\(B=\sqrt[3]{170-78\sqrt{3}}+\sqrt[3]{170+78\sqrt{3}}\)

\(\Rightarrow B^3=\left(\sqrt[3]{170-78\sqrt{3}}+\sqrt[3]{170+78\sqrt{3}}\right)^3\)

\(=170-78\sqrt{3}+170+78\sqrt{3}+3\sqrt[3]{170-78\sqrt{3}}.\sqrt[3]{170+78\sqrt{3}}\left(\sqrt[3]{170-78\sqrt{3}}+\sqrt[3]{170+78\sqrt{3}}\right)\)

\(=340+3B\sqrt[3]{\left(170-78\sqrt{3}\right)\left(170+78\sqrt{3}\right)}\)

\(=340+3B\sqrt[3]{28900-18252}\)

\(=340+3B\sqrt[3]{10648}\)

\(=340+3B.22\)

\(=340+66B\)

\(\Rightarrow B^3-66B-340=0\)

\(\Leftrightarrow\left(B-10\right)\left(B^2+10B+34\right)=0\)

\(\Leftrightarrow B-10=0\) (vì \(B^2+10B+34=\left(B^2+10B+25\right)+9=\left(B+5\right)^2+9>0\))

\(\Leftrightarrow B=10\)

Vậy \(B=10\)

-3x(x - 7) + 3(7x - 2) + 3x^2 = 5(2 - x)

<=> -3x^2 + 21x + 21x - 6 + 3x^2 = 10 - 5x

<=> 47x = 16

<=> x = 16/47

Vậy phương trình có tập nghiệm là: S = {16/47}.

30 - 2x = 10

2x = 30 - 10

2x = 20

x = 10

50 - x = 18

x = 50 - 18

x = 32

(x - 10) - 32 = 7

x - 10 = 39

x = 49

Ta có: \(\dfrac{2016}{2019}< \dfrac{2019}{2019}=1\)

Do đó: \(\dfrac{2016}{2019}< \dfrac{2016+1}{2019+1}=\dfrac{2017}{2020}\)

Vậy \(\dfrac{2016}{2019}< \dfrac{2017}{2020}\)

Điều kiện: \(x\ne0\)

\(\sqrt[3]{27x}-\sqrt[3]{125x}+x\sqrt[3]{\dfrac{1}{x^2}}=4\)

\(\Leftrightarrow\sqrt[3]{3^3.x}-\sqrt[3]{5^3.x}+\sqrt[3]{\dfrac{x^3}{x^2}}=4\)

\(\Leftrightarrow3\sqrt[3]{x}-5\sqrt[3]{x}+\sqrt[3]{x}=4\)

\(\Leftrightarrow-\sqrt[3]{x}=4\)

\(\Leftrightarrow\sqrt[3]{x}=-4\)

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

\(\Leftrightarrow x=-64\) (nhận)

Vậy phương trình có tập nghiệm là: \(S=\left\{-64\right\}.\)

\(\dfrac{\sqrt[3]{2}}{\sqrt[3]{2}-1}-\sqrt[3]{4}-\sqrt[3]{2}\)

\(=\dfrac{\sqrt[3]{2}\left(\sqrt[3]{4}+\sqrt[3]{2}+1\right)}{\left(\sqrt[3]{2}-1\right)\left(\sqrt[3]{4}+\sqrt[3]{2}+1\right)}-\sqrt[3]{4}-\sqrt[3]{2}\)

\(=\dfrac{\sqrt[3]{8}+\sqrt[3]{4}+\sqrt[3]{2}}{2-1}-\sqrt[3]{4}-\sqrt[3]{2}\)

\(=2+\sqrt[3]{4}+\sqrt[3]{2}-\sqrt[3]{4}-\sqrt[3]{2}\)

\(=2\)

\(\dfrac{7}{6},\dfrac{8}{5},\dfrac{9}{4},\dfrac{10}{3},\dfrac{11}{2}\)

Gọi b là thương của phép chia a cho 12, ta có: 

\(a=12b+4=2\left(6b+2\right)⋮2\)

Mặt khác, ta có:

\(a=12b+4=\left(12b+3\right)+1=3\left(4b+1\right)+1⋮̸3\)

Do đó: \(a⋮̸6\)

a) \(A=153^2-53^2=\left(153-53\right)\left(153+53\right)=100.206=20600\)

b) \(B=2020^2-2019^2+2018^2-2017^2+...+2^2-1^2\)

\(=\left(2020-2019\right)\left(2020+2019\right)+\left(2018-2017\right)\left(2018+2017\right)+...+\left(2-1\right)\left(2+1\right)\)

\(=\left(2020+2019\right)+\left(2018+2017\right)+...+\left(2+1\right)\)

\(=2020+2019+2018+2017+...+2+1\) (có \(\left(2020-1\right)+1=2020\) (số hạng))

\(=\left(2020+1\right)+\left(2019+2\right)+...+\left(1011+1010\right)\) (có \(2020:2=1010\) (cặp))

\(=2021+2021+...+2021\) (có 1010 (số hạng))

\(=1010.2021\)

\(=2041210\)

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

Thay \(x=101\) vào biểu thức C, ta được:

\(C=\left(101-1\right)^3=100^3=1000000\)

d) \(D=x^3+6x^2+12x+8=\left(x+2\right)^3\)

Thay \(x=8\) vào biểu thức D, ta được:

\(D=\left(8+2\right)^3=10^3=1000\)

e) Vì \(x=9\Rightarrow x+1=10,\) thay vào biểu thức E ta được:

\(E=x^{2013}-\left(x+1\right)x^{2012}+\left(x+1\right)x^{2011}+...-\left(x+1\right)x^2+\left(x+1\right)x-\left(x+1\right)\)

\(=x^{2013}-x^{2013}-x^{2012}+x^{2012}+x^{2011}+...-x^3-x^2+x^2+x-x-1\)

\(=-1\)

46. How far is it from our house to the mountain?

47. How long does the show last?

48. Why is watching too much TV not good?

49. What do both parents and their children enjoy?

50. What time did Mr Cuong do morning exercises yesterday?