Ta có :

\(B=\frac{x^2+15}{x^2+3}=\frac{x^2+3+12}{x^2+3}=1+\frac{12}{x^2+3}\)

vì x² \(\ge\)0 \(\Rightarrow\)x² + 3 \(\ge\)3 

\(\Rightarrow\frac{12}{x^2+3}\le4\)

\(\Rightarrow B\le1+4=5\)

Vậy GTLN của B là 5 khi x² + 3 = 3 hay x = 0

Ta có: \(B=1+\frac{12}{x^2+3}\)

Mà \(x^2+3\ne0\in Z\)

\(\Rightarrow\)Ta có 2 trường hợp

+) x²+3 nguyên dương

 \(\Rightarrow\frac{12}{x^2+3}\le12\Rightarrow B\le13\)(1)

+) x²+3 nguyên âm

\(\Rightarrow\frac{12}{x^2+3}< 0\Rightarrow B< 0\)(2)

Từ (1)(2) \(\Rightarrow B\le13\)

ADTCCDTSBN,TC :

\(\frac{2016c-a-b}{c}=\frac{2016b-a-c}{b}=\frac{2016a-b-c}{a}\)

\(=\frac{\left(2016c-a-b\right)+\left(2016b-a-c\right)+\left(2016a-b-c\right)}{c+b+a}=\frac{2014.\left(a+b+c\right)}{a+b+c}=2014\)

\(\frac{2016c-a-b}{c}=2014\Rightarrow2016c-a-b=2014c\Rightarrow2c=a+b\)( 1 )

\(\frac{2016b-a-c}{b}=2014\Rightarrow2016b-a-c=2014b\Rightarrow2b=a+c\)( 2 )

\(\frac{2016a-b-c}{a}=2014\Rightarrow2016a-b-c=2014a\Rightarrow2a=b+c\)( 3 )

Từ ( 1 ), ( 2 ) và ( 3 ) \(\Rightarrow\)a = b = c

\(\Rightarrow A=\left(1+\frac{a}{b}\right)\left(1+\frac{b}{c}\right)\left(1+\frac{c}{a}\right)=\left(1+1\right)\left(1+1\right)+\left(1+1\right)=2^3=8\)

thêm 1 câu nữa

d)F là trung điểm của BC

giúp mình với mình cần gắp

đặt \(A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+...+\frac{1}{3^{99}}\)

\(3A=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{98}}\)

\(3A-A=\left(1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{98}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+...+\frac{1}{3^{99}}\right)\)

\(2A=1-\frac{1}{3^{99}}\)

\(A=\frac{1-\frac{1}{3^{99}}}{2}\)

Đặt biểu thức là A, ta có:

\(A=\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}\)

\(\frac{1}{3}A=\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{100}}\)

\(A-\frac{1}{3}A=\left(\frac{1}{3^2}-\frac{1}{3^3}\right)+\left(\frac{1}{3^3}-\frac{1}{3^3}\right)+...+\left(\frac{1}{3}-\frac{1}{3^{100}}\right)\)

\(\frac{2}{3}A=\frac{1}{3}-\frac{1}{3^{100}}\)

\(\Rightarrow A=\frac{1}{3}:\frac{2}{3}=\frac{1}{2}\)

Đặt \(\frac{x}{a+2b+c}=\frac{y}{2a+b-c}=\frac{z}{4a-4b+c}=A\)

Áp dụng TC DTSBN ta có :

\(A=\frac{x+2y+z}{a+2b+c+2\left(2a+b-c\right)+4a-4b+c}=\frac{x+2y+z}{a+2b+c+4a+2b-2c+4a-4b+c}\)

\(=\frac{x+2y+z}{9a}=\frac{1}{9}.\frac{x+2y+z}{a}\) (1)

\(A=\frac{2x+y+z}{2\left(a+2b+c\right)+2a+b-c+4a-4b+c}=\frac{2x+y-z}{2a+4b+2c+2a+b-c-4a+4b-c}\)

\(=\frac{2x+y-z}{9b}=\frac{1}{9}.\frac{2x+y-z}{b}\) (2)

\(A=\frac{4x-4y+z}{4\left(a+2b+c\right)-4\left(2a+b-c\right)+4a-4b+c}=\frac{4x-4y+z}{4a+8b+4c-8a-4b+4c+4a-4b+c}\)

\(=\frac{4x-4y+z}{9c}=\frac{1}{9}.\frac{4x-4y+z}{c}\)(3)

Từ (1);(2);(3) \(\Rightarrow\frac{a}{x+2y+z}=\frac{b}{2x+y+z}=\frac{c}{4x-4y+z}\) (đpcm)

Gọi các góc của tam giác ABC là a,b,c

Ta có: \(\frac{a}{2}=\frac{b}{3}=\frac{c}{5}=\frac{a+b+c}{2+3+5}=\frac{180}{10}=18\)

=> a=36,b=54,c=90

Vậy góc A = 36 độ, góc B = 54 độ, góc C = 90 độ

xét \(\Delta\)ABC có : \(\widehat{A}+\widehat{B}+\widehat{C}=180^o\)

Vì A,B,C tỉ lệ thuận với 2,3,5

\(\Rightarrow\frac{\widehat{A}}{2}=\frac{\widehat{B}}{3}=\frac{\widehat{C}}{5}\)

áp dụng tính chất của dãy tỉ số bằng nhau, ta có :

\(\frac{\widehat{A}}{2}=\frac{\widehat{B}}{3}=\frac{\widehat{C}}{5}=\frac{\widehat{A}+\widehat{B}+\widehat{C}}{2+3+5}=\frac{180^o}{10}=18^o\)

\(\Rightarrow\widehat{A}=36;\widehat{B}=54;\widehat{C}=90\)

a: Xét ΔABD và ΔAED có

AB=AE
góc BAD=góc EAD

AD chung

Do đo: ΔABD=ΔAED
Suy ra: DB=DE
b:Ta có: AB=AE

DB=DE

Do đó: AD là đường trung trực của BE

1 \(-\)\(\frac{1}{3.5}\)\(-\)\(\frac{1}{5.7}\)\(-\)\(\frac{1}{7.9}\)\(-\)..... \(-\)\(\frac{1}{53.55}\)\(-\)\(\frac{1}{55.57}\)

= 1 \(-\)( \(\frac{1}{3.5}\)  + \(\frac{1}{5.7}\) + \(\frac{1}{7.9}\) + ..... + \(\frac{1}{53.55}\)  + \(\frac{1}{55.57}\)  )

= 1 \(-\)( \(\frac{1}{3}\)\(-\)\(\frac{1}{5}\)+ \(\frac{1}{5}\)\(-\)\(\frac{1}{7}\)+ \(\frac{1}{7}\)\(-\)\(\frac{1}{9}\)+....+ \(\frac{1}{53}\)\(-\)\(\frac{1}{55}\)+ \(\frac{1}{55}\)\(-\)\(\frac{1}{57}\)) . \(\frac{1}{2}\)

= 1 \(-\)( \(\frac{1}{3}\)\(-\)\(\frac{1}{57}\)) . \(\frac{1}{2}\)

= 1 \(-\) \(\frac{6}{19}\). \(\frac{1}{2}\)= 1 \(-\)\(\frac{3}{19}\)= \(\frac{16}{19}\)

\(1-\frac{1}{3.5}-\frac{1}{5.7}-\frac{1}{7.9}-...-\frac{1}{53.55}-\frac{1}{55.57}\)

đặt \(A=1-\frac{1}{3.5}-\frac{1}{5.7}-\frac{1}{7.9}-...-\frac{1}{53.55}-\frac{1}{55.57}\)

\(A=1-\left(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+....+\frac{1}{53.55}+\frac{1}{55.57}\right)\)

đặt \(B=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+.....+\frac{1}{53.55}+\frac{1}{55.57}\)

\(2B=2\left(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+....+\frac{1}{53.55}+\frac{1}{55.57}\right)\)

\(2B=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+....+\frac{2}{53.55}+\frac{2}{55.57}\)

\(2B=\frac{5-3}{3.5}+\frac{7-5}{5.7}+\frac{9-7}{7.9}+....+\frac{55-53}{53.55}+\frac{57-55}{55.57}\)

\(2B=\frac{5}{3.5}-\frac{3}{3.5}+\frac{7}{5.7}-\frac{5}{5.7}+\frac{9}{7.9}-\frac{7}{7.9}+...+\frac{55}{53.55}-\frac{53}{53.55}+\frac{57}{55.57}-\frac{55}{55.57}\)

\(2B=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{53}-\frac{1}{55}+\frac{1}{55}-\frac{1}{57}\)

\(2B=\frac{1}{3}-\frac{1}{57}\)

\(2B=\frac{54}{171}\)

\(\Rightarrow B=\frac{54}{171}:2\)

\(\Rightarrow B=\frac{9}{57}\)

mà \(A=1-B\)

\(\Rightarrow A=1-\frac{9}{57}\)

\(\Rightarrow A=\frac{48}{57}\)

chúc bạn học giỏi ^^