Có \(\sqrt{3}>\sqrt{1}\)

=> \(1-\sqrt{3}< 0\)

Có \(A=\sqrt{\left(1-\sqrt{3}\right)^2+\left(1-\sqrt{3}\right)^2}\)

\(=\sqrt{2\left(1-\sqrt{3}\right)^2}=\sqrt{2}.\sqrt{\left(1-\sqrt{3}\right)^2}=\sqrt{2}.\left(\sqrt{3}-1\right)=\sqrt{6}-\sqrt{2}\)

\(B=\sqrt{\left(1+\sqrt{3}\right)^2-\left(1-\sqrt{3}\right)^2}\)

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

Áp dụng 2 quy tác đơn giản: \(cosx=sin\left(90^0-x\right)\)

                                           và \(sin^2x+cos^2x=1\)

Xét \(cos^21^0+cos^22^0+...+cos^289^0-45.0,5\)

\(=\left(cos^21^0+sin^21^0\right)+\left(cos^22^0+sin^22^0\right)+...+\left(cos^244^0+sin^244^0\right)+cos^245^0-22,5\)

\(=1+1+...+1+\left(\frac{1}{\sqrt{2}}\right)^2-22,5\)

\(=44+\frac{1}{2}-22,5=22\)

a) Đề có lẽ là:

đk: \(x\ge0\)

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

\(\Leftrightarrow x+2\sqrt{x}+1+x\sqrt{x}+2x-x\sqrt{x}+\sqrt{x}-3=0\)

\(\Leftrightarrow3x+3\sqrt{x}-2=0\)

\(\Leftrightarrow3\left(x+\sqrt{x}+\frac{1}{4}\right)-\frac{11}{4}=0\)

\(\Leftrightarrow\left(\sqrt{x}+\frac{1}{2}\right)^2-\frac{11}{12}=0\)

\(\Leftrightarrow\left(\sqrt{x}+\frac{3+\sqrt{33}}{6}\right)\left(\sqrt{x}+\frac{3-\sqrt{33}}{6}\right)=0\)

Vì \(\sqrt{x}\ge0\left(\forall x\right)\)

=> \(\sqrt{x}=\frac{3-\sqrt{33}}{6}\Rightarrow x=\frac{7-\sqrt{33}}{6}\)

b) đk: \(x\ge1\)

Ta có: \(\sqrt{4\left(x^2-1\right)}-2\sqrt{15}=0\)

\(\Leftrightarrow\sqrt{x^2-1}=\sqrt{15}\)

\(\Leftrightarrow x^2-1=15\)

\(\Leftrightarrow x^2=16\Rightarrow\orbr{\begin{cases}x=4\\x=-4\end{cases}}\)

a) Ta có: Áp dụng định lý Pytago:

\(AC^2=AB^2+BC^2=5^2+12^2=169\)

\(\Rightarrow AC=13\left(cm\right)\)

Áp dụng định lý thứ 4 ta có:

\(\frac{1}{BH^2}=\frac{1}{AB^2}+\frac{1}{BC^2}=\frac{1}{5^2}+\frac{1}{12^2}\)

\(\Leftrightarrow BH^2=\frac{3600}{169}\Rightarrow BH=\frac{60}{13}\left(cm\right)\)

Ta có: ΔAHN ~ ΔKDN (g.g)

=> \(\frac{AN}{NH}=\frac{KN}{ND}\Leftrightarrow HN\cdot NK=AN\cdot ND\) (1)

Lại có: ΔAHN ~ ΔADC (g.g)

=> \(\frac{AN}{AH}=\frac{AC}{AD}\Leftrightarrow\frac{AN}{AH}=\frac{HC}{ND}\Rightarrow AN\cdot ND=AH\cdot HC\) (2)

Từ (1) và (2) => \(AH\cdot HC=HN\cdot NK\Leftrightarrow BH^2=HN.NK\)

=> đpcm

Ta có: \(n^3-n=n\left(n^2-1\right)=n\left(n-1\right)\left(n+1\right)=\left(n-1\right)\left(n+1\right)\)

Vì \(n\in Z\Rightarrow\left(n-1\right)n\left(n+1\right)\)là tích của ba số nguyên liên tiếp \(A'⋮3!\)

Hay \(n^3-n⋮6\). Nên \(\left(11^3-11\right)+\left(12^3-12\right)+...+\left(1945^3-1945\right)⋮6\)

\(\Rightarrow\left(11^3+12^3+...+1945^3\right)-\left(11+12+...+1945\right)⋮6\)

Mà

\(11+12+...+1945=\frac{1935\left(1945+11\right)}{2}=\frac{1935.1956}{2}=1935.978=1935.163.6⋮6\)

Do đó, suy ra \(11^3+12^3+...+1945^3⋮6\left(\text{đ}pcm\right)\)

ĐKXĐ: \(-5\le x\le5\)

Đề bài tương đương: \(\sqrt{\left(5-x\right)\left(5+x\right)}+x\sqrt{\left(5-x\right)\left(5+x\right)}-\left(5-x\right)=0\)

\(\Leftrightarrow\sqrt{5-x}\left(\sqrt{5+x}+x\sqrt{5+x}-\sqrt{5-x}\right)=0\)

+) \(\sqrt{5-x}=0\Leftrightarrow x=5\)

+) \(\sqrt{5+x}+x\sqrt{5+x}=\sqrt{5-x}\ge0\Rightarrow\sqrt{5+x}\left(1+x\right)\ge0\Leftrightarrow x\ge-1\)

Bình phương 2 vế của phương trình:

\(\Rightarrow\left(5+x\right)+x^2\left(5+x\right)+2x\left(5+x\right)=5-x\)

\(\Leftrightarrow x^3+7x^2+12x=0\Leftrightarrow x\left(x+3\right)\left(x+4\right)=0\)

\(\Leftrightarrow x=0;x=-3;x=-4\)Chỉ  nhận \(x=0\)vì \(x\ge-1\)

\(\Rightarrow S=\left\{0;5\right\}\)

ĐK: x>0, x \(\ne1;4\)

Rút gọn :

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

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

\(A>1\Leftrightarrow\frac{2\left(x+1\right)}{x-1}>1\Leftrightarrow\frac{2\left(x+1\right)}{x-1}-1>0\)

\(\Leftrightarrow\frac{2x+2-x+1}{x-1}>0\)

\(\Leftrightarrow\frac{x+3}{x-1}>0\)(theo đk x>0=>x+3>0)

\(\Rightarrow x-1>0\Rightarrow x>1\)

Kết hợp điều kiện x>0, x khác 1;4

=> x>1, x khác 4 thì P>1