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1)
\(\frac{a}{b}=\frac{a\left(b+c\right)}{b\left(b+c\right)}=\frac{ab+ac}{b\left(b+c\right)}\)
\(\frac{a+c}{b+c}=\frac{b\left(a+c\right)}{b\left(b+c\right)}=\frac{ab+bc}{b\left(b+c\right)}\)
mà ab = ab; ac > bc ( vì a > b )
=> \(\frac{a}{b}>\frac{a+c}{b+c}\left(đpcm\right)\)
\(\text{A = }\frac{\text{-1}}{\text{2011}}-\frac{\text{3}}{\text{11}^2}-\frac{\text{5}}{\text{11}^2.\text{11}}-\frac{\text{7}}{\text{11}^2.\text{11}^2}=\text{ }\frac{\text{-1}}{\text{2011}}-\frac{\text{1}}{\text{11}^2}.\left(3-\frac{\text{5}}{\text{11}}-\frac{\text{7}}{\text{11}^2}\right)\)
\(\text{B = }\text{ }\frac{\text{-1}}{\text{2011}}-\frac{7}{\text{11}^2}-\frac{5}{\text{11}^2.\text{11}}-\frac{3}{\text{11}^2.\text{11}^2}=\frac{\text{-1}}{\text{2011}}-\frac{\text{1}}{\text{11}^2}.\left(7-\frac{5}{\text{11}}-\frac{3}{\text{11}^2}\right)\)
\(\text{Vì }3-\frac{\text{5}}{\text{11}}-\frac{\text{7}}{\text{11}^2}< 7-\frac{5}{\text{11}}-\frac{3}{\text{11}^2}\)
\(\Rightarrow\frac{\text{-1}}{\text{2011}}-\frac{\text{1}}{\text{11}^2}.\left(3-\frac{\text{5}}{\text{11}}-\frac{\text{7}}{\text{11}^2}\right)>\frac{\text{-1}}{\text{2011}}-\frac{\text{1}}{\text{11}^2}.\left(7-\frac{5}{\text{11}}-\frac{3}{\text{11}^2}\right)\)
=> A > B
Vậy A > B
\(\text{A = }\frac{\text{-1}}{\text{2011}}-\frac{\text{3}}{\text{11}^2}-\frac{\text{5}}{\text{11}^2.\text{11}}-\frac{\text{7}}{\text{11}^2.\text{11}^2}=\text{ }\frac{\text{-1}}{\text{2011}}-\frac{\text{1}}{\text{11}^2}.\left(3-\frac{\text{5}}{\text{11}}-\frac{\text{7}}{\text{11}^2}\right)\)
\(\text{B = }\frac{\text{-1}}{\text{2011}}-\frac{7}{\text{11}^2}-\frac{5}{\text{11}^2.\text{11}}-\frac{3}{\text{11}^2.\text{11}^2}=\frac{\text{-1}}{\text{2011}}-\frac{\text{1}}{\text{11}^2}.\left(7-\frac{5}{\text{11}}-\frac{3}{\text{11}^2}\right)\)
\(\text{Vì }3-\frac{\text{5}}{\text{11}}-\frac{\text{7}}{\text{11}^2}< 7-\frac{5}{\text{11}}-\frac{3}{\text{11}^2}\)
\(\Rightarrow\frac{\text{-1}}{\text{2011}}-\frac{\text{1}}{\text{11}^2}.\left(3-\frac{\text{5}}{\text{11}}-\frac{\text{7}}{\text{11}^2}\right)>\frac{\text{-1}}{\text{2011}}-\frac{\text{1}}{\text{11}^2}.\left(7-\frac{5}{\text{11}}-\frac{3}{\text{11}^2}\right)\)
=> A > B
Vậy A > B
Bài 1:
a) \(A=\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{2012.2015}\)
\(A=\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{2012}-\frac{1}{2015}\right)\)
\(A=\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{2015}\right)\)
\(A=\frac{1}{3}\cdot\frac{2013}{4030}=\frac{671}{4030}\)
Bài 2:
ta có: \(\frac{a}{b+c}=\frac{b}{a+c}=\frac{c}{a+b}\)
\(\Rightarrow\frac{b+c}{a}=\frac{a+c}{b}=\frac{a+b}{c}=\frac{b+c+a+c+a+b}{a+b+c}=\frac{2a+2b+2c}{a+b+c}\)
\(=\frac{2.\left(a+b+c\right)}{a+b+c}=2\)
\(\Rightarrow\frac{b+c}{a}=\frac{a+c}{b}=\frac{a+b}{c}=2\)
\(\Rightarrow A=\frac{b+c}{a}+\frac{a+c}{b}+\frac{a+b}{c}=2+2+2=6\)
Bài 3:
a) f(1) = 4/1 = 4
=> f(1) = 4
g(-1) = (-1)^2 = 1
=> g(-1) = 1
h(-5) = -2.(-5)^2 - 5/(-5) = -2.25 + 1 = -50 + 1 = -49
=> h(-5) = -49
b) ta có: k(x)=f(x)+g(x)+h(x)
=> k(x) = 4/x + x^2 -2x^2 - 5/x
k(x) = - (5/x - 4/x) - (2x^2-x^2)
k(x) = -1/x - x
\(k_{\left(x\right)}=\frac{-1}{x}-\frac{x.x}{x}=\frac{-1-x^2}{x}\)
c) Để k(x) = 0
=> -1-x^2/x = 0 ( x khác 0)
=> -1-x^2 = 0
=> x^2 = -1
=> không tìm được x
Bài 4:
a) Xét tam giác ABC vuông tại A
có: góc B + góc C = 90 độ ( 2 góc phụ nhau)
thay số: 60 độ + góc C = 90 độ
góc C = 90 độ - 60 độ
góc C = 30 độ
=> AB = BC/2 ( cạnh đối diện với góc 30 độ)
thay số: 5 = BC/2
=> BC = 5.2
=> BC = 10 cm
Xét tam giác ABC vuông tại A
có: AC^2 + AB^2 = BC^2 ( py - ta - go)
thay số: AC^2 + 5^2 = 10^2
AC^2 + 25 = 100
AC^2 = 75
\(\Rightarrow AC=\sqrt{75}\) cm
Ta có : \(\frac{20}{60.63}+\frac{20}{63.66}+.....+\frac{20}{117.120}+\frac{20}{2011}\)
\(=\left(\frac{20}{60.63}+\frac{20}{63.66}+.....+\frac{20}{117.120}\right)+\frac{20}{2011}\)
\(=\frac{20}{3}\left(\frac{3}{60.63}+\frac{3}{63.66}+.....+\frac{3}{117.120}\right)+\frac{20}{2011}\)
\(=\frac{20}{3}\left(\frac{1}{60}-\frac{1}{63}+\frac{1}{63}-\frac{1}{66}+.....+\frac{1}{117}-\frac{1}{120}\right)+\frac{20}{2011}\)
\(=\frac{20}{3}.\left(\frac{1}{60}-\frac{1}{120}\right)+\frac{20}{2011}\)
\(=\frac{20}{3}.\frac{1}{120}+\frac{20}{2011}\)
\(=\frac{1}{18}+\frac{20}{2011}\)
Ta có:
\(A=\frac{20}{60.63}+\frac{20}{63.66}+...+\frac{20}{117.120}+\frac{20}{2011}\)
\(\Rightarrow A=\left(\frac{20}{60.63}+\frac{20}{63.66}+...+\frac{20}{117.120}\right)+\frac{20}{2011}\)
\(\Rightarrow A=\frac{20}{3}.\left(\frac{1}{60}-\frac{1}{63}+\frac{1}{63}-\frac{1}{66}+...+\frac{1}{117}-\frac{1}{120}\right)+\frac{20}{2011}\)
\(\Rightarrow A=\frac{20}{3}.\left(\frac{1}{60}-\frac{1}{120}\right)+\frac{20}{2011}\)
\(\Rightarrow A=\frac{20}{3}.\left(\frac{1}{60}-\frac{1}{120}\right)+\frac{20}{2011}\)
\(\Rightarrow A=\frac{20}{3}.\frac{1}{120}+\frac{20}{2011}=\frac{1}{18}+\frac{20}{2011}\)
\(B=\frac{5}{40.44}+\frac{5}{44.48}+\frac{5}{48.52}+...+\frac{5}{76.80}+\frac{5}{2011}\)
\(\Rightarrow B=\left(\frac{5}{40.44}+\frac{5}{44.48}+\frac{5}{48.52}+...+\frac{5}{76.80}\right)+\frac{5}{2011}\)
\(\Rightarrow B=\frac{5}{4}.\left(\frac{1}{40}-\frac{1}{44}+\frac{1}{44}-\frac{1}{48}+\frac{1}{48}-\frac{1}{52}+...+\frac{1}{76}-\frac{1}{80}\right)+\frac{5}{2011}\)
\(\Rightarrow B=\frac{5}{4}.\left(\frac{1}{40}-\frac{1}{80}\right)+\frac{5}{2011}\)
\(\Rightarrow B=\frac{5}{4}.\frac{1}{80}+\frac{5}{2011}=\frac{1}{64}+\frac{5}{2011}\)
Ta có \(A=\frac{1}{18}+\frac{20}{2011}\) và \(B=\frac{1}{64}+\frac{5}{2011}\)
So sánh từng số hạng: \(\frac{1}{18}>\frac{1}{64};\frac{20}{2011}>\frac{5}{2011}\)
\(\Rightarrow A>B\)
1)\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{b}+1=\frac{c}{d}+1\Leftrightarrow\frac{a+b}{b}=\frac{c+d}{d}\)
\(\frac{a}{b}=\frac{c}{d}\Rightarrow ad=bc\Rightarrow ac-ad=ac-bc\Leftrightarrow a\left(c-d\right)=c\left(a-b\right)\Rightarrow\frac{a}{a-b}=\frac{c}{c-d}\)
2) Gọi độ dài các cạnh của tam giác đó là a,b,c thì a : b : c = 3 : 4 : 5 ; a + b + c = 36
\(\Rightarrow\frac{a}{3}=\frac{b}{4}=\frac{c}{5}=\frac{a+b+c}{3+4+5}=\frac{36}{12}=3\Rightarrow\hept{\begin{cases}a=3.3=9\\b=3.4=12\\c=3.5=15\end{cases}}\).Vậy tam giác đó có 3 cạnh là 9 cm ; 12 cm ; 15 cm
3)\(\hept{\begin{cases}a:b:c:d=3:4:5:6\\a+b+c+d=3,6\end{cases}\Rightarrow\frac{a}{3}=\frac{b}{4}=\frac{c}{5}=\frac{d}{6}=\frac{a+b+c+d}{3+4+5+6}=\frac{3,6}{18}=0,2}\)
=> a = 0,2.3 = 0,6 ; b = 0,2.4 = 0,8 ; c = 0,2.5 = 1 ; d = 0,2.6 = 1,2
4)\(\frac{x}{3}=\frac{y}{2}\Rightarrow\frac{x}{3}:5=\frac{y}{2}:5\Leftrightarrow\frac{x}{15}=\frac{y}{10}\)
\(\frac{y}{5}=\frac{z}{7}\Rightarrow\frac{y}{5}:2=\frac{z}{7}:2\Leftrightarrow\frac{y}{10}=\frac{z}{14}\)
\(\Rightarrow\frac{x}{15}=\frac{y}{10}=\frac{z}{14}=\frac{x+y+z}{15+10+14}=\frac{184}{39}=4\frac{28}{39}\Rightarrow\hept{\begin{cases}x=4\frac{28}{39}.15=70\frac{10}{13}\\y=4\frac{28}{39}.10=47\frac{7}{39}\\z=4\frac{28}{39}.14=66\frac{2}{39}\end{cases}}\)