ĐKXĐ : \(x\ne2;3;4;5;6\)

(...) \(\Leftrightarrow\dfrac{1}{\left(x-2\right)\left(x-3\right)}+\dfrac{1}{\left(x-3\right)\left(x-4\right)}+\dfrac{1}{\left(x-4\right)\left(x-5\right)}+\dfrac{1}{\left(x-5\right)\left(x-6\right)}=\dfrac{1}{3}\)

\(\Leftrightarrow-1\left(\dfrac{1}{x-2}-\dfrac{1}{x-3}+\dfrac{1}{x-3}-\dfrac{1}{x-4}+\dfrac{1}{x-4}-\dfrac{1}{x-5}+\dfrac{1}{x-5}-\dfrac{1}{x-6}\right)=\dfrac{1}{3}\)

\(\Leftrightarrow\dfrac{1}{x-2}-\dfrac{1}{x-6}=-\dfrac{1}{3}\)  \(\Leftrightarrow\dfrac{-4}{x^2-8x+12}=-\dfrac{1}{3}\) 

\(\Leftrightarrow x^2-8x+12=12\Leftrightarrow x\left(x-8\right)=0\) \(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=8\end{matrix}\right.\)  

Vậy ... 

a. \(\Delta=5^2-4.3.2=1>0\) => P/t luôn có no (đpcm)

b. \(Theo\) Viet ta có : \(S=\dfrac{5}{3};P=\dfrac{2}{3}\)

c. Ta có : \(\dfrac{x_1-2}{x_2+2}+\dfrac{x_2-2}{x_1+2}=\dfrac{x_1^2-4+x_2^2-4}{\left(x_1+2\right)\left(x_2+2\right)}=\dfrac{\left(x_1+x_2\right)^2-2x_1x_2-8}{x_1x_2+2\left(x_1+x_2\right)+4}\)  

\(=\dfrac{\left(\dfrac{5}{3}\right)^2-\dfrac{4}{3}-8}{\dfrac{2}{3}+2.\dfrac{5}{3}+4}=-\dfrac{59}{72}\)

Thấy : \(a^2+2b^2+3=\left(a^2+b^2\right)+\left(b^2+1\right)+2\ge2ab+2b+2\) 

Suy ra : \(\dfrac{1}{a^2+2b^2+3}\le\dfrac{1}{2\left(ab+b+1\right)}\)  

\(CMTT:\dfrac{1}{b^2+2c^2+3}\le\dfrac{1}{2\left(bc+c+1\right)};\dfrac{1}{c^2+2a^2+3}\le\dfrac{1}{2\left(ac+a+1\right)}\)

Suy ra : (...) \(\le\dfrac{1}{2}\left(\dfrac{1}{ab+b+1}+\dfrac{1}{bc+c+1}+\dfrac{1}{ac+a+1}\right)\)  = \(\dfrac{1}{2}A\)

\(A=\dfrac{1}{ab+b+1}+\dfrac{ab}{abc.b+abc+ab}+\dfrac{b}{abc+ab+b}=\dfrac{1}{ab+b+1}+\dfrac{ab}{b+1+ab}+\dfrac{b}{1+ab+b}=1\)

Suy ra : \(\left(...\right)\le\dfrac{1}{2}\) 

" = " \(\Leftrightarrow a=b=c=1\)

\(a.\dfrac{2x+5}{2x}=\dfrac{x}{x+5}\left(x\ne0;-5\right)\Leftrightarrow\left(2x+5\right)\left(x+5\right)=2x^2\Leftrightarrow2x^2+15x+25=2x^2\Leftrightarrow3x+5=0\Leftrightarrow x=-\dfrac{5}{3}\)

\(b.\left|x+2\right|=3x+5\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x+2=3x+5\\x+2=-3x-5\end{matrix}\right.\\x\ge-\dfrac{5}{3}\end{matrix}\right.\)  \(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=-\dfrac{7}{4}\end{matrix}\right.\\x\ge-\dfrac{5}{3}\end{matrix}\right.\)   \(\Rightarrow x=-\dfrac{3}{2}\) 

c. \(\left(2x+1\right)\left(x-3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=3\end{matrix}\right.\)

a. P(x) là đa thức bậc 4 ; Q(x) là đa thức bậc 2 

b. \(P\left(x\right)+Q\left(x\right)=x^4+x^3+2x^2-3x-4\)

\(P\left(x\right)-Q\left(x\right)=x^4+x^3-2x^2-x+6\)

Gọi giá 1 kg thịt bò ; 1 kg cá ; 1 kg rau quả lần lượt là a ; b ; c (a > b > c > 0)(đồng)   

Ta có : \(\left\{{}\begin{matrix}0,5a+0,5b+c=290000\\a=1,5b\\b=5c\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=300000\\b=200000\\c=40000\end{matrix}\right.\)      

Vậy ...

Thể tích của khối gỗ là : \(17.10.3-\left(17-8\right).\left(10-6\right).3=402\left(cm^3\right)\) 

Giả sử cd ; cr sân trường lần lượt là a ; b ( a > b > 0 ) ( m ) 

S1 là diện tích sân trường ; S2 là diện tích sân khấu HV

Ta có : \(S_1=ab;S_2=\left(a-20\right)\left(b-5\right)\)

\(S_1-S_2=225\left(m^2\right)\) 

Suy ra : \(ab-\left(a-20\right)\left(b-5\right)=225\)

\(\Rightarrow ab-\left(ab-20b-5a+100\right)=225\) 

\(\Rightarrow5a+20b=325\) 

Có : \(a-20=b-5\Rightarrow a=b+15\)

Suy ra : \(5\left(b+15\right)+20b=325\Rightarrow25b=250\Rightarrow b=10\left(m\right)\)

\(\Rightarrow a=25\) (m)

Cạnh sân khấu : 25 - 20 = 5 (m)

Diện tích sân trường : \(25.10=250\left(m^2\right)\)

Đ/s : ...