Câu hỏi của Hiền Nguyễn Thị

sao ko làm đi :v

ĐKXĐ : \(x\ge1;x\ne2;x\ne3\)

\(P=\left[\frac{\sqrt{x}+\sqrt{x-1}}{1}-\frac{\left(x-3\right)\left(\sqrt{x-1}+\sqrt{2}\right)}{x-3}\right].\frac{2\sqrt{x}-\sqrt{x}-\sqrt{2}}{\sqrt{x}\left(\sqrt{2}-\sqrt{x}\right)}\)

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

\(x=3-2\sqrt{2}=\left(\sqrt{2}-1\right)^2\Rightarrow\sqrt{x}=\sqrt{2}-1\)

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

\(\frac{x+2}{x-3}=9+\frac{6}{2-x}ĐKXĐ:\orbr{\begin{cases}x\ne3\\x\ne2\end{cases}}\)

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

\(\Leftrightarrow\left(2+x\right)\left(2-x\right)=9\left(2-x\right)\left(x-3\right)+6\left(x-3\right)\)

\(\Leftrightarrow4-x^2=\left(18-9x\right)\left(x-3\right)+6x-18\)

\(\Leftrightarrow4-x^2=18x-54-9x^2+27x+6x-18\)

\(\Leftrightarrow4-x^2=51x-72-9x^2\)

\(\Leftrightarrow51x-72-9x^2+x^2-4=0\)

\(\Leftrightarrow-8x^2+51x-76=0\)

\(\Leftrightarrow\left(x-4\right)\left(x-\frac{19}{8}\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-4=0\\x-\frac{19}{8}=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=4\\x=\frac{19}{8}\end{cases}}\)

\(\frac{x+2}{x-3}=9+\frac{6}{2-x}\)

\(\Leftrightarrow\left(x+2\right)\left(x-2\right)=9\left(x-3\right)\left(2-x\right)+6\left(x-3\right)\)

\(\Leftrightarrow4-x^2=51x-9x^2-72\)

\(\Leftrightarrow4-x^2-51x+9x^2+72=0\)

\(\Leftrightarrow76+8x^2-51x=0\)

\(\Leftrightarrow8x^2-19x-32x+76=0\)

\(\Leftrightarrow x\left(8x-19\right)-4\left(8x-19\right)=0\)

\(\Leftrightarrow\left(8x-19\right)\left(x-4\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}8x-19=0\\x-4=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{19}{8}\\x=4\end{cases}}\)

Vậy nghiệm phương trình là: \(\left\{\frac{19}{8};4\right\}\)

3x² - 5x + m = 0 là PT bậc 2

Áp dụng hệ thức Vi-et, ta có : \(\hept{\begin{cases}x_1x_2=\frac{m}{3}\\x_1+x_2=\frac{5}{3}\end{cases}}\)

\(x_1^2-x_2^2=\left(x_1-x_2\right)\left(x_1+x_2\right)=\pm\left(x_1+x_2\right)\sqrt{\left(x_1+x_2\right)^2-4x_1x_2}=\frac{5}{9}\)

+) Xét \(x_1-x_2\ge0\) thì : \(\frac{5}{9}=\frac{5}{3}.\sqrt{\left(\frac{5}{3}\right)^2-4.\frac{m}{3}}\Rightarrow\frac{1}{3}=\sqrt{\frac{25}{9}-\frac{4m}{3}}\Rightarrow m=2\)

+) Xét \(x_1-x_2< 0\)thì : \(\frac{5}{9}=-\frac{5}{3}.\sqrt{\left(\frac{5}{3}\right)^2-4.\frac{m}{3}}\)rồi giải đc m

k nguyên dương => \(k\ge1\)\(\Leftrightarrow\)\(a^k\ge a\)\(\Leftrightarrow\)\(\frac{a^k}{b+c}\ge\frac{a}{b+c}\)

Tương tự với 2 phân thức còn lại, cộng 3 bđt ta thu đc bđt Nesbit 3 ẩn => đpcm 

Ủa bất đẳng thức \(a^k\ge a\)chỉ đúng với a>1 thôi

a,BC^2 = AB^2 + AC^2.
AB^2= AH^2 + HB^2= AH^2 + HE^2 + BE^2
AC^2= AH^2 + CH^2 = AH^2 + CF^2 + FH^2
Cộng AB^2 và AC^2 rồi ghép HE^2 + FH^2 = AH^2.

ta de co tu giac AEHF la hinh chu nhat
=>AH=EF
ma EF^2=HE^2+HF^2(chu vi tam giac HEFvuông)
=>AH^2=HE^2+HF^2
ap dung dinh ly pytago cho cac tam giac ABC AHC AHB ta co
AB^2=AH^2+BH^2
AC^2=AH^2+HC^2
=>AB^2+AB^2=BH^2+CH^2+2AH^2
ma BH^2=BE^2+HE^2 ; CF^2+HF^2=CH^2;AB^2+AC^2=BC^2
=>BC^2=BE^2+CF^2+2AH^2+HE^2+HF^2=3AH^2+CF^2+BE^2