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trả lời:
p(-1)=5(-1)^5+3(-1)-4(-1)^4-2(-1)^3+6+4(-1)^2
=-5-3-4+2+6+4=0
q(1)=2.1^4-1+3.1^2-2.1^3+1/4-1^4
=2+3-2+1/4-1=9/4>>4.q(1)=4.9/4=9
Vì \(\left\{{}\begin{matrix}\left|x^2-4\right|\ge0\\\left|y^2-9\right|\ge0\end{matrix}\right.\)=>|x2 - 4 | + | y2 - 9 | \(\ge\) 0
Dấu "=" xảy ra khi |x2 - 4 | = | y2 - 9 | = 0
|x2 - 4 |=0 => x2-4=0 => x2=4 => \(x=\pm2\)
| y2 - 9 |=0 =>y2-9=0=>y2=9=>\(y=\pm3\)
Vậy \(x=\pm2\) và \(y=\pm3\)
Ta có :
\(\left[{}\begin{matrix}\left|x^2-4\right|\ge0\\\left|y^2-9\right|\ge0\end{matrix}\right.\) \(\Rightarrow\left|x^2-4\right|+\left|y^2-9\right|\ge0\)
\(x^2-4=0\Rightarrow x=\pm2\)
\(y^2-9=0\Rightarrow y=\pm3\)
Vậy......................
\(3x^2y^4\)-\(5xy^3\)-\(\dfrac{3}{2}x^2y^4\)+\(3xy^3\)+\(2xy^3\)+1=1,5\(x^2y^4\)+1>0
\(A=\dfrac{4^2}{1.3}+\dfrac{4^2}{3.5}+\dfrac{4^2}{5.8}+...+\dfrac{4^2}{45.47}.\dfrac{1-3-5-...-49}{8}\)
\(A=4\left(\dfrac{4}{1.3}+\dfrac{4}{3.5}+\dfrac{4}{5.8}+...+\dfrac{4}{45.47}\right).\dfrac{1-3-5-...-49}{8}\)\(A=4\left[2\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+...+\dfrac{1}{45}-\dfrac{1}{47}\right)\right].\dfrac{1-3-5-...-49}{8}\)\(A=8\left(1-\dfrac{1}{47}\right).\dfrac{1-3-5-...-49}{8}\)
\(A=8\left(1-\dfrac{1}{47}\right).\dfrac{-623}{8}\)
\(A=\dfrac{368}{47}.\dfrac{-623}{8}=\dfrac{-28658}{47}\)
\(\dfrac{4^5\cdot9^4-2\cdot6^9}{2^{10}\cdot3^8+6^8\cdot20}\)=\(\dfrac{\left(2^2\right)^5\cdot\left(3^2\right)^4-2\cdot6^9}{2^{10}\cdot3^8+6^8\cdot2\cdot10}=\dfrac{2^{10}\cdot3^8-2\cdot6^9}{2^{10}\cdot3^8+6^8\cdot2\cdot10}=\dfrac{6}{10}=\dfrac{3}{5}\)
mk ko chép đề mà tách luôn nha
M = x2x2 + x2x2 + x2y2 + x2y2 + x2y2 + y2y2 + y2
= ( x2x2 + x2y2 ) + ( x2x2 + x2y2 ) + ( x2y2 + y2y2 ) + y2
= x2( x2 + y2 ) + x2( x2 + y2 ) + y2( x2 + y2 ) + y2
= ( x2 + y2 ) (x2 + x2 + y2 ) + y2
= 1( x2 + 1) + y2
= x2 + y2 +1 = 2
Ta có: \(1^2+2^2+3^2+...+10^2=358\)
\(S=2^2+4^2+6^2+...+20^2\)
\(=\left(1.2\right)^2+\left(2.2\right)^2+\left(2.3\right)^2+...+\left(2.10\right)^2\)
\(=1^2.2^2+2^2.2^2+3^2.2^2+...+10^2.2^2\)
\(=2^2\left(1^2+2^2+3^2+...+10^3\right)\)
\(=2^2.385\)
\(=4.385=1540\)
1. a, Ta có: \(2^{24}=2^{3^8}=8^8\)
Lại có: \(3^{16}=3^{2^8}=9^8\)
Vì \(8^8< 9^8\Rightarrow2^{24}< 3^{16}\)
b, Ta có: \(5^{300}=5^{3^{100}}=125^{100}\)
Lại có: \(3^{500}=3^{5^{100}}=243^{100}\)
Vì \(125^{100}< 243^{100}\Rightarrow5^{300}< 3^{500}\)
c, Ta có: \(2^{700}=2^{7^{100}}=128^{100}\)
Lại có: \(5^{300}=5^{3^{100}}=125^{100}\)
Vì \(128^{100}>125^{100}\Rightarrow2^{700}>5^{300}\)
d, Ta có: \(2^{400}=2^{2^{200}}=4^{200}\)
\(\Rightarrow2^{400}=4^{200}\)
e, Ta có: \(99^{20}=99^{2^{10}}=9801^{10}\)
Vì \(9801^{10}< 9999^{10}\Rightarrow99^{20}< 9999^{10}\)
Bài 1:
a) Ta có: 224 = (23)8 = 88 ; 316 = (32)8 = 98
Vì 8 < 9 nên 88 < 98
Vậy 224 < 316.
b) Ta có: 5300 = (53)100 =125100 ; 3500 = (35)100 = 243100
Vì 125 < 243 nên 125100 < 243100
Vậy 5300 < 3500.
c) Ta có: 2700 = (27)100 = 128100; 5300 = (53)100 = 125100
Vì 128 > 125 nên 128100 > 125100
Vậy 2700 > 5300.
d) (làm tương tự)
Vậy 2400 = 4200.
e) (tương tự)
Vậy 9920 < 999910.
f) Ta có: 321 = 320. 3 = 910. 3 ; 231 = 230. 3 = 810. 2
Vì 910 > 810 ; 3 > 2
Nên 910. 3 > 810. 2
Vậy 321 > 231.
Bài 2: phương trình dễ ợt :v
Theo bài ra, ta có: \(C=75\left(4^{2001}+4^{2000}+4^{1999}+...+4^2+4+1\right)+25\)
Đặt \(S=4^{2001}+4^{2000}+4^{1999}+...+4^2+4+1\)
\(\Rightarrow4S=4^{2002}+4^{2001}+4^{2000}+...+4^3+4^2+4\)
\(\Rightarrow4S-S=4^{2002}+4^{2001}+4^{2000}+...+4^3+4^2+4-4^{2001}-4^{2000}-4^{1999}-...4^2-4-1\)
\(\Rightarrow3S=4^{2002}-1\)
\(\Rightarrow S=\dfrac{4^{2002}-1}{3}\)
Khi đó \(C=75.\dfrac{4^{2002}-1}{3}+25=\dfrac{75}{3}.\left(4^{2002}-1\right)+25=25\left(4^{2002}-1\right)+25=25\left(4^{2002}-1+1\right)=25.4^{2002}⋮4^{2002}\)
Vậy \(C⋮4^{2002}\left(đpcm\right)\)