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Ta có \(-A=\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)...\left(1-\frac{1}{2014^2}\right)\)
\(=\left(\frac{2^2-1}{2^2}\right)\left(\frac{3^2-1}{3^2}\right)...\left(\frac{2014^2-1}{2014^2}\right)\)
\(=\frac{\left(2-1\right)\left(2+1\right)}{2^2}.\frac{\left(3-1\right)\left(3+1\right)}{3^2}...\frac{\left(2014-1\right)\left(2014+1\right)}{2014^2}\)
\(=\frac{1.3}{2.2}.\frac{2.4}{3.3}...\frac{2013.2015}{2014.2014}\)
\(=\frac{1.2...2013}{2.3...2014}.\frac{3.4...2015}{2.3...2014}\)
\(=\frac{1}{2014}.\frac{2015}{2}\)
\(=\frac{2015}{2014.2}>\frac{1}{2}\)hay -A>1/2
=>\(A< \frac{-1}{2}\)hay A<B
Ta có:
\(\left(\frac{1}{2}\right)^{225}=\left[\left(\frac{1}{2}\right)^9\right]^{25}=\left(\frac{1}{516}\right)^{25}\)
\(\left(\frac{1}{3}\right)^{100}=\left[\left(\frac{1}{3}\right)^4\right]^{25}=\left(\frac{1}{81}\right)^{25}\)
\(\frac{1}{516}< \frac{1}{81}\Rightarrow\left(\frac{1}{516}\right)^{25}< \left(\frac{1}{81}\right)^{25}\Rightarrow\left(\frac{1}{2}\right)^{225}< \left(\frac{1}{3}\right)^{100}\)
a/
\(x-y=\frac{a}{b}-\frac{c}{d}=\frac{ad-cb}{bd}=\frac{1}{bd}.\) (1)
\(y-z=\frac{c}{d}-\frac{e}{h}=\frac{ch-de}{dh}=\frac{1}{dh}\)(2)
+ Nếu d>0 => (1)>0 và (2)>0 => x>y; y>x => x>y>z
+ Nếu d<0 => (1)<0 và (2)<0 => x<y; y<z => x<y<z
b/
\(m-y=\frac{a+e}{b+h}-\frac{c}{d}=\frac{ad+de-cb-ch}{d\left(b+h\right)}=\frac{\left(ad-cb\right)-\left(ch-de\right)}{d\left(b+h\right)}=\frac{1-1}{d\left(b+h\right)}=0\)
=> m=y
+
cảm ơn bn nha Nguyễn Ngoc Anh Minh mk k cho bn r đó kb vs mk nha
a) (x + 1/2)^2 = 1/16
=> (x + 1/2)^2 = (1/4)^2 hoặc (x + 1/2)^2 = (-1/4)^2
=> x + 1/2 = 1/4 hoặc x + 1/2 = -1/4
* x + 1/2 = 1/4
x = 1/4 - 1/2
x = -1/4
* x + 1/2 = -1/4
x = -1/4 - 1/2
x = -3/4
Vậy x = -1/4 hoặc x = -3/4
b) 2^x+2 - 2^x = 9^6
=> 2^x . 2^2 - 2^x = 9^6
=> 2^x . (2^2 - 1) = 9^6
=> 2^x . (4 - 1) = 9^6
=> 2^x . 3 = (3^2)^6
=> 2^x . 3 = 3^12
=> 2^x = 3^12 : 3
=> 2^x = 3^11
Vì 3^11 không chia hết cho 2
=> Không có giá trị nào của x thõa mãn đề bài
c) (3^x)^2 : 3^3 = 1/243
=> 3^2x = 1/243 . 3^3
=> 3^2x = 1/243 . 27
=> 3^2x = 1/9
=> 3^2x . 9 = 1
=> 3^2x . 3^2 = 1
=> 3^2x+2 = 1
=> 3^2x+2 = 3^0
=> 2x + 2 = 0
=> 2x = 0 - 2
=> 2x = -2
=> x = -2 : 2
=> x = -1
a\(\left(x+\frac{1}{2}\right)^2=\left(\frac{1}{4}\right)^2
\)
\(\Rightarrow x+\frac{1}{2}=\frac{1}{4}\)
\(\Leftrightarrow x=\frac{1}{4}-\frac{1}{2}\)
\(\Rightarrow x=\frac{-1}{4}\)
Ta có :
\(A+3=\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}+3\)
\(=\left(\frac{a}{b+c}+1\right)+\left(\frac{b}{a+c}+1\right)+\left(\frac{c}{a+b}+1\right)\)
\(=\frac{a+b+c}{b+c}+\frac{a+b+c}{a+c}+\frac{a+b+c}{a+b}\)
\(=\left(a+b+c\right)\left(\frac{1}{b+c}+\frac{1}{a+c}+\frac{1}{a+b}\right)\)
\(=2017.\frac{1}{2017}=1\)
\(\Rightarrow A=1-3=-2\)
\(A=\left|x-\frac{1}{3}\right|+\frac{1}{4}\ge\frac{1}{4}>\frac{1}{5}\)