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1.
-15 - 12 = -15 + (-12) = -27
0+ (-7)= 0-7 = -7
0-(-5) = 0 + 5 = 5
đap số 1 . - 27
2. -7
3. 5
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\(A=1+2\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{98.99}+\frac{1}{99.100}\right)\)
\(A=1+2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\right)\)
\(A=1+2\left(\frac{1}{2}-\frac{1}{100}\right)=1+2.\frac{49}{100}=1+\frac{49}{50}\)
\(A=\frac{99}{50}\)
Vậy \(A=\frac{99}{50}\)
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Ta có: A=2009.2011=2009.(2010+1)=2009.2010+2009
B=20102=2010.2010=(2009+1).2010=2009.2010+2010
Vì 2009<2010 => A<B
(Nhớ k cho tui á nha boss)
A = 2009.(2010+1) = 2009.2010+2009 = (2009.2010+2010)-1 = 2010.(2009+1)-1 = 2010^2-1 = B-1 < B
=> A < B
k mk nha
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a.
\(\left|x+10\right|=15\Rightarrow\orbr{\begin{cases}x+10=15\\x+10=-15\end{cases}\Rightarrow\orbr{\begin{cases}x=5\\x=-25\end{cases}}}\)
b.
\(\left|x-3\right|+5=7\Rightarrow\left|x-3\right|=2\Rightarrow\orbr{\begin{cases}x-3=2\\x-3=-2\end{cases}}\Rightarrow\orbr{\begin{cases}x=5\\x=1\end{cases}}\)
c.
\(\left|x-3\right|+12=6\Rightarrow\left|x-3\right|=-6\Rightarrow x=\Phi\)
Phương trình vô nghiệm
d.
\(\left(2x+4\right)\left(3x-9\right)=0\Rightarrow\orbr{\begin{cases}2x+4=0\\3x-9=0\end{cases}\Rightarrow}\orbr{\begin{cases}2x=-4\\3x=9\end{cases}}\Rightarrow\orbr{\begin{cases}x=-2\\x=3\end{cases}}\)
e.
\(x^2-5x=0\Rightarrow x\left(x-5\right)=0\Rightarrow\orbr{\begin{cases}x=0\\x-5=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=5\end{cases}}\)
f.
\(\left(x+3\right)\left(4-2x\right)=70\Rightarrow4x-2x^2+7-6x=70\Rightarrow2x^2+2x+63=0\Rightarrow2\left(x+\frac{1}{2}\right)^2+\frac{123}{2}=0\)(vô lí)
Vậy phương trình vô nghiệm
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\(75\%+1,1:\left(\frac{2}{5}-1\frac{1}{2}\right)-\left(\frac{1}{3}\right)^2\)
\(=\frac{3}{4}+\frac{11}{10}:\left(\frac{2}{5}-\frac{3}{2}\right)-\frac{1}{9}\)
=\(\frac{3}{4}+\frac{11}{10}:\frac{-11}{10}-\frac{1}{9}\)
\(=\frac{3}{4}+\left(-1\right)-\frac{1}{9}\)
\(=\frac{27}{36}+\left(\frac{-36}{36}\right)-\frac{4}{36}\)
\(=\frac{-13}{36}\)
chúc bạn học tốt
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B=\(\frac{12}{\left(2.4\right)^2}\)+\(\frac{20}{\left(4.6\right)^2}\)+.....+\(\frac{396}{\left(98.100\right)^2}\)
B=\(\frac{4^2-2^2}{2^2.4^2}\)+ \(\frac{6^2-4^2}{4^2.6^2}\)+....+\(\frac{100^2-98^2}{\left(98^2.100^2\right)}\)
B=\(\frac{1}{2^2}\)-\(\frac{1}{4^2}\)+\(\frac{1}{4^2}\)-\(\frac{1}{6^2}\)+....+\(\frac{1}{98^2}\)-\(\frac{1}{100^2}\)
B=\(\frac{1}{2^2}\)-\(\frac{1}{100^2}\)< \(\frac{1}{2^2}\)=\(\frac{1}{4}\)
Vậy B<\(\frac{1}{4}\)
\(\Rightarrow2A=2+2^2+...+2^{1011}\)
\(\Rightarrow2A-A=\left(2+2^2+..+2^{1011}\right)-\left(1+2+...+2^{2010}\right)\)
\(\Rightarrow A=2^{1011}-1\)
Ta có: \(A=2^0+2^1+2^2+...+2^{2010}\)
\(\Rightarrow2A=2^1+2^2+2^3+...+2^{2010}+2^{2011}\)
\(\Rightarrow2A-A=\left(2^1+2^2+2^3+...+2^{2010}+2^{2011}\right)-\left(2^0+2^1+2^2+...+2^{2010}\right)\)
\(\Rightarrow A=2^{2011}-1\)