\(Cho:a+b+c=1;a;b;c>0.timMax:P=ab-ac\)
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30 tháng 11 2021
Câu 2:
a: \(\Leftrightarrow x+2\in\left\{3;9\right\}\)
hay \(x\in\left\{1;7\right\}\)
6 tháng 1 2016
b + c - (a + c) = 3 - 2 = 1
b + c - a - c = 1
b -a = 1
a = (1+1)/2 = 1
b = 1 - 1 = 0
c = 3- 0 = 3
Vậy (a,b,c)= (1, 0 ,3)
ML
0
QM
1
29 tháng 11 2021
a) (x - 1)2 = 1.
<=> x - 1 = 1 hoặc x - 1 = -1.
<=> x = 2 hoặc x = 0.
b) 72x - 6 = 49.
<=> 72x - 6 = 72.
<=> 2x - 6 = 2.
<=> x = 4.
c) (2x - 16)7 = 128.
<=> (2x - 16)7 = 27.
<=> 2x - 16 = 2.
<=> x = 9.
SB
1
PN
4
24 tháng 1 2021
a) Ta có:\(n-6⋮n-1\)
\(\Leftrightarrow n-1-5⋮n-1\)
mà \(n-1⋮n-1\)
nên \(-5⋮n-1\)
\(\Leftrightarrow n-1\inƯ\left(-5\right)\)
\(\Leftrightarrow n-1\in\left\{1;-1;5;-5\right\}\)
hay \(n\in\left\{2;0;6;-4\right\}\)
Vậy: \(n\in\left\{2;0;6;-4\right\}\)
b) Ta có: \(3n+2⋮n-1\)
\(\Leftrightarrow3n-3+5⋮n-1\)
mà \(3n-3⋮n-1\)
nên \(5⋮n-1\)
\(\Leftrightarrow n-1\inƯ\left(5\right)\)
\(\Leftrightarrow n-1\in\left\{1;-1;5;-5\right\}\)
hay \(n\in\left\{2;0;6;-4\right\}\)
Vậy: \(n\in\left\{2;0;6;-4\right\}\)
c) Ta có: \(n^2+5⋮n+1\)
\(\Leftrightarrow n^2+2n+1-2n+4⋮n+1\)
\(\Leftrightarrow\left(n+1\right)^2-2n-2+6⋮n+1\)
mà \(\left(n+1\right)^2⋮n+1\)
và \(-2n-2⋮n+1\)
nên \(6⋮n+1\)
\(\Leftrightarrow n+1\inƯ\left(6\right)\)
\(\Leftrightarrow n+1\in\left\{1;-1;2;-2;3;-3;6;-6\right\}\)
hay \(n\in\left\{0;-2;1;-3;2;-4;5;-7\right\}\)
Vậy: \(n\in\left\{0;-2;1;-3;2;-4;5;-7\right\}\)
Sửa đề: \(a;b;c\ge0\) (nếu không thì không có max đâu cu!)
Ta có: \(P=a\left(b-c\right)\le ab\le\frac{\left(a+b\right)^2}{4}=\frac{1}{4}\)
Đẳng thức xảy ra khi \(a=b=\frac{1}{2};c=0\)
Vậy..