-11/abc 

dạng này dễ mà bạn 

bạn tìm ĐK, đối chiếu giá trị với ĐK thấy thỏa mãn rồi thay vô 

toàn SCP nên tính cũng đơn giản:)

1) Thay x = 64 (TMĐK ) vào A, có :

           A = \(\frac{\sqrt{64}}{\sqrt{64}-2}\)=\(\frac{4}{3}\)

     Vậy A = \(\frac{4}{3}\)khi x = 64

2)  Thay x = 36 ( TMĐK ) vào A, có

        A =\(\frac{\sqrt{36}+4}{\sqrt{36}+2}\)=\(\frac{5}{4}\)

     Vậy A =\(\frac{5}{4}\)khi x = 36

3)   Thay x=9 (TMĐK  ) vào A, có :

         A= \(\frac{\sqrt{9}-5}{\sqrt{9}+5}\)=  \(\frac{-1}{4}\)

     Vậy A=\(\frac{-1}{4}\)khi x = 9

4)   Thay x = 25( TMĐK ) vào A có:

         A =\(\frac{2+\sqrt{25}}{\sqrt{25}}\)=\(\frac{7}{5}\)

      Vậy A=\(\frac{7}{5}\) khi x = 25

P₁ = (\(\frac{1}{\sqrt{x}}+\frac{\sqrt{x}}{\sqrt{x}+1}\)) : \(\frac{\sqrt{x}}{x+\sqrt{x}}\)= \(\frac{\sqrt{x}+1+x}{\sqrt{x}\left(\sqrt{x}+1\right)}\):\(\frac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\)=\(\frac{x+\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}\).

(\(\sqrt{x}+1\)) =\(\frac{x+\sqrt{x}+1}{\sqrt{x}}\)(ĐKXĐ : x > 0 )

P₂ =\(\frac{\sqrt{x}}{\sqrt{x}-1}+\frac{3}{\sqrt{x}+1}-\frac{6\sqrt{x}-4}{x-1}\)=\(\frac{\sqrt{x}\left(\sqrt{x}+1\right)+3\left(\sqrt{x}-1\right)-6\sqrt{x}+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)= \(\frac{x+\sqrt{x}+3\sqrt{x}-3-6\sqrt{x}+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)= \(\frac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)=\(\frac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)=\(\frac{\sqrt{x}-1}{\sqrt{x}+1}\)

(ĐKXĐ: x\(\ge\)0,  x\(\ne\)1)

a, \(P=\frac{a^3-a+2b-\frac{b^2}{a}}{\left(1-\sqrt{\frac{a+b}{a^2}}\right)\left(a+\sqrt{a+b}\right)}:\left[\frac{a^2\left(a+b\right)+a\left(a+b\right)}{\left(a-b\right)\left(a+b\right)}+\frac{b}{a-b}\right]\)

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

\(=\frac{a^4-a^2-2ab-b^2}{a^2-a-b}:\frac{a^2+a+b}{a-b}\)

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

\(=\frac{\left(a^4-a^2-2ab-b^2\right).\left(a-b\right)}{a^4-\left(a+b\right)^2}=\frac{\left[a^4-\left(a+b\right)^2\right].\left(a-b\right)}{a^4-\left(a+b\right)^2}=a-b\)

b, Có \(P=a-b=1\)\(\Rightarrow a=1+b\)

\(a^3-b^3=7\Leftrightarrow\left(a^2+ab+b^2\right)\left(a-b\right)=7\)

\(\Rightarrow a^2+ab+b^2=7\)

\(\Leftrightarrow\left(1+b\right)^2+\left(1+b\right)b+b^2=7\)

\(\Leftrightarrow b^2+2b+1+b^2+b+b^2=7\)

\(\Leftrightarrow3b^2+3b-6=0\)

Bạn tự giải phương trình tìm b => a

Bài 2 :

\(a,y=\left(m+1\right)x-2m-5\) \(\Leftrightarrow\left(m+1\right)x-2m-5-y=0\)

\(\Leftrightarrow mx+x-2m-5-y=0\)\(\Leftrightarrow m\left(x-2\right)+x-y-5=0\)

Có y luôn qua điểm A cố định với A( x₀ ; y₀ ) \(\orbr{\begin{cases}x_0-2=0\\x_0-y_0-5=0\end{cases}}\Rightarrow\orbr{\begin{cases}x_0=2\\y_0=-3\end{cases}}\)

=> A( 2;-3)

Gọi H là chân đường vuông góc hạ từ O xuống d => \(OH\le OA\)

\(OH_{max}=OA\)khi \(H\equiv A\)\(\left(d\perp OA\right)\)

=> đường thẳng OA qua O( 0;0 ) và A( 2;-3 ) => \(y=-\frac{3}{2}x\)

\(\Rightarrow d\perp OA\)=> hệ số góc \(m.\) \(-\frac{3}{2}=-1\Rightarrow m=\frac{2}{3}\)

b, \(y=0\Rightarrow\left(m+1\right)x-2m-5=0\)\(\Rightarrow x=\frac{2m+5}{m+1}\)\(\Rightarrow A\left(\frac{2m+5}{m+1};0\right)\)

\(x=0\Rightarrow y=-2m-5\Rightarrow B\left(0;-2m-5\right)\)

\(\Rightarrow OA=\sqrt{\frac{2m+5}{m+1}};OB=\sqrt{-2m-5}\)

\(\Rightarrow\frac{1}{2}.OA.OB=\frac{3}{2}\Rightarrow OA.OB=3\)

\(\Rightarrow\left(OA.OB\right)^2=9\Rightarrow\frac{\left(2m+5\right)^2}{m+1}=9\)

\(\Rightarrow4m^2+20m+25-9m-9=\)

\(\Rightarrow4m^2+11m+16=0\)

a, \(P=\frac{\sqrt{x}}{\sqrt{x}-1}+\frac{3}{\sqrt{x}+1}-\frac{6\sqrt{x}-4}{x-1}\)ĐK : \(x\ge0;x\ne1\)

\(=\frac{x+\sqrt{x}+3\sqrt{x}-3-6\sqrt{x}+4}{x-1}=\frac{x-2\sqrt{x}+1}{x-1}=\frac{\sqrt{x}-1}{\sqrt{x}+1}\)

b, \(B=\frac{3x-4}{x-2\sqrt{x}}-\frac{\sqrt{x}+2}{\sqrt{x}}+\frac{\sqrt{x}-1}{2-\sqrt{x}}\)ĐK : \(x>0;x\ne4\)

\(=\frac{3x-4-\left(x-4\right)-\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\)

\(=\frac{3x-4-x+4-x+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}=\frac{x+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}=\frac{\sqrt{x}+1}{\sqrt{x}-2}\)

c, \(Q=\frac{3}{\sqrt{a}-3}+\frac{2}{\sqrt{a}+3}+\frac{a-5\sqrt{a}-3}{a-9}\)ĐK : \(a\ge0;a\ne9\)

\(=\frac{3\sqrt{a}+9+2\sqrt{a}-6+a-5\sqrt{a}-3}{a-9}=\frac{a}{a-9}\)

d, \(B=\frac{x}{x-4}-\frac{1}{2-\sqrt{x}}+\frac{1}{\sqrt{x}+2}\)ĐK : \(x\ge0;x\ne4\)

\(=\frac{x}{x-4}+\frac{\sqrt{x}+2}{x-4}+\frac{\sqrt{x}-2}{x-4}=\frac{x+2\sqrt{x}}{x-4}=\frac{\sqrt{x}}{\sqrt{x}-2}\)