Tính nhanh :
a) \(\frac{3x4x7}{5x3x4}\) b) \(\frac{2x5x6x8}{6x2x8x9}\) c) \(\frac{4x5x6}{3x10x8}\)
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\(\dfrac{3\times4\times7}{3\times10\times8}=\)\(\dfrac{3\times4\times7}{3\times10\times2\times4}=\dfrac{3\times7}{10\times2}=\dfrac{21}{20}\)
\(\frac{x}{5}x\frac{5}{6}x9=\frac{4x5x6}{5x6}\)
\(\frac{x}{6}x9=4\)
\(\frac{x}{6}=\frac{4}{9}\)
\(x=4x6:9=\frac{8}{3}\)
học tốt ~~~
x/5 x 5/6 x 9 = 120/30 = 4
x/5 x 5/6 = 4 : 9 = 4/9
x/5 = 4/9 : 5/6 = 8/15
=> x/5 = 8/15 => x = 8/3 chắc chắn
Áp dụng tính chất của dãy tỉ số = nhau ta có:
\(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=\frac{d}{a}=\frac{a+b+c+d}{b+c+d+a}=1\)
=> a = b = c = d
=> \(D=\frac{2a-a}{2a-a}+\frac{2a-a}{2a-a}+\frac{2a-a}{2a-a}+\frac{2a-a}{2a-a}\)
D = 1 + 1 + 1 + 1 = 4
a) Sử dụng phương pháp dãy tỉ số bằng nhau
=> \(\frac{a+b-c}{c}\)= \(\frac{b+c-a}{a}\)=\(\frac{c+a-b}{b}\)=\(\frac{\left(a+b-c\right)+\left(b+c-a\right)+\left(c+a-b\right)}{a+b+c}\)=\(\frac{a+b+c}{a+b+c}\)=1
=>a+b=2c , b+c=2a , c+a=2b (*)
b)P=(1+\(\frac{b}{a}\))(1+\(\frac{c}{b}\))(1+\(\frac{a}{c}\))=1+ (\(\frac{b}{a}\)+\(\frac{c}{b}+\frac{a}{c}\)) + \(\frac{abc}{abc}\)+(\(\frac{c}{a}+\frac{a}{b}+\frac{b}{c}\)) (Tách ra )
=\(\frac{\left(b+c\right)bc+\left(c+a\right)ca+\left(a+b\right)ab}{abc}\)+ 2 = \(\frac{\left(a+b+c\right)\left(ab+bc+ca\right)}{abc}-\frac{3abc}{abc}\)+ 2
=\(\frac{\left(a+b\right)\left(b+c\right)\left(c+a\right)+abc}{abc}-1\)
Từ (*) =>P=\(\frac{8abc+abc}{abc}\)- 1 =8
Ta có : \(\frac{a}{b+c+d}=\frac{b}{c+d+a}=\frac{c}{d+a+b}=\frac{d}{a+b+c}\)
\(\Rightarrow\frac{a}{b+c+d}+1=\frac{b}{c+d+a}+1=\frac{c}{d+a+b}+1=\frac{d}{a+b+c}+1\)
\(\Rightarrow\frac{a+b+c+d}{b+c+d}=\frac{a+b+c+d}{c+d+a}=\frac{a+b+c+d}{d+a+b}=\frac{a+b+c+d}{a+b+c}\)
Nếu a + b + c + d = 0
=> a + b = - c - d
b + c = - a - d
c + d = - b - a
d + a = - b - c
Khi đó \(P=\frac{-\left(c+d\right)}{c+d}+\frac{-\left(a+d\right)}{d+a}+\frac{-\left(b+a\right)}{b+a}=\frac{-\left(b+c\right)}{b+c}\)
\(=-1+\left(-1\right)+\left(-1\right)+\left(-1\right)=-4\)
Nếu a + b + c + d \(\ne\)0
\(\Rightarrow\frac{1}{c+d}=\frac{1}{d+a}=\frac{1}{b+a}=\frac{1}{b+c}\)
\(\Rightarrow c+d=d+a=b+a=b+c\)
\(\Rightarrow a=b=c=d\)
Khi đó \(P=1+1+1+1=4\)
Vậy nếu a + b + c + d = 0 thì P = - 4
nếu a + b + c + d \(\ne\)0 thì P = 4
Ta có : \(\left(a+b+c\right)\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)=2015.5\)
\(\Leftrightarrow\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}+\frac{a+b}{a+b}+\frac{a+c}{c+a}+\frac{b+c}{b+c}=2015.5\)
\(\Leftrightarrow Q+3=2015.5\Rightarrow Q=2015.5-3=10072\)
Ta có :
\(A+3=\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}+3\)
\(=\left(\frac{a}{b+c}+1\right)+\left(\frac{b}{a+c}+1\right)+\left(\frac{c}{a+b}+1\right)\)
\(=\frac{a+b+c}{b+c}+\frac{a+b+c}{a+c}+\frac{a+b+c}{a+b}\)
\(=\left(a+b+c\right)\left(\frac{1}{b+c}+\frac{1}{a+c}+\frac{1}{a+b}\right)\)
\(=2017.\frac{1}{2017}=1\)
\(\Rightarrow A=1-3=-2\)
a)
\(\frac{3\times4\times7}{5\times3\times4}=\frac{7}{5}\)
b)
\(\frac{2\times5\times6\times8}{6\times2\times8\times9}=\frac{5}{9}\)
c)
\(\frac{4\times5\times6}{10\times3\times8}=\frac{4\times5\times3\times2}{5\times2\times3\times4\times2}=\frac{1}{2}\)
\(a,\frac{3\times4\times7}{5\times3\times4}=\frac{7}{5}\)
\(b,\frac{2\times5\times6\times8}{6\times2\times8\times9}=\frac{5}{9}\)