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Xét tam giác ABH vuông tại H( AH là đường cao) có:

\(AH=AB.sinB\Rightarrow AB=\frac{AH}{sinB}=\frac{2,5}{sin60^o}=\frac{5\sqrt{3}}{3}\left(cm\right)\)

Xét tam giác ACH vt H (AH là đường cao) có:

\(AH=AC.sinC\Rightarrow AC=\frac{AH}{sinC}=\frac{2,5}{sin40^o}\approx3,9\left(cm\right)\)

Lại có:

+) \(\Delta ABH\) vt H => BH=AH.cot B = 2,5 . cot 60^o=\(\frac{5\sqrt{3}}{6}\)(cm)

+) \(\Delta ACH\) vt H => CH=AH.cot C = 2,5 . cot 40^{o\(\approx3\)(cm)}

^{=> \(BC=BH+CH\approx\frac{5\sqrt{3}}{6}+3\approx4,44\)}(cm)

hfgfvfmyd

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

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

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

b) Để P = -1/2 

=> \(\frac{x}{\sqrt{x}-1}=-\frac{1}{2}\)

=> \(2x+\sqrt{x}-1=0\)

<=> \(\left(2\sqrt{x}-1\right)\left(\sqrt{x}+1\right)=0\)

<=> \(2\sqrt{x}-1=0\)

<=> \(x=\frac{1}{4}\)(tm)

Vậy x = 1/4 

c) Nhận thấy \(\sqrt{P}\)đạt MIN khi P \(>0\Rightarrow x>1\)

Khi đó \(P=\frac{x}{\sqrt{x}-1}=\frac{4\sqrt{x}-4+x-4\sqrt{x}+4}{\sqrt{x}-1}=4+\frac{\left(\sqrt{x}-2\right)^2}{\sqrt{x}-1}\ge4\)

=> Min \(\sqrt{P}=2\)

Dấu "=" xảy ra <=> x = 4 

Vậy Min \(\sqrt{P}\)= 2 khi x = 4

\(P=\left(\frac{\sqrt{x}}{x-1}+\frac{\sqrt{x}}{\sqrt{x}-1}\right):\left(\frac{2}{x}-\frac{2-x}{x\sqrt{x}+x}\right)\)\(\left(x>0;x\ne1\right)\)

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

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

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

\(P=\frac{x}{\sqrt{x}-1}\)

b) Để \(P=-\frac{1}{2}\)thì \(\frac{x}{\sqrt{x}-1}=\frac{-1}{2}\Rightarrow2x=1-\sqrt{x}\Leftrightarrow2x+\sqrt{x}-1=0\Leftrightarrow2x+2\sqrt{x}-\sqrt{x}-1=0\)

\(\Leftrightarrow2\sqrt{x}\left(\sqrt{x}+1\right)-\left(\sqrt{x}+1\right)=0\Leftrightarrow\left(\sqrt{x}+1\right)\left(2\sqrt{x}-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x}+1=0\\2\sqrt{x}-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}\sqrt{x}=-1\left(vôlí\right)\\\sqrt{x}=\frac{1}{2}\end{cases}\Leftrightarrow}x=\frac{1}{4}\left(nhận\right)}\)

Vậy để P = -1/2 thì x = 1/4

ta có :

\(B=\sqrt{\frac{16x}{x+4}}=\sqrt{16-\frac{64}{x+4}}\)ta có 

\(\frac{64}{x+4}\text{ nguyên khi x+4 là ước của 64}\)

\(\Rightarrow x+4\in\left\{4,8,16,32,64\right\}\Rightarrow x\in\left\{0,4,12,28,60\right\}\)

thay lại ta có các giá trị x =0 là giá trị duy nhất thỏa mãn

Thế \(x=2,x=\frac{1}{2}\)thì được

\(\hept{\begin{cases}f\left(2\right)+3f\left(\frac{1}{2}\right)=4\\f\left(\frac{1}{2}\right)+3f\left(2\right)=\frac{1}{4}\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}f\left(2\right)=-\frac{13}{32}\\f\left(\frac{1}{2}\right)=\frac{47}{32}\end{cases}}\)