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Bài 1 :
Ta có : \(S=\frac{a+b}{c}+\frac{b+c}{a}+\frac{c+a}{b}\)
\(=\frac{a}{c}+\frac{b}{c}+\frac{b}{a}+\frac{c}{a}+\frac{c}{b}+\frac{a}{b}\)
\(=\left(\frac{a}{b}+\frac{b}{a}\right)+\left(\frac{a}{c}+\frac{c}{a}\right)+\left(\frac{b}{c}+\frac{c}{b}\right)\)
Ta chứng minh BĐT \(\frac{x}{y}+\frac{y}{x}\ge2,\forall x,y>0\)
Thật vậy : BĐT \(\Leftrightarrow\frac{x}{y}+\frac{y}{x}-2=\frac{\left(x-y\right)^2}{xy}\ge0\) ( đúng )
Vậy \(\frac{x}{y}+\frac{y}{x}\ge2,\forall x,y>0\)
Áp dụng vào bài toán ta có : \(S\ge2+2+2=6\)
Dấu "=" xảy ra \(\Leftrightarrow a=b=c\)
Vậy min \(S=6\) tại \(a=b=c\)
Bài 1:
c/
\(\left(2x-7\right)^2=18:2\)
\(\left(2x-7\right)^2=9=3^2\)
=>\(2x-7=3\)
=>\(2x=10\)
=>\(x=5\)
Bài 1:
|2x+3|=5
=>2x+3=5 hoặc (-5)
- Với 2x+3=5
=>2x=2
=>x=1
- Với 2x+3=-5
=>2x=-8
=>x=-4
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2019.2020}\)và 1
gọi
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2019.2020}\)
\(A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2019}-\frac{1}{2020}\)
\(A=\frac{1}{1}-\frac{1}{2020}=\frac{2019}{2020}\)
VÌ \(\frac{2019}{2020}< 1\Rightarrow A< 1\)
VẬY \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2019.2020}< 1\)
1. a) P = 4 - ( x - 2 )32
( x - 2 )32 ≥ 0 ∀ x => - ( x - 2 )32 ≤ 0 ∀ x
=> 4 - ( x - 2 )32 ≤ 4 ∀ x
Dấu bằng xảy ra <=> x - 2 = 0 => x = 2
Vậy PMax = 4 khi x = 2
b) Q = 20 - | 3 - x |
| 3 - x | ≥ 0 ∀ x => - | 3 - x | ≤ 0 ∀ x
=> 20 - | 3 - x | ≤ 20 ∀ x
Dấu bằng xảy ra <=> 3 - x = 0 => x = 3
Vậy QMax = 20 khi x = 3
c) C = \(\frac{5}{\left(x-3\right)^2+1}\)
Để C có GTLN => ( x - 3 )2 + 1 nhỏ nhất dương
=> ( x - 3 )2 + 1 = 1
=> ( x - 3 )2 = 0
=> x - 3 = 0
=> x = 3
=> CMax = \(\frac{5}{\left(3-3\right)^2+1}=\frac{5}{1}=5\)khi x = 3
3n+2/ n-1 =3n-3+5/n-1=3 + 5/ n-1
Để phân số a nguyên
=>n-1 thuộc Ư(5)
=>n-1 thuoc {-5 ;-1 ;1 ;5 }
n thuộc {-4 ; 0 :2 :6}
Chú ý : Vì là lớp 6 nên giải zậy chứ lớp 9 là cách lm này là k chuẩn........( vì n không thuộc Z)
b,2B=1=1/2 +......+1/22015
2B-B=(1 +1/2 +.....+1/22015) - (1/2 +1/22+......+1/22016)
B=1 -1/22016
Vi 1-1/22016<1
=>B<1
a)
\(A=\frac{3n+2}{n-1}=\frac{3\left(n-1\right)+5}{n-1}=3+\frac{5}{n-1}\)
Để A nguyên thì 5 chia hết cho n-1
\(\Rightarrow n-1\in U\left(5\right)=+-1;+-5\)
lập bảng nhé!
b)
\(B=\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2016}}\)
\(\Rightarrow\frac{1}{2}B=\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2017}}\)
\(\Rightarrow B=\left(B-\frac{1}{2}B\right).2=\left(\frac{1}{2}-\frac{1}{2^{2017}}\right).2\)
\(\Rightarrow B=1-\frac{1}{2^{2016}}< 1\)
Sao nhiều quá vại??
mk lm k nổi đâu
Dài quá nhìn lòi bảng họng lun ak
Bài : 4
a/ \(\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+....+\frac{1}{24\cdot25}\)
\(=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+....+\frac{1}{24}-\frac{1}{25}\)
\(=\frac{1}{5}-\frac{1}{25}\)
\(=\frac{4}{25}\)
b/ \(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+....+\frac{2}{99\cdot101}\)
\(=\frac{3-1}{1\cdot3}+\frac{5-3}{3\cdot5}+\frac{7-5}{5\cdot7}+...+\frac{101-99}{99\cdot101}\)
\(=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....+\frac{1}{99}-\frac{1}{101}\)
\(=\frac{1}{1}-\frac{1}{101}\)
\(=\frac{100}{101}\)
c/ \(\frac{5^2}{1\cdot6}+\frac{5^2}{6\cdot11}+\frac{5^2}{11\cdot16}+\frac{5^2}{16\cdot21}+\frac{5^2}{21\cdot26}+\frac{5^2}{26\cdot31}\)
\(=\frac{25}{1\cdot6}+\frac{25}{6\cdot11}+\frac{25}{11\cdot16}+\frac{25}{16\cdot21}+\frac{25}{21\cdot26}+\frac{25}{26\cdot31}\)
\(=\frac{6-1}{1\cdot6}+\frac{11-6}{6\cdot11}+....+\frac{31-26}{26\cdot31}\)
\(=\frac{25}{5}\cdot\left(\frac{1}{1}-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+....+\frac{1}{26}-\frac{1}{31}\right)\)
\(=\frac{25}{5}\cdot\left(\frac{1}{1}-\frac{1}{31}\right)\)
\(=\frac{25}{5}\cdot\frac{30}{31}\)
\(=\frac{150}{31}\)
d/ \(\frac{3}{1\cdot3}+\frac{3}{3\cdot5}+\frac{3}{5\cdot7}+....+\frac{3}{49\cdot51}\)
\(=\frac{3-1}{1\cdot3}+\frac{5-3}{3\cdot5}+\frac{7-5}{5\cdot7}+....+\frac{51-49}{49\cdot51}\)
\(=\frac{3}{2}\cdot\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....+\frac{1}{49}-\frac{1}{51}\right)\)
\(=\frac{3}{2}\cdot\left(\frac{1}{1}-\frac{1}{51}\right)\)
\(=\frac{3}{2}\cdot\frac{50}{51}\)
\(=\frac{25}{17}\)
e/ \(\frac{1}{7}+\frac{1}{91}+\frac{1}{247}+\frac{1}{475}+\frac{1}{775}+\frac{1}{1147}\)
\(=\frac{1}{1\cdot7}+\frac{1}{7\cdot13}+\frac{1}{13\cdot19}+\frac{1}{19\cdot25}+\frac{1}{25\cdot31}+\frac{1}{31\cdot37}\)
\(=\frac{7-1}{1\cdot7}+\frac{13-7}{7\cdot13}+....+\frac{37-31}{31\cdot37}\)
\(=\frac{1}{6}\cdot\left(1-\frac{1}{7}+\frac{1}{7}-\frac{1}{13}+....+\frac{1}{31}-\frac{1}{37}\right)\)
\(=\frac{1}{6}\cdot\left(1-\frac{1}{37}\right)\)
\(=\frac{1}{6}\cdot\frac{36}{37}\)
\(=\frac{6}{37}\)
\(C=1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2012^2}\)
\(< 1+\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2011.2012}\)
\(=1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2011}-\frac{1}{2012}\)
\(=2-\frac{1}{2012}< 2\)
\(C=1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2012^2}\)
\(>1+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{2012.2013}\)
\(=1+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{2012}-\frac{1}{2013}\)
\(=1+\frac{1}{2}-\frac{1}{2013}>1\)
=> \(1< C< 2\)
=> Số tự nhiên bé nhất mà lớn hơn C là 2
Ta có \(\frac{1}{2^2}< \frac{1}{1.2};\frac{1}{3^2}< \frac{1}{2.3};...;\frac{1}{2012^2}< \frac{1}{2011.2012}\)
\(C< 1+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2011.2012}\)
\(C< 1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2011}-\frac{1}{2012}\)
\(C< 2-\frac{1}{2012}< 2\)
Vậy giá trị nguyên nhỏ nhất lớn hơn C là 2
_Kudo_