Câu 1: Tìm giá trị của x để A=x+2y-\(\sqrt{2x-1}\)- 5\(\sqrt{4y-3}\)+20 đạt GTNN. Tìm GTNN
Câu 2: Tìm x, biết
a) \(\sqrt{3x+2}\)=5
b) \(\sqrt{\left(1-x\right)^2}\) = 2x+1
Câu 3: Cho biểu thức M= (\(\dfrac{\sqrt{x}}{\sqrt{x}-2}+\dfrac{\sqrt{x}}{\sqrt{x}+2}\)) \(\dfrac{\sqrt{x}+2}{\sqrt{4x}}\)
a) Rút gọn M
b) Tìm giá trị của M khi x=6+4\(\sqrt{2}\)
c) Tìm các giá trị của x để M<1
Câu 3
a, ĐKXĐ: x>0, x\(\ne\)4
M=( \(\dfrac{\sqrt{x}}{\sqrt{x}-2}+\dfrac{\sqrt{x}}{\sqrt{x}+2}\)). \(\dfrac{\sqrt{x}+2}{\sqrt{4x}}\)
M= \(\left(\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\right)\). \(\dfrac{\sqrt{x}+2}{\sqrt{4x}}\)
M= \(\dfrac{x+2\sqrt{x}+x-2\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\). \(\dfrac{\sqrt{x}+2}{\sqrt{4x}}\)
M= \(\dfrac{2x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}.\dfrac{\sqrt{x}+2}{\sqrt{4x}}\)
M= \(\dfrac{\sqrt{x}}{\sqrt{x}-2}\)
b, Thay x= \(6+4\sqrt{2}\) ( x>0, x\(\ne\)4) ta có:
M= \(\dfrac{\sqrt{6+4\sqrt{2}}}{\sqrt{6+4\sqrt{2}}-2}\)
= \(\dfrac{\sqrt{\left(\sqrt{2}+2\right)^2}}{\sqrt{\left(\sqrt{2}+2\right)^2-2}}\) = \(\dfrac{\sqrt{2}+2}{\sqrt{2}+2-2}\)
= \(\dfrac{\sqrt{2}\left(1+\sqrt{2}\right)}{\sqrt{2}}\) = \(1+\sqrt{2}\)
Vậy khi x= \(6+4\sqrt{2}\) thì M= \(1+\sqrt{2}\)
c, Để M<1 <=> \(\dfrac{\sqrt{x}}{\sqrt{x}-2}< 1\)
<=> \(\dfrac{\sqrt{x}}{\sqrt{x}-2}-\dfrac{\sqrt{x}-2}{\sqrt{x}-2}< 0\)
<=> \(\dfrac{2}{\sqrt{x}-2}< 0\)
Vì 2>0 <=> \(\sqrt{x}-2< 0\)
<=> \(\sqrt{x}< 2\)
<=> x<4
Vậy để M<1 thì 0<x<4
<=>
Câu 2
a, \(\sqrt{3x+2}=5\) (x\(\ge\dfrac{-2}{3}\))
<=> \(\sqrt{3x+2}=\sqrt{25}\)
<=> 3x+2=25
<=> 3x= 23
<=> x=\(\dfrac{23}{3}\)
Vậy S= \(\left\{\dfrac{23}{3}\right\}\)
Câu 2
b, \(\sqrt{\left(1-x\right)^2}=2x+1\) (x\(\ne1\) ; x\(\ge\dfrac{1}{2}\))
<=> \(|1-x|=2x+1\)
<=> \(\left[{}\begin{matrix}1-x=2x+1\\1-x=-1-2x\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}x=0\left(tmđk\right)\\x=-2\left(loai\right)\end{matrix}\right.\)
Vậy S=\(\left\{0\right\}\)