\(x:y:z=3:5:\left(-2\right)\)

\(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{-2}=\dfrac{5x}{15}=\dfrac{y}{5}=\dfrac{3z}{-6}=\dfrac{5x+y-3z}{15-5-6}=-\dfrac{16}{4}=-4\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{3}=-4\Rightarrow x=\left(-4\right).3=-12\\\dfrac{y}{5}=-4\Rightarrow y=\left(-4\right).5=-20\\\dfrac{z}{-2}=-4\Rightarrow z=\left(-4\right).\left(-2\right)=8\end{matrix}\right.\)

a)gọi CTPT Z :\(C_xH_yO\)

giả sủ có 1(mol) Z

ta có:\(n_{CO_2}=xn_{C_xH_yO}=x\left(mol\right)\)

\(n_{H_2O}=\dfrac{y}{2}n_{C_xH_yO}=\dfrac{y}{2}\left(mol\right)\)

\(\rightarrow\dfrac{m_{CO_2}}{m_{H_2O}}=\dfrac{44x}{9y}=\dfrac{88}{45}\\ \rightarrow\dfrac{x}{y}=\dfrac{2}{5}\)

\(\rightarrow CTPTZ:C_2H_5O\)

b)z có tính chất hóa học giống rựu etylic +Z có nhóm-OH

\(\rightarrow CTCTZ:C_2H_4OH\)

                \(\rightarrow CH_2=CH_2-OH\)

vì ABCD là hình vuông nên AB=BC=DC=AD=7cm

26.\(x=\dfrac{4}{7}y\Rightarrow3\left(\dfrac{4}{7}y\right)^2-4y^2=100\Rightarrow\\ -\dfrac{148}{49}y^2=100\left(vôlí\right)\Rightarrow vôn_{\delta}\)

28.\(x=\dfrac{3}{2}y\Rightarrow\left(\dfrac{3}{2}y\right)^3+y^3=-35\Rightarrow y=-2\Rightarrow x=-3\)

30.\(x=\dfrac{3}{4}y\Rightarrow\left(\dfrac{3}{4}y\right)^3-y^3=37\Rightarrow y=-4\Rightarrow x=-3\)

32.\(x=\dfrac{3}{2}y\Rightarrow2\left(\dfrac{3}{2}y\right)^3+3y^3=78\Rightarrow y=2\Rightarrow x=3\)

34.\(x=\dfrac{3}{4}y\Rightarrow2\left(\dfrac{3}{4}y\right)^2+y^2=100\Rightarrow\left[{}\begin{matrix}y=6\\y=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\\x=-8\end{matrix}\right.\)

36.\(x=\dfrac{2}{5}y\Rightarrow\left(\dfrac{2}{5}y\right)^3+y^3=133\Rightarrow y=5\Rightarrow x=2\)

\(4\le n!+\left(n+1\right)!< 50\\ \Leftrightarrow4\le n!+n!\left(n+1\right)< 50\)

\(\Leftrightarrow4\le n!\left(n+2\right)< 50\)

\(\Rightarrow\left\{{}\begin{matrix}n!\left(n+2\right)\ge4\Rightarrow n\ge2\\n!\left(n+2\right)< 50\left(\cdot\right)\end{matrix}\right.\)

Giải(*) \(n!\left(n+2\right)< 50\)

*)xét n=4

\(\Rightarrow4!\left(4+2\right)=144\left(loại\right)\)

*)xét n=3

\(\Rightarrow3!\left(3+2\right)=30\left(T/m\right)\)

\(\Rightarrow2\le n\le3\Rightarrow n=\left\{2;3\right\}\)

ta có \(a-b|P\left(a\right)-P\left(b\right).màP\left(b\right)=-1\) nên suy ra \(\left[{}\begin{matrix}a-b=1\\a-b=-1\end{matrix}\right.\)

tương tự ta cũng được \(\left[{}\begin{matrix}c-b=1\\c-b=-1\end{matrix}\right.\) rõ ràng a≠c(do P(a)≠P(a)) nên a-b≠c-b

từ đây ta được

\(\left[{}\begin{matrix}a-b=1\\c-b=-1\end{matrix}\right.V\left[{}\begin{matrix}a-b=-1\\c-b=1\end{matrix}\right.\)

suy ra \(a+c=2b\) 

vậy ta được đpcm

a)M,N là trung điểm AB,AC

\(\Rightarrow MN\) là đường trung bình

\(\Rightarrow MN//BC\)

b) M là trung điểm \(AB\Rightarrow MB=\dfrac{AB}{2}màAB=AC\)

N_____\(AC\Rightarrow NC=\dfrac{AC}{2}\Rightarrow MB=NC\)         

\(BNC=CMB\left(C-g-c\right)\Rightarrow CM=BN\)

\(\dfrac{\sqrt{16a^4b^6}}{\sqrt{128a^6b^6}}=\dfrac{4a^2b^3}{8\sqrt{2}a^3b^3}=\dfrac{1}{2\sqrt{2}a}\)

\(ab\left(a+b\right)⋮ab+2\)

\(\Leftrightarrow\left(ab+2\right)\left(a+b\right)-2\left(a+b\right)⋮ab+2\)

\(\rightarrow2\left(a+b\right)⋮ab+2\rightarrow2\left(a+b\right)\ge ab+2\)

giả sử \(a\ge b\ge1\)

+,nếu \(b\ge4\rightarrow ab+2\ge ab\ge4a\ge2\left(a+b\right)\left(vô-lý\right)\)

+,nếu\(b\ge4\)

      +,\(b=1\rightarrow2\left(a+1\right)⋮a+2\) 

                     \(\rightarrow2a+4-2⋮a+2\) 

                       \(\rightarrow2⋮a+2\left(vô-lí-vìa+2.2\right)\)

+,

\(b=2\rightarrow2\left(a+2\right)⋮2a+2\\ \rightarrow2⋮2a+2\left(vô-lí\right)\)

\(+,b=3\rightarrow2\left(a+3\right)⋮3a+2\\ \rightarrow2a+6⋮3a+2\)

\(\rightarrow2a+6\ge⋮3a+2\\ \rightarrow4\ge a\)

\(\rightarrow\left[{}\begin{matrix}a=3\\a=4\end{matrix}\right.\)→ thứ tùng TH

đề có phải là:Tìm tất cả các cặp số nguyên dương (a;b) sao cho\(\dfrac{ab\left(a+b\right)}{ab+2}\) là số nguyên không bạn