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Câu 2
(a+3)(b-4)-(a-3)(b+4)=0
=>ab-4a+3b-12-ab-4a+3b+12=0
=>-8a=-6b
=>a/b=3/4
=>a/3=b/4
Bài 2:
x=13 nên x+1=14
\(f\left(x\right)=x^{14}-x^{13}\left(x+1\right)+x^{12}\left(x+1\right)-...+x^2\left(x+1\right)-x\left(x+1\right)+14\)
\(=x^{14}-x^{14}-x^{13}+x^{13}-...+x^3+x^2-x^2-x+14\)
=14-x=1
x=13 nên x+1=14
f(x)=x14−x13(x+1)+x12(x+1)−...+x2(x+1)−x(x+1)+14f(x)=x14−x13(x+1)+x12(x+1)−...+x2(x+1)−x(x+1)+14
=x14−x14−x13+x13−...+x3+x2−x2−x+14=x14−x14−x13+x13−...+x3+x2−x2−x+14
=14-x=1
\(\left(\frac{3}{4}-81\right)\left(\frac{3^2}{5}-81\right)\left(\frac{3^3}{6}-81\right)....\left(\frac{3^{2000}}{2003}-81\right)\)
\(=\left(\frac{3}{4}-81\right)\left(\frac{3^2}{5}-81\right)\left(\frac{3^3}{6}-81\right)...\left(\frac{3^6}{9}-81\right)...\left(\frac{3^{2000}}{2003}-81\right)\)
\(=\left(\frac{3}{4}-81\right)\left(\frac{3^2}{5}-81\right)\left(\frac{3^3}{6}-81\right)....\left(81-81\right)...\left(\frac{3^{2000}}{2003}-81\right)\)
\(=\left(\frac{3}{4}-81\right)\left(\frac{3^2}{5}-81\right)....0....\left(\frac{3^{2000}}{2003}-81\right)\)
\(=0\)
(3/4 -81 )(3^2/5 -81 )(3^3/6 -81)......(3^2000/2003 -81)
ta viết tiếp dãy số (3/4 -81 )(3^2/5 -81 )(3^3/6 -81)(3^4/7 - 81 ) (3^5/8 -81)(3^6/9 -81).........(3^2000/2003 -81) thì thấy 3^6/9=81 ->3^6/9 -81=0 -> dãy số bằng 0 -> (3/4 -81 )(3^2/5 -81 )(3^3/6 -81)......(3^2000/2003 -81) =0
Minh k hiểu cho lắm. Bạn viết theo công thức toán olm cho sẵn đi cho dễ đọc
a/ \(2016\dfrac{1}{6}:\dfrac{-2}{5}-16\dfrac{1}{6}:\dfrac{-2}{5}\)
\(=2016\dfrac{1}{6}.\dfrac{-5}{2}-16\dfrac{1}{6}.\dfrac{-5}{2}\)
\(=\dfrac{-5}{2}\left(2016\dfrac{1}{6}-16\dfrac{1}{6}\right)\)
\(=\dfrac{-5}{2}.2000\)
\(=-5000\)
b/ \(\left(\dfrac{4}{3}-\dfrac{3}{2}\right)^2-2.\left|-\dfrac{1}{9}\right|+\sqrt{\dfrac{4}{81}}\)
\(=\left(\dfrac{8}{6}-\dfrac{9}{6}\right)^2-2.\dfrac{1}{9}+\dfrac{2}{9}\)
\(=\dfrac{1}{4}-\dfrac{2}{9}+\dfrac{2}{9}\)
\(=\dfrac{1}{36}+\dfrac{2}{9}\)
\(=\dfrac{1}{4}\)
\(a,=\dfrac{9}{4}-\dfrac{9}{4}+\dfrac{5}{6}=\dfrac{5}{6}\\ b,=\dfrac{4}{9}+1=\dfrac{13}{9}\)
b. \(\left(\dfrac{3^2}{9}.\dfrac{3^3}{81}\right)^{12}:\left(\dfrac{3^6}{81^2}\right)^{10}\)
\(=\left(1.\dfrac{1}{3}\right)^{12}:\left(\dfrac{1}{9}\right)^{10}\)
\(=\left(\dfrac{1}{3}\right)^{12}:\left(\dfrac{1}{9}\right)^{10}\)
\(=\left[\left(\dfrac{1}{3}\right)^2\right]^6:\left(\dfrac{1}{9}\right)^{10}\)
\(=\left(\dfrac{1}{9}\right)^6:\left(\dfrac{1}{9}\right)^{10}\)
\(=\left(\dfrac{1}{9}\right)^{-4}=6561\)
Đặt \(A=\left(\dfrac{3}{4}-81\right)\left(\dfrac{3^2}{5}-81\right)\left(\dfrac{3^3}{6}-81\right)\cdot...\cdot\left(\dfrac{3^{2000}}{2003}-81\right)\)
\(=\left(\dfrac{3^6}{9}-81\right)\cdot\left(\dfrac{3}{4}-81\right)\left(\dfrac{3^2}{5}-81\right)\cdot...\cdot\left(\dfrac{3^{2000}}{2003}-81\right)\)
\(=\left(81-81\right)\cdot\left(\dfrac{3}{4}-81\right)\left(\dfrac{3^2}{5}-81\right)\cdot...\cdot\left(\dfrac{3^{2000}}{2003}-81\right)\)
=0