Cách 1:đá vào tường cách chôc mk đá 3 m là đc

Cách 2:Đá lên trời cao đủ 3m là đc

Cậu bé nên đá quả bóng lên trời cao 3 mét sau đos quả bóng sẽ rớt xuống lại với cậu bé

\(A=x+2\sqrt{x}+3=\left(\sqrt{x}+1\right)^2+2\ge3\)

Dấu "=" xảy ra \(\Leftrightarrow\sqrt{x}=0\Leftrightarrow x=0\)

\(B=\left|2x+1\right|+3\ge3\)

Dấu "=" xảy ra \(\Leftrightarrow\left|2x+1\right|=0\)

\(\Leftrightarrow x=\frac{-1}{2}\)

\(C=\left(\sqrt{x}+10\right)^2\ge100\)

Dấu "=" xảy ra \(\Leftrightarrow\sqrt{x}=0\Leftrightarrow x=0\)

\(\sqrt{\left(1-\sqrt{2006}\right)^2}\sqrt{2017+2\sqrt{2016}}\)

\(\left|1-\sqrt{2006}\right|\sqrt{\left(\sqrt{2016}+1\right)^2}\)

\(=\left(\sqrt{2006}-1\right)\left(\sqrt{2016}+1\right)\)

\(=2016-1=2015\)

\(\sqrt{\left(1-\sqrt{2016}\right)^2}.\sqrt{2017+2\sqrt{2016}}\)

\(=\left(\sqrt{2016}-1\right)\left(\sqrt{1+2\sqrt{2016}+2016}\right)\)

\(=\left(\sqrt{2016}-1\right)\left(\sqrt{\left(\sqrt{2016}+1\right)^2}\right)\)

\(=\left(\sqrt{2016}-1\right)\left(\sqrt{2016}+1\right)=2016-1=2015\)

Giải từ câu 8 trở đi thôi

\(8,\sqrt{17-12\sqrt{2}}+\sqrt{8}\)

\(\sqrt{3^2-12\sqrt{2}+\left(2\sqrt{2}\right)^2}+\sqrt{8}\)

\(\sqrt{\left(3-2\sqrt{2}\right)^2}+\sqrt{8}\)

\(\left|3-2\sqrt{2}\right|+\sqrt{8}\)

\(=3-\sqrt{8}+\sqrt{8}=3\)

\(9,\sqrt{7+4\sqrt{21-12\sqrt{3}}}\)

\(\sqrt{7+4\sqrt{\left(2\sqrt{3}\right)^2-12\sqrt{3}+3^2}}\)

\(\sqrt{7+4\sqrt{\left(2\sqrt{3}-3\right)^2}}\)

\(\sqrt{7+4.\left|2\sqrt{3}-3\right|}\)

\(\sqrt{7+8\sqrt{3}-12}\)

\(=\sqrt{8\sqrt{3}-5}\)

\(10,\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}\)

\(\sqrt{\sqrt{5}-\sqrt{3-\sqrt{\left(2\sqrt{5}-3\right)^2}}}\)

\(\sqrt{\sqrt{5}-\sqrt{6-2\sqrt{5}}}\)

\(=\sqrt{\sqrt{5}-\sqrt{\left(\sqrt{5}-1\right)^2}}\)

\(=\sqrt{\sqrt{5}-\sqrt{5}+1}=\sqrt{1}=1\)

\(B=\frac{3\sqrt{x}+1}{x+2\sqrt{x}-3}-\frac{2}{\sqrt{x}+3}\)

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

\(=\frac{\sqrt{x}+3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\frac{1}{\sqrt{x}-1}\)

b) \(\frac{A}{B}=\frac{\sqrt{x}+4}{\sqrt{x-1}}:\frac{1}{\sqrt{x}-1}=\sqrt{x}+4\)

Để \(\frac{A}{B}\ge\frac{x}{4}+5\)

\(\Leftrightarrow\sqrt{x}+4\ge\frac{x}{4}+5\)

\(\Leftrightarrow4\sqrt{x}+16\ge x+20\)

\(\Leftrightarrow x-4\sqrt{x}+4\le0\)

\(\Leftrightarrow\left(\sqrt{x}-2\right)^2\le0\)

Mà \(\left(\sqrt{x}-2\right)^2\ge0;\forall x\ge0\)

\(\Rightarrow\left(\sqrt{x}-2\right)^2=0\)

\(\Leftrightarrow x=4\)

Vậy ...

bổ sung thêm đề bài là \(x\ge0;x\ne25\) nha

\(a,B=\left(\frac{15-\sqrt{x}}{x-25}+\frac{2}{\sqrt{x}+5}\right):\frac{\sqrt{x}+1}{\sqrt{x}-5}\)

\(B=\left(\frac{15-\sqrt{x}+2\sqrt{x}-10}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}\right).\frac{\sqrt{x}-5}{\sqrt{x}+1}\)

\(B=\frac{5+\sqrt{x}}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}\frac{\sqrt{x}-5}{\sqrt{x}+1}\)

\(B=\frac{1}{\sqrt{x}+1}\)

\(b,P=A.B=\frac{4\left(\sqrt{x}+1\right)}{25-x}.\frac{1}{\sqrt{x}+1}\)

\(P=\frac{4}{25-x}\)

bổ sung điều kiện cho câu b là x nguyên

\(TH1:x>25< =>P< 0\left(KTM\right)\)

\(TH2:x< 25< =>P>0\)mà x nguyên

\(\frac{4}{25-x}\le4\)

dấu "=" xảy ra khi \(x=24\)

\(< =>MAX:P=4\)

\(2\sqrt{\left(-3\right)^6}+3\sqrt{\left(-2\right)^4}\)

\(2\sqrt{\left(-3^3\right)^2}+3\sqrt{\left(-2^2\right)^2}\)

\(2.\left|-27\right|+3.\left|4\right|\)

\(54+12=66\)

\(2,-4\sqrt{\left(-3\right)^6}+\sqrt{\sqrt{\left(-2\right)^8}}\)

\(-4\sqrt{\left(-3^3\right)^2}+\sqrt{\sqrt{\left(-2^2\right)^{2^2}}}\)

\(-4.\left|-27\right|+\left|4\right|\)

\(=-104\)