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\(1990\times1990-1992\times1998\)
\(=1990\times1990-\left(1992-2\right)\times\left(1998+2\right)\)
\(=1990\times1990-1990\times1990\)
\(=0\)
Ta có: \(\frac{1+5y}{5x}=\frac{1+7y}{4x}=\frac{1+5y-1-7y}{5x-4x}=\frac{-2y}{x}\)
\(\Rightarrow\frac{1+5y}{5x}=\frac{-2y}{x}\)\(\Rightarrow\frac{1+5y}{5}=-2y\)\(\Rightarrow1+5y=-10y\)\(\Rightarrow15y=-1\)\(\Rightarrow y=\frac{-1}{15}\)
Ta có: \(\frac{1+3y}{12}=\frac{1+5y}{5x}\)\(\Rightarrow\frac{1+3.\frac{-1}{15}}{12}=\frac{1+5.\frac{-1}{15}}{5x}\)\(\Rightarrow\frac{\frac{4}{5}}{12}=\frac{\frac{2}{3}}{5x}\)\(\Rightarrow5x=\frac{\frac{2}{3}.12}{\frac{4}{5}}=10\)\(\Rightarrow x=2\)
\(\dfrac{0,2-0,375+\dfrac{5}{11}}{-0,3+\dfrac{9}{16}-\dfrac{15}{22}}=\dfrac{\dfrac{2}{10}-\dfrac{6}{16}+\dfrac{5}{11}}{-\dfrac{3}{10}+\dfrac{9}{16}-\dfrac{15}{22}}=\dfrac{2.\left(\dfrac{1}{10}-\dfrac{3}{16}+\dfrac{5}{22}\right)}{-3.\left(\dfrac{1}{10}-\dfrac{3}{16}-\dfrac{5}{22}\right)}=\dfrac{-2}{3}\)
\(\dfrac{0,2-0,375+\dfrac{5}{11}}{-0,3+\dfrac{9}{16}-\dfrac{15}{22}}=\dfrac{\dfrac{2}{10}-\dfrac{6}{16}+\dfrac{5}{11}}{-\dfrac{3}{10}+\dfrac{9}{16}-\dfrac{15}{22}}=\dfrac{2.\left(\dfrac{1}{10}-\dfrac{3}{16}+\dfrac{5}{22}\right)}{-3.\left(\dfrac{1}{10}-\dfrac{3}{16}-\dfrac{5}{22}\right)=\dfrac{-2}{3}}\)
( chú ý vì x/5 = y/7 = z/3 =>x;y;z cùng dấu )
x/5 = y/7 = z/3 =>(x/5)^2= (y/7)^2 = (z/3)^2 hay x^2/25 = y^2/49 =z^2 /9
x^2/25 = y^2/49 =z^2 /9 = (x^2 + y^2 - z^2) /(25+49 -9)=585/65 =9=3^2
=> (x/5)^2=3^2 =>x/5 =+-3 =>x=+-15
(y/7)^2=3^2 =>y/7 =+-3 =>y=+-21
(z/3)^2 =3^2 =>z/3 =+-3 =>z=+-9
vậy có 2 cặp (x;y;z) là: (15;21;9) và (-15;-21;-9)
= (2.3)^3+ 3. ( 2.3)^2 + 3^3 / -13
= 2^3 .3^3 + 3. 3^2 . 2^2 + 3^3 / -13
= 3^3. ( 2^3 + 2^2 + 1) /-13
= 27.13/-13
= -27
\(\frac{6^3+3\cdot6^2+3^3}{-13}=\frac{6^2\left(6+3\right)+3^2\cdot3}{-13}=\frac{36\cdot9+9\cdot3}{-13}=\frac{39\cdot9}{-13}=-3\cdot9=-27\)
\(2017.2018\left(\frac{2016}{2017}-\frac{2016}{2018}\right)\)
\(=2017.2018\left(\frac{2016.2018}{2017.2018}-\frac{2016.2017}{2018.2017}\right)\)
\(=2017.2018\left(\frac{2016\left(2018-2017\right)}{2018.2017}\right)\)
\(=\frac{2017.2018.2016}{2017.2018}=2016\)