a, \(\frac{a^3\times b}{4}\)+ \(\frac{5c^2}{2}\)+ d với a = 2 , b = a , c = 6 , d = 9
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B
2 tháng 11 2017
b) Đặt \(\hept{\begin{cases}\frac{a}{b}=k\Rightarrow a=kb\\\frac{c}{d}=k\Rightarrow c=kd\end{cases}}\)
VT : \(\frac{5a+3b}{5a-3b}\Rightarrow\frac{5kb+3b}{5ka-3b}=\frac{b\left(5k+3\right)}{b\left(5k-3\right)}=\frac{5k+3}{5k-3}\) (1)
VP : \(\frac{5c+3d}{5c-3d}=\frac{5kd+3d}{5kd-3d}=\frac{d\left(5k+3\right)}{d\left(5k-3\right)}=\frac{5k+3}{5k-3}\) (2)
Từ (1) và (2) => đpcm
LM
27 tháng 8 2016
a/ Ta có: \(b^2=ac\Rightarrow\frac{a}{b}=\frac{b}{c};c^2=bd\Rightarrow\frac{b}{c}=\frac{c}{d}\)\(\Rightarrow\frac{a}{b}=\frac{b}{c}=\frac{c}{d}\)
Đặt \(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=k\Rightarrow\left(\frac{a}{b}\right)^3=\left(\frac{b}{c}\right)^3=\left(\frac{c}{d}\right)^3=k^3\Rightarrow\frac{a^3}{b^3}=\frac{b^3}{c^3}=\frac{c^3}{d^3}=k^3\)
Áp dụng tính chất của tỉ lệ thức ta có:\(\frac{a^3}{b^3}=\frac{b^3}{c^3}=\frac{c^3}{d^3}=\frac{a^3+b^3+c^3}{b^3+c^3+d^3}=k^3\)
Mặt khác: \(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=k\Rightarrow\frac{a+b+c}{b+c+d}=k\Rightarrow\left(\frac{a+b+c}{b+c+d}\right)^3=k^3\)
\(\Rightarrow\frac{a^3+b^3+c^3}{b^3+c^3+d^3}=\left(\frac{a+b+c}{b+c+d}\right)^3\left(=k^3\right)\)
DL
15 tháng 2 2020
Có
\(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=\frac{d}{a}\\ \Rightarrow\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=\frac{d}{a}=\frac{a+b+c+d}{b+c+d+a}=1\\ \Rightarrow a=b=c=d\)
Vậy
\(M=\frac{2a-b}{c+d}+\frac{2b-c}{d+a}+\frac{2c-d}{a+b}+\frac{2d-a}{b+c}\\ =\frac{2a-a}{a+a}+\frac{2a-a}{a+a}+\frac{2a-a}{a+a}+\frac{2a-a}{a+a}\\ =\frac{a}{2a}+\frac{a}{2a}+\frac{a}{2a}+\frac{a}{2a}\\ =\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}\\ =\frac{1+1+1+1}{2}\\ =\frac{4}{2}=2\)
Vậy M=2
DP
18 tháng 6 2017
a, \(A=\frac{2}{5}+\frac{-1}{6}-\frac{3}{4}-\frac{-2}{3}\)
\(A=\left(\frac{2}{5}-\frac{3}{4}\right)+\left(\frac{-1}{6}-\frac{-2}{3}\right)\)
\(A=\left(\frac{8}{20}-\frac{15}{20}\right)+\left(\frac{-3}{18}-\frac{-12}{18}\right)\)
\(A=\frac{-7}{20}+\frac{1}{2}\)
\(\Rightarrow A=\frac{-7}{20}+\frac{10}{20}=\frac{3}{20}\)
DP
18 tháng 6 2017
b, \(B=\frac{7}{10}-\frac{-3}{4}+\frac{-5}{6}-\frac{1}{5}+\frac{-2}{3}\)
\(B=\left(\frac{7}{10}-\frac{1}{5}\right)+\left(\frac{-5}{6}+\frac{-2}{3}\right)-\frac{-3}{4}\)
\(B=\left(\frac{7}{10}-\frac{2}{10}\right)+\left(\frac{-5}{6}+\frac{-4}{6}\right)-\frac{-3}{4}\)
\(B=\frac{1}{2}+\frac{-3}{2}-\frac{-3}{4}\)
\(B=\frac{2}{4}+\frac{-6}{4}-\frac{-3}{4}\)
\(\Rightarrow B=\frac{2+-6+3}{4}=\frac{-1}{4}\)
c, \(C=\frac{\left(\frac{1}{2}-0,75\right)\times\left(0,2-\frac{2}{5}\right)}{\frac{5}{9}-1\frac{1}{12}}\)
\(C=\frac{\left(\frac{1}{2}-\frac{3}{4}\right)\times\left(\frac{1}{5}-\frac{2}{5}\right)}{\frac{5}{9}-\frac{1\times12+1}{12}}\)
\(C=\frac{\left(\frac{2}{4}-\frac{3}{4}\right)\times\left(\frac{-1}{5}\right)}{\frac{5}{9}-\frac{13}{12}}\)
\(C=\frac{\left(\frac{-1}{4}\right)\times\left(\frac{-1}{5}\right)}{\frac{60}{108}-\frac{117}{108}}\)
\(C=\frac{\frac{1}{20}}{\frac{-19}{36}}=\frac{1}{20}\div\frac{-19}{36}=\frac{1}{20}\times\frac{36}{-19}\)
\(\Rightarrow C=\frac{36}{-380}=\frac{-9}{95}\)
d, \(D=\frac{\frac{2}{3}+\frac{2}{7}-\frac{1}{4}}{-1-\frac{3}{7}+\frac{3}{28}}\)
\(D=\frac{\frac{56}{84}+\frac{24}{84}-\frac{21}{84}}{\frac{-10}{7}+\frac{3}{28}}\)
\(D=\frac{\frac{59}{84}}{\frac{-40}{28}+\frac{2}{28}}=\frac{59}{84}\div\frac{-37}{28}=\frac{59}{84}\times\frac{28}{-37}\)
\(\Rightarrow D=\frac{1652}{-3108}=\frac{-59}{111}\)
ND
10 tháng 8 2019
1.
C/m bổ đề: \(a^3-b^3\ge\frac{1}{4}\left(a^3-b^3\right)\) với \(\forall a,b\in R,a\ge b\)
\(\Leftrightarrow4a^3-4b^3-\left(a^3-3a^2b+3ab^2-b^3\right)\ge0\)
\(\Leftrightarrow3a^3+3a^2b-3ab^2-3b^3\ge0\)
\(\Leftrightarrow3\left(a^2-b^2\right)\left(a+b\right)\ge0\)
\(\Leftrightarrow3\left(a+b\right)^2\left(a-b\right)\ge0\)(đúng)
Theo bài ra: \(a^3-b^3\ge3a-3b-4\)
\(\Leftrightarrow\) Cần c/m: \(\left(a-b\right)^3\ge12a-12b-16\)(1)
Thật vậy:
\(\left(1\right)\)\(\Leftrightarrow\left(a-b\right)^3-12\left(a-b\right)+16\ge0\)
\(\Leftrightarrow\left[\left(a-b\right)^3-8\right]-12\left(a-b-2\right)\ge0\)
\(\Leftrightarrow\left(a-b-2\right)\left[\left(a-b\right)^2+2\left(a-b\right)+4\right]-12\left(a-b-2\right)\ge0\)
\(\Leftrightarrow\left(a-b-2\right)\left[\left(a-b\right)^2+2\left(a+b\right)-8\right]\ge0\)
\(\Leftrightarrow\left(a-b-2\right)^2\left(a-b+4\right)\ge0\) (đúng với mọi a,b thỏa mãn \(a,b\in R,a\ge b\))
ND
10 tháng 8 2019
2.
\(BĐT\Leftrightarrow\frac{1}{\frac{a+b}{ab}}+\frac{1}{\frac{c+d}{cd}}\le\frac{1}{\frac{a+b+c+d}{\left(a+c\right)\left(b+d\right)}}\)
\(\Leftrightarrow\frac{ab}{a+b}+\frac{cd}{c+d}\le\frac{\left(a+c\right)\left(b+d\right)}{a+b+c+d}\)
\(\Leftrightarrow\frac{ab\left(c+d\right)+cd\left(a+b\right)}{\left(a+b\right)\left(c+d\right)}\le\)\(\frac{ab+ad+bc+cd}{a+b+c+d}\)
\(\Leftrightarrow\frac{abc+abd+acd+bcd}{ac+ad+bc+bd}\le\frac{ab+ad+bc+cd}{a+b+c+d}\)
\(\Leftrightarrow\left(ad+ab+bc+cd\right)\left(ac+ad+bc+bd\right)\ge\)\(\left(a+b+c+d\right)\left(abc+abd+acd+bcd\right)\)
\(\Leftrightarrow\left(ad\right)^2-2abcd+\left(bc\right)^2\ge0\)
\(\Leftrightarrow\left(ad-bc\right)^2\ge0\) (đúng với mọi a,b,c,d>0)