a) Đầu tiên bạn tự đi chứng minh hai công thức sau, do quá dài nên bạn có thể lên mạng tham khảo cách chứng minh:

\(\sin2a=2\sin a.\cos a\)

\(cos2a=cos^2a-sin^2a\)

Áp dụng hai công thức trên ta có:

\(sin30^o=2sin15^ocos15^o\Leftrightarrow sin15^ocos15^o=\frac{1}{4}\Leftrightarrow cos15^o=\frac{1}{4sin15^o}\)

\(cos30^o=cos^215^o-sin^215^o\)

\(\Leftrightarrow\frac{\sqrt{3}}{2}=cos^215^o-sin^215^o\)

\(\Leftrightarrow\left(\frac{1}{4sin^215^o}\right)^2-sin^215^o=\frac{\sqrt{3}}{2}\)

\(\Leftrightarrow\frac{1}{16sin^415^o}-sin^215^o=\frac{\sqrt{3}}{2}\)

\(\Leftrightarrow-32sin^415^o-16sin^215^o\sqrt{3}+2=0\)

\(\Leftrightarrow sin^215^o=\frac{2-\sqrt{3}}{4}\left(sin^215^o\ge0\right)\)

\(\Leftrightarrow sin15^o=\sqrt{\frac{2-\sqrt{3}}{4}}=\sqrt{\frac{\left(\sqrt{3}-1\right)^2}{4\sqrt{2}}}=\frac{\sqrt{3}-1}{2\sqrt{2}}=\frac{\sqrt{6}-\sqrt{2}}{4}\left(đpcm\right)\)

a) ta có với n nguyên dương n²+n+1=n²+2n+1-n=(n+1)²-n

như vậy có n²<n²+n+1<n²+2n+1 hay n²<n²+n+1<(n+1)²

mà n² và (n+1)² là 2 số chính phương liên tiếp

=> n²+n+1 không là số chính phương với mọi n nguyên dương (đpcm)

dùng bunhia cho phần mẫu số là ra 

Ta có:

\(x^4+y^4+y^4+16\ge4\sqrt[4]{16x^4y^8}=8xy^2\)

Tương tự:

\(y^4+z^4+z^4+16\ge8yz^2\)

\(z^4+x^4+x^4+16\ge8zx^2\)

Cộng vế với vế ta được: \(3\left(x^4+y^4+z^4\right)+48\ge8xy^2+8yz^2+8zx^2\)

\(\Leftrightarrow24\ge xy^2+yz^2+xz^2\)

Dấu = xảy ra khi x = y = z = 2

:))

\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=3\Leftrightarrow xy+yz+xz=3xyz\)

\(\Rightarrow3xyz=xy+yz+xy\ge3\sqrt[3]{x^2y^2z^2}\)

\(\Rightarrow x^3y^3z^3\ge x^2y^2z^2\Leftrightarrow\left(x^2y^2z^2\right)\left(xyz-1\right)\ge0\)

\(\Leftrightarrow xyz\ge1\left(x^2y^2z^2>0\right)\)

\(\Rightarrow P=x+\frac{y^2}{2}+\frac{z^3}{3}\)

\(=\frac{x}{6}+\frac{x}{6}+\frac{x}{6}+\frac{x}{6}+\frac{x}{6}+\frac{x}{6}+\frac{y^2}{6}+\frac{y^2}{6}+\frac{y^2}{6}+\frac{z^3}{6}+\frac{z^3}{6}\)

\(\ge11\sqrt[11]{\frac{x^6y^6z^6}{6^{11}}}\ge\frac{11}{6}\)

Dấu "=" xảy ra khi \(x=y=z=1\)

a) ta có \(\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}\Rightarrow\frac{a}{\sin A}=\frac{b+c}{\sin B+\sin C}=\frac{2a}{\sin B+\sin C}\)

do đó \(2a\cdot\sin A=a\left(\sin B+\sin C\right)\)

\(\Rightarrow2\sin A=\sin B+\sin C\)

b) ta có \(\frac{2}{h_a}=\frac{2a}{h_a\cdot a}=\frac{2a}{2S_{ABC}}=\frac{a}{S_{ABC}}\left(1\right)\)

\(\frac{1}{h_b}+\frac{1}{h_c}=\frac{b}{h_b\cdot b}+\frac{c}{h_c\cdot c}=\frac{b}{2S_{ABC}}+\frac{c}{2S_{ABC}}=\frac{b+c}{2S_{ABC}}=\frac{2a}{2S_{ABC}}=\frac{a}{S_{ABC}}\left(2\right)\)

từ (1) và (2) \(\Rightarrow\frac{2}{h_a}=\frac{1}{h_b}+\frac{1}{h_c}\)

a) ĐKXĐ: \(\hept{\begin{cases}x-9\ne0\\\sqrt{x}\ge0\\\sqrt{x}\ne0\end{cases}\Leftrightarrow\hept{\begin{cases}x\ne9\\x\ge0\\x\ne0\end{cases}\Leftrightarrow}\hept{\begin{cases}x\ne9\\x>0\end{cases}}}\)

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

\(\Leftrightarrow A=\left(\frac{x+3}{x-9}+\frac{\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\right).\frac{\sqrt{x}-3}{\sqrt{x}}\)

\(\Leftrightarrow A=\left(\frac{x+3}{x-9}+\frac{\sqrt{x}-3}{x-9}\right).\frac{\sqrt{x}-3}{\sqrt{x}}\)

\(\Leftrightarrow A=\frac{x+\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}.\frac{\sqrt{x}-3}{\sqrt{x}}\)

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

b) \(x=\sqrt{6+4\sqrt{2}}-\sqrt{3+2\sqrt{2}}\)

\(\Leftrightarrow x=\sqrt{4+4\sqrt{2}+2}-\sqrt{2+2\sqrt{2}+1}\)

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

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

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

Vậy với x= 1 thì giá trị của biểu thức \(A=\frac{\left(1+1\right)\left(1-3\right)}{1-9}=\frac{2.\left(-2\right)}{-8}=\frac{-4}{-8}=\frac{1}{2}\)

c)

Ta có :

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

+)  \(\frac{1}{A}\)nguyên 

\(\Leftrightarrow1+\frac{2}{\sqrt{x}+1}\)nguyên

\(\Leftrightarrow\sqrt{x}+1\inƯ\left(2\right)\)

\(\Leftrightarrow x=1\)

Vậy ..............

ĐKXĐ: x \(\ge\)0; x \(\ne\)1

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

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

P = \(\frac{2\sqrt{x}+4-5}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\cdot\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}{\sqrt{x}+1}\)

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

b) P = \(\frac{1}{\sqrt{x}}\) <=> \(\frac{2\sqrt{x}+1}{\sqrt{x}+1}=\frac{1}{\sqrt{x}}\)

=> \(\sqrt{x}\left(2\sqrt{x}+1\right)-\sqrt{x}-1=0\)

<=> \(2x+\sqrt{x}-\sqrt{x}-1=0\)

<=> \(x=\frac{1}{2}\)(tm)

c)Với đk: x \(\ge\)0 và x \(\ne\)1

 \(x-2\sqrt{x-1}=0\) (đk: \(x\ge1\))

<=> \(x-1-2\sqrt{x-1}+1=0\)

<=> \(\left(\sqrt{x-1}-1\right)^2=0\)

<=> \(\sqrt{x-1}-1=0\)

<=> \(\sqrt{x-1}=1\)

<=> \(\left(\sqrt{x-1}\right)^2=1\)

<=> \(\left|x-1\right|=1\)

<=> \(\orbr{\begin{cases}x=0\left(ktm\right)\\x=2\left(tm\right)\end{cases}}\)

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

a) P = \(\frac{2\sqrt{x}-1}{\sqrt{x}+1}\)(sửa lại)

b)  \(\frac{2\sqrt{x}-1}{\sqrt{x}+1}=\frac{1}{\sqrt{x}}\) => \(2x-\sqrt{x}-\sqrt{x}-1=0\)

<=> \(2x-2\sqrt{x}-1=0\)<=> \(2\left(x-\sqrt{x}+\frac{1}{4}\right)-\frac{3}{4}=0\)

<=>  \(2\left(\sqrt{x}-\frac{1}{2}\right)^2=\frac{3}{4}\) <=> \(\left(\sqrt{x}-\frac{1}{2}\right)^2=\frac{3}{8}\)....(tiếp tự lm)