Bài 2:

a) \(2^n-64=0\)

\(2^n=64\)

\(2^n=2^6\)

\(n=6\)

b) \(5.3^{n-3}-405=0\)

\(5.3^{n-3}=405\)

\(3^{n-3}=405:5\)

\(3^{n-3}=81\)

\(n-3=4\)

\(n=4+3\)

\(n=7\)

c) \(4^n.8=2^{15}\)

\(\left(2^2\right)^n.2^3=2^{15}\)

\(2^{2n}.2^3=2^{15}\)

\(2^{2n+3}=2^{15}\)

\(2n+3=15\)

\(2n=15-3\)

\(2n=12\)

\(n=12:2\)

\(n=6\)

d) \(3.2^{n+1}+2^{n+2}=160\)

\(2^{n+1}.\left(3+2\right)=160\)

\(2^{n+1}.5=160\)

\(2^{n+1}=160:5\)

\(2^{n+1}=32\)

\(2^{n+1}=2^5\)

\(n+1=5\)

\(n=5-1\)

\(n=4\)

Bài 1

a) \(2^{11}.64=2^{11}.2^6=2^{17}\)

Do \(16< 17\Rightarrow2^{16}< 2^{17}\)

Vậy \(2^{16}< 2^{11}.64\)

b) Do \(18>17\Rightarrow9^{18}>9^{17}\)   (1)

\(9^{18}=\left(3^2\right)^{18}=3^{36}\)

Do \(36< 37\Rightarrow3^{36}< 3^{37}\)

\(\Rightarrow9^{18}< 3^{37}\)  (2)

Từ (1) và (2) \(\Rightarrow9^{17}< 3^{37}\)

c) \(2^{333}=\left(2^3\right)^{111}=8^{111}\)

\(3^{222}=\left(3^2\right)^{111}=9^{111}\)

Do \(8< 9\Rightarrow8^{111}< 9^{111}\)

Vậy \(2^{333}< 3^{222}\)

d) \(3^{50}=\left(3^2\right)^{25}=9^{25}\)

Do \(9< 11\Rightarrow9^{25}< 11^{25}\)

Vậy \(3^{50}< 11^{25}\)

e) \(37< 38\Rightarrow3^{37}< 3^{38}\) (1)

Lại có: \(3^{38}=3^{2.19}=\left(3^2\right)^{19}=9^{19}\)

Do \(9< 10\Rightarrow9^{19}< 10^{19}\)

\(\Rightarrow3^{38}< 10^{19}\) (2)

Từ (1) và (2) \(\Rightarrow3^{37}< 10^{19}\)

f) Do \(17>16\Rightarrow17^{14}>16^{14}\)   (1)

Do \(32>31\Rightarrow32^{11}>31^{11}\)   (2)

Lại có:

\(16^{14}=\left(2^4\right)^{14}=2^{56}\)

\(32^{11}=\left(2^5\right)^{11}=2^{55}\)

Do \(56>55\Rightarrow2^{56}>2^{55}\)

\(\Rightarrow16^{14}>32^{11}\)   (3)

Từ (1), (2) và (3) \(\Rightarrow17^{14}>31^{11}\)

Mình mắc câu này 

\(\Leftrightarrow\left\{{}\begin{matrix}2x-3y=3\sqrt{5}-1\\4x+\left(2\sqrt{5}+2\right)y=-4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}4x-6y=6\sqrt{5}-2\\4x+\left(2\sqrt{5}+2\right)y=-4\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(-8-2\sqrt{5}\right)y=6\sqrt{5}+2\\2x-3y=3\sqrt{5}-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=1-\sqrt{5}\\x=\dfrac{3\sqrt{2}-3\sqrt{5}+2}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}4x-6y=6\sqrt{5}-2\\3.\left(\sqrt{5}-1\right)x+6y=3-3\sqrt{5}\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}\left(3\sqrt{5}+1\right)x=1+3\sqrt{5}\\y=\dfrac{3\sqrt{5}-1-2x}{-3}\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=1\\y=\dfrac{3\sqrt{5}-1-2.1}{-3}\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=1\\y=\dfrac{-3.\left(1-\sqrt{5}\right)}{-3}=1-\sqrt{5}\end{matrix}\right.\\ \Rightarrow\left(x;y\right)=\left(1;1-\sqrt{5}\right)\)

\(ĐK:x\ge\dfrac{1}{3}\\ PT\Leftrightarrow-2x^2+14x-10+\left(4x-3\right)\left(x-2-\sqrt{3x-1}\right)=0\\ \Leftrightarrow-2\left(x^2-7x+5\right)+\dfrac{\left(4x-3\right)\left(x^2-7x+5\right)}{x-2+\sqrt{3x-1}}=0\\ \Leftrightarrow\left(x^2-7x+5\right)\left(\dfrac{4x-3}{x-2+\sqrt{3x-1}}-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x^2-7x+5=0\\\dfrac{4x-3}{x-2+\sqrt{3x-1}}=2\left(1\right)\end{matrix}\right.\\ \left(1\right)\Leftrightarrow4x-3=2x-4+2\sqrt{3x-1}\\ \Leftrightarrow2x+1=2\sqrt{3x-1}\\ \Leftrightarrow4x^2+4x+1=12x-4\\ \Leftrightarrow4x^2-8x+5=0\left(\text{vô nghiệm}\right)\\ \Leftrightarrow x^2-7x+5=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7+\sqrt{29}}{2}\left(tm\right)\\x=\dfrac{7-\sqrt{29}}{2}\left(tm\right)\end{matrix}\right.\)

oh no I can't do it

i can't its verry diffcult

do you think

after (shout) shouting for help, she (take) took off her shoes and (jump) jumped in (save)to save him.

Động từ đầu tiên đứng sau giới từ nên thêm ing, còn các đt phía sau là các chuỗi hành động trong quá khứ

After (shout)..shouting.. for help, she (take)...took. off her shoes and (jump)..jumped.. in (save).to save... him

giải thik nè: sau after(là giới từ) ta chia v_ing; mik nghĩ đây là QKĐ nên 2 cái tiếp +ed; còn in save chia là to save(cái này cô dạy rồi)