Bài 1:

ta có: 333<3333; 444<4444

=> 333⁴⁴⁴<3333⁴⁴⁴⁴

Bài 2:

\(A=\frac{21^5}{81}=\frac{\left(3.7\right)^5}{3^4}=\frac{3^5.7^5}{3^4}=3.7^5=50421\)

\(B=\frac{3^3.\left(0,5\right)^5}{\left(1,5\right)^5}=\frac{3^3.\left(0,5\right)^5}{\left(3.0,5\right)^5}=\frac{3^3.\left(0,5\right)^5}{3^5.\left(0,5\right)^5}=\frac{1}{3^2}=\frac{1}{9}\)

\(C=2^2.\frac{1}{128}.45.2^{-6}=\frac{2^2.45}{128.64}=\frac{2^2.45}{2^7.2^6}=\frac{45}{2^{11}}=\frac{45}{2048}\)

\(D=\frac{6^3+3.6^2+3^3}{-13}=\frac{2^3.3^3+2^2.3^3+3^3}{-13}=\frac{3^3.\left(2^3+2^2+1\right)}{-13}=\frac{3^3.13}{-13}\)\(=3^3.\left(-1\right)=-27\)

Trả lời:

\(\frac{2}{\sqrt{5}+\sqrt{3}}-\sqrt{\frac{2}{4-\sqrt{15}}}+6\sqrt{\frac{1}{3}}\)

\(=\frac{2.\left(\sqrt{5}-\sqrt{3}\right)}{5-3}-\sqrt{\frac{2\times2}{2\times\left(4-\sqrt{15}\right)}}+6\times\frac{1}{\sqrt{3}}\)

\(=\frac{2.\left(\sqrt{5}-\sqrt{3}\right)}{2}-\sqrt{\frac{4}{8-2\sqrt{15}}}+6\times\frac{\sqrt{3}}{3}\)

\(=\sqrt{5}-\sqrt{3}-\sqrt{\frac{4}{5-2\sqrt{15}+3}}+2\sqrt{3}\)

\(=\sqrt{5}-\sqrt{3}-\sqrt{\frac{4}{\left(\sqrt{5}-\sqrt{3}\right)^2}}+2\sqrt{3}\)

\(=\sqrt{5}-\sqrt{3}-\frac{2}{\sqrt{5}-\sqrt{3}}+2\sqrt{3}\)

\(=\sqrt{5}+\sqrt{3}-\frac{2}{\sqrt{5}-\sqrt{3}}\)

\(=\frac{\left(\sqrt{5}-\sqrt{3}\right).\left(\sqrt{5}+3\right)-2}{\sqrt{5}-\sqrt{3}}\)

\(=\frac{5-3-2}{\sqrt{5}-\sqrt{3}}\)

\(=0\)

Học tốt 

-2.6314...

2/7 nhân 5/1/4 - 2/7 nhân 3/1/4 

ai giúp mình với xin người đi mà nhớ ghé qua trả lời cho mình với huhuhuhuhu 

5 k cho ai trả lời đc yêu các bạn nếu trả lời đc câu hỏi của mình

\(\frac{11}{\frac{3}{13}}-\left(\frac{2}{\frac{4}{7}}+\frac{5}{\frac{3}{13}}\right)\)

\(=11:\frac{3}{13}-\left(2:\frac{4}{7}+5:\frac{3}{13}\right)\)

\(=11.\frac{13}{3}-\left(2.\frac{7}{4}+5.\frac{13}{3}\right)\)

\(=11.\frac{13}{3}-2.\frac{7}{4}-5.\frac{13}{3}\)

\(=\frac{13}{3}.\left(11-5\right)-2.\frac{7}{4}\)

\(=\frac{13}{3}.6-\frac{7}{2}\)

\(=13.2-\frac{7}{2}\)

\(=26-\frac{7}{2}\)

\(=\frac{52}{2}-\frac{7}{2}=\frac{45}{2}\)

1. \(\frac{\left(0,6\right)^5}{\left(0.2\right)^5}=\left(\frac{0.6}{0.2}\right)^5=\left(3\right)^5=243\)

2.\(\frac{6^3+3.6^2+3^3}{-13}=\frac{6^2\left(6+3\right)+3^3}{-13}=\frac{6^2.9+3^2.3}{-13}\)

\(\Leftrightarrow\frac{6^2.9+3^2.3}{-13}=\frac{3^2.39}{-13}=3^2.\left(-3\right)=-27\)

Bạn ashura nhầm mũ mẫu dưới rồi

\(2\frac{1}{3}+\left(x-\frac{3}{2}\right)=\left(3-\frac{3}{2}\right)x\)

\(\Rightarrow\frac{7}{3}+x-\frac{3}{2}=\frac{3}{2}x\)

\(\Rightarrow x-\frac{3}{2}x=\frac{3}{2}-\frac{7}{3}\)

\(\Rightarrow\frac{-1}{2}x=-\frac{5}{6}\)

\(\Rightarrow x=\frac{5}{3}\)

\(ĐKXĐ:x\ne\pm3\)

\(pt\Leftrightarrow\frac{\left(x+3\right)^2-\left(x-3\right)^2}{x^2-9}=\frac{17}{x^2-9}\)

\(\Leftrightarrow\left(x+3\right)^2-\left(x-3\right)^2=17\)

Tự dừng bấm Gửi tl

\(\Leftrightarrow x^2+6x+9-x^2+6x-9=17\)

\(\Leftrightarrow12x=17\Leftrightarrow x=\frac{17}{12}\)

\(\frac{x+3}{x-3}-\frac{17}{x^2-9}=\frac{x-3}{x+3}\)

\(\frac{x+3}{x-3}-\frac{17}{x^2-3^2}=\frac{x-3}{x+3}\)

\(\frac{x+3}{x-3}-\frac{17}{\left(x+3\right)\left(x-3\right)}=\frac{x-3}{x+3}\)

\(\left(x+3\right)^2-17=\left(x-3\right)^2\)

\(x^2+6x-8=x^2-6x+9\)

\(6x-8=-6x+9\)

\(6x=-6x+9+8\)

\(6x+6x=17\)

\(12x=17\Leftrightarrow x=\frac{17}{12}\)

\(x=\frac{\left[6+3^2.2-6.3^2\right]^2}{\left[3^2+3.3^2-3^4\right]^2}=\frac{\left[6\left[1+3-9\right]\right]^2}{\left[9+27-81\right]^2}=\frac{\left[-30\right]^2}{\left[-45\right]^2}=\frac{900}{2025}=\frac{4}{9}\)

hay doi hon so thanh phan  so de tinh cho de nhe

Bạn trả lời rõ đi

tính ak?

7-12/13=79/13

188/45

Còn lại tự tính nha!