e) Vì \(\left\{{}\begin{matrix}\sqrt{x+5}\ge0\\\left(2x+10\right)^4\ge0\end{matrix}\right.\Rightarrow\sqrt{x+5}+\left(2x+10\right)^4\ge0\)

Dấu bằng xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x+5}=0\\\left(2x+10\right)^4=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+5=0\\2x+10=0\end{matrix}\right.\Leftrightarrow x=-5\)

Vậy \(x=-5\)

\(2006^2-2005^2+2004^2-2003^2+...+2^2-1^2\)

\(=\left(2006-2005\right)\left(2006+2005\right)+\left(2004-2003\right)\left(2004+2003\right)+...+\left(2-1\right)\left(2+1\right)\)

\(=2006+2005+2004+2003+...+2+1\) (có \(\left(2006-1\right)+1=2006\) (số hạng))

\(=\left(2006+1\right)+\left(2005+2\right)+...+\left(1004+1003\right)\) (có \(2006:2=1003\) (cặp))

\(=2007+2007+...+2007\) (có 1003 số hạng)

\(=2007.1003\)

\(=2013021\)

56) We will sell the bike for 20 Euros if Ron repairs it.
57) If you use a pencil, the drawing will be perfect.
58) The children will be happy if he teaches them English.
59) If Ireen visits us, we will go out tonight.
60) They will come again if he plans a second stay.
61) If you check the car, it won't break down in the middle of the desert.

1) If you help your grandma, I will do the shopping.
2) Andrew will water the flowers if he stays at home.
3) If she has 5 pounds more, she will buy herself this T-shirt.
4) If they offer me the job, I will take it.
5) You will have summer holidays from June till August if you live in the 

a) \(4^8.2^{20}=\left(2^2\right)^8.2^{20}=2^{16}.2^{20}=2^{36}\)

b) \(25^{20}.125^4=\left(5^2\right)^{20}.\left(5^3\right)^4=5^{40}.5^{12}=5^{52}\)

c) \(9^{12}.27^5.81^4=\left(3^2\right)^{12}.\left(3^3\right)^5.\left(3^4\right)^4=3^{24}.3^{15}.3^{16}=3^{55}\)

d) \(x^7.x^4.x^3=x^{14}\)

e) \(27^6:9^3=\left(3^3\right)^6:\left(3^2\right)^3=3^{18}:3^6=3^{12}\)

f) \(4^{20}:2^{15}=\left(2^2\right)^{20}:2^{15}=2^{40}:2^{15}=2^{25}\)

g) \(4^{10}:64^3=4^{10}:\left(4^3\right)^3=4^{10}:4^9=4\)

h) \(5^9:25^3=5^9:\left(5^2\right)^3=5^9:5^6=5^3\)

i) \(3^6.4^6=\left(3.4\right)^6=12^6\)

k) \(18^{10}:3^{10}=\left(18:3\right)^{10}=6^{10}\)

m) \(48^n:4^{2n}=48^n:\left(4^2\right)^n=48^n:16^n=\left(48:16\right)^n=3^n\)

n) \(64^4.16^5:4^{20}=\left(4^3\right)^4.\left(4^2\right)^5:4^{20}=4^{12}.4^{10}:4^{20}=4^2\)

p) \(\left(ab\right)^6:b^6=a^6.b^6:b^6=a^6\)

q) \(27^3.9^4.243=\left(3^3\right)^3.\left(3^2\right)^4.3^5=3^9.3^8.3^5=3^{22}\)

o) \(8^2.32^4.\left(2^4\right)^2=\left(2^3\right)^2.\left(2^5\right)^4.\left(2^4\right)^2=2^6.2^{20}.2^8=2^{34}\)

\(x^2+4y^2+z^2+14\ge2x+12y+4z\)

\(\left(x^2-2x+1\right)+\left(4y^2-12y+9\right)+\left(z^2-4z+4\right)\ge0\)

\(\left(x-1\right)^2+\left(2y-3\right)^2+\left(z-2\right)^2\ge0\) (luôn đúng)

Dấu bằng xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}\left(x-1\right)^2=0\\\left(2y-3\right)^2=0\\\left(z-2\right)^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-1=0\\2y-3=0\\z-2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=\dfrac{3}{2}\\z=2\end{matrix}\right.\)

Ta có: điều phải chứng minh

\(x^2+x=72\)

\(x^2+x-72=0\)

\(\left(x^2-8x\right)+\left(9x-72\right)=0\)

\(x\left(x-8\right)+9\left(x-8\right)=0\)

\(\left(x+9\right)\left(x-8\right)=0\)

\(\Rightarrow x-8=0\) (do \(x+9>0\) vì x thuộc N)

\(x=8\)

\(\dfrac{12}{13}-\dfrac{21}{26}=\dfrac{12\times2}{13\times2}-\dfrac{21}{26}=\dfrac{24}{26}-\dfrac{21}{26}=\dfrac{24-21}{26}=\dfrac{3}{26}\)

a) Vì \(x=9\) thỏa mãn điều kiện xác định, nên thay \(x=9\) vào biểu thức \(A\) ta được:

\(A=\dfrac{9-2\sqrt{9}+1}{\sqrt{9}}=\dfrac{9-2.3+1}{3}=\dfrac{9-6+1}{3}=\dfrac{4}{3}\)

b) Với \(x>0,\) \(x\ne4,\) ta có:

\(B=\dfrac{\sqrt{x}}{\sqrt{x}+2}-\dfrac{1}{\sqrt{x}-2}-\dfrac{5\sqrt{x}+2}{4-x}\)

\(=\dfrac{\sqrt{x}}{\sqrt{x}+2}-\dfrac{1}{\sqrt{x}-2}+\dfrac{5\sqrt{x}+2}{x-4}\)

\(=\dfrac{\sqrt{x}}{\sqrt{x}+2}-\dfrac{1}{\sqrt{x}-2}+\dfrac{5\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)

\(=\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)-\left(\sqrt{x}+2\right)+\left(5\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)

\(=\dfrac{x-2\sqrt{x}-\sqrt{x}-2+5\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)

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

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

\(=\dfrac{\sqrt{x}}{\sqrt{x}-2}\) (điều phải chứng minh)

c) \(P=A.B=\dfrac{x-2\sqrt{x}+1}{\sqrt{x}}.\dfrac{\sqrt{x}}{\sqrt{x}-2}=\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}-2}\)

Để \(P< 0\) thì \(\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}-2}< 0\Rightarrow\left\{{}\begin{matrix}\sqrt{x}-1\ne0\\\sqrt{x}-2< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x}\ne1\\\sqrt{x}< 2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne1\\0\le x< 4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne1\\0< x< 4\end{matrix}\right.\)

Mà \(x\) nguyên nên \(x\in\left\{2;3\right\}.\)

Vậy với \(x\in\left\{2;3\right\}\) thì \(P< 0.\)

(1) 90%

(2) chính trị

(3) kinh tế

(4) ruộng đất

(5) nông dân