`a, y xx 2,8 + 47,8 = 61,24`

`=> y xx 2,8=61,24 - 47,8`

`=> y xx 2,8=13,44`

`=> y=13,44 : 2,8`

`=>y=4,8`
`b, 13,9 + y : 5,7 = 26,23`

`=>  y : 5,7 = 26,23-13,9`

`=> y : 5,7 =12,33`

`=> y= 12,33 xx 5,7`

`=>y= 70,281`
`c, 68,5 - y xx 2,8 = 49,18`

`=> y xx 2,8 =  68,5 - 49,18`

`=>  y xx 2,8 =19,32`

`=>y=19,32 : 2,8`

`=>y=6,9`
`d, y : 5,7 - 3,6 = 5,8`

`=> y : 5,7 = 5,8 + 3,6`

`=> y : 5,7=9,4`

`=>y=9,4 xx 5,7`

`=>y= 53,58`

Em cảm ơn ạ!

Chọn đáp án C

Gọi công thức của amino aixt là C n H 2 n + 1 N O 2

Công thức của Y là C 4 n H 8 n - 2 N 4 O 5

C 4 n H 8 n - 2 N 4 O 5 + O₂ → 4n C O 2 + (4n-1) H 2 O   + N 2

có m C O 2 + m H 2 O = 47, 8 → 0,1.4n. 44 + 0,1.( 4n-1) . 18 = 47,8 → n = 2

Đốt cháy 0,1 mol X có công thức C 6 H 11 N 3 O 4

C 6 H 11 N 3 O 4 + 6,75 O 2 → 6 C O 2 + 5,5 H 2 O   + 1,5 N 2

Có n O 2 = 6,75.0,1 = 0,675 mol.

y*97+y*4 - y*1=376

y*(97+4 - 1)=376

y*100=376

y=376/100=3,76

\(y\times97+y:\frac{1}{4}-y=47,8\)

\(\Leftrightarrow y\times97+y\times4-y=47,8\)

\(\Leftrightarrow y\times\left[97+4-1\right]=47,8\)

\(\Leftrightarrow y\times100=47,8\)

\(\Leftrightarrow y=47,8:100=0,748\)

1.
\(\left(\frac{3}{1\times3}+\frac{3}{3\times5}+\frac{3}{5\times7}+...+\frac{3}{97\times99}\right)-x:\frac{3}{2}=\frac{7}{3}\\ \left(\frac{2}{1\times3}+\frac{2}{3\times5}+\frac{2}{5\times7}+...+\frac{2}{97\times99}\right):\frac{3}{2}-x:\frac{3}{2}=\frac{7}{3}\\\left[\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\right)-x\right]:\frac{3}{2}=\frac{7}{3}\\ \left(1-\frac{1}{99}\right)-x=\frac{7}{3}\times\frac{3}{2}\\ \frac{98}{99}-x=\frac{7}{2}\\ x=\frac{98}{99}-\frac{7}{2}=\frac{-497}{198}\)

2.\(\frac{x}{y}=\frac{4}{3}\Rightarrow\hept{\begin{cases}x=4a\\y=3a\\x-y=4a-3a=a\end{cases}}\\ \left(x-y\right)^{2015}=5^{2015}\Rightarrow x-y=5\\ \Rightarrow a=5\Rightarrow\hept{\begin{cases}x=4\times5=20\\y=3\times5=15\end{cases}}\)

1.

a/ \(\left\{{}\begin{matrix}xy+1+x+y=10\\\left(x+y\right)\left(xy+1\right)=1\end{matrix}\right.\)

Đặt \(\left\{{}\begin{matrix}x+y=a\\xy+1=b\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a+b=10\\ab=1\end{matrix}\right.\)

Theo Viet đảo, a và b là nghiệm:

\(t^2-10t+1=0\) \(\Rightarrow\left[{}\begin{matrix}t=5+2\sqrt{6}\\t=5-2\sqrt{6}\end{matrix}\right.\)

TH1: \(\left\{{}\begin{matrix}x+y=5+2\sqrt{6}\\xy=4-2\sqrt{6}\end{matrix}\right.\)

Theo Viet đảo, x và y là nghiệm:

\(t^2-\left(5+2\sqrt{6}\right)t+4-2\sqrt{6}=0\) (bấm máy, số xấu quá)

TH2: \(\left\{{}\begin{matrix}x+y=5-2\sqrt{6}\\xy=4+2\sqrt{6}\end{matrix}\right.\)

Ta có \(\left(5-2\sqrt{6}\right)^2-4\left(4+2\sqrt{6}\right)=33-28\sqrt{6}< 0\) nên vô nghiệm

b/ \(\left\{{}\begin{matrix}x^4+y^4=97\\xy\left(x^2+y^2\right)=78\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(x^2+y^2\right)^2-2x^2y^2=97\\xy\left(x^2+y^2\right)=78\end{matrix}\right.\)

Đặt \(\left\{{}\begin{matrix}x^2+y^2=a>0\\xy=b\end{matrix}\right.\) với \(a\ge2b\) hệ trở thành:

\(\left\{{}\begin{matrix}a^2-2b^2=97\\ab=78\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a^2-2b^2=97\\b=\frac{78}{a}\end{matrix}\right.\)

\(\Rightarrow a^2-2\left(\frac{78}{a}\right)^2=97\)

\(\Leftrightarrow a^4-97a^2-12168=0\Rightarrow\left[{}\begin{matrix}a^2=169\\a^2=-72\left(l\right)\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}a=13\Rightarrow b=6\\a=-13< 0\left(l\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x^2+y^2=13\\xy=6\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x^2+y^2=13\\y=\frac{6}{x}\end{matrix}\right.\)

\(\Rightarrow x^2+\frac{36}{x^2}=13\Leftrightarrow x^4-13x^2+36=0\) \(\Rightarrow\left[{}\begin{matrix}x^2=9\\x^2=4\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=3\Rightarrow y=2\\x=-3\Rightarrow y=-2\\x=2\Rightarrow y=3\\x=-2\Rightarrow y=-3\end{matrix}\right.\)