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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
1: \(A=\dfrac{15-4+1}{10}+\dfrac{18-8+1}{12}\)
\(=\dfrac{12}{10}+\dfrac{11}{12}\)
\(=\dfrac{6}{5}+\dfrac{11}{12}=\dfrac{72+55}{60}=\dfrac{127}{60}\)
\(1,\\ a,=\left(\dfrac{1}{4}\right)^3\cdot32=\dfrac{1}{64}\cdot32=\dfrac{1}{2}\\ b,=\left(\dfrac{1}{8}\right)^3\cdot512=\dfrac{1}{512}\cdot512=1\\ c,=\dfrac{2^6\cdot2^{10}}{2^{20}}=\dfrac{1}{2^4}=\dfrac{1}{16}\\ d,=\dfrac{3^{44}\cdot3^{17}}{3^{30}\cdot3^{30}}=3\\ 2,\\ a,A=\left|x-\dfrac{3}{4}\right|\ge0\\ A_{min}=0\Leftrightarrow x=\dfrac{3}{4}\\ b,B=1,5+\left|2-x\right|\ge1,5\\ A_{min}=1,5\Leftrightarrow x=2\\ c,A=\left|2x-\dfrac{1}{3}\right|+107\ge107\\ A_{min}=107\Leftrightarrow2x=\dfrac{1}{3}\Leftrightarrow x=\dfrac{1}{6}\)
\(d,M=5\left|1-4x\right|-1\ge-1\\ M_{min}=-1\Leftrightarrow4x=1\Leftrightarrow x=\dfrac{1}{4}\\ 3,\\ a,C=-\left|x-2\right|\le0\\ C_{max}=0\Leftrightarrow x=2\\ b,D=1-\left|2x-3\right|\le1\\ D_{max}=1\Leftrightarrow x=\dfrac{3}{2}\\ c,D=-\left|x+\dfrac{5}{2}\right|\le0\\ D_{max}=0\Leftrightarrow x=-\dfrac{5}{2}\)
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 \(B=x+x^2+x^3+...+x^{100}\)
\(\Rightarrow B=\left(-11\right)+\left(-11\right)^2+\left(-11\right)^3+...+\left(-11\right)^{100}\)
\(\Rightarrow-11B=\left(-11\right)^2+\left(-11\right)^3+\left(-11\right)^4+...+\left(-11\right)^{101}\)
\(\Rightarrow-11B-B=\left(-11\right)^{101}-\left(-11\right)\)
\(\Rightarrow-12B=\left(-11\right)^{101}+11\Rightarrow B=\frac{\left(-11\right)^{101}+11}{-12}\)
\(\Rightarrow A=1+B=\frac{\left(-11\right)^{101}+11}{-12}+1\)
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