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`a,`\(2^x -15= 2^4+1\)
`-> 2^x-15=17`
`-> 2^x=17+15`
`-> 2^x=32`
`-> 2^x=2^5`
`-> x=5`
`b,` Có phải đề là \(\dfrac{x+1}{65}+\dfrac{x+2}{64}=\dfrac{x+3}{63}+\dfrac{x+4}{62}\) ?
`=>`\(\dfrac{x+1}{65}+1+\dfrac{x+2}{64}+1=\dfrac{x+3}{63}+1+\dfrac{x+4}{62}+1\)
`=>`\(\dfrac{x+1+65}{65}+\dfrac{x+2+64}{64}-\dfrac{x+3+63}{63}-\dfrac{x+4+62}{62}=0\)
`=>`\(\dfrac{x+66}{65}+\dfrac{x+66}{64}-\dfrac{x+66}{63}-\dfrac{x+66}{62}=0\)
`=>`\(\left(x+66\right)\left(\dfrac{1}{65}+\dfrac{1}{64}-\dfrac{1}{63}-\dfrac{1}{62}\right)=0\)
Mà `1/65+1/64-1/63-1/62 \ne 0`
`-> x+66=0`
`-> x=-66`
a: =>2^x=2^4+16=32
=>x=5
b: Sửa đề: \(\dfrac{x+1}{65}+\dfrac{x+2}{64}=\dfrac{x+3}{63}+\dfrac{x+4}{62}\)
=>\(\left(\dfrac{x+1}{65}+1\right)+\left(\dfrac{x+2}{64}+1\right)=\left(\dfrac{x+3}{63}+1\right)+\left(\dfrac{x+4}{62}+1\right)\)
=>x+66=0
=>x=-66
1.
\(A\le0,5\)
Dấu "=" xảy ra khi x-3,5 = 0
<=> x = 3,5
Vậy max A = 0,5 khi x = 3,5
\(B\le-2\)
Dấu "=" xảy ra khi 1,4 -x =0
<=> x = 1,4
Vậy max B = -2 khi x =1,4
1.
A nhỏ hơn hoặc bằng 0,5 suy ra GTLN của A là 0,5.
B sẽ nhơ hơn hoặc bằng 2 suy ra GTLN
\(\frac{x}{2}+\frac{x}{2.3}+\frac{x}{3.4}+.....+\frac{x}{2015.2016}=\frac{2015}{4032}\)
\(x.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{2015.2016}\right)=\frac{2015}{4032}\)
\(x.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{2015}-\frac{1}{2016}\right)=\frac{2015}{4032}\)
\(x.\left(1-\frac{1}{2016}\right)=\frac{2015}{4032}\)
\(x.\frac{2015}{2016}=\frac{2014}{4032}\)
\(x=\frac{2015}{4032}:\frac{2015}{2016}\)
\(x=\frac{1}{2}\)
\(=\frac{x}{1}-\frac{x}{2}+\frac{x}{2}-\frac{x}{3}+...+\frac{x}{2015}-\frac{x}{2016}=\frac{2015}{4023}\)
\(=\frac{x}{1}-\frac{x}{2016}=\frac{2015}{4023}\)
\(=\frac{2015}{2016}x=\frac{2015}{4023}\)
=> x = \(\frac{2015}{4023}\cdot\frac{2016}{2015}\)= 2016/4023
\(2^x+3.4^2=64\\ \Rightarrow2^x+48=64\\ \Rightarrow2^x=16\\ \Rightarrow2^x=2^4\\ \Rightarrow x=4\)