a: \(B=\left(4\cdot2^5\right):\left(2^3\cdot\frac{1}{16}\right)\)

\(=\left(4\cdot32\right):\left(\frac{8}{16}\right)\)

\(=128:\frac12=128\cdot2=256\)

b: \(B=3^2\cdot\frac{1}{243}\cdot81^2\cdot\frac{1}{3^3}\)

\(=3^2\cdot\frac{1}{3^5}\cdot3^8\cdot\frac{1}{3^3}=\frac{3^8}{3^8}\cdot3^2=3^2=9\)

c: \(D=\left\lbrace\left(0,1\right)^2\right\rbrace^0+\left\lbrack\left(\frac17\right)^1\right\rbrack^2:\frac{1}{49}\cdot\left\lbrack\left(2^2\right)^3:2^5\right\rbrack\)

\(=1+\left(\frac17\right)^2\cdot49\cdot2^6:2^5\)

\(=1+49\cdot\frac{1}{49}\cdot2=1+2=3\)

d: \(C=\left(-0,5\right)^5:\left(-0,5\right)^3-\left(\frac{17}{2}\right)^7:\left(\frac{17}{2}\right)^6\)

\(=\left(-0,5\right)^{5-3}-\left(\frac{17}{2}\right)\)

\(=\left(-0,5\right)^2-\frac{17}{2}=0,25-\frac{17}{2}=\frac14-\frac{34}{4}=-\frac{33}{4}\)

a: \(A=\left(1\frac34\right)^3-\left(1\frac34\right)^2+\left(-1,031\right)^0\)

\(=\left(\frac74\right)^3-\left(\frac74\right)^2+1\)

\(=\frac{343}{64}-\frac{49}{16}+1=\frac{343}{64}-\frac{33}{16}=\frac{343-132}{64}=\frac{211}{64}\)

b: \(B=\left(\frac23\right)^3-4\cdot\left(-1\frac34\right)^2+\left(-\frac23\right)^3\)

\(=\frac{8}{27}-4\cdot\left(-\frac74\right)^2-\frac{8}{27}\)

\(=-4\cdot\frac{49}{16}=-\frac{49}{4}\)

Bài 1:

|\(x\)| = 1 ⇒ \(x\) \(\in\) {-\(\dfrac{1}{3}\); \(\dfrac{1}{3}\)}

A(-1) = 2(-\(\dfrac{1}{3}\))² - 3.(-\(\dfrac{1}{3}\)) + 5

A(-1) = \(\dfrac{2}{9}\) + 1 + 5

A (-1) = \(\dfrac{56}{9}\)

A(1) = 2.(\(\dfrac{1}{3}\) )²- \(\dfrac{1}{3}\).3 + 5

A(1) = \(\dfrac{2}{9}\) - 1 + 5

A(1) = \(\dfrac{38}{9}\)

|y| = 1 ⇒ y \(\in\) {-1; 1} 

⇒ (\(x;y\)) = (-\(\dfrac{1}{3}\); -1); (-\(\dfrac{1}{3}\); 1); (\(\dfrac{1}{3};-1\)); (\(\dfrac{1}{3};1\))

B(-\(\dfrac{1}{3}\);-1) = 2.(-\(\dfrac{1}{3}\))² - 3.(-\(\dfrac{1}{3}\)).(-1) + (-1)²

B(-\(\dfrac{1}{3}\); -1) = \(\dfrac{2}{9}\) - 1 + 1

B(-\(\dfrac{1}{3}\); -1) = \(\dfrac{2}{9}\)

B(-\(\dfrac{1}{3}\); 1) = 2.(-\(\dfrac{1}{3}\))² - 3.(-\(\dfrac{1}{3}\)).1 + 1²

B(-\(\dfrac{1}{3};1\)) = \(\dfrac{2}{9}\) + 1 + 1

B(-\(\dfrac{1}{3}\); 1) = \(\dfrac{20}{9}\) 

B(\(\dfrac{1}{3};-1\)) = 2.(\(\dfrac{1}{3}\))² - 3.(\(\dfrac{1}{3}\)).(-1) + (-1)²

B(\(\dfrac{1}{3}\); -1) = \(\dfrac{2}{9}\) + 1 + 1

B(\(\dfrac{1}{3}\); -1) = \(\dfrac{20}{9}\)

B(\(\dfrac{1}{3}\); 1) = 2.(\(\dfrac{1}{3}\))² - 3.(\(\dfrac{1}{3}\)).1 + (1)²

B(\(\dfrac{1}{3}\); 1) = \(\dfrac{2}{9}\) - 1 + 1

B(\(\dfrac{1}{3}\);1) = \(\dfrac{2}{9}\)

a. x = 2

b. x = -1

c. y = 2

d. x = 1

e. y= -2018

a)\(\left(x-2\right)\left(x-3\right)=0\)

Hoặc \(x-2=0\Leftrightarrow x=2\)(nhận)

Hoặc \(x-3=0\Leftrightarrow x=3\)(nhận)

b)\(\left(x+1\right)\left(x^2+1\right)=0\)

Hoặc \(x+1=0\Leftrightarrow x=-1\)(nhận)

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

c)\(5.y^2-20=0\)

\(\Rightarrow5.y^2=20\)

\(\Rightarrow y^2=4\)

\(\Rightarrow\hept{\begin{cases}y=2\\y=-2\end{cases}}\)

d)\(|x-2|-1=0\)

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

\(\Rightarrow\hept{\begin{cases}x-2=1\\x-2=-1\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}x=3\\x=1\end{cases}}\)

e)\(|y-1|-2019=0\)

\(\Rightarrow|y-1|=2019\)

\(\Rightarrow\hept{\begin{cases}y-1=2019\\y-1=-2019\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}y=2020\\y=-2018\end{cases}}\)

                                                                             HOK TOT                                                                                

f) \(\left(1:\frac{1}{7}\right)^2\left[\left(2^2\right)^3:2^5\right]\cdot\frac{1}{49}\)

\(=\left(1\cdot7\right)^2:\left(2^6:2^5\right)\cdot\frac{1}{49}=7^2\cdot\frac{1}{2}\cdot\frac{1}{49}=49\cdot\frac{1}{49}\cdot\frac{1}{2}=\frac{1}{2}\)

g) \(\frac{4^6\cdot3^5-2^{12}\cdot3^6}{2^{12}\cdot9^3+8^4\cdot3^5}=\frac{\left(2^2\right)^6\cdot3^5-2^{12}\cdot3^6}{2^{12}\cdot\left(3^2\right)^3+\left(2^3\right)^4\cdot3^5}\)

\(=\frac{2^{12}\cdot3^5-2^{12}\cdot3^6}{2^{12}\cdot3^6+2^{12}\cdot3^5}=\frac{2^{12}\left(3^5-3^6\right)}{2^{12}\left(3^6+3^5\right)}=\frac{2^{12}\left[3^5\left(1-3\right)\right]}{2^{12}\left[3^5\left(3+1\right)\right]}=\frac{2^{12}\cdot3^5\cdot\left(-2\right)}{2^{12}\cdot3^5\cdot4}=\frac{-2}{4}=-\frac{1}{2}\)

                                                               Bài giải

\(f,\text{ }\left(1\text{ : }\frac{1}{7}\right)^2\left[\left(2^2\right)^3\text{ : }2^5\right]\cdot\frac{1}{49}\)

\(=7^2\left(2^6\text{ : }2^5\right)\cdot\frac{1}{7^2}\)

\(=2\)

\(g,\text{ }\frac{4^6\cdot3^5-2^{12}\cdot3^6}{2^{12}\cdot9^3+8^4\cdot3^5}=\frac{2^{12}\cdot3^5-2^{12}\cdot3^6}{2^{12}\cdot3^6+2^{12}\cdot3^5}=\frac{2^{12}\cdot3^5\cdot\left(1-3\right)}{2^{12}\cdot3^5\cdot\left(3+1\right)}=-\frac{2}{4}=-\frac{1}{2}\)