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Đặt 3759=a; 5741=b
Theo đề, ta có: \(E=4\dfrac{7}{b}\cdot\dfrac{1}{a}-\dfrac{4}{a}\cdot\left(1+\dfrac{2}{b}\right)+\dfrac{1}{a}+\dfrac{1}{ab}\)
\(=\dfrac{4b+7}{b}\cdot\dfrac{1}{a}-\dfrac{4}{a}\cdot\dfrac{b+2}{b}+\dfrac{b+1}{ab}\)
\(=\dfrac{4b+7-4b-8+b+1}{ab}=\dfrac{b}{ab}=\dfrac{1}{a}=\dfrac{1}{3759}\)
\(xy\left(x-y\right)+yz\left(y-z\right)+xz\left(z-x\right)\\ =xy\left(x-y\right)+yz\left[-\left(x-y\right)-\left(z-x\right)\right]+xz\left(z-x\right)\\ =xy\left(x-y\right)-yz\left(x-y\right)-yz\left(z-x\right)+xz\left(z-x\right)\\ =\left(x-y\right)\left(xy-yz\right)+\left(z-x\right)\left(xz-yz\right)\\ =y\left(x-y\right)\left(x-z\right)+z\left(z-x\right)\left(x-y\right)\\ =\left(x-y\right)\left(x-z\right)\left(y-z\right)\)
\(\frac{\frac{1}{3}-\frac{1}{7}-\frac{1}{13}}{\frac{2}{3}-\frac{2}{7}-\frac{2}{13}}\cdot\frac{\frac{3}{4}-\frac{3}{16}-\frac{3}{64}-\frac{3}{264}}{1-\frac{1}{4}-\frac{1}{16}-\frac{1}{64}}+\frac{5}{8}\)
\(=\frac{\frac{1}{3}-\frac{1}{7}-\frac{1}{13}}{2\left(\frac{1}{3}-\frac{1}{7}-\frac{1}{13}\right)}\cdot\frac{\frac{3}{4}\left(1-\frac{1}{4}-\frac{1}{16}-\frac{1}{64}\right)}{1-\frac{1}{4}-\frac{1}{16}-\frac{1}{64}}\)\(+\frac{5}{8}\)
\(\frac{1}{2}\cdot\frac{3}{4}+\frac{5}{8}=\frac{3}{8}+\frac{5}{8}=1\)
\(\frac{1-\frac{1}{\sqrt{49}}+\frac{1}{49}-\frac{1}{\left(7\sqrt{7}\right)^2}}{\frac{\sqrt{64}}{2}-\frac{4}{7}+\left(\frac{2}{7}\right)^2-\frac{4}{343}}\)
\(=\frac{1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}}{4-\frac{4}{7}+\frac{4}{49}-\frac{4}{343}}\)
\(=\frac{1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}}{4\left(1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}\right)}\)
\(=\frac{1}{4}\)
f: =>35x-5=96-6x
=>41x=101
hay x=101/41
g: =>3(x-3)=90-5(1-2x)
=>3x-9=90-5+10x
=>3x-9=10x+85
=>-7x=94
hay x=-94/7
h.3x - 2/6 - 5 = 3 - 2(x + 7)/4
<=> 3x - 2 - 30/6 = 3 - 2(x + 7)/4
<=> 3x - 32/6 = 3 - 2x - 14/4
<=> 3x - 32/6 = -2x - 11/4
<=> 6x - 64/12 = -6x - 33/12
<=> 6x - 64 = -6x - 33 <=> 12x = 31 <=> x = 31/12
