a, \(B=\left(\frac{9-3x}{x^2+4x-5}-\frac{x+5}{1-x}-\frac{x+1}{x+5}\right):\frac{7x-14}{x^2-1}\)

\(=\left(\frac{9-3x}{\left(x-1\right)\left(x+5\right)}+\frac{\left(x+5\right)^2}{\left(x-1\right)\left(x+5\right)}-\frac{\left(x-1\right)\left(x+1\right)}{\left(x-1\right)\left(x+5\right)}\right):\frac{7\left(x-2\right)}{\left(x-1\right)\left(x+1\right)}\)

\(=\frac{9-3x+x^2+10x+25-x^2+1}{\left(x-1\right)\left(x+5\right)}.\frac{\left(x-1\right)\left(x+1\right)}{7\left(x-2\right)}\)

\(=\frac{35+7x}{x+5}\frac{x+1}{7\left(x-2\right)}=\frac{7\left(x+5\right)\left(x+1\right)}{7\left(x+5\right)\left(x-2\right)}=\frac{x+1}{x-2}\)

b, Ta có : \(\left(x+5\right)^2-9x-45=0\)

\(\Leftrightarrow x^2+10x+25-9x-45=0\Leftrightarrow x^2+x-20=0\)

\(\Leftrightarrow\left(x-4\right)\left(x-5\right)=0\Leftrightarrow\orbr{\begin{cases}x=4\\x=5\end{cases}}\)

TH1 : Thay x = 4 vào biểu thức ta được : \(\frac{4+1}{4-2}=\frac{5}{2}\)

TH2 : THay x = 5 vào biểu thức ta được : \(\frac{5+1}{5-2}=\frac{6}{3}=2\)

c, Để B nhận giá trị nguyên khi \(\frac{x+1}{x-2}\inℤ\Rightarrow x-2+3⋮x-2\)

\(\Leftrightarrow3⋮x-2\Rightarrow x-2\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)

d, Ta có : \(B=-\frac{3}{4}\Rightarrow\frac{x+1}{x-2}=-\frac{3}{4}\)ĐK : \(x\ne2\)

\(\Rightarrow4x+4=-3x+6\Leftrightarrow7x=2\Leftrightarrow x=\frac{2}{7}\)( tmđk )

e, Ta có B < 0 hay \(\frac{x+1}{x-2}< 0\)

TH1 : \(\hept{\begin{cases}x+1< 0\\x-2>0\end{cases}\Rightarrow\hept{\begin{cases}x< -1\\x>2\end{cases}}}\)( ktm )

TH2 : \(\hept{\begin{cases}x+1>0\\x-2< 0\end{cases}}\Rightarrow\hept{\begin{cases}x>-1\\x< 2\end{cases}\Rightarrow-1< x< 2}\)

bạn ơi đề bài có sai không :)) sao mình với Tú ra cùng 1 kết quả mà đề bài cho khác vậy :v xem lại đề bài đi bạn

g) \(B=\frac{x^2+x+1}{x-2}=\frac{x^2-2x+3x-6+7}{x-2}=\frac{x\left(x-2\right)+3\left(x-2\right)+7}{x-2}=x+3+\frac{7}{x-2}\)

\(=\left[\left(x-2\right)+\frac{7}{x-2}\right]+5\)

Vì x > 2, áp dụng bất đẳng thức AM-GM ta có :

\(\left(x-2\right)+\frac{7}{x-2}\ge2\sqrt{\left(x-2\right)\cdot\frac{7}{x-2}}=2\sqrt{7}\)

=> \(\left[\left(x-2\right)+\frac{7}{x-2}\right]+5\ge2\sqrt{7}+5\)

Đẳng thức xảy ra <=> ( x - 2 ) = 7/(x-2) [ bạn tự giải nốt ]

Vậy ...

Ta có:

\(2P=\frac{2x^2}{y^2}+\frac{2y^2}{x^2}-6\left(\frac{x}{y}+\frac{y}{x}\right)+10\)

\(=\left(\frac{x^2}{y^2}+2+\frac{y^2}{x^2}\right)-4\left(\frac{x}{y}+\frac{y}{x}\right)+4+\left(\frac{x^2}{y^2}-2\frac{x}{y}+1\right)+\left(\frac{y^2}{x^2}-2\frac{y}{x}+1\right)+2\)

\(=\left(\frac{x}{y}+\frac{y}{x}-2\right)^2+\left(\frac{x}{y}-1\right)^2+\left(\frac{y}{x}-1\right)^2+2\)

\(\ge2\)

\(\Rightarrow P\ge1\)

Dấu = xảy ra khi x = y

gì mà  đánh võng kính thế

1/a/
\(A=\frac{2}{xy}+\frac{3}{x^2+y^2}=\left(\frac{1}{xy}+\frac{1}{xy}+\frac{4}{x^2+y^2}\right)-\frac{1}{x^2+y^2}\)

\(\ge\frac{\left(1+1+2\right)^2}{\left(x+y\right)^2}-\frac{1}{\frac{\left(x+y\right)^2}{2}}=16-2=14\)

Dấu = xảy ra khi \(x=y=\frac{1}{2}\)

b/

\(4B=\frac{4}{x^2+y^2}+\frac{8}{xy}+16xy=\left(\frac{4}{x^2+y^2}+\frac{1}{xy}+\frac{1}{xy}\right)+\left(\frac{1}{xy}+16xy\right)+\frac{5}{xy}\)

\(\ge\frac{\left(1+1+2\right)^2}{\left(x+y\right)^2}+2\sqrt{\frac{1}{xy}.16xy}+\frac{5}{\frac{\left(x+y\right)^2}{4}}\)

\(=16+8+20=44\)

\(\Rightarrow B\ge11\)

Dấu = xảy ra khi \(x=y=\frac{1}{2}\)

2/

\(A=\left(x+\frac{1}{x}\right)^2+\left(y+\frac{1}{y}\right)^2\)

\(\ge\frac{\left(x+\frac{1}{x}+y+\frac{1}{y}\right)^2}{2}\ge\frac{\left(1+\frac{4}{x+y}\right)^2}{2}=\frac{25}{2}\)

Dấu = xảy ra khi \(x=y=\frac{1}{2}\)

1.(√x -2)^2 ≥ 0 --> x -4√x +4 ≥ 0 --> x+16 ≥ 12 +4√x --> (x+16)/(3+√x) ≥4 
--> Pmin=4 khi x=4

2. Đặt \(\sqrt{x^2-4x+5}=t\ge1\)1

=> M=2x²-8x+\(\sqrt{x^2-4x+5}\)+6=2(t²-5)+t+6

<=> M=2t²+t-4\(\ge\)2.1²+1-4=-1

M_min=-1 khi t=1 hay x=2