a)  \(0,75+\left(\dfrac{-1}{3}\right)-\dfrac{5}{18}=\dfrac{3}{4}+\left(\dfrac{-1}{3}\right)-\dfrac{5}{18}=\dfrac{5}{12}-\dfrac{5}{18}=\dfrac{5}{36}\)

c) \(\dfrac{4}{15}\cdot\dfrac{1}{3}\cdot\dfrac{15}{20}=\dfrac{4}{15}\cdot\dfrac{1}{3}\cdot\dfrac{3}{4}=\dfrac{5}{45}\cdot\dfrac{3}{4}=\dfrac{15}{180}=\dfrac{1}{12}\)

d) \(\left(\dfrac{-1}{9}\right)\cdot\left(\dfrac{15}{22}\right):\left(\dfrac{-25}{9}\right)=\dfrac{-5}{66}:\left(\dfrac{-25}{9}\right)=\dfrac{-5}{66}\cdot\left(\dfrac{9}{-25}\right)=\dfrac{-3}{-110}=\dfrac{3}{110}\)

a) \(0,75\) + \(\dfrac{-1}{3}\) - \(\dfrac{5}{18}\)= \(\dfrac{5}{12}\) - \(\dfrac{5}{18}\) = \(\dfrac{5}{36}\)

b) \(\dfrac{4}{15}\)x \(\dfrac{1}{3}\)x \(\dfrac{15}{20}\)= 4/45 x 15/20 = 1/15

c) -1/9 x 15/22 : -25/9 = -5/66 : -25/9 = 3/110

\(a.0,75+\left(\dfrac{-1}{3}\right)-\dfrac{5}{18}=\dfrac{75}{100}+\left(\dfrac{-6}{18}\right)-\dfrac{5}{18}=\dfrac{3}{4}-\dfrac{11}{18}=\dfrac{5}{36}\\ b.\dfrac{4}{15}.\dfrac{1}{3}.\dfrac{15}{20}=\dfrac{4}{15}.\dfrac{1}{3}.\dfrac{3}{4}=\dfrac{4.1.3}{15.3.4}=\dfrac{1.1.1}{15.1.1}=\dfrac{1}{15}\\ c.\dfrac{-1}{9}.\dfrac{15}{22}:\left(\dfrac{-25}{9}\right)=\dfrac{-1}{9}.\dfrac{15}{22}.\dfrac{-9}{25}=\dfrac{\left(-1\right).15.\left(-9\right)}{9.22.25}=\dfrac{\left(-1\right).3.\left(-1\right)}{1.22.5}=\dfrac{3}{110}\)

\(x^8+x^7+1\)

\(=x^8+x^7+x^6-x^6+x^5-x^5+x^4-x^4+x^3-x^3+x^2-x^2+x-xx+1\)

\(=\left(x^8-x^6+x^5-x^3+x^2\right)\)

\(+\left(x^7-x^5+x^4-x^2+x\right)\)

\(+\left(x^6-x^4+x^3-x+1\right)\)

\(=\left(x^2+x+1\right)\left(x^6-x^4+x^3-x+1\right)\)

\(x^5+x+1\)

\(=x^5-x^2+x^2+x+1\)

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

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

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

\(x^4+2x^2-24\)

\(=x^4+2x^2+1-25\)

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

\(=\left(x^2+6\right)\left(x^2-4\right)\)

\(=\left(x^2+6\right)\left(x+2\right)\left(x-2\right)\)

2^x+3.16=64

2^x+3=64:16

2^x+3=4

2^x=1

x=0

2^x+3.16=64

2^x+3=64:16

2^x+3=4

2^x=1

x=0 .

a.

\(2x-x^2+7=-\left(x^2-2x+1\right)+8=-\left(x-1\right)^2+8\le8\)

\(\Rightarrow2+\sqrt{2x-x^2+7}\le2+\sqrt{8}=2+2\sqrt{2}\)

\(\Rightarrow\dfrac{3}{2+\sqrt{2x-x^2+7}}\ge\dfrac{3}{2+2\sqrt{2}}=\dfrac{3\sqrt{2}-3}{2}\)

\(A_{min}=\dfrac{3\sqrt{2}-3}{2}\) khi \(x=1\)

b. ĐKXĐ: \(x\le1\)

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

\(B=-\left(1-x-\sqrt{2\left(1-x\right)}+\dfrac{1}{2}\right)+\dfrac{3}{2}\)

\(B=-\left(\sqrt{1-x}-\dfrac{\sqrt{2}}{2}\right)^2+\dfrac{3}{2}\le\dfrac{3}{2}\)

\(B_{max}=\dfrac{3}{2}\) khi\(x=\dfrac{1}{2}\)

\(\dfrac{5}{7}=\dfrac{30}{42};\dfrac{6}{7}=\dfrac{36}{42}. Vay5phansocantimla:\dfrac{31}{42};\dfrac{32}{42};\dfrac{33}{42};\dfrac{34}{42};\dfrac{35}{42}\)

Qui đồng 2 phân số lên ta có: 

5/7 = 30/42 và 6/7 = 36/42 

Vậy phân số nằm giữa là: 31/42 ; 32/42; 33/42; 34/42; 35/42

\(lim\left(\sqrt[3]{n^3+4}-\sqrt[3]{n^3-1}\right)\)

\(=lim\left(\sqrt[3]{1+\dfrac{4}{n^3}}-\sqrt[3]{1-\dfrac{1}{n^3}}\right)=\sqrt[3]{1}-\sqrt[3]{1}=0\)

c: \(=\dfrac{x^3+2x+2x^2+2x+x^2-x+1}{\left(x+1\right)\left(x^2-x+1\right)}\)

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