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b)\(\dfrac{a+b}{c}=\dfrac{b+c}{a}=\dfrac{c+a}{b}\)
Ta có:
\(\dfrac{a+b}{c}=\dfrac{b+c}{a}\) và \(\dfrac{b+c}{a}=\dfrac{c+a}{b}\)
\(\Rightarrow1+\dfrac{a+b}{c}=1+\dfrac{b+c}{a}\)và \(1+\dfrac{b+c}{a}=1 +\dfrac{c+a}{b}\)
\(\Rightarrow\dfrac{c}{c}+\dfrac{a+b}{c}=\dfrac{a}{a}+\dfrac{b+c}{a}\)và \(\dfrac{a}{a}+\dfrac{b+c}{a}=\dfrac{b}{b}+\dfrac{c+a}{b}\)
\(\Rightarrow\dfrac{a+b+c}{c}=\dfrac{a+b+c}{a}\)và \(\dfrac{a+b+c}{a}=\dfrac{a+b+c}{b}\)
\(\Rightarrow\dfrac{a+b+c}{c}-\dfrac{a+b+c}{a}=0\) \(\Rightarrow\left(a+b+c\right)\cdot\left(\dfrac{1}{c}-\dfrac{1}{a}\right)=0\)
và \(\dfrac{a+b+c}{a}-\dfrac{a+b+c}{b}=0\)
\(\Rightarrow\left(a+b+c\right)\cdot\left(\dfrac{1}{a}-\dfrac{1}{b}\right)=0\)
+) Vì a,b,c đôi một khác 0
\(\Rightarrow a+b+c=0\)
\(\rightarrow a+b=\left(-c\right)\)
\(\rightarrow a+c=\left(-b\right)\)
\(\rightarrow b+c=\left(-a\right)\)
+) Ta có:
\(M=\left(1+\dfrac{a}{b}\right)\cdot\left(1+\dfrac{b}{c}\right)\cdot\left(1+\dfrac{c}{a}\right)\)
\(=\left(\dfrac{a+b}{b}\right)\cdot\left(\dfrac{b+c}{a}\right)\cdot\left(\dfrac{c+a}{c}\right)\)
\(=\dfrac{-c}{b}\cdot\dfrac{-a}{c}\cdot\dfrac{-b}{a}\)
\(=\left(-1\right)\)
Bài 1:
Áp dụng t.c của dãy tỉ số bằng nhau, ta có:
\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{a+b+c}{b+c+d}\\ =\left(\dfrac{a+b+c}{b+c+d}\right)^3=\dfrac{a^3}{b^3}=\dfrac{a.b.c}{b.c.d}=\dfrac{a}{d}\left(dpcm\right)\)
Đặt \(\frac{a}{2020}=\frac{b}{2021}=\frac{c}{2022}=k\Rightarrow\hept{\begin{cases}a=2020k\\b=2021k\\c=2022k\end{cases}}\)
Khi đó M = 4(a - b)(b - c) - (c - a)2
= 4(2020k - 2021k)(2021k - 2022k) - (2022k - 2020k)2
= 4(-k)(-k) - (2k)2
= 4k2 - 4k2 = 0
Vậy M = 0
Đặt \(\frac{a}{2020}=\frac{b}{2021}=\frac{c}{2022}=k\)( \(k\ne0\))
\(\Rightarrow a=2020k\); \(b=2021k\); \(c=2022k\)
Thay a, b, c vào biểu thức M ta có:
\(M=4\left(a-b\right)\left(b-c\right)-\left(c-a\right)^2\)
\(=4\left(2020k-2021k\right)\left(2021k-2022k\right)-\left(2022k-2020k\right)^2\)
\(=4.\left(-k\right).\left(-k\right)-\left(2k\right)^2=4k^2-4k^2=0\)
Vậy \(M=0\)
Bài 1:
a) \(\left(\dfrac{1}{9}-1\right)\left(\dfrac{1}{10}-1\right)......\left(\dfrac{1}{2004}-1\right)\left(\dfrac{1}{2005}-1\right)\)
= \(\dfrac{-8}{9}.\dfrac{-9}{10}.......\dfrac{-2003}{2004}.\dfrac{-2004}{2005}\) = \(\dfrac{-8}{2005}\)
b) \(-2+\dfrac{1}{-2+\dfrac{1}{-2+\dfrac{1}{-2+3}}}\) = \(-2+\dfrac{1}{-2+\dfrac{1}{-2+\dfrac{1}{1}}}\)
= \(-2+\dfrac{1}{-2+\dfrac{1}{-1}}\) = \(-2+\dfrac{1}{-3}\) = \(\dfrac{-7}{3}\)
\(\text{Câu 1 : }\) Tính
\(\text{a) }\left(\dfrac{1}{9}-1\right)\left(\dfrac{1}{10}-1\right)...\left(\dfrac{1}{2004}-1\right)\left(\dfrac{1}{2005}-1\right)\\ =\left(1-\dfrac{9}{9}\right)\left(\dfrac{1}{10}-\dfrac{10}{10}\right)...\left(\dfrac{1}{2004}-1\right)\left(\dfrac{1}{2005}-\dfrac{2005}{2005}\right)\\ =\dfrac{-8}{9}\cdot\dfrac{-9}{10}\cdot...\cdot\dfrac{-2003}{2004}\cdot\dfrac{-2004}{2005}\\ =\dfrac{\left(-8\right)\cdot\left(-9\right)\cdot..\cdot\left(-2003\right)\cdot\left(-2004\right)}{9\cdot10\cdot...\cdot2004\cdot2005}\\ =-\dfrac{8\cdot9\cdot...\cdot2003\cdot2004}{9\cdot10\cdot...\cdot2004\cdot2005}\\ =-\dfrac{8}{2005}\)
\(-2+\dfrac{1}{-2+\dfrac{1}{-2+\dfrac{1}{-2+3}}}\\ =-2+\dfrac{1}{-2+\dfrac{1}{-2+\dfrac{1}{1}}}\\ =-2+\dfrac{1}{-2+\dfrac{1}{-1}}\\ =-2+\dfrac{1}{-3}\\ =-2+\dfrac{-1}{3}=-\dfrac{7}{3}\)
1)\(B=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot\dfrac{3}{4}\cdot...\cdot\dfrac{2017}{2018}\)
\(B=\dfrac{1}{2018}\)
2)a)\(x^2-2x-15=0\)
\(\Leftrightarrow x^2-2x+1-16=0\)
\(\Leftrightarrow\left(x-1\right)^2-16=0\)
\(\Leftrightarrow\left(x-5\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-3\end{matrix}\right.\)
3)\(\dfrac{a}{b}=\dfrac{d}{c}\)
\(\Rightarrow\dfrac{a^2}{b^2}=\dfrac{d^2}{c^2}=\dfrac{a}{b}\cdot\dfrac{d}{c}=\dfrac{ad}{bc}\)
Lại có:\(\dfrac{a^2}{b^2}=\dfrac{d^2}{c^2}=\dfrac{a^2+d^2}{b^2+c^2}\)
\(\Rightarrow\dfrac{a^2+d^2}{b^2+c^2}=\dfrac{ad}{bc}\)
4)Ta có:\(g\left(x\right)=-x^{101}+x^{100}-x^{99}+...+x^2-x+1\)
\(g\left(x\right)=-x^{101}+\left(x^{100}-x^{99}+...+x^2-x+1\right)\)
\(g\left(x\right)=-x^{101}+f\left(x\right)\)
\(\Rightarrow f\left(x\right)-g\left(x\right)=f\left(x\right)+x^{101}-f\left(x\right)=x^{101}\)
Tại x=0 thì f(x)-g(x)=0
Tại x=1 thì f(x)-g(x)=1