a) \(P=\dfrac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{2x+\sqrt{x}}{\sqrt{x}}+\dfrac{2\left(x-1\right)}{\sqrt{x}-1}\left(x>0,x\ne1\right)\)

\(=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{x+\sqrt{x}+1}-\dfrac{\sqrt{x}\left(2\sqrt{x}+1\right)}{\sqrt{x}}+\dfrac{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}-1}\)

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

b) \(P=x-\sqrt{x}+1=\left(\sqrt{x}\right)^2-2.\sqrt{x}.\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+\dfrac{3}{4}\)

\(=\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)

\(\Rightarrow P_{min}=\dfrac{3}{4}\) khi \(x=\dfrac{1}{4}\)

c) \(Q=\dfrac{2\sqrt{x}}{P}=\dfrac{2\sqrt{x}}{x-\sqrt{x}+1}\)

Ta có: \(\left\{{}\begin{matrix}2\sqrt{x}>0\left(x>0\right)\\x+\sqrt{x}+1>0\end{matrix}\right.\Rightarrow Q>0\)

Lại có: \(3x-5\sqrt{x}+3=3\left(\left(\sqrt{x}\right)^2-2.\sqrt{x}.\dfrac{5}{6}+\left(\dfrac{5}{6}\right)^2\right)+\dfrac{11}{12}\)

\(=3\left(\sqrt{x}-\dfrac{5}{6}\right)^2+\dfrac{11}{12}>0\)

\(\Rightarrow3x-5\sqrt{x}+3>0\Rightarrow3x-3\sqrt{x}+3>2\sqrt{x}\Rightarrow3\left(x-\sqrt{x}+1\right)>2\sqrt{x}\)

\(\Rightarrow3>\dfrac{2\sqrt{x}}{x-\sqrt{x}+1}\Rightarrow Q< 3\Rightarrow0< Q< 3\)

mà \(Q\in Z\Rightarrow Q\in\left\{1;2\right\}\)

Từ\(Q\) tính ta x thôi

a, \(P=\dfrac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{2x+\sqrt{x}}{\sqrt{x}}+\dfrac{2\left(x-1\right)}{\sqrt{x}-1}\)ĐK : \(x>0;x\ne1\)

\(=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{x+\sqrt{x}+1}-\dfrac{2\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}}+\dfrac{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}-1}\)

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

\(=x-\sqrt{x}\)

b, Ta có : \(x-\sqrt{x}+\dfrac{1}{4}-\dfrac{1}{4}=\left(\sqrt{x}-\dfrac{1}{2}\right)^2-\dfrac{1}{4}\ge-\dfrac{1}{4}\)

Dấu ''='' xảy ra khi \(x=\dfrac{1}{4}\)

Vậy GTNN P là -1/4 khi x = 1/4 

c, Ta có : \(G=\dfrac{2\sqrt{x}}{P}\Rightarrow G=\dfrac{2\sqrt{x}}{x-\sqrt{x}}=\dfrac{2}{\sqrt{x}-1}\)

\(\Rightarrow\sqrt{x}-1\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)

a) Ta có: \(P=\dfrac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{2x+\sqrt{x}}{\sqrt{x}}+\dfrac{2\left(x-1\right)}{\sqrt{x}-1}\)

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

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

\(=x-\sqrt{x}+1\)

Bài 3: 

Hình a: x=30 độ

Hình b: x=32 độ

2 mountainous 

3 picks 

4 pray

5 thanks 

II

1 F

2 F

3 T

4 T

5 F

?

lỗi bn !

Lag Lag

\(=\dfrac{3^{57}\cdot5^{30}\cdot2^{32}}{5^{30}\cdot3^{30}\cdot2^{33}}=\dfrac{3^{27}}{2}\)

D

Câu 24 : 

$n_{OH^-\ pư} = n_{H^+} = 0,1.10^{-1} = 0,01(mol)$
$n_{OH^-\ dư} = 0,2.(10^{-14} : 10^{-12}) = 0,002(mol)$
$\Rightarrow n_{OH^-} = 0,01 + 0,002 = 0,012(mol)$
$\Rightarrow a = \dfrac{0,012}{0,1} = 0,12M$

Đáp án D

Câu 25 : 

$2Na + 2H_2O \to 2NaOH + H_2$
$n_{Na} = n_{NaOH} = 0,1.(10^{-14} : 10^{-13}) = 0,01(ml)$

$m_{Na} = 0,01.23 = 0,23(gam)$

D  A