\(\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)=1\) chứng minh \(\dfrac{1}{a^{2005}}+\dfrac{1}{b^{2005}}+\dfrac{1}{c^{2005}}=\dfrac{1}{a^{2005}+b^{2005}+c^{2005}}\)
Cho a, b, c là các số thực khác 0 và \(\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)=1\). Tính giá trị của biểu thức \(P=\left(a^{2004}-b^{2004}\right)\left(b^{2005}+c^{2005}\right)\left(c^{2006}-a^{2006}\right)\).
Cho a,b,c là các số thực khác 0 và \(\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)=1\)
Tính giá trị của biểu thức: \(P=\left(a^{2004}-b^{2004}\right)\left(b^{2005}+c^{2005}\right)\left(c^{2006}-a^{2006}\right)\)
\(\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)=1\)
\(\Leftrightarrow\left(a+b+c\right)\left(ab+ac+bc\right)=abc\)
\(\Leftrightarrow a\left(ab+ac+bc\right)+\left(b+c\right)\left(ab+ac+bc\right)-abc=0\)
\(\Leftrightarrow a\left(ab+ac+bc-bc\right)+\left(b+c\right)\left(ab+ac+bc\right)=0\)
\(\Leftrightarrow a^2\left(b+c\right)+\left(b+c\right)\left(ab+ac+bc\right)=0\)
\(\Leftrightarrow\left(a^2+ab+ac+bc\right)\left(b+c\right)=0\)
\(\Leftrightarrow\left(a+b\right)\left(a+c\right)\left(b+c\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}a=-c\\a=-b\\b=-c\end{matrix}\right.\)
- Nếu \(a=-c\Rightarrow a^{2006}=c^{2006}\Rightarrow c^{2006}-a^{2006}=0\Rightarrow P=0\)
- Nếu \(a=-b\Rightarrow a^{2004}=b^{2004}\Rightarrow a^{2004}-b^{2004}=0\Rightarrow P=0\)
- Nếu \(b=-c\Rightarrow b^{2005}=-c^{2005}\Rightarrow b^{2005}+c^{2005}=0\Rightarrow P=0\)
Vậy \(P=0\)
1. Cho \(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}\). Chứng minh rằng \(\left(\dfrac{a+b+c}{b+c+d}\right)^3=\dfrac{a}{d}\)
2. Cho \(\dfrac{a}{2003}=\dfrac{b}{2004}=\dfrac{c}{2005}\). Chứng minh rằng \(4\left(a-b\right)\left(b-c\right)=\left(c-a\right)^2\)
Bài 1:
Áp dụng t.c của dãy tỉ số bằng nhau, ta có:
\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{a+b+c}{b+c+d}\\ =\left(\dfrac{a+b+c}{b+c+d}\right)^3=\dfrac{a^3}{b^3}=\dfrac{a.b.c}{b.c.d}=\dfrac{a}{d}\left(dpcm\right)\)
Cho a; b; c là các số thực khác 0 và \(\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)=1\).
Tính giá trị của biểu thức P = (a2004 - b2004)(b2005 + c2005)(c2006-a2006)
Từ giả thiết suy ra: \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{1}{a+b+c}\)
\(\Rightarrow\left(\dfrac{1}{a}+\dfrac{1}{b}\right)+\left(\dfrac{1}{c}-\dfrac{1}{a+b+c}\right)=0\)
\(\Rightarrow\dfrac{a+b}{ab}+\dfrac{a+b}{c\left(a+b+c\right)}=0\)
\(\Rightarrow\) (a + b)[c(a + b + c) + ab] = 0
\(\Rightarrow\) (a + b)(ac + ab + bc + c2) = 0
\(\Rightarrow\) (a + b)(b + c)(a + c) = 0
P = (a2004 - b2004)(b2005 + c2005)(c2006 - a2006)
= (a + b)(b + c)(a + c) = 0
So Sánh:
A=\(\dfrac{10^{11}-1}{10^{12}-1}\) và B=\(\dfrac{10^{10}+1}{10^{11}+1}\)
C=\(\dfrac{2005^{2005}+1}{2005^{2006}+1}\) và D=\(\dfrac{2005^{2004}+1}{2005^{2005}+1}\)
Cho \(\dfrac{a}{b} = \dfrac{c}{d}\) . Chứng minh :
a, \(\dfrac{a^{2005}}{b^{2005}} = \dfrac{(a-c)^{2005}}{(b-d)^{2005}}\)
b, \(\dfrac{(a^2+b^2)^3}{(c^2+d^2)^3}\) =\(\dfrac{a^3+b^3)^2}{(c^3+d^3)^2}\)
c, \((\dfrac{a-b}{c-d})^{2005}\) = \(\dfrac{2.a^{2005}-b^{2005}}{2.c^{2005}-d^{2005}}\)
d, \(\dfrac{(a^2-b^2)^5}{(c^2-d^2)^5} = \) \(\dfrac{a^{10}+b^{10}}{c^{10}+d^{10}}\)
e, \(\dfrac{2.a^{2005}+5.b^{2005}}{2.c^{2005}+5.d^{2005}}\) = \(\dfrac{(a+b)^{2005}}{(c+d)^{2005}}\)
f, \(\dfrac{(a^{2004}+b^{2004})^{2005}}{(c^{2004}+d^{2004})^{2005}}\) = \(\dfrac{(a^{2005} -b^{2005})^{2004}}{(c^{2005}-d^{2005})^{2004}}\)
cho hỏi chút
\(\frac{a}{b}=\frac{c}{d}\)
trong đó
\(a=c\) hay \(a\ne c\)
\(b=d\) hay \(b\ne d\)
( bài có thiếu điều kiện ko vậy )
a) Ta có: \(\dfrac{2x+1}{6}-\dfrac{x-2}{4}=\dfrac{3-2x}{3}-x\)
\(\Leftrightarrow\dfrac{2\left(2x+1\right)}{12}-\dfrac{3\left(x-2\right)}{12}=\dfrac{4\left(3-2x\right)}{12}-\dfrac{12x}{12}\)
\(\Leftrightarrow4x+2-3x+6=12-8x-12x\)
\(\Leftrightarrow x+8-12+20x=0\)
\(\Leftrightarrow21x-4=0\)
\(\Leftrightarrow21x=4\)
\(\Leftrightarrow x=\dfrac{4}{21}\)
Vậy: \(S=\left\{\dfrac{4}{21}\right\}\)
Hình như em viết công thức bị lỗi rồi. Em cần chỉnh sửa lại để được hỗ trợ tốt hơn!
a)
PT \(\Leftrightarrow \frac{4x+2}{12}-\frac{3x-6}{12}=\frac{12-8x}{12}-\frac{12x}{12}\)
\(\Leftrightarrow 4x+2-3x+6=12-8x-12x\)
\(\Leftrightarrow 21x=4\Leftrightarrow x=\frac{4}{21}\)
b)
PT \(\Leftrightarrow \frac{30x+15}{20}-\frac{100}{20}-\frac{6x+4}{20}=\frac{24x-12}{20}\)
\(\Leftrightarrow 30x+15-100-6x-4=24x-12\Leftrightarrow -89=-12\) (vô lý)
Vậy pt vô nghiệm.
So sánh
a, \(\dfrac{2005^{2014}+1}{2005^{2015}+1}\&\dfrac{2005^{2016}+1}{2005^{2017}+1}\)
b, \(\dfrac{19}{10}\&\dfrac{49}{40}\)
c, \(\dfrac{13}{20}\&\dfrac{33}{40}\)
ồ, lâu h ms gặp
a,
Dễ thấy \(\dfrac{2005^{2016}+1}{2005^{2017}+1}< 1\)
Áp dụng khi \(\dfrac{a}{b}< 1\Rightarrow\dfrac{a}{b}< \dfrac{a+n}{b+n}\left(n\in N^{\circledast}\right)\)
Ta có:
\(\dfrac{2005^{2016}+1}{2005^{2017}+1}< \dfrac{2005^{2016}+1+\left(2005^2-1\right)}{2005^{2017}+1+\left(2005^2-1\right)}=\dfrac{2005^{2016}+2005^2}{2005^{2017}+2005^2}=\dfrac{2005^2\left(2005^{2014}+1\right)}{2005^2\left(2005^{2015}+1\right)}=\dfrac{2005^{2014}+1}{2005^{2015}+1}\)
Vậy \(\dfrac{2005^{2016}+1}{2005^{2017}+1}< \dfrac{2005^{2014}+1}{2005^{2015}+1}\)
b,
\(\dfrac{19}{10}=\dfrac{10+9}{10}=\dfrac{10}{10}+\dfrac{9}{10}=1+\dfrac{9}{10}\\ \dfrac{49}{40}=\dfrac{40+9}{40}=\dfrac{40}{40}+\dfrac{9}{40}=1+\dfrac{9}{40}\)
Vì \(10< 40\Rightarrow\dfrac{9}{10}>\dfrac{9}{40}\Rightarrow1+\dfrac{9}{10}>1+\dfrac{9}{40}\Leftrightarrow\dfrac{19}{10}>\dfrac{49}{40}\)Vậy \(\dfrac{19}{10}>\dfrac{49}{40}\)
c,
\(\dfrac{13}{20}=\dfrac{20-7}{20}=\dfrac{20}{20}-\dfrac{7}{20}=1-\dfrac{7}{20}\\ \dfrac{33}{40}=\dfrac{40-7}{40}=\dfrac{40}{40}-\dfrac{7}{40}=1-\dfrac{7}{40}\)
Vì \(20< 40\Rightarrow\dfrac{7}{20}>\dfrac{7}{40}\Rightarrow1-\dfrac{7}{20}< 1-\dfrac{7}{40}\Leftrightarrow\dfrac{13}{20}< \dfrac{33}{40}\)
Vậy \(\dfrac{13}{20}< \dfrac{33}{40}\)
Áp dụng tính chất:
\(\dfrac{a}{b}< 1\Rightarrow\dfrac{a+m}{b+m}< 1\left(m\in N\right)\)
\(\)Đặt: \(B=\dfrac{2005^{2016}+1}{2005^{2017}+1}< 1\)
\(\Rightarrow B< \dfrac{2005^{2016}+1+4020024}{2005^{2017}+1+4020024}\)
\(B< \dfrac{2005^{2016}+4020025}{2005^{2017}+4020025}\)
\(B< \dfrac{2005^2\left(2005^{2014}+1\right)}{2005^2\left(2005^{2015}+1\right)}\)
\(B< \dfrac{2005^{2014}+1}{2005^{2015}+1}=A\)
\(B< A\)
A=\(\dfrac{1}{4.9}+\dfrac{1}{9.14}+\dfrac{1}{14.19}+...+\dfrac{1}{44.49}+\left(\dfrac{1-3-5-7-...-49}{89}\right)\)
B=\(\dfrac{212.3^5.4^6.9^2}{\left(2^2.3\right)^6+8^4.3^5}-\dfrac{5^{10}.7^3-25^4.49^2}{\left(125.71^3+59.14^3\right)}\)
C=\(\dfrac{\dfrac{1}{2003}+\dfrac{1}{2004}-\dfrac{1}{2005}}{\dfrac{5}{2003}+\dfrac{5}{2004}-\dfrac{5}{2005}}-\dfrac{\dfrac{2}{2002}+\dfrac{2}{2003}-\dfrac{2}{2004}}{\dfrac{3}{2002}+\dfrac{3}{2003}-\dfrac{3}{2004}}\)
D=\(\left(\dfrac{1,5+1-0,75}{2,5+\dfrac{5}{3}-1,25}\right)+\left(\dfrac{0,375-0,3+\dfrac{3}{11}+\dfrac{3}{12}}{-0,625+0,5-\dfrac{5}{11}-\dfrac{5}{12}}\right):\dfrac{1890}{2005}+115\)
E=13+23+...+103=3025
Tính F=23+42+63+...+203=?