\(a-b=2\Leftrightarrow a=b+2\)

\(P=3a^2+b^2+8\\ P=3\left(b+2\right)^2+b^2+8\\ P=3b^2+12b+12+b^2+8\\ P=4b^2+12b+20\\ P=\left(4b^2+12b+9\right)+11\\ P=\left(2b+3\right)^2+11\ge11\forall a;b\)

Dấu "=" xảy ra \(\Leftrightarrow b=\dfrac{-3}{2}\)

Pmin = 11

\(\dfrac{2A}{2A+16.5}=\dfrac{43,66}{100}\)

=> \(200A=43,66.\left(2A+16.5\right)\)

=> \(200A-87,32A=3492,8\)

=> \(112,68A=3492,8\)

=> A= 31 

Cái đó là tìm ra A là bn ạ

a, thay x=25 vào A ta có:

\(A=\dfrac{\sqrt{x}}{\sqrt{x}-1}=\dfrac{\sqrt{25}}{\sqrt{25}-1}=\dfrac{5}{5-1}=\dfrac{5}{4}\)

b, \(P=\dfrac{\sqrt{x}}{\sqrt{x}-1}\left(\dfrac{3x+3}{x\sqrt{x}-1}-\dfrac{2}{\sqrt{x}-1}\right)\)

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

\(\Rightarrow P=\dfrac{\sqrt{x}}{\sqrt{x}-1}\left(\dfrac{3x+3}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\dfrac{2x+2\sqrt{x}+2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right)\)

\(\Rightarrow P=\dfrac{\sqrt{x}}{\sqrt{x}-1}.\dfrac{3x+3-2x-2\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)

\(\Rightarrow P=\dfrac{\sqrt{x}\left(x-2\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)^2\left(x+\sqrt{x}+1\right)}\)

\(\Rightarrow P=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)^2\left(x+\sqrt{x}+1\right)}\)

\(\Rightarrow P=\dfrac{\sqrt{x}}{x+\sqrt{x}+1}\)

\(=2\sqrt{3}-10\sqrt{3}-\sqrt{3}+\dfrac{5\sqrt{3}}{2}=\dfrac{5\sqrt{3}}{2}-9\sqrt{3}=\dfrac{5\sqrt{3}-18\sqrt{3}}{2}=\dfrac{-13\sqrt{3}}{2}\)

\(=\dfrac{1}{2}.4\sqrt{3}-2.5\sqrt{3}-\sqrt{3}+5.\dfrac{\sqrt{3}}{2}\)

\(=2\sqrt{3}-10\sqrt{3}-\sqrt{3}+\dfrac{5\sqrt{3}}{2}\)

\(=-9\sqrt{3}+\dfrac{5\sqrt{3}}{2}=\dfrac{-18\sqrt{3}+5\sqrt{3}}{2}=-\dfrac{13\sqrt{3}}{2}\)

Giải hpt:

Đặt: \(\left[{}\begin{matrix}\sqrt{x-1}=a\\y+1=b\end{matrix}\right.\)

Ta có: \(\left\{{}\begin{matrix}3a-2b=-1\\5a-9b=-13\end{matrix}\right.< =>\left\{{}\begin{matrix}15a-10b=-5\\15a-27b=-39\end{matrix}\right.< =>\left\{{}\begin{matrix}b=2\\15a-27\cdot2=-39\end{matrix}\right.< =>\left\{{}\begin{matrix}b=2\\a=1\end{matrix}\right.\)

Thay: \(\left[{}\begin{matrix}\sqrt{x-1}=1\\y+1=2\end{matrix}\right.< =>\left[{}\begin{matrix}x=2\\y=1\end{matrix}\right.\)

10.

\(H\left(x\right)=-5x^4+10x^3-15x+1\)

\(=-5x\left(x^3-2x^2+3\right)+1\)

\(=-5x.0+1\)

\(=1\)

9.

\(P\left(x\right)-Q\left(x\right)=\left(1-a\right)x^3+x^2+x-6\)

\(P\left(x\right)-Q\left(x\right)\) là đa thức bậc 3 khi và chỉ khi \(1-a\ne0\)

\(\Rightarrow a\ne1\)

\(a,ĐK:x+y\ne0;x\ne y\\ HPT\Leftrightarrow\left\{{}\begin{matrix}\dfrac{2}{x+y}+\dfrac{4}{x-y}=\dfrac{14}{3}\left(1\right)\\\dfrac{3}{x+y}+\dfrac{4}{x-y}=5\left(2\right)\end{matrix}\right.\\ \left(2\right)-\left(1\right)=\dfrac{1}{x+y}=\dfrac{1}{3}\\ \Leftrightarrow x+y=3\\ \Leftrightarrow x=3-y\\ \text{Thay vào }\left(1\right)\Leftrightarrow\dfrac{2}{3}+\dfrac{4}{3-2y}=\dfrac{14}{3}\\ \Leftrightarrow\dfrac{4}{3-2y}=4\\ \Leftrightarrow3-2y=1\\ \Leftrightarrow y=1\Leftrightarrow x=2\)

Vậy hệ có nghiệm \(\left(x;y\right)=\left(2;1\right)\)

\(b,ĐK:y\ne-\dfrac{1}{2};x-2y\ne0\\ HPT\Leftrightarrow\left\{{}\begin{matrix}\dfrac{6}{x-2y}+\dfrac{y}{1+2y}=3\left(1\right)\\\dfrac{6}{x-2y}-\dfrac{8}{1+2y}=-2\left(2\right)\end{matrix}\right.\\ \left(1\right)-\left(2\right)=\dfrac{y+8}{2y+1}=5\\ \Leftrightarrow y+8=10y+5\Leftrightarrow y=\dfrac{1}{3}\\ \text{Thay vào }\left(1\right)\Leftrightarrow\dfrac{6}{x-\dfrac{2}{3}}+\dfrac{\dfrac{1}{3}}{\dfrac{5}{3}}=3\\ \Leftrightarrow\dfrac{6}{x-\dfrac{2}{3}}=\dfrac{14}{5}\\ \Leftrightarrow x-\dfrac{2}{3}=\dfrac{15}{7}\Leftrightarrow x=\dfrac{59}{21}\)

Vậy hệ có nghiệm \(\left(x;y\right)=\left(\dfrac{59}{21};\dfrac{1}{3}\right)\)

\(c,HPT\Leftrightarrow\left\{{}\begin{matrix}xy-5x+3y-15=xy\\2xy+30x-9y-135=2xy\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}-5x+3y=15\\10x-3y=45\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=12\\y=25\end{matrix}\right.\)