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Đặt các cặp 1+1/3+1/5+..+1/4025 của A ra so sánh (1/2+1/4+..+1/4026)/B với 2013/2014
thấy A/B<1+2013/2014
1/3^2+1/5^2+1/7^2+...+1/(2n+1)^2 < 1/1.3+1/3.5+1/5.7+...+1/(2n-1)(2n+1)
= 1/2(1-1/3+1/3-1/5+1/5-1/7+...+1/(2n-1)-1/(2n+1)
= 1/2(1-1/(2n+1))
= 1/2 . 2n/(2n+1)
= 2n/2(2n+1)
3:
a: =>x=0 hoặc x+5=0
=>x=0 hoặc x=-5
b: =>x^2=4
=>x=2 hoặc x=-2
c: =>(x-5)(2x+1+x+6)=0
=>(x-5)(3x+7)=0
=>x=5 hoặc x=-7/3
1.
a. 2x - 6 > 0
\(\Leftrightarrow\) 2x > 6
\(\Leftrightarrow\) x > 3
S = \(\left\{x\uparrow x>3\right\}\)
b. -3x + 9 > 0
\(\Leftrightarrow\) - 3x > - 9
\(\Leftrightarrow\) x < 3
S = \(\left\{x\uparrow x< 3\right\}\)
c. 3(x - 1) + 5 > (x - 1) + 3
\(\Leftrightarrow\) 3x - 3 + 5 > x - 1 + 3
\(\Leftrightarrow\) 3x - 3 + 5 - x + 1 - 3 > 0
\(\Leftrightarrow\) 2x > 0
\(\Leftrightarrow\) x > 0
S = \(\left\{x\uparrow x>0\right\}\)
d. \(\dfrac{x}{3}-\dfrac{1}{2}>\dfrac{x}{6}\)
\(\Leftrightarrow\dfrac{2x}{6}-\dfrac{3}{6}>\dfrac{x}{6}\)
\(\Leftrightarrow2x-3>x\)
\(\Leftrightarrow2x-3-x>0\)
\(\Leftrightarrow x-3>0\)
\(\Leftrightarrow x>3\)
\(S=\left\{x\uparrow x>3\right\}\)
2.
a.
Ta có: a > b
3a > 3b (nhân cả 2 vế cho 3)
3a + 7 > 3b + 7 (cộng cả 2 vế cho 7)
b. Ta có: a > b
a > b (nhân cả 2 vế cho 1)
a + 3 > b + 3 (cộng cả 2 vế cho 3) (1)
Ta có; 3 > 1
b + 3 > b + 1 (nhân cả 2 vế cho 1b) (2)
Từ (1) và (2) \(\Rightarrow\) a + 3 > b + 1
c.
5a - 1 + 1 > 5b - 1 + 1 (cộng cả 2 vế cho 1)
5a . \(\dfrac{1}{5}\) > 5b . \(\dfrac{1}{5}\) (nhân cả 2 vế cho \(\dfrac{1}{5}\) )
a > b
3.
a. 2x(x + 5) = 0
\(\Leftrightarrow\left[{}\begin{matrix}2x=0\\x+5=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)
\(S=\left\{0,-5\right\}\)
b. x2 - 4 = 0
\(\Leftrightarrow x\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)
\(S=\left\{0,4\right\}\)
d. (x - 5)(2x + 1) + (x - 5)(x + 6) = 0
\(\Leftrightarrow\left(x-5\right)\left(2x+1+x+6\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(3x+7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\3x+7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=\dfrac{-7}{3}\end{matrix}\right.\)
\(S=\left\{5,\dfrac{-7}{3}\right\}\)
A=2012x2014=2012x(2012+2)=2012^2+4024
B=2013^2=(2012+1)^2=2012^2+2x2012+1=2012^2+2025
=>A<B
chúc bạn học tốt~~~
Bài 1 :
\(a)\)\(A=2012.2014=\left(2013-1\right)\left(2013+1\right)=2013^2-1< 2013^2=B\)
Vậy \(A< B\)
\(b)\)\(A=\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(2A=\left(3-1\right)\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(2A=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(2A=\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(2A=\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(2A=\left(3^{16}-1\right)\left(3^{16}+1\right)\)
\(2A=3^{32}-1\)
\(A=\frac{3^{32}-1}{2}< 3^{32}-1=B\)
\(c)\)\(A=2017^2-17^2=\left(2017-17\right)\left(2017+17\right)=2000.2034>2000.2000=2000^2=B\)
Vậy \(A>B\)
