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Đề bài : \(\dfrac{97}{20}\times\left(\dfrac{25}{8}-\dfrac{221}{200}\right)< x< \dfrac{91}{10}\times\left(\dfrac{137}{20}+\dfrac{11}{4}\right)\)
\(\dfrac{97}{20}\times\left(\dfrac{25}{8}-\dfrac{221}{200}\right)=\dfrac{97}{20}\times\dfrac{101}{50}=\dfrac{9797}{1000}\)
\(\dfrac{91}{10}\times\left(\dfrac{137}{20}+\dfrac{11}{4}\right)=\dfrac{91}{10}\times\dfrac{48}{5}=\dfrac{2184}{25}=\dfrac{2184\times40}{25\times40}=\dfrac{87360}{1000}\)
\(\Rightarrow\dfrac{9797}{1000}< x< \dfrac{87360}{1000}\)
Vậy \(x\in\left(\dfrac{9797}{1000};\dfrac{87360}{1000}\right)\)
1.
a. Ta có: \(A=2^{300}=2^{3.100}=\left(2^3\right)^{100}=8^{100}\)
\(B=3^{200}=3^{2.100}=\left(3^2\right)^{100}=9^{100}\)
Mà \(8^{100}< 9^{100}\)
\(\Rightarrow A< B\)
b. Ta có: \(A=2^{332}< 2^{333}=2^{3.111}=\left(2^3\right)^{111}=8^{111}\)
\(B=3^{223}>3^{222}=3^{2.111}=\left(3^2\right)^{111}=9^{111}\)
Mà \(8^{111}< 9^{111}\)
\(\Rightarrow A< B\)
c. Ta có: \(A=2^{91}=2^{13.7}=\left(2^{13}\right)^7=8192^7\)
\(B=5^{35}=5^{5.7}=\left(5^5\right)^7=3125^7\)
Mà \(8192^7>3125^7\)
\(\Rightarrow A>B\)
Câu 2:
a: =>(x-6)(x-7)=0
=>x=6 hoặc x=7
b: =>\(x^8\left(x^2-25\right)=0\)
\(\Leftrightarrow x^8\left(x-5\right)\left(x+5\right)=0\)
hay \(x\in\left\{0;5;-5\right\}\)
a) \(\left(\frac{7}{8}-\frac{3}{4}\right)\cdot\frac{1}{3}-\frac{2}{7}\cdot\left(3,5\right)=\left(\frac{7}{8}-\frac{3}{4}\right)\cdot\frac{1}{3}-\frac{2}{7}\cdot\frac{7}{2}\)
\(=\left(\frac{7}{8}-\frac{6}{8}\right)\cdot\frac{1}{3}-1=\frac{1}{8}\cdot\frac{1}{3}-1=\frac{1}{24}-\frac{24}{24}=-\frac{23}{24}\)
b) \(\left(\frac{3}{5}+0,415-\frac{3}{200}\right)\cdot2\frac{2}{3}\cdot0,25\)
\(=\left(\frac{3}{5}+\frac{83}{200}-\frac{3}{200}\right)\cdot\frac{8}{3}\cdot\frac{1}{4}\)
\(=\left(\frac{3}{5}+\frac{80}{200}\right)\cdot\frac{8}{3}\cdot\frac{1}{4}=\left(\frac{3}{5}+\frac{2}{5}\right)\cdot\frac{8}{3}\cdot\frac{1}{4}=1\cdot\frac{8}{3}\cdot\frac{1}{4}=1\cdot\frac{2}{3}\cdot\frac{1}{1}=\frac{2}{3}\)
c) \(\frac{5}{16}:0,125-\left(2\frac{1}{4}-0,6\right)\cdot\frac{10}{11}\)
\(=\frac{5}{16}:\frac{1}{8}-\left(\frac{9}{4}-\frac{3}{5}\right)\cdot\frac{10}{11}\)
\(=\frac{5}{16}\cdot8-\frac{33}{20}\cdot\frac{10}{11}=\frac{5}{2}-\frac{3}{2}=1\)
d) \(0,25:\left(10,3-9,8\right)-\frac{3}{4}=\frac{1}{4}:\left(\frac{103}{10}-\frac{98}{10}\right)-\frac{3}{4}\)
\(=\frac{1}{4}:\frac{1}{2}-\frac{3}{4}=\frac{1}{4}\cdot2-\frac{3}{4}=\frac{2}{4}-\frac{3}{4}=-\frac{1}{4}\)
Câu cuối tương tự
phần mik gửi vừa nãy là vt nhầm , nên mik gửi lại