\(b,=1^2-\left(x-y\right)^2=\left(1+x-y\right)\left(1-x+y\right)\)

\(c,=\left(x^2+1\right)^2-\left(2x\right)^2=\left(x^2+2x+1\right)\left(x^2-2x+1\right)=\left(x+1\right)^2\left(x-1\right)^2\)

\(a,=\left(2a\right)^3-\left(3b\right)^3-2a\left[\left(2a\right)^2-\left(3b\right)^2\right]\)

\(=\left(2a\right)^3-\left(3b\right)^3-\left(2a\right)^3+18ab^2\)

\(=18ab^2-27b^3=9b^2\left(2a-3b\right)\)

mn giúp em vs ạ em cần gấp cảm ơn ạ 

3a+9=11

3b-a=4 phần 3

Em cần làm gì với dữ liệu này?

Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\dfrac{a}{5}=\dfrac{b}{2}=\dfrac{2a-3b}{2\cdot5-3\cdot2}=\dfrac{12}{4}=3\)

Do đó: a=15; b=6

d) Áp dụng t/c dtsbn:

\(\dfrac{a}{5}=\dfrac{b}{2}=\dfrac{2a}{10}=\dfrac{3b}{6}=\dfrac{2a-3b}{10-6}=\dfrac{12}{4}=3\)

\(\Rightarrow\left\{{}\begin{matrix}a=3.5=15\\b=3.2=6\end{matrix}\right.\)

f) \(\Rightarrow\dfrac{x}{5}=\dfrac{y}{3}=-\dfrac{z}{2}\)

Áp dụng t/c dtsbn:

\(\dfrac{x}{5}=\dfrac{y}{3}=\dfrac{-z}{2}=\dfrac{x+y-z}{5+3+2}=\dfrac{2}{10}=\dfrac{1}{5}\)

\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1}{5}.5=1\\y=\dfrac{1}{5}.3=\dfrac{3}{5}\\z=\dfrac{1}{5}.\left(-2\right)=-\dfrac{2}{5}\end{matrix}\right.\)

g) \(\dfrac{x}{4}=\dfrac{y}{5}=k\)\(\Rightarrow\left\{{}\begin{matrix}x=4k\\y=5k\end{matrix}\right.\)

\(\Rightarrow xy=20k^2=500\Rightarrow k=\pm5\)

\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=20\\y=25\end{matrix}\right.\\\left\{{}\begin{matrix}x=-20\\y=-25\end{matrix}\right.\end{matrix}\right.\)

a)

$CH_3COOH + KOH \to CH_3COOK + H_2O$

n CH3COOH = n KOH = 0,1(mol)

$CH_3COOH + 2O_2 \xrightarrow{t^o} 2CO_2 + 2H_2O$

$C_2H_5OH + 3O_2 \xrightarrow{t^o} 2CO_2 + 3H_2O$
$CO_2 +C a(OH)_2 \to CaCO_3 + H_2O$

2n C2H5OH + 2n CH3COOH = n CO2 = n CaCO3 = 60/100 = 0,6(mol)

=> n C2H5OH = (0,6 - 0,1.2)/2 = 0,2(mol)

=> n O2 = 2n CH3COOH + 3n C2H5OH = 0,1.2 + 0,2.3 = 0,8(mol)

=> V O2 = 0,8.22,4 = 17,92 lít

b) m = 0,1.60 + 0,2.46 = 15,2 gam

%m CH3COOH = 0,1.60/15,2  .100% = 39,47%
%m C2H5OH = 100%- 39,47% = 60,53%