Question

giúp mình nhé, mình cần gấp

P=(\(\frac{\sqrt{x}}{\sqrt{x}-2}+\frac{4x}{2\sqrt{x}-x}\)) : \(\frac{\sqrt{x}+3}{\sqrt{x}-2}\) với x>0, x#4

a) rút gọn P

b) Tìm x để P= -1

c) so sánh P với 1

d) tính P tại x=11-4\(\sqrt{6}\)

Answer 1

a, ta có

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

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

P=\(\frac{-3\sqrt{x}}{\sqrt{x}-2}:\frac{\sqrt{x}+3}{\sqrt{x}-2}\)

P=\(\frac{-3\sqrt{x}}{\sqrt{x}-2}.\frac{\sqrt{x}-2}{\sqrt{x}+3}\)

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

b, để P=-1 thì \(\frac{-3\sqrt{x}}{\sqrt{x}+3}=-1\)

\(-3\sqrt{x}=-\sqrt{x}-3\)

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

\(\left(\sqrt{x}\right)^2=\left(\frac{3}{2}\right)^2\)

\(x=\frac{9}{4}\)

Answer 2

a, P={(√X/√X-2)+(4X/2√X-X}:√X+3/√-2

={(√X/√X-2)+(4X/√X[2-√X])}* √X-2/√X+3

={(√X/√X-2)+(4√X/2-√X)}* √X-2/√X+3

={(√X/√X-2)-(4√X/√X-2)}* √X-2/√X+3

=-3√X/√X-2 *√X-2/√X+3

=-3√X/√X+3

b,P=-1

<=>-3√X/√X+3=-1

<=>-3√X=-(√X+3)

<=>3√X=√X+3

<=>9X=X+6√X+9

<=>8X-6√X-9=0

<=>8X+6√X-12√X-9=0

<=>2√X(4√X+3)-3(4√X+3)=0

<=>(4√X+3)(2√X-3)=0

==>2√X-3=0

==>2√X=9

==>√X=9/2

==>X=9/4 (TM X>0,X#4)

c,Ta có P=-3√X/√X+3

=(-3√X-3+3)/√X+3

=-3+(3/√X+3)

Mà 3/√X+3 đạt GTLN=1 khi X=0

==>GTLN của P=-3+1=-2<1

Vậy p<1 (đpcm)

d,Ta có X=11-4√6 TM X>0,X#4

==> p=-3√(11-4√6)/ √(11-4√6)+3

=-3(2√6-1)/(2√6-1)+3

=-6√6+3/2√6+2

=-3(2√6+2)/2√6+2

=-3