\(P=\frac{\sqrt{x}-1}{\sqrt{x}+9}=\frac{\sqrt{x}+9-10}{\sqrt{x}+9}=1-\frac{10}{\sqrt{x}+9}\)

Để \(\sqrt{p}< \frac{1}{3}\)thì\(P< \frac{1}{9}\)hay\(1-\frac{10}{\sqrt{x}+9}< \frac{1}{9}\Leftrightarrow\frac{8}{9}< \frac{10}{\sqrt{x}+9}\Leftrightarrow\frac{10}{11,25}< \frac{10}{\sqrt{x}+9}\Leftrightarrow\sqrt{x}+9>11,25\)

\(\Leftrightarrow\sqrt{x}>2,25\Leftrightarrow x>\frac{81}{16}\)

\(A=\frac{\left(x-9\right)+25}{\sqrt{x}+3}=\frac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)+25}{\sqrt{x}+3}=\sqrt{x}-3+\frac{25}{\sqrt{x}+3}\)\(=\left(\sqrt{x}+3\right)+\frac{25}{\sqrt{x}+3}-6\ge2\sqrt{\left(\sqrt{x}+3\right).\frac{25}{\sqrt{x}+3}}-6=2.5-4=6\)

Dấu'=' xảy ra khi và chỉ khi \(\sqrt{x}+3=\frac{25}{\sqrt{x}+3}\)

\(\Rightarrow\left(\sqrt{x}+3\right)^2=25\Rightarrow\sqrt{x}+3=5\left(do\sqrt{x}+3>0\right)\Rightarrow\sqrt{x}=2\Rightarrow x=4\)

Vậy MinA=4 khi và chỉ khi x=4

+) \(2^{2n}=4^n=4\left(4^{n-1}-1\right)+4\)với \(n\inℕ^∗\)

+) \(4^{n-1}-1\equiv1-1\equiv0\)(mod 3)

\(\Rightarrow4\left(4^{n-1}-1\right)⋮12\)

Vậy \(2^{2n}\)chia 12 dư 4

Ta chứng minh: \(\sqrt{a^2+b^2}+\sqrt{c^2+d^2}\ge\sqrt{\left(a+c\right)^2+\left(b+d\right)^2}\)

\(\Leftrightarrow a^2+b^2+c^2+d^2+2\sqrt{\left(a^2+b^2\right)\left(c^2+d^2\right)}\ge\left(a+c\right)^2+\left(b+d\right)^2\)

\(\Leftrightarrow\sqrt{\left(a^2+b^2\right)\left(c^2+d^2\right)}\ge ac+bd\)

\(\Leftrightarrow\left(a^2+b^2\right)\left(c^2+d^2\right)\ge\left(ac+bd\right)^2\)

\(\Leftrightarrow a^2b^2-2abcd+c^2d^2=\left(ab-cd\right)^2\ge0\)(luôn đúng)

Tương tự cho \(\sqrt{\left(a+c\right)^2+\left(b+d\right)}^2,\sqrt{m^2+n^2}\), chứng minh được:

\(\sqrt{a^2+b^2}+\sqrt{c^2+d^2}+\sqrt{m^2+n^2}\ge\sqrt{\left(a+c+n\right)^2}+\sqrt{\left(b+d+n\right)^2}\)(BDT Minkowski)

đề là dì

ĐKXĐ : \(\hept{\begin{cases}x\ge0\\y\ge0\end{cases}}\)

Ta có :

\(\frac{x\sqrt{x}+y\sqrt{y}}{\sqrt{x}+\sqrt{y}}-\left(\sqrt{x}-\sqrt{y}\right)^2\)

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

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

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

\(=\frac{x\sqrt{x}+y\sqrt{y}-x\sqrt{x}-y\sqrt{y}+y\sqrt{x}+x\sqrt{y}}{\sqrt{x}+\sqrt{y}}\)

\(=\frac{\sqrt{xy}\left(\sqrt{y}+\sqrt{x}\right)}{\sqrt{x}+\sqrt{y}}=\sqrt{xy}\)

a) \(-0.8\sqrt{\left(-0.125\right)^2}=-0.8\left|-0.125\right|=-0.8\times0.125=0,1\)

b) \(\sqrt{\left(-2\right)^6}=\sqrt{2^6}=\sqrt{\left(2^3\right)^2}=\left|8\right|=8\)

Đề có nhầm không bạn ??

ình gửi l;ại câu hỏi bạn giải hộ mình nhé

ĐKXĐ : \(x\ge0\)

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

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

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

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

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