số viên phấn có trong 7 thùng là : 

\(305\times7=2135\left(vien\right)\)

đáp số : 2135 viên phấn

\(\left(3x-4\right)^3=7+1^{2019}\)

\(\Leftrightarrow\left(3x-4\right)^3=7+1\)

\(\Leftrightarrow\left(3x-4\right)^3=8\)

\(\Leftrightarrow\left(3x-4\right)^3=2^3\)

\(\Leftrightarrow3x-4=2\)

\(\Leftrightarrow3x=6\)

\(\Leftrightarrow x=2\)

\(a,195.63+37.195\)

\(=195\left(63+37\right)\)

\(=195.100\)

\(=19500\)

\(b,7^{12}:7^{10}-36:3^2\)

\(=7^2-2^2.3^2:3^2\)

\(=7^2-2^2\)

\(=49-4=44\)

\(c,180-\left[130-\left(12-10\right)^3\right]+2023^0\)

\(=180-\left[130-2^3\right]+1\)

\(=180-\left(130-8\right)+1\)

\(=180-122+1\)

\(=59\)

Có gì k hiểu thì ib nhá

\(480:\left(75+7^2-8.3:5\right)+2021^0\)

\(=480:\left(75+49-\dfrac{24}{5}\right)+1\)

\(=480:\left(124-\dfrac{24}{5}\right)+1\)

\(=480:\dfrac{596}{5}+1\)

\(=\dfrac{600}{149}+1=\dfrac{749}{149}\)

\(a,32< 2^n< 512\)

\(\Leftrightarrow2^5< 2^n< 2^9\)

\(\Rightarrow n\in\left\{6;7;8\right\}\)

\(b,\dfrac{3^{10}.11+3^{10}.5}{3^9.2^4}\)

\(=\dfrac{3^{10}\left(11+5\right)}{3^9.2^4}\)

\(=\dfrac{3^{10}.16}{3^9.2^4}\)

\(=\dfrac{3^{10}.2^4}{3^9.2^4}=3\)

\(a,\dfrac{3^{43}+3^4}{3^{39}+1}\)

\(=\dfrac{3^4\cdot\left(3^{39}+1\right)}{3^{39}+1}\)

\(=3^4\)

\(=81\)

\(\left(x-2\right)^8=\left(x-2\right)^3\)

\(\Leftrightarrow\left(x-2\right)^8-\left(x-2\right)^3=0\)

\(\Leftrightarrow\left(x-2\right)^3\left[\left(x-2\right)^5-1\right]=0\)

\(\Rightarrow\left(x-2\right)^3=0\) hoặc \(\left(x-2\right)^5-1=0\)

     \(x-2=0\)              \(\left(x-2\right)^5=1\)

     \(x=2\)                     \(x-2=1\)

                                    \(x=3\)

Vậy ...

\(\dfrac{x-2}{x+1}=\dfrac{x^2-3x+2}{x^2-1}\)

\(\dfrac{x-2}{x+1}=\dfrac{\left(x^2-2x+1\right)-\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}\)

\(\dfrac{x-2}{x+1}=\dfrac{\left(x-1\right)^2-\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}\)

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

\(\dfrac{x-2}{x+1}=\dfrac{x}{x+1}\)

\(\Leftrightarrow x\left(x+1\right)=\left(x-2\right)\left(x+1\right)\)

\(\Leftrightarrow x^2+x=x^2+x-2x+1\)

\(\Leftrightarrow x^2+x-x^2+x+1=0\)

\(\Leftrightarrow2x=1\)

\(\Leftrightarrow x=\dfrac{1}{2}\left(tm\right)\)

Vậy ...