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a) \(\approx9,31\)
b) \(\approx4,77\)
c) \(\approx289,57\)
d) \(\approx23,73\)
\(\frac{1}{9}\). 3\(^4\). 3\(^x\)= 3\(^7\)
<=> \(\frac{3^4.3^x}{3^2}\)= 3\(^7\)
<=> 3\(^4\)- 3\(^x\)= 3\(^7\). 3\(^2\)
<=> 3\(^x\)= 3\(^9\): 3\(^4\)
<=> 3\(^x\)= 3\(^5\)
<=> x = 5
Vay x = 5
1) a.Từ\(\frac{x}{y}=\frac{11}{7}\Rightarrow\frac{x}{11}=\frac{y}{7}\)
Áp dụng t/c của dãy tỉ số bằng nhau, ta có:
\(\frac{x}{11}=\frac{y}{7}=\frac{x-y}{11-7}=\frac{12}{4}=3\)
\(\Rightarrow x=3.11=33;y=3.7=21\)
b) \(\sqrt{2x-3}=5\)
\(2x-3=25\)
\(2x=28\)
\(x=14\)
2) a) \(\frac{3}{2}-\frac{5}{6}:\left(\frac{1}{2}\right)^2+\sqrt{4}=\frac{3}{2}-\frac{5}{6}:\frac{1}{4}+2\)
\(=\frac{3}{2}-\frac{10}{3}+2\)
\(=\frac{1}{6}\)
_Học tốt nha_
1. a, \(\frac{x}{y}=\frac{11}{7}\)và x-y=12
\(\Rightarrow\frac{x}{11}=\frac{y}{7}\)và x-y=12
Áp dung tính chất của dãy tỉ số bằng nhau ta có:
\(\frac{x}{11}=\frac{y}{7}=\frac{x-y}{11-7}=\frac{12}{4}=3\)
\(\Rightarrow\hept{\begin{cases}\frac{x}{11}=3\\\frac{y}{7}=3\end{cases}}\)\(\Rightarrow\hept{\begin{cases}x=33\\y=21\end{cases}}\)
Vậy
b,\(\sqrt{2x-3}\)=5
\(\Rightarrow2x-3=25\)
\(\Rightarrow2x=28\)
\(\Rightarrow x=14\)
c,\(\frac{3}{2}-\frac{5}{6}:\left(\frac{1}{2}\right)^2+\sqrt{4}\)
\(=\frac{3}{2}-\frac{5}{6}:\frac{1}{4}+2\)
\(=\frac{3}{2}-\frac{10}{3}+2\)
\(=\frac{9}{6}-\frac{20}{6}+2\)
\(=\frac{-11}{6}+2\)
\(=\frac{1}{6}\)
\(125^7-25^{10}+5^{19}\)
\(=\left(5^3\right)^7-\left(5^2\right)^{10}+5^{19}\)
\(=5^{21}-5^{20}+5^{19}\)
\(=5^{19}.\left(5^2-5+1\right)\)
\(=5^{19}.21\)
\(=5^{18}.5.21\)
\(=5^{18}.105\)
Ta có: \(105⋮105\)
\(\Rightarrow5^{18}.105⋮105\)
\(\Rightarrow125^7-25^{10}+5^{19}⋮105\)
đpcm
\(125^7-25^{10}+5^{19}\)
\(=\left(5^3\right)^7-\left(5^2\right)^{10}+5^{19}\)
\(=5^{21}-5^{20}+5^{19}\)
\(=5^{19}.\left(5^2-5+1\right)\)
\(=5^{19}.21\)
\(=5^{18}.5.21=5^{18}.105⋮105\)
Vậy ......
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\frac{\left(1+2y\right)}{18}=\frac{\left(1+4y\right)}{24}\)\(=\frac{\left(1+4y-1-2y\right)}{24-18}\)\(=\frac{2y}{6}=\frac{y}{3}\)
\(\Rightarrow\frac{\left(1+2y\right)}{18}=\frac{y}{3}\)
\(\Rightarrow3.\left(1+2y\right)=18y\)
\(\Rightarrow3+6y=18y\)
\(\Rightarrow18y-6y=3\)
\(\Rightarrow12y=3\)
\(\Rightarrow y=\frac{1}{4}\)
Thay \(y=\frac{1}{4}\)vào \(\frac{\left(1+2y\right)}{18}=\frac{\left(1+6y\right)}{6x}\)ta có:
\(6x.\left(1+2y\right)=18.\left(1+6y\right)\)
\(\Rightarrow6x.\left(\frac{3}{2}\right)=18.\left(\frac{5}{2}\right)\)
\(\Rightarrow6x=30\)
\(\Rightarrow x=5\)
Vậy \(y=\frac{1}{4};x=5\)
\(-21,3\cdot15,4+8,3\cdot4,6-21,3\cdot4,6+8,3\cdot15,4\)
\(=\left(-21,3\cdot15,4-21,3\cdot4,6\right)+\left(8,3\cdot4,6+8,3\cdot15,4\right)\)
\(=-21,3\cdot\left(15,4+4,6\right)+8,3\left(4,6+15,4\right)\)
\(=20\cdot\left(8,3-21,3\right)\)
\(=-13\cdot20\)
\(=-260\)