\(B=\sqrt{x+2017}+2018\)

a) Đề biểu thức trên có nghĩa thì:

\(x+2017\ge0\Rightarrow x\ge-2017\)

b) Với mọi \(x\ge-2017\) ta có:

\(\sqrt{x+2017}\ge0\)

\(\Rightarrow\sqrt{x+2017}+2018\ge2018\)

Dấu "=" xảy ra khi:

\(\sqrt{x+2017}=0\Rightarrow x=-2017\)

\(\Rightarrow MIN_B=2018\) khi \(x=-2017\)

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

a: B(căn x+3)=10 căn x

=>x+16-10 căn x=0

=>(căn x-2)(căn x-8)=0

=>x=4 hoặc x=64

b: \(B=\dfrac{x-9+25}{\sqrt{x}+3}=\sqrt{x}-3+\dfrac{25}{\sqrt{x}+3}\)

=>\(B=\sqrt{x}+3+\dfrac{25}{\sqrt{x}+3}-6>=2\cdot\sqrt{25}-6=2\cdot5-6=4\)

Dấu = xảy ra khi (căn x+3)^2=25

=>căn x+3=5

=>căn x=2

=>x=4

ĐK : x >= 0

\(B=\frac{\sqrt{x}-2}{\sqrt{x}+1}=\frac{\sqrt{x}+1-3}{\sqrt{x}+1}=1-\frac{3}{\sqrt{x}+1}\)

Để B nguyên thì \(\frac{3}{\sqrt{x}+1}\)nguyên hay \(\sqrt{x}+1\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\Leftrightarrow x\in\left\{0;4\right\}\)

Ta có : \(\sqrt{x}+1\ge1\forall x\ge0\Rightarrow\frac{3}{\sqrt{x}+1}\le3\Leftrightarrow1-\frac{3}{\sqrt{x}+1}\ge-2\)

Dấu "=" xảy ra <=> x = 0 . Vậy MinB = -2

\(a,ĐK:x>0;x\ne1\\ b,B=\dfrac{x+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\\ c,B=\dfrac{\sqrt{x}-1+2}{\sqrt{x}-1}=1+\dfrac{2}{\sqrt{x}-1}\in Z\\ \Leftrightarrow\sqrt{x}-1\inƯ\left(2\right)=\left\{-2;-1;1;2\right\}\\ \Leftrightarrow\sqrt{x}\in\left\{2;3\right\}\left(x>0\right)\Leftrightarrow x\in\left\{4;9\right\}\left(tm\right)\)