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a)|x+0,573|=2
=>x+0,573=2 hoặc -2
Xét x+0,573=2
=>x=1,427
Xét x+0,573=-2
=>x=-2,573
a) | x + 0,573 | = 2
\(\Rightarrow\)x + 0,573 = 2 hoặc x + 0,573 = -2
+) x + 0,573 = 2\(\Rightarrow\)x = 1,427
+) x + 0,573 = -2\(\Rightarrow\)x = -2,573
Vậy x = 1,427 hoặc -2,573
b) \(\left|x+\frac{1}{3}\right|-4=-1\)
\(\Rightarrow\left|x+\frac{1}{3}\right|=3\)
\(\Rightarrow x+\frac{1}{3}=3\) hoặc \(x+\frac{1}{3}=-3\)
+) \(x+\frac{1}{3}=3\Rightarrow x=\frac{8}{3}\)
+) \(x+\frac{1}{3}=-3\Rightarrow x=\frac{-10}{3}\)
Vậy \(x=\frac{8}{3}\) hoặc \(x=\frac{-10}{3}\)
Các phần khác làm tương tự nhé bạn
\(\frac{1}{2}S=\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+...+\left(\frac{1}{2}\right)^{21}\)
\(\Rightarrow\left(\frac{1}{2}S\right)-S=\left(\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+...+\left(\frac{1}{2}\right)^{21}\right)-\left(\left(\frac{1}{2}\right)+\left(\frac{1}{2}\right)^2+...+\left(\frac{1}{2}\right)^{20}\right)\)
\(\Rightarrow-\frac{1}{2}S=\left(\frac{1}{2}\right)^{21}-\left(\frac{1}{2}\right)\)
\(\Rightarrow S=\frac{\left(\left(\frac{1}{2}\right)^{21}-\frac{1}{2}\right)}{-\frac{1}{2}}\)
Đặt \(A=\frac{1}{2}+\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+...+\left(\frac{1}{2}\right)^{20}\)(1)
\(\Rightarrow2A=1+\frac{1}{2}+\left(\frac{1}{2}\right)^2+...+\left(\frac{1}{2}\right)^{19}\)(2)
Lấy (2) trừ đi (1) ta có :
\(\Rightarrow2A-A=1-\left(\frac{1}{2}\right)^{20}\)
\(\Rightarrow A=1-\left(\frac{1}{2}\right)^{20}\)
a. \(\frac{20^5.5^{10}}{100^5}\)
\(=\frac{20^5.\left(5^2\right)^5}{100^5}\)
\(=\frac{20^5.25^5}{100^5}\)
\(=\frac{500^5}{100^5}\)
\(=\left(\frac{500}{100}\right)^5\)
\(=5^5=3125\)
b. \(\frac{\left(0,9\right)^5}{\left(0,3\right)^6}\)
\(=\frac{\left(0,9\right)^5}{\left(0,3\right)^5.0,3}\)
\(=\left(\frac{0,9}{0,3}\right)^5.\frac{1}{0,3}\)
\(=3^5.\frac{1}{0,3}\)
\(=810\)
c. \(\frac{6^3+3.6^2+3^3}{-13}\)
\(=\frac{\left(3.2\right)^3+3.\left(3.2\right)^2+3^3}{-13}\)
\(=\frac{3^3\left(2^3+2^2+1\right)}{-13}\)
\(=\frac{3^3.13}{-13}\)
\(=\left(-3\right)^3\)
\(=-27\)
\(B=1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)\)
\(+...+\frac{1}{20}\left(1+2+3+...+20\right)\)
\(=1+\frac{1}{2}.\frac{2.3}{2}+\frac{1}{3}.\frac{3.4}{2}\)\(+...+\frac{1}{20}.\frac{20.21}{2}\)
\(=\frac{2}{2}+\frac{3}{2}+\frac{4}{2}+...+\frac{21}{2}\)
\(=\frac{\frac{\left(21+2\right)\left[\left(21-2\right)+1\right]}{2}}{2}=115\)
Đặt \(A=\frac{1}{2}+\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+\left(\frac{1}{2}\right)^4+...+\left(\frac{1}{2}\right)^{20}\)
\(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{20}}\)
\(2A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{19}}\)
\(2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{19}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{20}}\right)\)
\(A=1-\frac{1}{2^{20}}\)
\(A=\frac{2^{20}}{2^{20}}-\frac{1}{2^{20}}\)
\(A=\frac{2^{20}-1}{2^{20}}\)
Vậy chọn câu a)