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=>\(\frac{4}{x}\)=2 =>x=2
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học tốt!!!!!!!!!!!!!!!!
\(1< \frac{4}{x}\le2\)
\(\Rightarrow\frac{4}{x}=2\)
\(\Rightarrow x=2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a)\(5\le2.x< 13\)''
\(\frac{5}{2}\le x< \frac{13}{2}\)( chia cho 2)
b)\(5< 4.x+1\le17\)
\(5-1< 4x\le17-1\)( trừ cho 1 )
\(4< 4x\le16\)
\(\frac{4}{4}< x\le\frac{16}{4}\)
\(1< x\le4\)
c) \(x+\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}=\frac{1}{8}\)
\(x+\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}=8\)
\(x+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}=8\)( Áp dụng \(\frac{1}{a.\left(a+1\right)}=\frac{1}{a}-\frac{1}{a+1}\))
\(x+1-\frac{1}{6}=8\)
\(x=8-1+\frac{1}{6}\)
\(x=\frac{43}{6}\)
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\(1\) = \(1\)
\(\frac{1}{2^2}\)< \(\frac{1}{1.2}\)
\(\frac{1}{3^2}\) < \(\frac{1}{2.3}\)
.........
\(\frac{1}{100^2}\) < \(\frac{1}{99.100}\)
\(\Rightarrow1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{100^2}\) < \(1+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)
Ta có: \(1+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)
\(=1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1+1-\frac{1}{100}\)
\(=2-\frac{1}{100}\)
\(\Rightarrow2-\frac{1}{100}\le2\)
Nên \(1+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\le2\)
=>\(1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{100^2}\le2\)
Vậy S \(\le2\)
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b: |x-5|+|y+6|<=0
=>x-5=0 và y+6=0
=>x=5 và y=-6
c: (x-5)(y+2)=5
nên \(\left(x-5;y+2\right)\in\left\{\left(1;5\right);\left(5;1\right);\left(-1;-5\right);\left(-5;-1\right)\right\}\)
hay \(\left(x,y\right)\in\left\{\left(6;3\right);\left(10;-1\right);\left(4;-7\right);\left(0;-3\right)\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Do \(a,b,c\in Z^+\)=> \(\frac{a}{a+b}>\frac{a}{a+b+c}\)\(\frac{b}{b+c}>\frac{b}{a+b+c}\)và \(\frac{c}{c+a}>\frac{c}{a+b+c}\)
\(\Rightarrow\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}>\frac{a}{a+b+c}+\frac{b}{a+b+c}+\frac{c}{a+b+c}=1\)
Giả sử \(a\ge b\ge c\)Ta có \(a,b,c\in Z^+\)và \(a\ge b\)\(\Rightarrow\)\(c+a\ge c+b\)\(\Rightarrow\frac{c}{c+a}\le\frac{c}{c+b}\Rightarrow\frac{b}{b+c}+\frac{c}{c+a}\le\frac{b}{b+c}+\frac{c}{c+b}=1\)
Do \(a,b,c\in Z^+\)\(\Rightarrow\frac{a}{a+b}< 1\Rightarrow\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}< 2\)
Vậy \(\frac{a}{a+b}+\frac{c}{b+c}+\frac{a}{c+a}\le2\)
\(x=3\)
=4 nha