1, Với x > 0 

Ta có : \(A=\frac{\sqrt{x}-2}{\sqrt{x}}=\frac{2}{3}\Rightarrow3\sqrt{x}-6=2\sqrt{x}\Leftrightarrow\sqrt{x}=6\Leftrightarrow x=36\)

2, Với \(x\ge0;x\ne1\)

\(B=\frac{\sqrt{x}}{\sqrt{x}-1}-\frac{2}{\sqrt{x}+1}-\frac{2}{x-1}=\frac{x+\sqrt{x}-2\sqrt{x}+2-2}{x-1}\)

\(=\frac{x-\sqrt{x}}{x-1}=\frac{\sqrt{x}}{\sqrt{x}+1}\)

3, Ta có : \(\frac{\sqrt{x}}{\sqrt{x}+1}=\frac{\sqrt{x}+1-1}{\sqrt{x}+1}=1-\frac{1}{\sqrt{x}+1}\)

\(\Rightarrow\sqrt{x}+1\inƯ\left(1\right)=\left\{1\right\}\)

CM \(\sqrt{9-4\sqrt{5}}-\sqrt{5}=-2\)chứ nhể

Ta có : \(\sqrt{9-4\sqrt{5}}-\sqrt{5}=\sqrt{\left(2-\sqrt{5}\right)^2}-\sqrt{5}=\left|2-\sqrt{5}\right|-\sqrt{5}=\sqrt{5}-2-\sqrt{5}=-2\)

ddeef sai

Để A = 2/3 thì \(\frac{\sqrt{x}-2}{\sqrt{x}}=\frac{2}{3}\Rightarrow3\sqrt{x}-6=2\sqrt{x}\Leftrightarrow\sqrt{x}=6\Leftrightarrow x=36\left(tm\right)\)

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

\(AB=\frac{\sqrt{x}-2}{\sqrt{x}}\cdot\frac{\sqrt{x}}{\sqrt{x}+1}=\frac{\sqrt{x}+1-3}{\sqrt{x}+1}=1-\frac{3}{\sqrt{x}+1}\)

Để AB nguyên thì \(\frac{3}{\sqrt{x}+1}\in Z\Leftrightarrow\sqrt{x}+1\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)đến đây bạn tự xét giá trị:)

\(1,A=\frac{\sqrt{x}-2}{\sqrt{x}}=\frac{2}{3}\Leftrightarrow3\sqrt{x}-6=2\sqrt{x}\)

\(\Leftrightarrow\sqrt{x}=6\Leftrightarrow x=36\left(tm\right)\)

\(2,B=\frac{\sqrt{x}}{\sqrt{x}-1}-\frac{2}{\sqrt{x}+1}-\frac{2}{x-1}\)

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

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

\(3,A\cdot B=\frac{\sqrt{x}-2}{\sqrt{x}}\cdot\frac{\sqrt{x}}{\sqrt{x}+1}=\frac{\sqrt{x}-2}{\sqrt{x}+1}\) mà AB nguyên

\(\Rightarrow\sqrt{x}-2⋮\sqrt{x}+1\)

\(\Leftrightarrow\sqrt{x}+1-3⋮\sqrt{x}+1\)

\(\Rightarrow3⋮\sqrt{x}+1\) 

\(\Rightarrow\sqrt{x}+1\inƯ\left(3\right)\) mà căn x + 1 > 0

\(\Rightarrow\sqrt{x}+1\in\left\{1;3\right\}\)

\(\Rightarrow\sqrt{x}\in\left\{0;2\right\}\Leftrightarrow x\in\left\{0;4\right\}\) mà x khác 0

=> x = 4

\(1,A=\frac{\sqrt{x}+1}{\sqrt{x}-2}+\frac{2\sqrt{x}}{\sqrt{x}+2}+\frac{5\sqrt{x}+2}{4-x}\)

\(A=\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\frac{2\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\frac{5\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)

\(A=\frac{x+3\sqrt{x}+2+2x-4\sqrt{x}-5\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)

\(A=\frac{3x-6\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\frac{3\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\frac{3\sqrt{x}}{\sqrt{x}+2}\)

\(2,x=\sqrt{9}\Rightarrow x=3\left(thỏamãn\right)\)

\(A=\frac{3\cdot3}{3+2}=\frac{9}{5}\)

\(A=\frac{3\sqrt{x}}{\sqrt{x}+2}=2\)

\(\Leftrightarrow3\sqrt{x}=2\sqrt{x}+4\)

\(\Leftrightarrow\sqrt{x}=4\)

\(\Leftrightarrow x=16\left(tm\right)\)

Gọi M( x₀ ; y₀ ) là điểm cố định mà y luôn đi qua với mọi m

Khi đó : y₀ = ( 2m - 3 )x₀ + 2m + 1 ∀ m

<=> y₀ - 2mx₀ + 3x₀ - 2m - 1 = 0 ∀ m

<=> -2m( x₀ + 1 ) + ( y₀ - 1 ) = 0 ∀ m

<=> x₀ + 1 = 0 và y₀ - 1 = 0 <=> x₀ = -1 ; y₀ = 1

Vậy y luôn đi qua điểm cố định M( -1 ; 1 )

a, y = (m-1)x + m cắt trục tung tại điểm có tung độ bằng 2 

<=> m = 2 

b, y = (m-1)x + m cắt trục hoành tại điểm có hoành độ bằng -3 

<=> -3(m-1) + m = 0 <=> -3m + 3 + m = 0 

<=> -2m + 3 = 0 <=> m = 3/2 

c, bạn tự vẽ 

X x X + Y x Y = 2 x X x X x Y x Y 

ko có chuyện x²+ y²= 2x²y²

nó có khác gì nhau kh ạ:)))) b nói hay ghê