\(a,P=\dfrac{\sqrt{x}+2+\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\cdot\dfrac{2-\sqrt{x}}{\sqrt{x}}=\dfrac{-2\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)}=\dfrac{-2}{\sqrt{x}+2}\\ P=-\dfrac{3}{5}\Leftrightarrow\dfrac{2}{\sqrt{x}+2}=\dfrac{3}{5}\\ \Leftrightarrow3\sqrt{x}+6=10\Leftrightarrow\sqrt{x}=\dfrac{4}{3}\Leftrightarrow x=\dfrac{16}{9}\left(tm\right)\)

`B=(x+sqrtx+5)/(sqrtx+1)=(sqrtx(sqrtx+1)+4)/(sqrtx+1)=sqrtx+4/(sqrtx+1)=[(sqrtx+1)+4/(sqrtx+1)]-1>=2\sqrt((sqrtx+1). 4/(sqrtx+1))-1=3`

Dấu "=" xảy ra `<=>x=1`

Vậy `B_(min)=3<=>x=1`

ĐK: \(x\ge2009\)

Khi đó :

\(C=x-2009-2.\sqrt{x-2009}.\frac{1}{2}+\frac{1}{4}-\frac{1}{4}+2009\)

\(=\left(\sqrt{x-2009}-\frac{1}{2}\right)^2+\left(2009-\frac{1}{4}\right)\)

\(=\left(\sqrt{x-2009}-\frac{1}{2}\right)^2+\frac{8035}{4}\ge\frac{8035}{4}\)

Dấu "=" xảy ra <=> \(\sqrt{x-2009}-\frac{1}{2}=0\)

<=> \(x-2009=\frac{1}{4}\)

<=> \(x=2009+\frac{1}{4}=\frac{8037}{4}\)( tm).

Vật min C = 8035/4 đạt tại x = 8037/4 .

ĐK: \(x\ge2009\)

Xét a > 0. Ta có:

\(C=x-\frac{1}{2\sqrt{a}}.2\sqrt{a\left(x-2009\right)}\ge\frac{2\sqrt{a}.x-a-x+2009}{2\sqrt{a}}\)(cô si xong rồi quy đồng)

\(=\frac{\left(2\sqrt{a}-1\right)x-a+2009}{2\sqrt{a}}\). Ta tìm a sao cho \(2\sqrt{a}-1=0\Leftrightarrow a=\frac{1}{4}\)

Giờ thay ngược cái a vào bên trên là ra:D

P/s: Is that true?

giúp mk với

giúp mk với

a. ĐKXĐ \(x\ge0\)và \(x\ne9\)

Ta có \(K=\left(\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}}{\sqrt{x}-3}-\frac{3\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\right):\left(\frac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\)

\(=\frac{2\sqrt{x}\left(\sqrt{x}-3\right)+\sqrt{x}\left(\sqrt{x}+3\right)-3\sqrt{x}-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}:\frac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\)

\(=\frac{2x-6\sqrt{x}+x+3\sqrt{x}-3\sqrt{x}-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}:\frac{\sqrt{x}+1}{\sqrt{x}-3}\)

\(=\frac{3x-6\sqrt{x}-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}.\frac{\sqrt{x}-3}{\sqrt{x}+1}=\frac{3\left(x-2\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}.\frac{\sqrt{x}-3}{\sqrt{x}+1}\)

\(=\frac{3\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}.\frac{\sqrt{x}-3}{\sqrt{x}+1}=\frac{3\left(\sqrt{x}-3\right)}{\sqrt{x}+3}\)

b. Để \(K< -1\Rightarrow\frac{3\sqrt{x}-9+\sqrt{x}+3}{\sqrt{x}+3}< 0\Rightarrow\frac{4\sqrt{x}-6}{\sqrt{x}+3}< 0\Rightarrow4\sqrt{x}-6< 0\)vì \(\sqrt{x}+3\ge3\)

\(\Rightarrow0\le x< \frac{9}{4}\left(tm\right)\)

Vậy với \(0\le x< \frac{9}{4}\)thì K<-1

c. \(K=\frac{3\sqrt{x}-9}{\sqrt{x}+3}=3+\frac{-18}{\sqrt{x}+3}\)

Ta có \(\sqrt{x}+3\ge3\Rightarrow\frac{1}{\sqrt{x}+3}\le\frac{1}{3}\Rightarrow-\frac{18}{\sqrt{x}+3}\ge-6\Rightarrow3+\frac{-18}{\sqrt{x}+3}\ge-3\)

\(\Rightarrow K\ge-3\)

Vậy \(MinK=-3\Leftrightarrow\sqrt{x}=0\Leftrightarrow x=0\)

Điều kiên (x<>1,X>0) xong rút gọn đi :)))

TRẢ LỜI HẾT MAU :(

\(Dk:x\ge0;x\ne1\) 

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

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

làmtp

a/ ĐKXĐ : \(0\le x\ne4\)

\(B=\frac{x\sqrt{x}+15\sqrt{x}-35}{x-\sqrt{x}-2}-\frac{\sqrt{x}+2}{\sqrt{x}+1}-\frac{\sqrt{x}-1}{\sqrt{x}-2}\)

\(=\frac{x\sqrt{x}+15\sqrt{x}-35-\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)-\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\)

\(=\frac{x\sqrt{x}+15\sqrt{x}-35-x+4-x+1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\)

\(=\frac{x\sqrt{x}-2x+15\sqrt{x}-30}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}=\frac{\left(\sqrt{x}-2\right)\left(x+15\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}=\frac{x+15}{\sqrt{x}+1}\)

c/ \(x=21-4\sqrt{5}=\left(2\sqrt{5}-1\right)^2\) thay vào B được

\(B=\frac{21-4\sqrt{5}+15}{2\sqrt{5}-1+1}=\frac{36-4\sqrt{5}}{2\sqrt{5}}=\frac{-10+18\sqrt{5}}{5}\)

d/ Đặt \(t=\sqrt{x},t\ge0\) thì \(B=\frac{t^2+15}{t+1}=6\Leftrightarrow t^2+15=6\left(t+1\right)\Leftrightarrow t^2-6t+9=0\Leftrightarrow t=3\)

=> x = 9

e/ \(B=\frac{t^2+15}{t+1}=\frac{6\left(t+1\right)+\left(t^2-6t+9\right)}{t+1}=\frac{\left(t-3\right)^2}{t+1}+6\ge6\)

Đẳng thức xảy ra khi t = 3 <=> x = 9

Vậy B đạt giá trị nhỏ nhất bằng 6 khi x = 9

a/ ĐKXĐ : 0≤x≠4

B=x√x+15√x−35x−√x−2 −√x+2√x+1 −√x−1√x−2 

=x√x+15√x−35−(√x+2)(√x−2)−(√x+1)(√x−1)(√x+1)(√x−2) 

=x√x+15√x−35−x+4−x+1(√x+1)(√x−2) 

=x√x−2x+15√x−30(√x+1)(√x−2) =(√x−2)(x+15)(√x+1)(√x−2) =x+15√x+1 

c/ x=21−4√5=(2√5−1)2 thay vào B được

B=21−4√5+152√5−1+1 =36−4√52√5 =−10+18√55 

d/ Đặt t=√x,t≥0 thì B=t2+15t+1 =6⇔t2+15=6(t+1)⇔t2−6t+9=0⇔t=3

=> x = 9

e/ B=t2+15t+1 =6(t+1)+(t2−6t+9)t+1 =(t−3)2t+1 +6≥6

Đẳng thức xảy ra khi t = 3 <=> x = 9

Vậy B đạt giá trị nhỏ nhất bằng 6 khi x = 9