b: \(\sqrt{8-2\sqrt{15}}-\sqrt{5}\)

\(=\sqrt{5}-\sqrt{3}-\sqrt{5}\)

\(=-\sqrt{3}\)

c: \(\sqrt{11-6\sqrt{2}}=3-\sqrt{2}\)

d: \(\sqrt{5-2\sqrt{6}}=\sqrt{3}-\sqrt{2}\)

a: \(A=x^2-40x+400+5=\left(x-20\right)^2+5\ge5\forall x\)

Dấu '=' xảy ra khi x=20

b: \(=-\left(x^2+30x+255\right)=-\left(x+15\right)^2-30\le-30\forall x\)

Dấu '=' xảy ra khi x=-15

\(\Delta'=4-\left(m+1\right)\ge0\Rightarrow m\le3\)

Theo hệ thức Viet: \(\left\{{}\begin{matrix}x_1+x_2=4\\x_1x_2=m+1\end{matrix}\right.\)

\(x_1^2+x_2^2=5\left(x_1+x_2\right)\)

\(\Leftrightarrow\left(x_1+x_2\right)^2-2x_1x_2=5\left(x_1+x_2\right)\)

\(\Leftrightarrow16-2\left(m+1\right)=20\)

\(\Leftrightarrow m=-3\) (thỏa mãn)

a. Ta có: \(x^2-4x+m+1=0\)

Thay m=2 ta được: \(x^2-4x+2+1=0\Leftrightarrow x^2-4x+3=0\Leftrightarrow\left[{}\begin{matrix}x=3\\x=1\end{matrix}\right.\)

b. Để phương trình có 2 nghiệm phân biệt thì \(\Delta=\left(-4\right)^2-4.1.\left(m+1\right)>0\)

\(\Leftrightarrow16-4\left(m+1\right)>0\Leftrightarrow16>4\left(m+1\right)\Leftrightarrow4>m+1\Leftrightarrow m< 3\)

Áp dụng định lí Vi-et ta có: \(\left\{{}\begin{matrix}x_1+x_2=4\\x_1x_2=m+1\end{matrix}\right.\)

Theo đề ta có: \(x_1^2+x_2^2=5\left(x_1+x_2\right)\)

\(\Leftrightarrow\left(x_1+x_2\right)^2-2x_1x_2=5\left(x_1+x_2\right)\)

\(\Leftrightarrow\left(4\right)^2-2\left(m+1\right)=5.4\)

\(\Leftrightarrow16-2m-2=20\Leftrightarrow m=-3\) (TM)

1. Yes,it is.

2. They watch TV about 5 hours (35(hours Americans watch TV for a week):7)

3. Because they think that there is a lot of violence on TV today, the programs are terrible and people don't get any exercise because they only sit and watch TV.

4. It brings news from around the world.

5. We learn many useful things.

A C C B C A B A A B 

Are running

Flies

A - C - C - B - C - A - B - A - A - B

are running - flies

\(a,=\dfrac{1}{64}\cdot64=1\\ b,=\left(\dfrac{3}{4}\cdot\dfrac{4}{3}\right)^3+\dfrac{1}{3}=1+\dfrac{1}{3}=\dfrac{4}{3}\\ c,=\left(\dfrac{3}{2}\right)^2=\dfrac{9}{4}\\ d,=\dfrac{1}{2^{2004}}\cdot9^{1002}\\ =\dfrac{9^{1002}}{4^{1002}}=\left(\dfrac{3}{2}\right)^{1002}\)

a. (0,125)² . 64

= \(\dfrac{1}{64}.\dfrac{64}{1}\)

= \(\dfrac{1.1}{1.1}=1\) 

\(\dfrac{x}{6}=\dfrac{7}{4}\Rightarrow x=\dfrac{6\cdot7}{4}=\dfrac{21}{2}\\ \dfrac{3}{x}=\dfrac{21}{17}\Rightarrow x=\dfrac{3\cdot17}{21}=\dfrac{17}{7}\)

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