K = (2x + 3)^2 + 3(4x - 1) + x - 2

= (4x^2 + 12x + 9) + (12x - 3) + x - 2

= 4x^2 + 12x + 9 + 12x - 3 + x - 2

= 4x^2 + 25x + 4

Ta có:

\(\left(7m-2\right)^2-\left(2m-7\right)^2\)

\(=\left(7m-2-2m+7\right)\left(7m-2+2m-7\right)\)

\(=\left(5m+5\right)\left(9m-9\right)\)

\(=5\left(m+1\right).9\left(m-1\right)\)

\(=45\left(m+1\right)\left(m-1\right)⋮45\)

Vậy \(\left(7m-2\right)^2-\left(2m-7\right)^2⋮45\)

Ta có: \(tanx.cotx=1\Rightarrow tanx=\dfrac{1}{cotx}\)

Khi đó: \(16cotx+\dfrac{1}{cotx}=8\)

\(\Rightarrow16cot^2x+1=8cotx\)

\(\Leftrightarrow16cot^2x-8cotx+1=0\)

\(\Leftrightarrow\left(4cotx-1\right)^2=0\)

\(\Leftrightarrow4cotx-1=0\)

\(\Leftrightarrow4cotx=1\)

\(\Leftrightarrow cotx=\dfrac{1}{4}\)

Ta có: \(cotx=\dfrac{cosx}{sinx}\Leftrightarrow\dfrac{1}{4}=\dfrac{cosx}{sinx}\Leftrightarrow sinx=4cosx\)

Do đó: \(P=sinx-4cosx=4cosx-4cosx=0\)

Điều kiện: \(\left\{{}\begin{matrix}a\ge0\\a\ne1\\a\ne9\end{matrix}\right.\)

\(B=\left[\dfrac{2\sqrt{a}\left(\sqrt{a}+3\right)}{\sqrt{a}+3}+\dfrac{\left(\sqrt{a}-3\right)^2}{\sqrt{a}-3}\right].\dfrac{2}{\sqrt{a}-1}\)

\(=\left[2\sqrt{a}+\left(\sqrt{a}-3\right)\right].\dfrac{2}{\sqrt{a}-1}\)

\(=\left[2\sqrt{a}+\sqrt{a}-3\right].\dfrac{2}{\sqrt{a}-1}\)

\(=\left[3\sqrt{a}-3\right].\dfrac{2}{\sqrt{a}-1}\)

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

\(=6\)

Điều kiện: \(\left\{{}\begin{matrix}x>0\\x\ne9\end{matrix}\right.\)

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

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

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

\(=\dfrac{-3\left(3+\sqrt{x}\right)}{\left(\sqrt{x}-3\right)\left(3+\sqrt{x}\right)}:\dfrac{2\left(2+\sqrt{x}\right)}{\sqrt{x}\left(\sqrt{x}-3\right)}\)

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

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

\(4b^2c^2-\left(b^2+c^2-a^2\right)^2\)

\(=\left(2bc-b^2-c^2+a^2\right)\left(2bc+b^2+c^2-a^2\right)\)

\(=\left[a^2-\left(b^2-2bc+c^2\right)\right]\left[\left(b^2+2bc+c^2\right)-a^2\right]\)

\(=\left[a^2-\left(b-c\right)^2\right]\left[\left(b+c\right)^2-a^2\right]\)

\(=\left(a-b+c\right)\left(a+b-c\right)\left(b+c-a\right)\left(b+c+a\right)\)

Áp dụng hệ thức \(sin^2\alpha+cos^2\alpha=1,\) ta có:

\(sin^2B+cos^2B=1\Rightarrow cos^2B=1-sin^2B=1-\dfrac{9}{25}=\dfrac{16}{25}\Rightarrow cosB=\dfrac{4}{5}\)

Áp dụng hệ thức \(tan\alpha=\dfrac{sin\alpha}{cos\alpha},\) ta có:

\(tanB=\dfrac{sinB}{cosB}=\dfrac{\dfrac{3}{5}}{\dfrac{4}{5}}=\dfrac{3}{4}\)

Vì ∆ABC vuông tại A => Hai góc B và C phụ nhau

\(\Rightarrow cotC=tanB=\dfrac{3}{4}\)

I. Pick out the word whose underlined part is pronounced differently from the others.

1.         A. hobby             B. honest            C. humor             D. hole

2.         A.  flood             B. typhoon          C. groom             D. balloon

3.         A.  fly                  B. hobby             C. energy             D. ordinary

4.         A. pleasure         B. leisure            C. sure                 D. music

5.         A. exchange       B. champagne    C. teacher           D. children

\(2MOH\rightarrow M_2O+H_2O\)

\(\Rightarrow n=1\)

Để \(6⋮\left(n+2\right)\Rightarrow\left(n+2\right)\inƯ\left(6\right)\)

Mà \(Ư\left(6\right)=\left\{1;2;3;6\right\}.\)

Vì n thuộc N \(\Rightarrow n+2\ge2\)

\(\Rightarrow\left[{}\begin{matrix}n+2=2\\n+2=3\\n+2=6\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}n=0\\n=1\\n=4\end{matrix}\right.\)

Vì \(\left(n+4\right)⋮2,\) \(4⋮2\Rightarrow n⋮2\)

\(\Rightarrow n\in\left\{0;4\right\}.\)

Vậy \(n\in\left\{0;4\right\}.\)