\(\frac{1}{9}\).27n = 3n +2
\(\frac{1}{9}\).34 . 3n = 37
\(\frac{1}{2}\). 4n + 4.4n = 9 . 2n + 1
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\(\frac{3x}{5x+5y}-\frac{x}{10x-10y}\)
= \(\frac{3x\left(x-y\right)}{5.2.\left(x+y\right)\left(x-y\right)}-\frac{x\left(x+y\right)}{10\left(x^2-y^2\right)}\)
= \(\frac{3x^2-3xy-x^2-xy}{10\left(x^2-y^2\right)}\)
= \(\frac{3x\left(x-y\right)}{10\left(x^2-y^2\right)}\)
= \(\frac{3x}{10\left(x+y\right)}\)
Tôi cũng là của FC Real Madrid ở Hà Nam.
Chúng mình kết bạn nhé.hihi.
Đặt \(A=\frac{4n+3}{7n+1}-\frac{3n-2}{7n+1}+\frac{2n-3}{7n+1}\) ta có :
\(A=\frac{4n+3-3n+2+2n-3}{7n+1}\)
\(A=\frac{3n+2}{7n+1}\)
Vậy \(A=\frac{3n+2}{7n+1}\)
Chúc bạn học tốt ~
a) lim \(\frac{\left(2n^2-3n+5\right)\left(2n+1\right)}{\left(4-3n\right)\left(2n^2+n+1\right)}\)
= lim \(\frac{\left(2-\frac{3}{n}+\frac{5}{n^2}\right)\left(2+\frac{1}{n}\right)}{\left(\frac{4}{n}-3\right)\left(2+\frac{1}{n}+\frac{1}{n^2}\right)}=\frac{4}{-6}=-\frac{2}{3}\)
b)lim ( \(\frac{\sqrt{n^4+1}}{n}-\frac{\sqrt{4n^6+2}}{n^2}\))
= lim ( \(\frac{n\sqrt{n^4+1}-\sqrt{4n^6+2}}{n^2}\) )
= lim \(\frac{\left(n^6+n^2\right)-\left(4n^6+2\right)}{n^2\left(n\sqrt{n^4+1}+\sqrt{4n^2+2}\right)}\)
= lim \(\frac{-3n^6+n^2+2}{n^3\sqrt{n^4+1}+n^2\sqrt{4n^2+2}}\)
= lim \(\frac{-3n\left(1-\frac{1}{n^4}-\frac{2}{n^6}\right)}{\sqrt{1+\frac{1}{n^4}}+\frac{1}{n^2}\sqrt{4+\frac{2}{n^2}}}\)
= lim \(-3n=-\infty\)
c) lim \(\frac{2n+3}{\sqrt{9n^2+3}-\sqrt[3]{2n^2-8n^3}}\)
= lim\(\frac{2+\frac{3}{n}}{\sqrt{9+\frac{3}{n^2}}-\sqrt[3]{\frac{2}{n}-8}}=\frac{2}{3+2}=\frac{2}{5}\)
\(A=\frac{5n+17}{n+3}+\frac{-3n}{2+3}+\frac{2n+9}{n+3}+\frac{-4n-23}{n+3}\)
\(=\frac{5n+17-3n+2n+9-4n-23}{n+3}\)
\(=\frac{3}{n+3}\)
Trl :
\(\frac{1}{9}.27^n=3^{n+2}\)
\(3^{-2}.\left(3^3\right)^n=3^{n+2}\)
\(3^{-2}.3^{3n}=3^{n+2}\)
\(\Rightarrow-2+3n=n+2\)
\(\Rightarrow3n=n+4\)
\(\Rightarrow2n=4\)\(\Rightarrow n=2\)
Hok tốt
Trl :
\(\frac{1}{9}3^4.3^n=3^7\)
\(3^{-2}.3^4.3^n=3^7\)
\(\Rightarrow-2+4+n=7\)
\(\Rightarrow2+n=7\)
\(\Rightarrow n=7-2\)
\(\Rightarrow n=5\)
Hok tốt !