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a. Vì \(\left|x-y-5\right|\ge0\forall x;y;2019\left|y-3\right|^{2020}\ge0\forall y\)
\(\Rightarrow\left|x-y-5\right|+2019\left|y-3\right|^{2020}\ge0\)
Dấu "=" xảy ra \(\Leftrightarrow\orbr{\begin{cases}\left|x-y-5\right|=0\\2019\left|y-3\right|^{2020}=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x-y-5=0\\y-3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x-y=5\\y=3\end{cases}}\)
b. \(2\left(x-5\right)^4\ge0\forall x;5\left|2y-7\right|^5\ge0\forall y\)
\(\Rightarrow2\left(x-5\right)^4+5\left|2y-7\right|^5\ge0\)
Dấu "=" xảy ra \(\Leftrightarrow\orbr{\begin{cases}2\left(x-5\right)^4=0\\5\left|2y-7\right|^5=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x-5=0\\2y-7=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=5\\y=\frac{7}{2}\end{cases}}\)
B = 5^0 + 5^1 + 5^2 + ... + 5^25 = 1 + 5 + 5^2 + ... + 5^25
5B = 5 + 5^2 + 5^3 + ... + 5^26
4B = ( 5 + 5^2 + 5^3 + ... + 5^26 ) - ( 1 + 5 + 5^2 + ... + 5^25 )
4B = 5^26 - 1
B = 5^26 - 1 / 4
Vậy B chắc chắn nhỏ hơn 5^26
10.
\(J=\dfrac{1\cdot2+2\cdot4+3\cdot6+4\cdot8+5\cdot10}{3\cdot4+6\cdot8+9\cdot12+12\cdot16+15\cdot20}\\ =\dfrac{1\cdot2+2\cdot1\cdot2\cdot2+3\cdot1\cdot3\cdot2+4\cdot1\cdot4\cdot2+5\cdot1\cdot5\cdot2}{3\cdot4+2\cdot3\cdot2\cdot4+3\cdot3\cdot3\cdot4+4\cdot3\cdot4\cdot4+5\cdot3\cdot4\cdot4}\\ =\dfrac{\left(1\cdot2\right)\cdot\left(1+2\cdot2+3\cdot3+4\cdot4+5\cdot5\right)}{\left(3\cdot4\right)\cdot\left(1+2\cdot2+3\cdot3+4\cdot4+5\cdot5\right)}\\ =\dfrac{1\cdot2}{3\cdot4}\\ =\dfrac{1\cdot1}{3\cdot2}\\ =\dfrac{1}{6}\)
11.
\(K=3^0+3^1+3^2+...+3^{100}\\ =1\cdot\left(3^0+3^1+3^2+...+3^{100}\right)\\ =\dfrac{3-1}{2}\cdot\left(3^0+3^1+3^2+...+3^{100}\right)\\ =\dfrac{\left(3-1\right)\cdot\left(3^0+3^1+3^2+...+3^{100}\right)}{2}\\ =\dfrac{3^1-3^0+3^2-3^1+3^3-3^2+...+3^{101}-3^{100}}{2}\\ =\dfrac{3^{100}-3^0}{2}=\dfrac{3^{100}-1}{2}\)
12.
\(L=1-5+5^2-5^3+...+5^{98}-5^{99}\\ =1\cdot\left(1-5+5^2-5^3+...+5^{98}-5^{99}\right)\\ =\dfrac{5+1}{6}\cdot\left(1-5+5^2-5^3+...+5^{98}-5^{99}\right)\\ =\dfrac{\left(5+1\right)\cdot\left(1-5+5^2-5^3+...+5^{98}-5^{99}\right)}{6}\\ =\dfrac{5+1-5^2-5+5^3+5^2-5^4-5^3+...+5^{99}+5^{98}-5^{100}-5^{99}}{6}\\ =\dfrac{1-5^{100}}{6}\)
a)\(\dfrac{6^2.6^3}{3^5}=\dfrac{2^2.3^2.2^3.3^3}{3^5}=2^5=32\)
b)\(\dfrac{25^2.4^2}{5^5\left(-2\right)^5}=\dfrac{5^2.5^2.2^2.2^2}{5^5.\left(-2\right)^5}=\dfrac{1}{-10}=-\dfrac{1}{10}\)
c)\(\dfrac{2^7.9^3}{8^2.3^6}=\dfrac{2^7.\left(3^2\right)^3}{\left(2^3\right)^2.3^6}=\dfrac{2^7.3^6}{2^6.3^6}=2\)
d)\(\dfrac{6^3+3.6^2+3^3}{-13}=\dfrac{2^3.3^3+3.2^2.3^2+3^3}{-13}=\dfrac{3^3\left(2^3+2^2+1\right)}{-13}\)
\(=\dfrac{3^3.13}{-13}=-3^3=-27\)
A<1
bạn tính phần mẫu ra rồi làm như dạng sai phân bình thường
a) \(\left(x^2-5\right)\left(x^2-25\right)< 0\)
Vì \(x^2-5>x^2-25\) nên \(\left\{{}\begin{matrix}x^2-5>0\\x^2-25< 0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x^2>5\\x^2< 25\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\sqrt{5}< x< -\sqrt{5}\left(vl\right)\\-5< x< 5\end{matrix}\right.\)
b) \(\left(x+5\right)\left(9+x^2\right)< 0\)
Vì \(9+x^2>0\) nên \(x+5< 0\Leftrightarrow x< -5\)
c) \(\left(x+3\right)\left(x^2+1\right)=0\)
Vì \(x^2+1>0\) nên \(x+3=0\Leftrightarrow x=-3\)
d) \(\left(x+5\right)\left(x^2-4\right)=0\)
\(\Rightarrow\left(x+5\right)\left(x+2\right)\left(x-2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=-5\\x=-2\\x=2\end{matrix}\right.\)
Biết câu b
C=(2.4)^20+4^20/4^25+(4^3)^5
=2^20.4^20+4^20/4^25+4^15
=4^20.(2^20+1)/4^15.(4^10+1)
=4^20.(2^20+1)/4^15.(2^20+1)
=4^20/4^15=4^5=1024.
k giúp mk nha! Chúc bạn học tốt.
=630:[25.(1+1)-2.16]
=630:[25.2-32]
=630:[ 50-32]
=630: 18
= 35
sửa đề nhanh thật, suýt nữa thì ấn gửi rồi