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Ta có: \(M=\dfrac{\dfrac{1}{99}+\dfrac{2}{98}+\dfrac{3}{97}+\dfrac{4}{96}+...+\dfrac{97}{3}+\dfrac{98}{2}+\dfrac{99}{1}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{100}}\)
\(=\dfrac{\left(1+\dfrac{1}{99}\right)+\left(1+\dfrac{2}{98}\right)+\left(1+\dfrac{3}{97}\right)+\left(1+\dfrac{4}{96}\right)+...+\left(1+\dfrac{98}{2}\right)+1}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{100}}\)
\(=\dfrac{\dfrac{100}{99}+\dfrac{100}{98}+\dfrac{100}{97}+...+\dfrac{100}{1}+\dfrac{100}{2}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{100}}\)
=100
Ta có: \(N=\dfrac{92-\dfrac{1}{9}-\dfrac{2}{10}-\dfrac{3}{11}-...-\dfrac{90}{98}-\dfrac{91}{99}-\dfrac{92}{100}}{\dfrac{1}{45}+\dfrac{1}{50}+\dfrac{1}{55}+...+\dfrac{1}{495}+\dfrac{1}{500}}\)
\(=\dfrac{\left(1-\dfrac{1}{9}\right)+\left(1-\dfrac{2}{10}\right)+\left(1-\dfrac{3}{11}\right)+...+\left(1-\dfrac{90}{98}\right)+\left(1-\dfrac{91}{99}\right)+\left(1-\dfrac{92}{100}\right)}{\dfrac{1}{5}\left(\dfrac{1}{9}+\dfrac{1}{10}+\dfrac{1}{11}+...+\dfrac{1}{99}+\dfrac{1}{100}\right)}\)
\(=\dfrac{\dfrac{8}{9}+\dfrac{8}{10}+\dfrac{8}{11}+...+\dfrac{8}{99}+\dfrac{8}{100}}{\dfrac{1}{5}\left(\dfrac{1}{9}+\dfrac{1}{10}+\dfrac{1}{11}+...+\dfrac{1}{99}+\dfrac{1}{100}\right)}\)
\(=\dfrac{8}{\dfrac{1}{5}}=40\)
\(\Leftrightarrow\dfrac{M}{N}=\dfrac{100}{40}=\dfrac{5}{2}\)
Do \(a,b,c\ne0\)
\(\dfrac{ab}{a+b}=\dfrac{bc}{b+c}=\dfrac{ac}{a+c}\Rightarrow\dfrac{a+b}{ab}=\dfrac{b+c}{bc}=\dfrac{a+c}{ac}\)
\(\Rightarrow\dfrac{a}{ab}+\dfrac{b}{ab}=\dfrac{b}{bc}+\dfrac{c}{bc}=\dfrac{a}{ac}+\dfrac{c}{ac}\)
\(\Rightarrow\dfrac{1}{a}+\dfrac{1}{b}=\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{1}{a}+\dfrac{1}{c}\) \(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{a}+\dfrac{1}{b}=\dfrac{1}{b}+\dfrac{1}{c}\\\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{1}{a}+\dfrac{1}{c}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{a}=\dfrac{1}{c}\\\dfrac{1}{b}=\dfrac{1}{a}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=c\\b=a\end{matrix}\right.\) \(\Rightarrow a=b=c\)
\(\Rightarrow M=\dfrac{a.a+a.a+a.a}{a^2+a^2+a^2}=\dfrac{3a^2}{3a^2}=1\)
a) \(\dfrac{x}{y}=\dfrac{1}{3}\Rightarrow y=3x\). Thay vào biểu thức N, ta có: \(N=\dfrac{x-3x}{x+9x}=\dfrac{-2x}{10x}=-\dfrac{1}{5}\)
b) \(x+y+1=0\Leftrightarrow x+y=-1\). Thay vào biểu thức M, ta có: \(M=\left(-1\right)^2-y^3\left(-1\right)+x^2-y^3+3\) \(=1+y^3+x^2-y^3+3\) \(=x^2+4\)
\(\text{Ta có : }\dfrac{x}{y+z}=\dfrac{y}{x+z}=\dfrac{z}{y+x}\\ \Rightarrow\dfrac{y+z}{x}=\dfrac{x+z}{y}=\dfrac{y+x}{z}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta được :
\(\dfrac{y+z}{x}=\dfrac{x+z}{y}=\dfrac{y+x}{z}\\ =\dfrac{\left(y+z\right)+\left(x+z\right)+\left(y+x\right)}{x+y+z}\\ =\dfrac{y+z+x+z+y+x}{x+y+z}\\ =\dfrac{\left(y+y\right)+\left(z+z\right)+\left(x+x\right)}{x+y+z}\\ =\dfrac{2y+2z+2x}{x+y+z}\\ =\dfrac{2\left(x+y+z\right)}{x+y+z}\\ =2\\ \)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{y+z}{x}=2\\\dfrac{x+z}{y}=2\\\dfrac{y+x}{z}=2\end{matrix}\right.\Rightarrow\dfrac{y+z}{x}+\dfrac{x+z}{y}+\dfrac{y+x}{z}=2+2+2=6\)
Vậy \(\dfrac{y+z}{x}+\dfrac{x+z}{y}+\dfrac{y+x}{z}=6\)
Câu 7:
x=2014 nên x-1=2013
\(A=x^{2014}-x^{2013}\left(x-1\right)-x^{2012}\left(x-1\right)-...-x\left(x-1\right)+1\)
\(=x^{2014}-x^{2014}+x^{2013}-x^{2013}+x^{2012}-...-x^2+x+1\)
=x+1
=2014+1=2015
Đặt \(\dfrac{x}{-4}=\dfrac{y}{-7}=\dfrac{z}{3}=k\)
\(\Rightarrow x=-4k;y=-7k;z=3k\) (1)
Thay (1) vào A , ta được
\(A=\dfrac{-2.\left(-4k\right)+\left(-7k\right)+5.3k}{2\left(-4k\right)-3\left(-7k\right)-6.3k}\)
\(\Rightarrow A=\dfrac{8k+\left(-7k\right)+15k}{-8k+21k+\left(-18k\right)}\)
\(\Rightarrow A=\dfrac{k[8+\left(-7\right)+15]}{k[-8+21+\left(-18\right)]}\)
\(\Rightarrow A=\dfrac{16k}{-5k}\)
\(\Rightarrow A=\dfrac{16}{5}\)
Vậy \(A=\dfrac{16}{5}\)
1/ Ta có: \(P=\frac{2}{6-m}\)\(\le2\left(\forall m\in Z\right)\)
Dấu "=" xảy ra \(\Leftrightarrow6-m=1\Rightarrow m=5\).
Vậy Max P =2 khi m = 5.
2/ Ta có: \(Q=\frac{8-n}{n-3}\ge0\left(\forall n\in Z\right)\)
Dấu "=" xảy ra \(\Leftrightarrow8-n=0\Rightarrow n=8.\)
Vậy Min Q = 0 khi n = 8.
Chúc bn hc tốt!^_^.
Nhớ kb và cho tớ nhé mọi người!
1/ta có :2/6-m max
suy ra:6-m>0,6-m min
suy ra:6-m=1
suy ra: m=5
Vậy ...
Từ m-n=3=>m=n+3
Ta có: \(\frac{m-8}{n-3}=\frac{\left(n+3\right)-8}{n-3}=\frac{n-5}{n-5}=1\) (1)
\(\frac{4m-n}{3m+3}=\frac{4.\left(n+3\right)-n}{3.\left(n+3\right)+3}=\frac{4n+12-n}{3n+9+3}=\frac{\left(4n-n\right)+12}{3n+12}=\frac{3n+12}{3n+12}=1\) (2)
Từ (1) và (2) \(\Rightarrow A=1-1=0\)
Vậy A=0
0
có m=3 +n -> thay m thành 3+n -> làm như bình thường ->ra