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a) \(\frac{325}{260}=\frac{650}{520}< \frac{650}{320}< \frac{876}{320}\)
\(\Rightarrow\frac{325}{260}< \frac{876}{320}.\)
b) \(\frac{13}{9}< \frac{24}{9}=\frac{8}{3}\)
\(\Rightarrow\frac{13}{9}< \frac{8}{3}.\)
c) Do \(18>-17;-39< 41\)
\(\Rightarrow\frac{18}{-39}>\frac{-17}{41}.\)
d) \(\frac{-151515}{232323}=\frac{\left(-15\right).10101}{23.10101}=\frac{-15}{23}\)
Mà \(\frac{-15}{25}< \frac{-15}{23}\)
\(\Rightarrow\frac{-15}{25}< \frac{-151515}{232323}.\)
a) Ta có : \(31^5< 32^5=\left(2^5\right)^5=2^{25}< 2^{28}=\left(2^4\right)^7=16^7< 17^7\)
\(\Rightarrow31^5< 17^7\)
b) Ta có : \(8^{12}=\left(2^3\right)^{12}=2^{36}>2^{32}=\left(2^4\right)^8=16^8>12^8\)
\(\Rightarrow8^{12}>12^8\)
c) \(A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}\)
\(3A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}\)
\(3A-A=\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}\right)\)
\(2A=1-\frac{1}{99}\)
\(A=\frac{1-\frac{1}{99}}{2}< \frac{1}{2}\)
\(\Rightarrow A< \frac{1}{2}\)
a) \(31^5< 34^5=2^5.17^5=32.17^5\)
\(17^7=17^2.17^5=289.17^5\)
\(\Rightarrow31^5< 17^7\)
b) \(12^8< 16^8=\left(2^4\right)^8=2^{32}\)
\(8^{12}=\left(2^3\right)^{12}=2^{36}\)
\(\Rightarrow8^{12}>12^8\)
c) \(A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}\)
\(3A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}\)
\(\Rightarrow3A-A=1+\left(\frac{1}{3}-\frac{1}{3}\right)+\left(\frac{1}{3^2}-\frac{1}{3^2}\right)+...+\left(\frac{1}{3^{98}}-\frac{1}{3^{98}}\right)-\frac{1}{3^{99}}\)
\(\Rightarrow2A=1-\frac{1}{3^{99}}< 1\Rightarrow A< \frac{1}{2}\)
a) \(2\frac{7}{9}\)và \(8\frac{1}{3}\)
Ta có:
\(2\frac{7}{9}=\frac{25}{9}\)
\(8\frac{1}{3}=\frac{25}{3}=\frac{25.3}{3.3}=\frac{75}{9}\)
Vì \(\frac{25}{9}< \frac{75}{9}\)nên \(2\frac{7}{9}< 8\frac{1}{3}\)
b) \(\frac{12}{7}\)và \(\frac{48}{28}\)
Ta có:
\(\frac{48}{28}=\frac{48:4}{28:4}=\frac{12}{7}\)
Mà \(\frac{12}{7}=\frac{12}{7}\)nên \(\frac{12}{7}=\frac{48}{28}\)
c) \(\frac{2^9}{\left(4^3\right)^8+45}\)và \(\frac{5^2}{\left(2^4\right)^3.12}\)
Ta có:
\(\frac{2^9}{\left(4^3\right)^8+45}=\frac{\left(2^2\right).2^7}{\left(2^5\right)^8+45}=\frac{\left(2^2\right).2^7}{2^{40}+45}=\frac{2^{31}}{45}\)
Tương tự với phân số kia
Phần d tương tự nha
a) Ta có 3/7 < 3,5/7 = 1/2 = 7,5/15 <11/15
Vậy 3/7 < 11/15
b) -11/6 < -1 < -8/9
Ai đủ điểm hỏi đáp đi ngang qua tk giùm với :(
a/b=x/y (=) a.y=b.x
A= x/12= (-1)/4
=) x.4=(-1).12
=)x.4=(-12)
=)x.=(-12);4
=)x=(-3)
a/b=x/y <=>a.y=b.x
a,=>x.4=12.-1
=>x.4=-12
=>x=-3
b,(1/3)^2x-1=1/243
=>(1/3)^2x-1=(1/3^5)
=>2x-1=1
=>2x=2
=>x=1
c,4/7.x-2/3=1/5
=>4/7x=13/15
=>x=13/15:4/7
=>x=91/60