Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
bài này dễ vào TH 0,5 điểm trong bài thi
nghe có vẻ khó nhưng chú ý 1 chút là có thể làm được
\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\Rightarrow\frac{a^{2016}}{c^{2016}}=\frac{b^{2016}}{d^{2016}}\)\(\Rightarrow\left(\frac{a^{2016}}{c^{2016}}\right)^{2017}=\left(\frac{b^{2016}}{d^{2016}}\right)^{2017}\)
áp dụng t/c dãy t/s = nhau
\(\Rightarrow\left(\frac{a^{2016}}{c^{2016}}\right)^{2017}=\left(\frac{b^{2016}}{d^{2016}}\right)^{2017}=\)\(\frac{\left(a^{2016}+b^{2016}\right)^{2017}}{\left(c^{2016}+d^{2016}\right)^{2017}}\)
biến đổi tiếp cái kia tương tự rồi suy ra chúng = nhau nhé
casi phần áp dụng tc thì phải bằng (a^2016)^2017+(b^2016)^2017 chớ nhỉ bạn hỏi đáp
Đặt \(\frac{a}{2016}=\frac{b}{2017}=\frac{c}{2018}=t\)
\(\Rightarrow a=2016t,b=2017t,c=2018t\)
Ta có: \(4\left(a-b\right)\left(b-c\right)=4\left(2016t-2017t\right)\left(2017t-2018t\right)=4.\left(-t\right).\left(-t\right)=4t^2\)
\(\left(c-a\right)^2=\left(2018t-2016t\right)^2=\left(2t\right)^2=4t^2\)
Vậy \(4\left(a-b\right)\left(b-c\right)=\left(c-a\right)^2\)
Áp dụng tính chất của dãy tỉ số bằng nhau ta có:
a/2016 = b/2017 = c/2018 = (a-b) / (2016-2017) = (b-c) / (2017-2018) = (c-a) / (2018-1026)
= (a-b) / (-1) = (b-c) / ( -1) = (c-a) / 2
Vì (a-b) / (-1) = (b-c) / ( -1) = (c-a) / 2 nên (a-b) / (-1) . (b-c) / (-1) =[ (c-a) / 2 ]2
=> (a-b)(b-c) / (-1).(-1) = (c-a)2 / 22
=> (a-b)(b-c).1= (c-a)2 / 4
=> (a-b)(b-c) =(c-a)2 / 4
=> 4(a-b)(b-c)= (c-a)2
TH1: Nếu a+b+c \(\ne0\)
áp dụng tính chất của dãy tỉ số bằng nhau ta có:
\(\frac{a+b-c}{c}=\frac{b+c-a}{a}=\frac{c+a-b}{b}=\frac{a+b-c+b+c-a+c+a-b}{a+b+c}=1\)
mà \(\frac{a+b-c}{c}+1=\frac{b+c-a}{a}+1=\frac{c+a-b}{b}+1=2\)
\(\Rightarrow\frac{a+b}{c}=\frac{b+c}{a}=\frac{c+a}{b}=2\)
Vậy \(B=\left(1+\frac{b}{a}\right)\left(1+\frac{a}{c}\right)\left(1+\frac{c}{b}\right)=\left(\frac{a+b}{a}\right)\left(\frac{a+c}{c}\right)\left(\frac{b+c}{b}\right)=8\)
TH2 : Nếu a+b+c = 0
áp dụng tính chất của dãy tỉ số bằng nhau ta có :
\(\frac{a+b-c}{c}=\frac{b+c-a}{a}=\frac{c+a-b}{b}=\frac{a+b-c+b+c-a+c+a-b}{a+b+c}=0\)
mà \(\frac{a+b-c}{c}+1=\frac{b+c-a}{a}+1=\frac{c+a-b}{b}+1=1\)
\(\Rightarrow\frac{a+b}{c}=\frac{b+c}{a}=\frac{c+a}{b}=1\)
vậy \(B=\left(1+\frac{b}{a}\right)\left(1+\frac{a}{c}\right)\left(1+\frac{c}{b}\right)=\left(\frac{a+b}{a}\right)\left(\frac{a+c}{c}\right)\left(\frac{b+c}{b}\right)=1\)
\(\frac{a+b-c}{c}+2=\frac{b+c-a}{a}+2=\frac{c+a-b}{b}+2\)
\(\Leftrightarrow\frac{a+b+c}{c}=\frac{a+b+c}{b}=\frac{a+b+c}{a}\)
TH1: a+b+c=0
\(\Rightarrow\hept{\begin{cases}a=-\left(b+c\right)\\b=-\left(a+c\right)\\c=-\left(a+b\right)\end{cases}}\Rightarrow B=\left(1-\frac{a+c}{a}\right).\left(1-\frac{b+c}{c}\right).\left(1-\frac{a+b}{b}\right)=-1\)
TH2: a+b+c khác 0
\(\Rightarrow a=b=c\Rightarrow B=\left(1+\frac{a}{a}\right).\left(1+\frac{a}{a}\right).\left(1+\frac{a}{a}\right)=2^3=8\)
A = \(\frac{\frac{3}{4}-\frac{3}{11}+\frac{3}{13}}{\frac{5}{4}-\frac{5}{11}+\frac{5}{13}}+\frac{\frac{1}{2}-\frac{1}{3}+\frac{1}{4}}{\frac{5}{4}-\frac{5}{6}+\frac{5}{8}}\)
\(=\frac{3.\left(\frac{1}{4}-\frac{1}{11}+\frac{1}{13}\right)}{5.\left(\frac{1}{4}-\frac{1}{11}+\frac{1}{13}\right)}+\frac{\frac{1}{2}-\frac{1}{3}+\frac{1}{4}}{\frac{5}{2}.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{4}\right)}\)
\(=\frac{3}{5}+\frac{1}{\frac{5}{2}}\)
\(=\frac{3}{5}+\frac{2}{5}=1\)
b) B = \(\frac{2^{12}.3^5-4^6.9^2}{\left(2^2.3\right)^6.8^4.3^5}-\frac{5^{10}.7^3:25^5.49}{\left(125.7\right)^3+5^9.14^3}\)
\(=\frac{2^{12}.3^5-\left(2^2\right)^6.\left(3^2\right)^2}{2^{12}.3^6+\left(2^3\right)^4.3^5}-\frac{5^{10}.7^3-\left(5^2\right)^5.7^2}{\left(5^3\right)^3.7^3+5^9.\left(7.2\right)^3}\)
\(=\frac{2^{12}.3^5-2^{12}.3^4}{2^{12}.3^6+2^{12}.3^5}-\frac{5^{10}.7^3-5^{10}-7^2}{5^9.7^3+5^9.7^3.2^3}\)
\(=\frac{2^{12}.3^4.\left(3-1\right)}{2^{12}.3^5\left(3+1\right)}-\frac{5^{10}.7^2.\left(7-1\right)}{5^9.7^3\left(1+2^3\right)}\)
\(=\frac{1}{3.2}-\frac{5.2}{7.3}\)
\(=\frac{7}{3.2.7}-\frac{5.2.2}{7.3.2}\)
\(=\frac{7}{42}-\frac{20}{42}\)
\(=-\frac{13}{42}\)
a) \(a^2+b^2+c^2=ab+bc+ac\)
\(\Leftrightarrow2\left(a^2+b^2+c^2\right)=2\left(ab+bc+ac\right)\)
\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2bc-2ac=0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(c-a\right)^2+\left(b-c\right)^2=0\)
Ta có : \(\hept{\begin{cases}\left(a-b\right)^2\ge0\\\left(c-a\right)^2\ge0\\\left(b-c\right)^2\ge0\end{cases}}\)
\(\Rightarrow\left(a-b\right)^2+\left(c-a\right)^2+\left(b-c\right)^2=0\)
\(\Leftrightarrow a=b=c\)
a. \(a^2+b^2+c^2=ab+bc+ca\)
\(\Leftrightarrow2a^2+2b^2+2c^2=2ab+2bc+2ca\)
\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2ab-2ca=0\)
\(\Leftrightarrow\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ca+a^2\right)=0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
\(\Leftrightarrow\hept{\begin{cases}a-b=0\\b-c=0\\c-a=0\end{cases}}\Leftrightarrow a=b=c\left(đpcm\right)\)
\(\frac{2a-b}{a+b}=\frac{2}{3}\)
\(\Leftrightarrow6a-3b=2a+2b\)
\(\Leftrightarrow6a-2a=2b+3b\)
\(\Leftrightarrow4a=5b\)
\(\frac{b-c+a}{2a-b}=\frac{2}{3}\)
\(\Leftrightarrow4a-2b=3b-3c+3a\)
\(\Leftrightarrow4a-3a=3b-3c+2b\)
\(\Leftrightarrow a=5b-3c\)
\(\Leftrightarrow a=4a-3c\)
\(\Leftrightarrow3a=3c\)
\(\Rightarrow a=c\)
\(\Rightarrow P=\frac{\left(4a+4a\right)^5}{\left(4a+4a\right)^2\left(a+3a\right)^3}=\frac{\left(8a\right)^5}{\left(8a\right)^2\left(4a\right)^3}=\frac{\left(8a\right)^3}{\left(4a\right)^3}=\frac{8^3}{4^3}=2^3=8\)
bạn giải thử để tham khảo đi
Nè Phong Đãng - Trang của Phong Đãng - Học toán với OnlineMath:
Đặt \(\frac{a}{2016}=\frac{b}{2017}=\frac{c}{2018}=k\left(k\in R\right)\)
\(\Rightarrow a=2016k,b=2017k,c=2018k\)
Thay vào biểu thức: \(4\left(a-b\right)\left(b-c\right)\), có:
\(=4\left(2016k-2017k\right)\left(2017k-2018k\right)\)
\(=4\cdot\left(-1\right)\cdot k\cdot\left(-1\right)\cdot k=4k^2\) (1)
Làm tương tự như vậy: \(\left(c-a\right)^2=\left(2018k-2016k\right)^2=\left(2k\right)^2=4k^2\)(2)
Từ (1)(2) suy ra: \(4\left(a-b\right)\left(b-c\right)=\left(c-a\right)^2\)