\(A=\dfrac{32}{x^2+2x+4}=\dfrac{32}{x^2+2x+1+3}=\dfrac{32}{\left(x^2+2x+1\right)+3}\)
= \(\dfrac{32}{\left(x+1\right)^2+3}\)
do (x+1)2 ≥ 0 ∀x
=> (x+1)2+3 ≥ 3
=> \(\dfrac{32}{\left(x+1\right)^2+3}\le\dfrac{32}{3}\)
=> A ≤ \(\dfrac{32}{3}\)
max A= \(\dfrac{32}{3}\) dấu "=" xảy ra khi
x+1=0
=> x=-1
vậy max A= \(\dfrac{32}{3}\) khi x=-1
Tìm các số nguyên x,y biết rằng
\(\dfrac{3-x}{y-4}\)= \(\dfrac{2}{5}\) và 2x+y= 0
Các bạn giải giúp mình nhé, ai nhanh nhất mình tick cho nha !!!
Tìm x,y
\(\dfrac{2x}{3}-\dfrac{2}{y}=\dfrac{1}{3}\) với \(x,y\in Z\)
a) Tính A = ( 1 - \(\dfrac{1}{2}\) )( 1 - \(\dfrac{1}{3}\) ) (1-\(\dfrac{1}{4}\) ) ....(1-\(\dfrac{1}{2014}\) ) (1-\(\dfrac{1}{2015}\) ) (1-\(\dfrac{1}{2016}\) )
b)Tìm x biết \(\dfrac{x-2}{12}\) + \(\dfrac{x-2}{20}\) + \(\dfrac{x-2}{30}\)+ \(\dfrac{x-2}{42}\) + \(\dfrac{x-2}{56}\) +\(\dfrac{x-2}{72}\) = \(\dfrac{16}{9}\)
Tìm GTLN của :
A = | 2x - 1 | + 2019 .
B = 2 - 3 . | 4 - 7x |.
C = 2 . ( x - 1 )2 + | 3y + 1| - 2018 .
D = 3 . ( x2 - 9 )4 + 4 . ( 2y2 - 32 )2 .
Tìm \(x,y\in N\):
a) 32x+1 . 7y = 9 . 21x
b) \(\dfrac{27^x}{3^{2x-y}}=243\) và \(\dfrac{25^x}{5^{x+y}}=125\)
Đề bài : Tìm các số tự nhiên x
1 , \(\dfrac{2x+5}{x+1}\) 4 , \(\dfrac{3x+5}{x-1}\)
2 , \(\dfrac{2x+4}{x}\) 5 , \(\dfrac{3x+6}{x-1}\)
3 , \(\dfrac{2x+7}{x+1}\)
Tìm x :
a) \(\left|x+\dfrac{11}{17}\right|+\left|x+\dfrac{2}{17}\right|+\left|x+\dfrac{4}{17}\right|=4x\)
b) \(\left|x+\dfrac{1}{2}\right|+\left|x+\dfrac{1}{6}\right|+\left|x+\dfrac{1}{12}\right|+\left|x+\dfrac{1}{20}\right|+..+\left|x+\dfrac{1}{110}\right|=11x\)
Tính tổng sau đây bằng 2 cách:
\(A=\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{1}{32}+\dfrac{1}{64}\)
1: \(\dfrac{2\cdot\left(x+2\right)}{3}-\dfrac{5\cdot\left(x-1\right)}{4}=\dfrac{3\cdot\left(5-x\right)}{2}-1\dfrac{1}{2}\cdot\left(2x+3\right)\)
