1)\(\left(2^5:2^3\right).2^x=64\)

\(\Rightarrow2^{5-3+x}=2^6\)

\(\Rightarrow2^{2+x}=2^6\)

\(\Rightarrow.2^22^x=2^6\)

\(\Rightarrow2^x=2^6:2^2\)

\(\Rightarrow2^x=2^4\Rightarrow x=4\)

2)Tính:

\(F=3^0+3^1+...+3^9\)

\(\Rightarrow3F=3\left(3^0+3^1+...+3^9\right)=3+3^2+3^3+...+3^{10}\)

\(3F-F=3+3^2+...+3^{10}-3^0-3^1-...-3^9\)

\(2F=3^{10}-3^0=3^{10}-1\)

\(F=\frac{3^{10}-1}{2}\)

2 

ta có : F = 1 + 3 + 3² + ..... + 3⁹

=> 3F = 3 + 3² + 3³ +..... + 3¹⁰ 

=> 3F - F = 3¹⁰ - 1

=> 2F = 3¹⁰ - 1

=> F = \(\frac{3^{10}-1}{2}\)

(2^5:2^3)*2^x=64

2^2. 2^x = 2^6

2^x= 2^6 : 2^2

2^x = 2^4

=> x= 4

\(\Rightarrow\dfrac{2}{3}:x=\dfrac{5}{3}\Rightarrow x=\dfrac{2}{3}:\dfrac{5}{3}=\dfrac{2}{5}\)

a) \(\sqrt{4x^2+4x+1}=6\)

\(\Leftrightarrow\sqrt{\left(2x+1\right)^2}=6\)

\(\Leftrightarrow\left(2x+1\right)^2=6^2\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+1=6\\2x+1=-6\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-\dfrac{7}{2}\end{matrix}\right.\)

b) \(\sqrt{4x^2-4\sqrt{7}x+7}=\sqrt{7}\)

\(\Leftrightarrow\sqrt{\left(2x-\sqrt{7}\right)^2}=\sqrt{7}\)

\(\Leftrightarrow\left(2x-\sqrt{7}\right)^2=\left(\sqrt{7}\right)^2\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-\sqrt{7}=\sqrt{7}\\2x-\sqrt{7}=-\sqrt[]{7}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{7}\\x=0\end{matrix}\right.\)

a) \(\sqrt{4x^2+4x+1}=6\)

\(\Leftrightarrow\sqrt{\left(2x+1\right)^2}=6\)

\(\Leftrightarrow\left|2x+1\right|=6\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+1=6\\2x+1=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-\dfrac{7}{2}\end{matrix}\right.\)

b) \(pt\Leftrightarrow\sqrt{\left(2x-\sqrt{7}\right)^2}=\sqrt{7}\)

\(\Leftrightarrow\left|2x-\sqrt{7}\right|=\sqrt{7}\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-\sqrt{7}=\sqrt{7}\\2x-\sqrt{7}=-\sqrt{7}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{7}\\x=0\end{matrix}\right.\)

c) \(PT\Leftrightarrow\sqrt{\left(x+\sqrt{3}\right)^2}=2\sqrt{3}\)

\(\Leftrightarrow\left|x+\sqrt{3}\right|=2\sqrt{3}\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\sqrt{3}=2\sqrt{3}\\x+\sqrt{3}=-2\sqrt{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{3}\\x=-3\sqrt{3}\end{matrix}\right.\)

d) \(pt\Leftrightarrow\left|x-3\right|=9\Leftrightarrow\left[{}\begin{matrix}x-3=-9\\x-3=9\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-6\\x=12\end{matrix}\right.\)

Ta có :

a, \(\frac{31}{12}-(\frac{2}{5}+x)=\frac{2}{3}\)

\(\Rightarrow\frac{2}{5}+x=\frac{31}{12}-\frac{2}{3}=\frac{23}{12}\)

\(\Rightarrow\frac{23}{12}-\frac{31}{12}=\frac{-8}{12}=\frac{-2}{3}\)

Câu b để mk làm sau

ờ bn , giup mk đik r mk kick cho

\(x+x\cdot3:\dfrac{2}{9}+x:\dfrac{2}{7}=252\)

\(\Leftrightarrow x+x\cdot3\cdot\dfrac{9}{2}+x\cdot\dfrac{7}{2}=252\)

\(\Leftrightarrow x\cdot18=252\)

hay x=14

\(a,50\%x-0,2+x=\dfrac{4}{5}\)

\(\Leftrightarrow\dfrac{1}{2}x-0,2+x=\dfrac{4}{5}\)

\(\Leftrightarrow\dfrac{1}{2}x+x=\dfrac{4}{5}+0,2\)

\(\Leftrightarrow\dfrac{3}{2}x=\dfrac{4}{5}+\dfrac{1}{5}\)

\(\Leftrightarrow\dfrac{3}{2}x=1\)

\(\Leftrightarrow x=\dfrac{2}{3}\)

\(b,\left(x-\dfrac{3}{4}\right):\dfrac{1}{2}+\dfrac{3}{2}=\dfrac{25}{2}\)

\(\Leftrightarrow\left(x-\dfrac{3}{4}\right).2=\dfrac{25}{2}-\dfrac{3}{2}\)

\(\Leftrightarrow\left(x-\dfrac{3}{4}\right).2=\dfrac{22}{2}\)

\(\Leftrightarrow x-\dfrac{3}{4}=11:2\)

\(\Leftrightarrow x=\dfrac{11}{2}+\dfrac{3}{4}\)

\(\Leftrightarrow x=\dfrac{25}{4}\)