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Bài 1 :
\(M=\dfrac{30-2^{20}}{2^{18}}=\dfrac{2.15-2^{20}}{2^{18}}=\dfrac{15}{2^{17}}-2^2=\dfrac{15}{2^{17}}-4< 0\left(\dfrac{15}{2^{17}}< 1\right)\)
\(N=\dfrac{3^5}{1^{2021}+2^3}=\dfrac{3^5}{9}=\dfrac{3^5}{3^2}=3^3=27\)
\(\Rightarrow M< N\)
Bài 3 :
a) \(t^2+5t-8\) khi \(t=2\)
\(=5^2+2.5-8\)
\(=25+10-8\)
\(=27\)
b) \(\left(a+b\right)^2-\left(b-a\right)^3+2021\left(1\right)\)
\(\left\{{}\begin{matrix}a=5\\b=a+1=6\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a+b=11\\b-a=1\end{matrix}\right.\)
\(\left(1\right)=11^2-1^3+2021=121-1+2021=2141\)
c) \(x^3-3x^2y+3xy^2-y^3=\left(x-y\right)^3\left(1\right)\)
\(\left\{{}\begin{matrix}x=3\\y=2\end{matrix}\right.\) \(\Rightarrow x-y=1\)
\(\left(1\right)=1^3=1\)
a, Ta có : \(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{199.200}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{199}-\frac{1}{200}\)
\(=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{199}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{200}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{199}+\frac{1}{200}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{200}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{199}+\frac{1}{200}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)\)
\(=\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}\)
=> \(\frac{\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{199.200}}{\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}}=1\)
=> đpcm
Study well ! >_<
Trả lời :
m2 . (n)3
Thay số : 22 . (-3)3 = 4 . (-27)
= - 108
Chúc học tốt nha !
tính giá trị biểu thức m2.(-n) 3 cho m=2 n=3.
Ta thay m ,n vào biểu thức ta đc:
22.(-3)3=4.(-9)=-36
học tốt
1/a,
-Ta có:
$B<1\Leftrightarrow B<\frac{10^{2005}+1+9}{10^{2006}+1+9}=\frac{10^{2005}+10}{10^{2006}+10}=\frac{10(10^{2004}+1)}{10(10^{2005}+1)}=\frac{10^{2004}+1}{10^{2005}+1}=A$
-Vậy: B<A
b,$A=1+(\frac{1}{2})^2+...+(\frac{1}{100})^2$
$\Leftrightarrow A=1+\frac{1}{2^2}+...+\frac{1}{100^2}$
$\Leftrightarrow A<1+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}$
$\Leftrightarrow A<1+\frac{1}{1}-\frac{1}{2}+...+\frac{1}{99}-\frac{1}{100}$
$\Leftrightarrow A<1+1-\frac{1}{100}\Leftrightarrow A<2-\frac{1}{100}\Leftrightarrow A<2(đpcm)$
2,
a.
-Ta có:$\Rightarrow \frac{3x+7}{x-1}=\frac{3(x-1)+16}{x-1}=\frac{3(x-1)}{x-1}+\frac{16}{x-1}=3+\frac{16}{x-1}
-Để: 3x+7/x-1 nguyên
-Thì: $\frac{16}{x-1}$ nguyên
$\Rightarrow 16\vdots x-1\Leftrightarrow x-1\in Ư(16)\Leftrightarrow ....$
b, -Ta có:
$\frac{n-2}{n+5}=\frac{n+5-7}{n+5}=1-\frac{7}{n+5}$
-Để: n-2/n+5 nguyên
-Thì: \frac{7}{n+5} nguyên
$\Leftrightarrow 7\vdots n+5\Leftrightarrow n+5\in Ư(7)\Leftrightarrow ...$
a,( 393+390) : (317. 373)
= (33+1). 390 : 390
= 33+1
=27+1
=28
b,(556+57) : (549+1)
=57. (549+1) : (549+1)
=57= 78125
c,(722+721+720) ; (25+24+32)
= 720. (72+71+1) : [24. (2+1)+32 ]
= 720. 57 : [ 24. 3 +32 ]
= 720. 57 : ( 24+3) . 3
= 720. 57 : 19 . 3
= 720. 57 : 57
= 720
M=1+2+22+23+...+299.
=> 2M = M= 2+2^2+2^3+...+2^99 + 2^100=>2M= M=2+22+23+...+299+2100
=> M = 2M-M = 2+2^2+2^3+...+2^99 + 2^100 - (1+2+2^2+2^3+...+2^99)=>M=2M−M= 2+22+23+...+299+2100−(1+2+22+23+...+299)
<=> M = 2^100-1 <2^100<=>M=2100−1<2100
<=>Vậy M<2^100