Em chưa học ạ

Hệ số biến dạng theo mỗi trục đo O'x', O'y', O'z' lần lượt là:

$p=\frac{O\text{'}A\text{'}}{OA}=\frac{2}{2}=1$;

$q=\frac{O\text{'}B\text{'}}{OB}=\frac{1}{3}$;

$r=\frac{O\text{'}C\text{'}}{OC}=\frac{4}{6}=\frac{2}{3}$.

TôToo Hân

 Tô Hân

\(=\lim\limits_{x\rightarrow0}\dfrac{2\left(\sqrt[]{2x+1}-1\right)+2-\sqrt[3]{x^2+x+8}}{x}\)

\(=\lim\limits_{x\rightarrow0}\dfrac{\dfrac{2.2x}{\sqrt[]{2x+1}+1}-\dfrac{x\left(x+1\right)}{\sqrt[3]{\left(x^2+x+8\right)^2}+2\sqrt[3]{x^2+x+8}+4}}{x}\)

\(=\lim\limits_{x\rightarrow0}\left(\dfrac{4}{\sqrt[]{2x+1}+1}-\dfrac{x+1}{\sqrt[3]{\left(x^2+x+8\right)^2}+2\sqrt[3]{x^2+x+8}+4}\right)\)

\(=\dfrac{23}{12}\)

\(=\lim\limits_{x\rightarrow0}\dfrac{\sqrt[]{1+2x}\left(\sqrt[3]{3x+1}-\left(x+1\right)\right)+\left(x+1\right)\left(\sqrt[]{1+2x}-\left(x+1\right)\right)+x^2+\left(2x+1-\sqrt[]{4x+1}\right)}{1+x-\sqrt[]{2x+1}}\)

\(=\lim\limits_{x\rightarrow0}\dfrac{\sqrt[]{1+2x}.\dfrac{x^2\left(-x-3\right)}{\sqrt[3]{\left(3x+1\right)^2}+\sqrt[3]{3x+1}+1}+\dfrac{x^2.\left(x+1\right)}{\sqrt[]{1+2x}+x+1}+x^2+\dfrac{x^2}{2x+1+\sqrt[]{4x+1}}}{\dfrac{x^2}{1+x+\sqrt[]{2x+1}}}\)

\(=\lim\limits_{x\rightarrow0}\dfrac{\dfrac{\left(-x-3\right)\sqrt[]{1+2x}}{\sqrt[3]{\left(3x+1\right)^2}+\sqrt[3]{3x+1}+1}+\dfrac{x+1}{\sqrt[]{1+2x}+x+1}+1+\dfrac{1}{2x+1+\sqrt[]{4x+1}}}{\dfrac{1}{1+x+\sqrt[]{2x+1}}}\)

\(=3\)

\(log_{140}98=\dfrac{log_298}{log_2140}=\dfrac{log_2\left(2.7^2\right)}{log_2\left(2^2.5.7\right)}=\dfrac{log_22+2log_27}{2log_22+log_25+log_27}=\dfrac{1+2log_27}{2+a+log_27}\)

Ta có:

\(log_27=\dfrac{log_57}{log_52}=log_25.log_57=a.\left(\dfrac{log_37}{log_35}\right)=\dfrac{a}{log_35.log_73}=\dfrac{a}{bc}\)

\(\Rightarrow log_{140}98=\dfrac{1+\dfrac{2a}{bc}}{2+a+\dfrac{a}{bc}}=\dfrac{2a+bc}{a+2bc+abc}\)

có 28 viên bi đỏ,đổi 2 viên bi đỏ lấy 9 viên bi xanh,đổi 2 viên bi vàng.hỏi đổi được bao nhiêu viên bi vàng?