Xét tam giác \(ABC\)có \(AB\le AC\le BC\), phân giác \(AD\)ứng với cạnh lớn nhất, đường cao \(CH\)ứng với cạnh nhỏ nhất. 

Kẻ \(DK\perp AB\).

\(\frac{AB}{BD}=\frac{AC}{DC}\Rightarrow BD\le DC\).

Suy ra \(BD\le\frac{1}{2}BC\Rightarrow DK\le\frac{1}{2}CH\).

Vì tam giác \(ABC\)có \(AB\le AC\le BC\)nên \(\widehat{BAC}\ge60^o\Leftrightarrow\widehat{BAD}\ge30^o\).

Suy ra tam giác \(BKD\)có \(DK\ge\frac{1}{2}AD\).

Suy ra \(CH\ge AD\).

Ta có đpcm. 

nko tồn tại

\(P=2\sqrt{x-3}+7\sqrt{4x-12}-3\sqrt{25x-75}+8\sqrt{\frac{x-3}{4}}+3\left(x\ge3\right)\)

\(P=2\sqrt{x-3}+14\sqrt{x-3}-15\sqrt{x-3}+4\sqrt{x-3}+3\)

\(P=5\sqrt{x-3}+3\ge3\)

Dấu = xảy ra

\(\Leftrightarrow x-3=0\)

\(\Leftrightarrow x=3\)

Vậy ...

a=2³⁰+2²⁰²⁰+4ⁿ=4¹⁵+4¹⁰¹⁰+4ⁿ=4¹⁵(1+4⁹⁹⁵+4^n-15) mà 4¹⁵ là số cp suy ra (1+4⁹⁹⁵+4^n-15)là số cp

ta có: 1+4⁹⁹⁵+4^n-15=2^2n-30+2.2¹⁹⁸⁹+1=(2^2n-30+1)²

đề 1+4⁹⁹⁵+4^n-15=(2^n-15)²+2.2¹⁹⁸⁹+1=(2^n-15+1)² là số cp thì n-15=1989 suy ra n=1974

nếu sai thì sorry bạn nha

\(P=\frac{3x+3\sqrt{x}-9}{x+\sqrt{x}-2}+\frac{\sqrt{x}+3}{\sqrt{x}+2}+\frac{\sqrt{x}-2}{1-\sqrt{x}}\)(ĐK: \(x\ge0,x\ne1\)) 

\(=\frac{3x+3\sqrt{x}-9+\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)-\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)

\(=\frac{3x+3\sqrt{x}-9+x+2\sqrt{x}-3-x+4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)

\(=\frac{3x+5\sqrt{x}-8}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)

\(=\frac{\left(3\sqrt{x}+8\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}=\frac{3\sqrt{x}+8}{\sqrt{x}+2}\)

\(P< \frac{15}{4}\)

\(\Leftrightarrow\frac{3\sqrt{x}+8}{\sqrt{x}+2}< \frac{15}{4}\)

\(\Leftrightarrow4\left(3\sqrt{x}+8\right)< 15\left(\sqrt{x}+2\right)\)

\(\Leftrightarrow3\sqrt{x}>2\)

\(\Leftrightarrow x>\frac{4}{9}\)

Vậy \(x>\frac{4}{9},x\ne1\)thì \(P< \frac{15}{4}\).

\(P=\frac{3\sqrt{x}+8}{\sqrt{x}+2}=3+\frac{2}{\sqrt{x}+2}\le3+\frac{2}{0+2}=4\)

Dấu \(=\)khi \(x=0\).

em

chịu 

nha 

anh

\(\frac{1-a\sqrt{a}}{1-\sqrt{a}}\)

\(\frac{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}+a\right)}{1-\sqrt{a}}\)

\(a+\sqrt{a}+1\)

Bổ sung: ĐKXĐ: \(a\ge0;a\ne1\)