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Bài 1:suy ra 5*(44-x)=3*(x-12)
220-5x=3x-36
-5x-3x=-36-220
-8x =-256
x=32
Bài 2 :Đặt a/3=b/4=k
suy ra a=3k ; b=4k
Ta có a*b=48
suy ra 3k*4k=48
12k =48
k=4
suy ra a=3*4=12
b=4*4 =16
Bài 3: áp dụng tính chất dãy số bằng nhau ta được
a+b+c+d/3+5+7+9 = 12/24=0,5
suy ra a=1,5; b=2,5; c=3,5; d=4,
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 :
Từ \(\frac{1}{4}< \frac{1}{3}\) suy ra \(\frac{1}{4}< \frac{1+1}{4+3}< \frac{1}{3}\) hay \(\frac{1}{4}< \frac{2}{7}< \frac{1}{3}\)
Từ \(\frac{1}{4}< \frac{2}{7}\)suy ra \(\frac{1}{4}< \frac{1+2}{4+7}< \frac{1}{3}\)hay \(\frac{1}{4}< \frac{3}{11}< \frac{1}{3}\)
Từ \(\frac{2}{7}< \frac{1}{3}\)suy ra \(\frac{2}{7}< \frac{2+1}{7+3}< \frac{1}{3}\)hay \(\frac{2}{7}< \frac{3}{10}< \frac{1}{3}\)
Vậy ta có : \(\frac{1}{4}< \frac{3}{11}< \frac{2}{7}< \frac{3}{10}< \frac{1}{3}\)
Chúc bạn học tốt ( -_- )
Bài 2 :
\(\frac{a}{a+b+c+d}< \frac{a}{a+b+c}< \frac{a}{a+c}\left(1\right)\)
\(\frac{b}{a+b+c+d}< \frac{b}{b+c+d}< \frac{b}{b+d}\left(2\right)\)
\(\frac{c}{a+b+c+d}< \frac{c}{c+d+a}< \frac{c}{c+a}\left(3\right)\)
\(\frac{d}{a+b+c+d}< \frac{d}{d+a+b}< \frac{d}{d+b}\left(4\right)\)
Cộng ( 1 ), ( 2 ) , (3 ) và ( 4 ) theo từng vế ta được :
\(1=\frac{a+b+c+d}{a+b+c+d}< \frac{a}{a+b+c}+\frac{b}{b+c+d}\)\(+\frac{c}{c+d+a}+\frac{d}{d+a+b}< \frac{a+c}{a+c}+\frac{b+d}{b+d}\)
Chúc bạn học tốt ( -_- )
\(\frac{a+1}{2}\)=\(\frac{b+2}{3}\)=\(\frac{c+2}{4}\)=>\(\frac{3a+3}{6}\)=\(\frac{-2b-4}{-6}\)=\(\frac{c+2}{4}\)=\(\frac{3a-2b+c+3-4+2}{6+-6+4}\)=\(\frac{105+3-4+2}{4}\)=26,5
=>a=26,5.2-1=52
b=26,5.3-2=77,5
c=26,5.4-2=104
chắc chắn đúng
-duc tuan nguyen-
=>( 3a + 3 ) / 6 = ( 2b + 4 ) / 6 =( c + 2 ) / 4
= ( 3a + 3 - 2b - 4 + c + 2 ) / ( 6 - 6 + 4 )
= 106 / 4
= 26,5
=> a + 1 = 53 => a = 52
=> b + 2 = 79,5 => b = 77,5
=> c + 2 = 106 => c = 104
Bài 1 :
\(A=\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{50-49}{49.50}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\)
\(=1-\frac{1}{50}< 1\left(1\right)\)
\(B=\frac{1}{10}+\left(\frac{1}{11}+\frac{1}{12}+...+\frac{1}{99}+\frac{1}{100}\right)\)\(>\frac{1}{10}+\frac{1}{100}.90=1\left(2\right)\)
Từ (1) và ( 2) ta có \(A< 1\) \(B>1\)NÊN \(A< B\)
Bài 2:
\(S=\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\)
\(=\frac{\left(a+b+c\right)-\left(b+c\right)}{b+c}+\)\(\frac{\left(a+b+c\right)-\left(c+a\right)}{c+a}\)\(+\frac{\left(a+b+c\right)-\left(a+b\right)}{a+b}\)
\(=\frac{7-\left(b+c\right)}{b+c}+\frac{7-\left(c+a\right)}{c+a}+\frac{7-\left(a+b\right)}{a+b}\)
\(=7.\left(\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{a+b}\right)-3\)
\(=7.\frac{7}{10}-3\)\(=\frac{49}{10}-3=\frac{19}{10}\)
\(S=\frac{19}{10}>\frac{19}{11}=1\frac{8}{11}\)
Chúc bạn học tốt ( -_- )
Bài 1:
ta có: \(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
\(A=1-\frac{1}{50}< 1\)
\(\Rightarrow A< 1\)(1)
ta có: \(\frac{1}{11}>\frac{1}{100};\frac{1}{12}>\frac{1}{100};...;\frac{1}{99}>\frac{1}{100}\)
\(\Rightarrow\frac{1}{11}+\frac{1}{12}+...+\frac{1}{99}+\frac{1}{100}>\frac{1}{100}+\frac{1}{100}+...+\frac{1}{100}+\frac{1}{100}\) ( có 90 số 1/100)
\(=\frac{90}{100}=\frac{9}{10}\)
\(\Rightarrow B=\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+...+\frac{1}{99}+\frac{1}{100}>\frac{1}{10}+\frac{9}{10}=1\)
\(\Rightarrow B>1\)(2)
Từ (1);(2) => A<B
Bài 3:
a,Đặt A = \(\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}+\frac{1}{32}-\frac{1}{64}\)
A = \(\frac{1}{2}-\frac{1}{2^2}+\frac{1}{2^3}-\frac{1}{2^4}+\frac{1}{2^5}-\frac{1}{2^6}\)
2A = \(1-\frac{1}{2}+\frac{1}{2^2}-\frac{1}{2^3}+\frac{1}{2^4}-\frac{1}{2^5}\)
2A + A = \(\left(1-\frac{1}{2}+\frac{1}{2^2}-\frac{1}{2^3}+\frac{1}{2^4}-\frac{1}{2^5}\right)+\left(\frac{1}{2}-\frac{1}{2^2}+\frac{1}{2^3}-\frac{1}{2^4}+\frac{1}{2^5}-\frac{1}{2^6}\right)\)
3A = \(1-\frac{1}{2^6}\)
=> 3A < 1
=> A < \(\frac{1}{3}\)(đpcm)
b, Đặt A = \(\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+...+\frac{99}{3^{99}}-\frac{100}{3^{100}}\)
3A = \(1-\frac{2}{3}+\frac{3}{3^2}-\frac{4}{4^3}+...+\frac{99}{3^{98}}-\frac{100}{3^{99}}\)
3A + A = \(\left(1-\frac{2}{3}+\frac{3}{3^2}-\frac{4}{4^3}+...+\frac{99}{3^{98}}-\frac{100}{3^{99}}\right)-\left(\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+...+\frac{99}{3^{99}}-\frac{100}{3^{100}}\right)\)
4A = \(1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+...+\frac{1}{3^{98}}-\frac{1}{3^{99}}-\frac{100}{3^{100}}\)
=> 4A < \(1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+...+\frac{1}{3^{98}}-\frac{1}{3^{99}}\) (1)
Đặt B = \(1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+...+\frac{1}{3^{98}}-\frac{1}{3^{99}}\)
3B = \(3-1+\frac{1}{3}-\frac{1}{3^2}+...+\frac{1}{3^{97}}-\frac{1}{3^{98}}\)
3B + B = \(\left(3-1+\frac{1}{3}-\frac{1}{3^2}+...+\frac{1}{3^{97}}-\frac{1}{3^{98}}\right)+\left(1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+...+\frac{1}{3^{98}}-\frac{1}{3^{99}}\right)\)
4B = \(3-\frac{1}{3^{99}}\)
=> 4B < 3
=> B < \(\frac{3}{4}\) (2)
Từ (1) và (2) suy ra 4A < B < \(\frac{3}{4}\)=> A < \(\frac{3}{16}\)(đpcm)
1. Ta có:\(\frac{a}{2}=\frac{b}{3}=\frac{c}{4}=\frac{a+2b-3c}{2+6-12}=\frac{-20}{-4}=5\)
\(\Rightarrow\hept{\begin{cases}\frac{a}{2}=5\\\frac{b}{3}=5\\\frac{c}{4}=5\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}a=10\\b=15\\c=20\end{cases}}\)
2. Ta có:\(\frac{a}{2}=\frac{b}{3}\Rightarrow\frac{a}{10}=\frac{b}{15}\)
\(\frac{b}{5}=\frac{c}{4}\Rightarrow\frac{b}{15}=\frac{c}{12}\)
\(\Rightarrow\frac{a}{10}=\frac{b}{15}=\frac{c}{12}=\frac{a-b+c}{10-15+12}=\frac{-49}{7}=-7\)
\(\Rightarrow\hept{\begin{cases}\frac{a}{10}=-7\\\frac{b}{15}=-7\\\frac{c}{12}=-7\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}a=-70\\b=-105\\c=-84\end{cases}}\)
1. Ta có:a2 =b3 =c4 =a+2b−3c2+6−12 =−20−4 =5
2. Ta có:a2 =b3 ⇒a10 =b15
b5 =c4 ⇒b15 =c12
⇒a10 =b15 =c12 =a−b+c10−15+12 =−497 =−7