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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

\(đkxđ\Leftrightarrow\hept{\begin{cases}a\ge0\\a\ne1\end{cases}}\)

\(A=\)\(\left(\frac{\sqrt{a}}{2}-\frac{1}{2\sqrt{a}}\right)^2\)\(\left(\frac{\sqrt{a}-1}{\sqrt{a}+1}-\frac{\sqrt{a}+1}{\sqrt{a}-1}\right)\)

\(=\left(\frac{\sqrt{a}.\sqrt{a}}{2\sqrt{a}}-\frac{1}{2\sqrt{a}}\right)^2\)\(\left(\frac{\left(\sqrt{a}-1\right)^2}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}-\frac{\left(\sqrt{a}+1\right)^2}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\right)\)

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

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

\(=\frac{\left(a-1\right)^2.-4\sqrt{a}}{4a\left(a-1\right)}=\frac{a-1}{\sqrt{a}}\)

\(b,A< 0\Rightarrow\frac{a-1}{\sqrt{a}}< 0\)

Mà \(\sqrt{a}\ge0\Rightarrow a-1\le0\Rightarrow a\le1\)

\(A=2\Rightarrow\frac{a-1}{\sqrt{a}}=2\)

\(\Rightarrow a-1=2\sqrt{a}\Rightarrow a-2\sqrt{a}-1=0\)

\(\Rightarrow a-2\sqrt{a}+1-2=0\)

\(\Rightarrow\left(\sqrt{a}-1\right)^2-\sqrt{2}^2=0\)

\(\Rightarrow\left(\sqrt{a}-1-\sqrt{2}\right)\left(\sqrt{a}-1+\sqrt{2}\right)=0\)

\(\Rightarrow\orbr{\begin{cases}\sqrt{a}=1+\sqrt{2}\\\sqrt{a}=1-\sqrt{2}\end{cases}\Rightarrow\orbr{\begin{cases}a=\left(1+\sqrt{2}\right)^2=3+2\sqrt{2}\\a=\left(1-\sqrt{2}\right)^2=3-2\sqrt{2}\end{cases}}}\)

\(\left(\frac{\sqrt{a}}{2}-\frac{1}{2\sqrt{a}}\right)^2\left(\frac{\sqrt{a}-1}{\sqrt{a}+1}-\frac{\sqrt{a}+1}{\sqrt{a}-1}\right)\)

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

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

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

\(=\frac{a-1}{4a}.\frac{2\sqrt{a}.\left(-2\right)}{1}\)

\(=\frac{a-1}{4a}.\frac{-4\sqrt{a}.}{1}\)

\(=\frac{1-a}{\sqrt{a}}\)

pt tương đương: x²-5x-xy+2+y=0

<=>x²-5x+2=y.(x-1)

<=>\(y=\frac{x^2-5x+2}{x-1}=x-4-\frac{2}{x-1}\)

y là số nguyên => x=0 hoặc x=3 hoặc x=-1

x=0 =>y=-2

x=3=>y=-2

x=-1=>y=-4

#)Giải :

Giả sử  \(p^3+\frac{p-1}{2}\) là tích của hai số tự nhiên liên tiếp 

\(\Rightarrow p^3+\frac{p-1}{2}=a\left(a+1\right)\Rightarrow2p\left(2p^2+1\right)=\left(2a+1\right)^2+1\)

Nếu \(p=3\Rightarrow p^3+\frac{p-1}{2}=3^3+\frac{3-1}{2}=27+1=28\left(ktm\right)\)

Nếu \(p\ne3\Rightarrow2p^2+1⋮3\Rightarrow\left(2a+1\right)^2+1⋮3\Rightarrow\left(2a+1\right)^2\div3\) dư 2 (mâu thuẫn)

\(\Rightarrowđpcm\)

cái cuối là chia 3 dư 1 chớ sao dư 2 vậy bạn

<=>\(\sqrt[3]{a}.b=\sqrt[3]{a.b^3}\)(luôn đúng)=>dpcm :v

\(\sqrt[3]{\frac{a}{b^2}}=\sqrt[3]{\frac{ab}{b^3}}=\frac{\sqrt[3]{ab}}{\sqrt[3]{b^3}}=\frac{\sqrt[3]{ab}}{b}\)

\(\sqrt{\frac{a}{b}}+\sqrt{ab}+\frac{a}{b}\sqrt{\frac{b}{a}}\)

\(=\sqrt{\frac{a}{b}}+\sqrt{ab}+\sqrt{\frac{a^2b}{b^2a}}\)

\(=\sqrt{\frac{a}{b}}+\sqrt{ab}+\sqrt{\frac{a}{b}}\)

\(=2\sqrt{\frac{a}{b}}+\sqrt{ab}\)