Giải hệ phương trình:

\(\left\{{}\begin{matrix}\left(\sqrt{5}\right)^{2x}-\left(\sqrt{5}\right)^{3y}=\left(3y-2x\right)\left(6xy+12\right)\\4x^2+9y^2=16\end{matrix}\right.\)

_{Chủ đề là gợi ý cho bài này, nhưng mình linh cảm có thể giải một cách đơn giản :v}

Xét hàm \(f\left(x\right)=\dfrac{x+m}{x+1}\) có \(f'\left(x\right)=\dfrac{\left(x+m\right)'\left(x+1\right)-\left(x+m\right)\left(x+1\right)'}{\left(x+1\right)^2}=\dfrac{1-m}{\left(x-1\right)^2}\)

Cho \(f'\left(x\right)=\dfrac{1-m}{\left(x-1\right)^2}=0\Leftrightarrow m=1\)

Khi đó \(f\left(x\right)=\dfrac{x+1}{x+1}=1\)

\(\Rightarrow max_{\left[0;1\right]}\left|f\left(x\right)\right|+min_{\left[0;1\right]}\left|f\left(x\right)\right|=1+1=2\) ( thỏa mãn )

Vậy \(m=1\) thỏa mãn bài toán.

Xét \(m\ne1\), ta thấy \(f\left(x\right)\) đơn điệu trên \(\left[0;1\right]\), xét các trường hợp:

*) \(f\left(0\right).f\left(1\right)\le0\Leftrightarrow\dfrac{m+1}{2}\cdot m\le0\) \(\Leftrightarrow-1\le m\le0\)

\(\Rightarrow\left\{{}\begin{matrix}min_{\left[0;1\right]}\left|f\left(x\right)\right|=0\\max_{\left[0;1\right]}\left|f\left(x\right)\right|=max\left\{\dfrac{\left|m+1\right|}{2};\left|m\right|\right\}\end{matrix}\right.\)

Khi đó: \(max_{\left[0;1\right]}\left|f\left(x\right)\right|+min_{\left[0;1\right]}\left|f\left(x\right)\right|=2\)

\(\Leftrightarrow0+\dfrac{\left|\dfrac{m+1}{2}+m\right|+\left|\dfrac{m+1}{2}-m\right|}{2}=2\)

\(\Leftrightarrow\left|\dfrac{3m+1}{2}\right|+\left|\dfrac{-m+1}{2}\right|=4\)

\(\Leftrightarrow\left|3m+1\right|+\left|m-1\right|=8\) (1)

Xét các trường hợp:

+) \(m\le\dfrac{-1}{3}\) : \(\left(1\right)\Leftrightarrow-3m-1-m+1=8\Leftrightarrow m=-2\) ( loại )

+) \(m\ge1\) : \(\left(1\right)\Leftrightarrow3m+1+m-1=8\Leftrightarrow m=2\) ( loại )

+) \(-\dfrac{1}{3}< m< 1\) : \(\left(1\right)\Leftrightarrow3m+1-m+1=8\Leftrightarrow m=3\) ( loại )

*) \(f\left(0\right)\cdot f\left(1\right)>0\Leftrightarrow\dfrac{m+1}{2}\cdot m>0\Leftrightarrow\left[{}\begin{matrix}m>0\\m< -1\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}min_{\left[0;1\right]}\left|f\left(x\right)\right|=min\left\{\dfrac{\left|m+1\right|}{2};\left|m\right|\right\}\\max_{\left[0;1\right]}\left|f\left(x\right)\right|=max\left\{\dfrac{\left|m+1\right|}{2};\left|m\right|\right\}\end{matrix}\right.\)

Khi đó: \(min_{\left[0;1\right]}\left|f\left(x\right)\right|+max_{\left[0;1\right]}\left|f\left(x\right)\right|=2\)

\(\Leftrightarrow\dfrac{\left|\left|\dfrac{m+1}{2}+m\right|-\left|\dfrac{m+1}{2}-m\right|\right|}{2}+\dfrac{\left|\left|\dfrac{m+1}{2}+m\right|\right|+\left|\left|\dfrac{m+1}{2}-m\right|\right|}{2}=2\)

\(\Leftrightarrow\dfrac{\left|\left|3m+1\right|-\left|m-1\right|\right|}{4}+\dfrac{\left|\left|3m+1\right|+\left|m-1\right|\right|}{4}=2\)

\(\Leftrightarrow\dfrac{2\left|3m+1\right|}{4}=2\)

\(\Leftrightarrow\left|3m+1\right|=4\)

\(\Leftrightarrow\left[{}\begin{matrix}m=1\\m=\dfrac{-5}{3}\end{matrix}\right.\)

Tóm lại ở cả 2 trường hợp thì ta có \(m\in\left\{1;\dfrac{-5}{3}\right\}\) thỏa mãn đề bài.

Vậy \(S=\left\{1;\dfrac{-5}{3}\right\}\) có \(2\) phần tử.

\(\lim\limits_{x\rightarrow-\infty}\dfrac{\sqrt{4x^2+2x-1}-x}{3x-2}=\lim\limits_{x\rightarrow-\infty}\dfrac{\sqrt{\dfrac{4x^2+2x-1}{x^2}}-\dfrac{x}{x}}{\dfrac{3x-2}{x}}\)

\(=\lim\limits_{x\rightarrow-\infty}\dfrac{\sqrt{4+\dfrac{2}{x}-\dfrac{1}{x^2}}-1}{3-\dfrac{2}{x}}=-\dfrac{4-1}{3}=-1\)

a) \(lim_{x\rightarrow3}\dfrac{9-x^2}{\sqrt{x+6}-3}=lim_{x\rightarrow3}\dfrac{\left(\sqrt{x+6}+3\right)\left(3-x\right)\left(3+x\right)}{\left(\sqrt{x+6}-3\right)\left(\sqrt{x+6}+3\right)}\)

\(=lim_{x\rightarrow3}\dfrac{\left(\sqrt{x+6}+3\right)\left(x-3\right)\left(-x-3\right)}{x+6-9}\)

\(=lim_{x\rightarrow3}\left(\sqrt{x+6}+3\right)\left(-x-3\right)\)

\(=\left(\sqrt{3+6}+3\right)\left(-3-3\right)=-36\)

b) \(lim_{x\rightarrow-\infty}\left(\sqrt{x^2+x}+x\right)=lim_{x\rightarrow-\infty}\dfrac{\left(\sqrt{x^2+x}+x\right)\left(\sqrt{x^2+x}-x\right)}{\sqrt{x^2+x}-x}\)

\(=lim_{x\rightarrow-\infty}\dfrac{x}{\sqrt{x^2+x}-x}=lim_{x\rightarrow-\infty}\dfrac{1}{\sqrt{1+\dfrac{1}{x}}-1}=\dfrac{-1}{2}\)

\(\left\{{}\begin{matrix}\sqrt{x}+\sqrt{xy}+\sqrt{y}=11+12\sqrt{13}\\x+y=134\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(\sqrt{x}+1\right)\left(\sqrt{y}+1\right)=12+12\sqrt{13}\\x+y=134\end{matrix}\right.\)

Đặt \(\left\{{}\begin{matrix}\sqrt{x}+1=a\\\sqrt{y}+1=b\end{matrix}\right.\) \(\left(a,b>0\right)\)

\(Hpt\Leftrightarrow\left\{{}\begin{matrix}ab=12+12\sqrt{13}\\a^2+b^2-2\left(a+b\right)+2=134\end{matrix}\right.\)

\(\Leftrightarrow a^2+b^2+2ab-2\left(a+b\right)+1=134+12+12\sqrt{13}-1\)

\(\Leftrightarrow\left(a+b\right)^2-2\left(a+b\right)+1=145+12\sqrt{13}\)

\(\Leftrightarrow\left(a+b-1\right)^2=145+12\sqrt{13}\)

\(\Leftrightarrow a+b=\sqrt{145+12\sqrt{13}}+1\)

\(Hpt\Leftrightarrow\left\{{}\begin{matrix}ab=12+12\sqrt{13}\\a+b=\sqrt{145+12\sqrt{13}}+1\end{matrix}\right.\)

Số xấu quá nên dừng tại đây :D

\(log_m\left(7\right)-log_m\left(5-y\right)=log_m\left(6\right)+log_m\left(4\right)\)

\(\Leftrightarrow log_m\left(\dfrac{7}{5-y}\right)=log_m\left(24\right)\)

\(\Leftrightarrow\dfrac{7}{5-y}=24\)

\(\Leftrightarrow y=\dfrac{113}{24}\)

Ý tưởng chung là chứng minh tử số và mẫu số nguyên tố cùng nhau.

a) Gọi \(\left(n;n+1\right)=d\Leftrightarrow\left\{{}\begin{matrix}n⋮d\\n+1⋮d\end{matrix}\right.\)

\(\Leftrightarrow n+1-n⋮d\)

\(\Leftrightarrow1⋮d\)

\(\Leftrightarrow d=1\) ( đpcm )

b) \(\left\{{}\begin{matrix}12n+1⋮d\\30n+2⋮d\end{matrix}\right.\)\(\Leftrightarrow5\left(12n+1\right)-2\left(30n+2\right)⋮d\)

\(\Leftrightarrow1⋮d\) \(\Leftrightarrow d=1\) ( đpcm )

c) \(\left\{{}\begin{matrix}5n+3⋮d\\3n+2⋮d\end{matrix}\right.\) \(\Leftrightarrow5\left(3n+2\right)-3\left(5n+3\right)⋮d\)

\(\Leftrightarrow1⋮d\) \(\Leftrightarrow d=1\) ( đpcm )

Áp dụng BĐT Cauchy:

\(\sqrt{\dfrac{x}{y+z}}+\sqrt{\dfrac{y}{z+x}}+\sqrt{\dfrac{z}{x+y}}\)

\(=\dfrac{x}{\sqrt{x\left(y+z\right)}}+\dfrac{y}{\sqrt{y\left(z+x\right)}}+\dfrac{z}{\sqrt{z\left(x+y\right)}}\)

\(\ge\dfrac{x}{\dfrac{x+y+z}{2}}+\dfrac{y}{\dfrac{x+y+z}{2}}+\dfrac{z}{\dfrac{x+y+z}{2}}\)

\(=\dfrac{2x}{x+y+z}+\dfrac{2y}{x+y+z}+\dfrac{2z}{x+y+z}\)

\(=\dfrac{2\left(x+y+z\right)}{x+y+z}=2\)

Dấu "=" không xảy ra nên \(\sqrt{\dfrac{x}{y+z}}+\sqrt{\dfrac{y}{z+x}}+\sqrt{\dfrac{z}{x+y}}>2\)