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4. a) Áp dụng tỉ dãy số bằng nhau:
\(\frac{a_1-1}{100}=\frac{a_2-2}{99}=...=\frac{a_{100}-100}{1}\)
\(=\frac{\left(a_1+a_2+...+a_{100}\right)-\left(1+2+...+100\right)}{100+99+...+2+1}=\frac{5050}{5050}=1\)
Suy ra: \(a_1-1=100\Leftrightarrow a_1=101\)
\(a_2-2=99\Leftrightarrow a_2=101\)
.......v.v...
\(a_{100}-100=1\Leftrightarrow a_{100}=101\)
Do đó: \(a_1=a_2=a_3=...=a_{100}=101\)
Bài 5/
Theo t/c dãy tỉ số bằng nhau,ta có: \(\frac{y+z-x}{x}=\frac{z+x-y}{y}=\frac{x+y-z}{z}=\frac{2\left(x+y+z\right)}{x+y+z}=2\)\(=\frac{2x}{x}\)
Suy ra:
\(\frac{y+z-x}{x}=\frac{2x}{x}\Leftrightarrow y+z-x=2x\Rightarrow x=y=z\) (vì nếu \(x\ne y\ne z\Rightarrow y+z-x\ne2x\) "không thỏa mãn")
Thay vào A,ta có: \(A=\left(1+\frac{x}{x}\right)\left(1+\frac{y}{y}\right)\left(1+\frac{z}{z}\right)=2.2.2=8\)
\(\left[\left(2+2\frac{1}{3}\right)-0,75\right]\left[3\frac{1}{2}-0,5:\left(\frac{3}{5}+\frac{1}{3}-\frac{1}{2}\right)\right]\)
\(=\left[\left(2+\frac{7}{3}\right)-\frac{75}{100}\right]\left[\frac{7}{2}-\frac{5}{10}:\left(\frac{3}{5}+\frac{1}{3}-\frac{1}{2}\right)\right]\)
\(=\left[\frac{2\cdot3+7}{3}-\frac{3}{4}\right]\left[\frac{7}{2}-\frac{1}{2}:\frac{13}{30}\right]\)
\(=\left[\frac{13}{3}-\frac{3}{4}\right]\left[\frac{7}{2}-\frac{1}{2}\cdot\frac{30}{13}\right]\)
\(=\left[\frac{13}{3}-\frac{3}{4}\right]\left[\frac{7}{2}-\frac{1}{1}\cdot\frac{15}{13}\right]\)
\(=\left[\frac{13}{3}-\frac{3}{4}\right]\left[\frac{7}{2}-\frac{15}{13}\right]\)
\(=\frac{43}{12}\cdot\frac{61}{26}=\frac{2623}{312}\)
\(\frac{2\left|2018x-2019\right|+2019}{\left|2018x-2019\right|+1}\)
\(=\frac{\left(2\left(\left|2018x-2019\right|+1\right)\right)+2017}{\left|2018x-2019\right|+1}\)
\(=2+\frac{2017}{\left|2018x-2019\right|+1}\)có giá trị lớn nhất
\(\Rightarrow\frac{2017}{\left|2018x-2019\right|+1}\)có giá trị lớn nhất
\(\Rightarrow\left|2018x-2019\right|+1\)có giá trị nhỏ nhất
Mà \(\left|2018x-2019\right|\ge0\)
\(\Rightarrow\left|2018x-2019\right|+1\ge1\)
Dấu "=" xảy ra khi và chỉ khi:
\(\left|2018x-2019\right|=0\)
\(\Leftrightarrow x=\frac{2019}{2018}\)
Vậy \(M_{MAX}=2019\)tại \(x=\frac{2019}{2018}\)
\(\frac{5^x+5^{x+1}+5^{x+2}}{31}=\frac{3^{2x}+3^{2x+1}+3^{2x+2}}{13}\)
\(\Rightarrow\frac{5^x\left(1+5+5^2\right)}{31}=\frac{3^{2x}\left(1+3+3^2\right)}{13}\)
\(\Rightarrow\frac{5^x\cdot31}{31}=\frac{3^{2x}\cdot13}{13}\)
\(\Rightarrow5^x=3^{2x}\)
Mà \(\left(5;3\right)=1\)
\(\Rightarrow x=2x=0\)
a) 273 : 32 = (33)3 : 32
= 39 : 32
= 37
b) (3/5)15 : (9/25)5 = (3/5)15 : [(3/5)2]5
= (3/5)15 : (3/5)10
= (3/5)2
a)\(x=\dfrac{-14\cdot52}{72}\\ x=\dfrac{-91}{9}\)
b)\(x=\dfrac{120\cdot7.2}{70}\\ x=\dfrac{432}{35}\)
c)\(x=\dfrac{2\dfrac{2}{3}\cdot8.5}{5}\\ x=\dfrac{68}{15}\)
d)\(x=\dfrac{4\dfrac{2}{5}\cdot9.5}{8}\\ x=\dfrac{209}{40}\)
a) \(\frac{790^4}{79^4}=\frac{79^4.10^4}{79^4}=10^4=10000\)
b) \(\frac{3^2}{0,375^2}=\frac{0,375^2.8^2}{0,375^2}=8^2=64\)
c) \(3^2.\frac{1}{243}.81^2.\frac{1}{3^3}=3^2.3^{-5}.3^8.3^{-3}=3^2=9\)
d) \(\left(4.2^5\right):\left(2^3.\frac{1}{16}\right)=2^7:\left(2^3.2^{-4}\right)=2^7:2^{-1}=2^7:\frac{1}{2}=2^8\)