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Lời giải:
\(t=\sqrt{2x-3}\Rightarrow t^2=2x-3\Rightarrow x=\frac{t^2+3}{2}\)
Khi đó:
\(P=x-2\sqrt{2x-3}=\frac{t^2+3}{2}-2t=\frac{t^2-4t+3}{2}\)
BT1.
a,Ta có :\(A^2=-5x^2+10x+11\)
\(=-5\left(x^2-2x+1\right)+16\)
\(=-5\left(x-1\right)^2+16\)
Vì \(\left(x-1\right)^2\ge0\Rightarrow-5\left(x-1\right)^2\le0\)
\(\Rightarrow A^2\le16\Rightarrow A\le4\)
Dấu ''='' xảy ra \(\Leftrightarrow x=1\)
Vậy Max A = 4 \(\Leftrightarrow x=1\)
Câu b,c tương tự nhé.
1 ) \(A=\sqrt{x-2}+\sqrt{4-x}\)
ĐKXĐ : \(2\le x\le4\)
\(\Rightarrow A^2=x-2+4-x+2\sqrt{\left(x-2\right)\left(4-x\right)}=2+2\sqrt{\left(x-2\right)\left(4-x\right)}\)
Áp dụng bđt AM - GM ta có :
\(2\sqrt{\left(x-2\right)\left(4-x\right)}\le x-2+4-x=2\)
\(\Rightarrow A^2\le2+2=4\Rightarrow-2\le A\le2\)
Mà A > 0 nên ko thể có min = - 2 nên \(2\le x\le4\) ta chọn x = 2
=> A = \(\sqrt{2}\)
Vậy \(\sqrt{2}\le A\le2\)
\(A=1+\sqrt{x-2}\)
Do \(\sqrt{x-2}\ge0\forall x>2\) nên \(A\ge1\forall x>2\)
Vậy \(minA=1\Leftrightarrow x=2\)
__________
\(B=5-\sqrt{2x-1}\)
Do \(\sqrt{2x-1}\ge0\forall x\ge\frac{1}{2}\)nên \(B\le5\forall x\ge\frac{1}{2}\)
Vậy \(maxB=5\Leftrightarrow x=\frac{1}{2}\)
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= \(\left\{\dfrac{23}{3}\right\}\)
ĐKXĐ:...
\(A=\frac{2\sqrt{x}\left(x+1\right)-3\left(x+1\right)}{2\sqrt{x}-3}=\frac{\left(2\sqrt{x}-3\right)\left(x+1\right)}{2\sqrt{x}-3}=x+1\)
\(B=\frac{2x\left(x-1\right)}{\sqrt{x}\left(x-1\right)}=\frac{2x}{\sqrt{x}}=2\sqrt{x}\)
\(A=x+1=\sqrt{4+\sqrt{7}}+1=\frac{\sqrt{8+2\sqrt{7}}}{\sqrt{2}}+1=\frac{\sqrt{\left(\sqrt{7}+1\right)^2}}{\sqrt{2}}+1=\frac{1+\sqrt{14}+\sqrt{2}}{2}\)
\(B< -x+3\Leftrightarrow2\sqrt{x}< -x+3\Leftrightarrow x+2\sqrt{x}-3< 0\)
\(\Leftrightarrow\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)< 0\Leftrightarrow\sqrt{x}-1< 0\Rightarrow x< 1\Rightarrow0< x< 1\)
Ta có:
\(A-B=x+1-2\sqrt{x}=\left(\sqrt{x}-1\right)^2\ge0\) \(\forall x\in TXĐ\)
Mà \(x\ne1\Rightarrow\) dấu "=" ko xảy ra
\(\Rightarrow A-B>0\Rightarrow A>B\)
\(t=\sqrt{2x-3}=>\frac{t^2+3}{2}=x\)
\(=>P=\frac{t^2+3}{2}-2t=\frac{t^2-4t+3}{2}=\frac{\left(t-2\right)^2-1}{2}=\frac{\left(t-2\right)^2}{2}-\frac{1}{2}\)
ta có \(\frac{\left(t-2\right)^2}{2}\ge0\left(\forall t\right)\)
\(=>\frac{\left(t-2\right)^2}{2}-\frac{1}{2}\ge-\frac{1}{2}\left(\forall t\right)\)
minP=-1/2
dấu = xảy ra khi x=7/2
a) \(t=\sqrt{2x-3}\ge0\)
<=> \(t^2=2x-3\)
<=> \(x=\frac{t^2+3}{2}\)
=> \(P=\frac{t^2+3}{2}-2t\)
b) khi đó: \(P=\frac{t^2+3}{2}-2t=\frac{t^2-4t+3}{2}=\frac{\left(t-2\right)^2-1}{2}\ge-\frac{1}{2}\)
Dấu "=" xảy ra <=> t = 2 khi đó: x = 7/2