23. \(\dfrac{1}{2^4}\) = 27: \(\dfrac{1}{2}\) = 27. 2= 28
c) 32. 25. \(\left(\dfrac{2}{3}\right)^2\) = 32. 25. \(\dfrac{2^2}{3^2}\) = 25. 22 = 27
d) \(\left(\dfrac{1}{3}\right)^2\) .\(\dfrac{1}{3}\) . 92 = \(\dfrac{1}{9}.\dfrac{1}{3}\). 92 = \(\dfrac{9}{3}\) = 31
b) Giải:
ĐK: \(a\ne-b\)
Ta có:
\(3a^2+b^2=4ab\)
\(\Leftrightarrow4a^2-4ab+b^2-a^2=0\)
\(\Leftrightarrow\left(2a-b\right)^2-a^2=0\)
\(\Leftrightarrow\left(3a-b\right)\left(a-b\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3a-b=0\\a-b=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}a=\dfrac{b}{3}\\a=b\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}a=\dfrac{b}{3}\Leftrightarrow P=\dfrac{\dfrac{b}{3}-b}{\dfrac{b}{3}+b}=\dfrac{-1}{2}\\a=b\Leftrightarrow P=\dfrac{a-a}{a+a}=\dfrac{0}{2a}=0\end{matrix}\right.\)
Vậy \(\left[{}\begin{matrix}P=\dfrac{-1}{2}\\P=0\end{matrix}\right.\)
1: \(A=\dfrac{15-4+1}{10}+\dfrac{18-8+1}{12}\)
\(=\dfrac{12}{10}+\dfrac{11}{12}\)
\(=\dfrac{6}{5}+\dfrac{11}{12}=\dfrac{72+55}{60}=\dfrac{127}{60}\)
Câu 3:
\(S=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2011}-\frac{1}{2012}+\frac{1}{2013}\)
\(\Rightarrow S=\left(1+\frac{1}{3}+...+\frac{1}{2013}\right)-\left(\frac{1}{2}-\frac{1}{4}-...-\frac{1}{2012}\right)\)
\(\Rightarrow S=\left(1+\frac{1}{2}+...+\frac{1}{2013}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2012}\right)\)
\(\Rightarrow S=\left(1+\frac{1}{2}+...+\frac{1}{2013}\right)-1-\frac{1}{2}-...-\frac{1}{2012}\)
\(\Rightarrow S=\frac{1}{1007}+\frac{1}{1008}+...+\frac{1}{2012}+\frac{1}{2013}\)
\(\Rightarrow S=P\)
\(\Rightarrow S-P=0\)
\(\Rightarrow\left(S-P\right)^{2013}=0^{2013}=0\)
Vậy \(\left(S-P\right)^{2013}=0\)
Bài 2.1
a: \(\dfrac{2012}{\left|x\right|+2013}\le\dfrac{2012}{2013}\)
Dấu '=' xảy ra khi x=0
b: \(\dfrac{\left|x\right|+2012}{-2013}\le-\dfrac{2012}{2013}\)
Dấu '=' xảy ra khi x=0