a: \(x^3 + 2x^2 + 3x + 2 = y^3\)

=>\(y^3-\left(x+1\right)^3=x^3+2x^2+3x+2-x^3-3x^2-3x-1=-x^2+1\)

Khi x<>1 và x<>-1 thì \(-x^2+1<0\)

=>\(y^3<\left(x+1\right)^3\)

=>y<x+1

\(y^3-x^3=(x^3+2x^2+3x+2)-x^3=2x^2+3x+2>0\forall x\)

=>\(y^3>x^3\)

=>y>x

=>x<y<x+1

mà x;x+1 là hai số nguyên liên tiếp

nên (x;y)∈∅

b: \(2x^4 + 3x^3 - 3x^2 + 3x + 2 = 0\) (1)

Khi x=0 thì (1) sẽ trở thành: \(2\cdot0^4+3\cdot0^3-3\cdot0^2+3\cdot0+2=0\)

=>2=0(vô lý)

(1) =>\(2x^2 + 3x - 3 + \frac{3}{x} + \frac{2}{x^2} = 0\)

=>\(2\left(x^2 + \frac{1}{x^2}\right) + 3\left(x + \frac{1}{x}\right) - 3 = 0\)

Đặt \(t=x+\frac{1}{x}\)

(1)=>\(2(t^2 - 2) + 3t - 3 = 0\)

=>\(2t^2+3t-7=0\) (2)

\(\Delta=3^2-4\cdot2\cdot\left(-7\right)=9+56=65>0\)

Do đó: (2) sẽ có hai nghiệm phân biệt là:

\(\left[\begin{array}{l}t=\frac{-3-\sqrt{65}}{2\cdot2}=\frac{-3-\sqrt{65}}{4}\\ t=\frac{-3+\sqrt{65}}{4}\end{array}\right.\)

=>\(x+\frac{1}{x}\in\left\lbrace\frac{-3-\sqrt{65}}{4};\frac{-3+\sqrt{65}}{4}\right\rbrace\)

mà x∈Z

nên x∈∅

a,\(\frac{x}{4}=\frac{y}{2}=\frac{z}{3}\Leftrightarrow\frac{3x}{12}=\frac{2y}{4}=\frac{4z}{12}\)

Áp dụng tính chất của dãy tỉ số bằng nhau:

\(\frac{3x}{12}=\frac{2y}{4}=\frac{4z}{12}=\frac{3x-2y+4z}{12-4+12}=\frac{20}{20}=1\)

Suy ra:\(\hept{\begin{cases}\frac{x}{4}=1\\\frac{y}{2}=1\\\frac{z}{3}=1\end{cases}}\Leftrightarrow\hept{\begin{cases}x=4\\y=2\\z=3\end{cases}}\)

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

\(\frac{x}{2}=\frac{y}{6}=\frac{x-y}{2-6}=\frac{10}{-4}=-\frac{5}{2}\)

Suy ra:\(\hept{\begin{cases}\frac{x}{2}=-\frac{5}{2}\\\frac{y}{6}=-\frac{5}{2}\end{cases}\Leftrightarrow\hept{\begin{cases}x=-5\\y=-15\end{cases}}}\)

a) \(\dfrac{x}{2}=\dfrac{y}{5}=\dfrac{z}{7};x+y+z=56\)

\(\dfrac{x}{2}=\dfrac{y}{5}=\dfrac{z}{7}=\dfrac{x+y+z}{2+5+7}=\dfrac{56}{14}=4\)

\(\Rightarrow\left\{{}\begin{matrix}x=4.2=8\\y=4.5=20\\z=4.7=28\end{matrix}\right.\)

b) \(\dfrac{x}{1,1}=\dfrac{y}{1,3}=\dfrac{z}{1,4}\left(1\right);2x-y=5,5\)

\(\left(1\right)\Rightarrow\dfrac{2x-y}{1,1.2-1,3}=\dfrac{5,5}{0,9}\)

\(\Rightarrow\left\{{}\begin{matrix}x=1,1.\dfrac{5,5}{0,9}=\dfrac{6,05}{0,9}\\y=1,3.\dfrac{5,5}{0,9}=\dfrac{7,15}{0,9}\\z=\dfrac{1,4}{1,1}.x=\dfrac{1,4}{1,1}.\dfrac{6,05}{0,9}=\dfrac{8,47}{0,99}\end{matrix}\right.\)

d) \(\dfrac{x}{2}=\dfrac{x}{3}=\dfrac{z}{5};xyz=-30\)

\(\dfrac{x}{2}=\dfrac{x}{3}=\dfrac{z}{5}=\dfrac{xyz}{2.3.5}=\dfrac{-30}{30}=-1\)

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

a) �2=�5=�7;�+�+�=562x​=5y​=7z​;x+y+z=56

�2=�5=�7=�+�+�2+5+7=5614=42x​=5y​=7z​=2+5+7x+y+z​=1456​=4

⇒{�=4.2=8�=4.5=20�=4.7=28⇒⎩⎨⎧​x=4.2=8y=4.5=20z=4.7=28​

b) �1,1=�1,3=�1,4(1);2�−�=5,51,1x​=1,3y​=1,4z​(1);2x−y=5,5

(1)⇒2�−�1,1.2−1,3=5,50,9(1)⇒<...

Ta có: \(\frac{x+2}{3}=\frac{y-1}{4}=\frac{z+5}{7}\)

\(\Rightarrow\frac{2\left(x+2\right)}{6}=\frac{y-1}{4}=\frac{z+5}{7}\)

\(\Rightarrow\frac{2x+4}{6}=\frac{y-1}{4}=\frac{z+5}{7}\)

Áp dụng tính chất dãy tỉ số bằng nhau được:

\(\frac{2x+4-\left(y-1\right)+z+5}{6-4+7}=\frac{2x+4-y+1+z+5}{6-4+7}=\frac{\left(2x-y+z\right)+\left(4+1+5\right)}{6-4+7}\)

                                                                                                     \(=\frac{17+10}{9}=\frac{27}{9}=3\)

Suy ra: \(2x+4=6.3\Rightarrow2x=14\Rightarrow x=7\)

            \(y-1=3.4\Rightarrow y=13\)

             \(z+5=3.7\Rightarrow z=16\)

Vậy x = 7 ; y = 13; z = 16

y*2+y*3+4*y+y=800*3

 y * ( 2 + 3 + 4 + 1 ) = 800 * 3

 y * 10 = 2400

    y = 2400 : 10

    y = 240

Thank bạn

Ta có

\(2\left(x^3+y^3\right)=2\left(x^3+3xy\left(x+y\right)+y^3\right)-6xy\left(x+y\right)\)

\(=2\left(x+y\right)^3-6xy=2-6xy\)

Vậy ta có 

\(B=2-6xy-3\left(x^2+y^2\right)=2-3\left(x+y\right)^2=-1\)