sao cô cho cả đáp án ra lun thế ạ @@
à ko em nhầm nhầm em xin lỗi cô
1) A = 2 khi x =9
3) 2 + \(\sqrt{x}\)- 3x = 0
-3x +\(\sqrt{x}\)+2 = 0
3x- \(\sqrt{x}\)- 2 =0
Đặt \(\sqrt{x}\)= t ; ĐK : t\(\ge\)0; t \(\ne\) 2
\(^{3t^2-t-2=0}\)
t1 =1 (TMĐK)
t2 = \(\dfrac{-2}{3}\)(Không TMĐK)
Với t = 1 => \(\sqrt{x}=1\Leftrightarrow x=1\)
1,thay x=9(TMĐKXĐ)vào A ta được:
A=\(\dfrac{2}{\sqrt{9}-2}\)=2
vậy khi x=9 thì A =2
2,với x≥0 ,x≠4 ta được:
B=\(\dfrac{\sqrt{x}}{\sqrt{x}+2}\)+\(\dfrac{4\sqrt{x}}{x-4}\)=\(\dfrac{x-2\sqrt{x}+4\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)=\(\dfrac{x+2\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)=\(\dfrac{\sqrt{x}}{\sqrt{x}-2}\)
vậy B=\(\dfrac{\sqrt{x}}{\sqrt{x}-2}\)
để A+B=\(\dfrac{3x}{\sqrt{x}-2}\)=
a, thay x=9(TMĐKXĐ) vào A ta được:
A=\(\dfrac{2}{\sqrt{9}-2}\)=2
vậy khi x=9 thì A=2
b,với x≥0,x≠4 ta được:
B=\(\dfrac{\sqrt{x}}{\sqrt{x}+2}\)+\(\dfrac{4\sqrt{x}}{x-4}\)=\(\dfrac{x-2\sqrt{x}+4\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)=\(\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)=\(\dfrac{\sqrt{x}}{\sqrt{x}-2}\)
vậy B=\(\dfrac{\sqrt{x}}{\sqrt{x}-2}\)
để A+B=\(\dfrac{3x}{\sqrt{x}-2}\)
⇒\(\dfrac{2}{\sqrt{x}-2}\)+\(\dfrac{\sqrt{x}}{\sqrt{x}-2}\)=\(\dfrac{3x}{\sqrt{x}-2}\)⇔2+\(\sqrt{x}\)=3x⇔3x-\(\sqrt{x}\)-2=0
⇔3x-\(3\sqrt{x}\)+\(2\sqrt{x}\)-2=0⇔\(3\sqrt{x}\)(\(\sqrt{x}\)-1)+2(\(\sqrt{x}-1\))=0
⇔(\(\sqrt{x}-1\))(\(3\sqrt{x}\)-2)=0⇔\(\left[{}\begin{matrix}\sqrt{x}-1=0\\3\sqrt{x}-2=0\end{matrix}\right.\)⇔ \(\left[{}\begin{matrix}x=1\left(TM\right)\\x=\sqrt{\dfrac{2}{3}}\\x=-\sqrt{\dfrac{2}{3}}\left(KTM\right)\end{matrix}\right.\left(TM\right)\)
vậy x=1,x=\(\sqrt{\dfrac{2}{3}}\)thì A+B=\(\dfrac{3x}{\sqrt{x}-2}\)
