\(A=a+b=12-2ab\ge12-2\frac{\left(a+b\right)^2}{4}=12-\frac{A^2}{2}\)

Vậy \(A^2+2A-24\le0\)

\(-6\le A\le4\)

Vậy \(A_{min}=-6\)

\(=\left(\frac{\sqrt{b}}{\sqrt{a}\left(\sqrt{a}-\sqrt{b}\right)}-\frac{\sqrt{a}}{\sqrt{b}\left(\sqrt{a}-\sqrt{b}\right)}\right)\left(a\sqrt{b}-b\sqrt{a}\right)\)

\(=\frac{b-a}{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)\)

=b-a

Gọi ban đầu số chỗ ngồi trong phòng được chia thành \(x\)dãy, \(x\inℕ^∗\).

Số ghế trong một dãy là: \(\frac{360}{x}\)(ghế) 

Theo bài ra ta có phương trình: 

\(\left(x-3\right)\left(\frac{360}{x}+4\right)=360\)

\(\Leftrightarrow\left(x-3\right)\left(360+4x\right)=360x\)

\(\Leftrightarrow4x^2-12x-1080=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=18\left(tm\right)\\x=-15\left(l\right)\end{cases}}\)

c,\(\left(\frac{\sqrt{1+a}}{\sqrt{1+a}-\sqrt{1-a}}+\frac{1-a}{\sqrt{1-a^2}-1+a}\right)\left(\sqrt{\frac{1}{a^2}-1}-\frac{1}{a}\right)\)

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

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

M chỉ làm tiếp thôi nha, ko chép lại đề với đk đâu

a,

\(=\frac{a+2\sqrt{ab}+b-4\sqrt{ab}}{\sqrt{a}-\sqrt{b}}-\)\(\frac{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{ab}}\)

\(=\frac{a-2\sqrt{ab}+b}{\sqrt{a}-\sqrt{b}}-\left(\sqrt{a}-\sqrt{b}\right)\)

\(=\frac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\sqrt{a}-\sqrt{b}}-\sqrt{a}+\sqrt{b}\)

\(=\sqrt{a}-\sqrt{b}-\sqrt{a}+\sqrt{b}\)

\(=0\)

b,

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

\(=\left(a-b\right)^2\left(\frac{a+b}{a-b}-1\right)\)

\(=\left(a-b\right)^2\cdot\frac{a+b-a+b}{a-b}\)

\(=\left(a-b\right)2b=2ab-2b^2\)

\(P=\frac{x+7}{\sqrt{x}+3}=\frac{x-9+16}{\sqrt{x}+3}=\frac{\left(\sqrt{x}+3\right)\left(\sqrt{x-3}\right)+16}{\sqrt{x}+3}\)\(=\sqrt{x}-3+\frac{16}{\sqrt{x}+3}\)

\(P+6=\sqrt{x}+3+\frac{16}{\sqrt{x}+3}\)

Theo Cô si ta có : \(\sqrt{x}+3+\frac{16}{\sqrt{x}+6}\ge2\sqrt{\sqrt{x}+3\times\frac{16}{\sqrt{x}+3}}\)=\(2\sqrt{16}=8\)

Vậy \(P+6\ge8\)\(=>P\ge2\)

Dấu bằng xảy ra \(< =>\left(\sqrt{x}+3\right)^2=16\)

\(x+6\sqrt{x}+9-16=0\)

\(x+6\sqrt{x}-7=0\)

\(\left(\sqrt{x}-1\right)\left(\sqrt{x}+7\right)=0\)

\(\orbr{\begin{cases}\sqrt{x}=1\left(tm\right)\\\sqrt{x}=-7\left(l\right)\end{cases}}\)

Vậy min P =2 \(< =>x=1\)

\(M=x-2\sqrt{x-3}+1=x-3-2\sqrt{x-3}+1+3=\left(\sqrt{x-3}-1\right)^2+3\ge3\)

Dấu \(=\)khi \(\sqrt{x-3}-1=0\Leftrightarrow x=4\). 

Để pt có nghiệm khi duy nhất khi \(\frac{1}{2}\ne-\frac{2}{1}\)* luôn đúng *

Ta có : \(\hept{\begin{cases}x-2y=m+3\\2x+y=2m+1\end{cases}\Leftrightarrow\hept{\begin{cases}2x-4y=2m+6\\2x+y=2m+1\end{cases}}\Leftrightarrow\hept{\begin{cases}-5y=5\\x-2y=m+3\end{cases}\Leftrightarrow}\hept{\begin{cases}y=-1\\x=m+1\end{cases}}}\)

Thay vào biểu thức trên ta có : \(3x+2y>3\Rightarrow3\left(m+1\right)-2>3\)

\(\Leftrightarrow3m+3-2>3\Leftrightarrow3m>2\Leftrightarrow m>\frac{2}{3}\)