2^2 . 2^3 - 3^8 : 3^5 + 2009^0 - 1^101
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a, \(-17.53-21.\left(-17\right)-17.17^0\)
\(=\)\(-17.\left(53-21\right)-17.1\)
\(=\) \(-17.32-17\)
\(=\)\(-544-17\)
\(=-561\)
b, \(\left|-2^2.2^3-3^5\right|+3^5+2009^0-\left(-1\right)^{101}\)
\(=\left|-4.8-243\right|+243+1+1^{101}\)
\(=\left|-32-243\right|+243+1+1\)
\(=\left|-275\right|+243+1+1\)
\(=275+243+1+1\)
\(=518+1+1\)
\(=519+1\)
\(=520\)
:)
Mình làm mẫu 1 bài rùi bạn tự giải những bài còn lại nha
1, 7A = 7+7^2+7^3+....+7^2008
6A = 7A - A = (7+7^2+7^3+....+7^2008)-(1+7+7^2+....+7^2007) = 7^2008-1
=> A = (7^2008-1)/6
Tk mk nha
\(A=1+7+7^2+7^3+...+7^{2007}\)
\(\Rightarrow7A=7+7^2+7^3+7^4+...+7^{2008}\)
\(\Rightarrow7A-A=\left(7+7^2+7^3+...+7^{2008}\right)-\left(1+7+7^2+...+7^{2007}\right)\)
\(\Rightarrow6A=7^{2008}-1\)
\(\Rightarrow A=\frac{7^{2008}-1}{6}\)
\(\frac{3x-7}{5}=\frac{2x-1}{3}\)
\(\Leftrightarrow9x-21=10x-5\)
\(\Leftrightarrow-x=16\Leftrightarrow x=-16\)
\(\frac{4x-7}{12}-x=\frac{3x}{8}\)
\(\Leftrightarrow\frac{4x-7-12x}{12}=\frac{3x}{8}\)
\(\Leftrightarrow\frac{-7-8x}{12}=\frac{3x}{8}\)
\(\Leftrightarrow-56-64x=36x\)
\(\Leftrightarrow-56=100x\Leftrightarrow x=\frac{-14}{25}\)
\(\frac{x-2009}{1234}+\frac{x-2009}{5678}-\frac{x-2009}{197}=0\)
\(\Leftrightarrow\left(x-2019\right)\left(\frac{1}{1234}+\frac{1}{5678}-\frac{1}{197}\right)=0\)
Vì \(\left(\frac{1}{1234}+\frac{1}{5678}-\frac{1}{197}\right)\ne0\)nên x - 2019 = 0
Vậy x = 2019
\(\frac{5x-8}{3}=\frac{1-3x}{2}\)
\(\Leftrightarrow10x-16=3-9x\)
\(\Leftrightarrow19x=19\Leftrightarrow x=1\)
Giúp vsssssssssssssssssssssssssssssssssssssssss nhaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa .........................
\(2^2.2^3-3^8:3^5+2009^0-1^{101}\)
\(=\)\(2^5-3^3+1-1\)
\(=\) \(32-27+1-1\)
\(=\) \(5+1-1\)
\(=\) \(6-1\)
\(=5\)