a³ + b³ + c³ = 3abc

⇔ ( a³ + b³ ) + c³ - 3abc = 0

⇔ ( a + b )³ - 3ab( a + b ) + c³ - 3abc = 0

⇔ [ ( a + b )³ + c³ ] - [ 3ab( a + b ) + 3abc ] = 0

⇔ ( a + b + c )[ ( a + b )² - ( a + b ).c + c² ] - 3ab( a + b + c ) = 0

⇔ ( a + b + c )( a² + 2ab + b² - ac - bc + c² - 3ab ) = 0

⇔ ( a + b + c )( a² + b² + c² - ab - bc - ac ) = 0

⇔ \(\orbr{\begin{cases}a+b+c=0\\a^2+b^2+c^2-ab-bc-ac=0\end{cases}}\)

Từ đây tự làm tiếp nhé :))

Ta có : \(a^3+b^3+c^3=3abc\)

\(\Rightarrow a^3+b^3+c^3-3abc=0\)

\(\Rightarrow\left(a+b\right)^3-3ab\left(a+b\right)+c^3-3abc=0\)

\(\Rightarrow\left(a+b+c\right)^3-3\left(a+b\right)c\left(a+b+c\right)-3ab\left(a+b+c\right)=0\)

\(\Rightarrow\left(a+b+c\right)[\left(a+b+c\right)^2-3ac-3bc-3ab]=0\)

\(\Rightarrow\left(a+b+c\right)\left(a^2+b^2+c^2+2ab+2bc+2ac-3ab-3bc-3ac\right)=0\)

\(\Rightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)=0\)

\(\Rightarrow\orbr{\begin{cases}a+b+c=0\\a^2+b^2+c^2-ab-bc-ac=0\end{cases}}\)

​Để \(N\)có nghĩa thì \(\left(a+b+c\right)^2\ne0\)

Hay \(a+b+c\ne0\)

\(\Rightarrow a^2+b^2+c^2-ab-bc-ac=0\)

\(\Rightarrow2a^2+2b^2+2c^2-2ab-2bc-2ac=0\)

\(\Rightarrow\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ac+a^2\right)=0\)

\(\Rightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)

Vì \(\hept{\begin{cases}\left(a-b\right)^2\ge0\forall a,b\\\left(b-c\right)^2\ge0\forall b,c\\\left(c-a\right)^2\ge0\forall c,a\end{cases}}\)\(\Rightarrow\hept{\begin{cases}\left(a-b\right)^2=0\\\left(b-c\right)^2=0\\\left(c-a\right)^2=0\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}a-b=0\\b-c=0\\c-a=0\end{cases}}\)\(\Rightarrow a=b=c\)

Thay \(a=b=c\)vào \(N\), ta có : \(N=\frac{3a^2}{\left(3a\right)^2}=\frac{3a^2}{9a^2}=\frac{1}{3}\)

Vậy \(N=\frac{1}{3}\)

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

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

Xét : \(\left(x-y\right)^2=x^2+y^2-2xy\)

Thay \(\hept{\begin{cases}x-y=-7\\xy=-6\end{cases}\left(3\right)}\)vào , ta được :

\(x^2+y^2=49-12=37\left(2\right)\)

Thay \(\left(2\right)\),\(\left(3\right)\)vào \(\left(1\right)\)vào , ta có giá trị của biểu thức tương đương với :

\(-7\left(37-6\right)-\left(-7^2\right)=-7.31-49=-266\)

a, \(\frac{5x}{x-1}+\frac{-5}{x-1}=\frac{5x-5}{x-1}=\frac{5\left(x-1\right)}{x-1}=5\)

b, \(\frac{1}{x-3}+\frac{2}{x+3}+\frac{9-x}{x^2-9}=\frac{x+3}{\left(x-3\right)\left(x+3\right)}+\frac{2\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}+\frac{9-x}{\left(x-3\right)\left(x+3\right)}\)

\(=\frac{x+3+2x-6+9-x}{\left(x-3\right)\left(x+3\right)}=\frac{2x+6}{\left(x-3\right)\left(x+3\right)}=\frac{2}{x-3}\)

Tương tự 

a) Điều kiện: \(x\ne0;x\ne1\)

b) \(A=\left(\frac{x}{x-1}-\frac{1}{x^2-x}\right):\frac{x^2+2x+1}{x}\)

\(A=\left(\frac{x}{x-1}-\frac{1}{x.\left(x-1\right)}\right):\frac{\left(x+1\right)^2}{x}\)

\(A=\left(\frac{x^2}{\left(x-1\right).x}-\frac{1}{x.\left(x-1\right)}\right):\frac{\left(x+1\right)^2}{x}\)

\(A=\frac{\left(x-1\right)\left(x+1\right)}{\left(x-1\right).x}.\frac{x}{\left(x+1\right)^2}\)

\(A=\frac{x+1}{x}.\frac{x}{\left(x+1\right)^2}=\frac{1}{x+1}\)

c) Thay: \(x=2\)vào \(\frac{1}{x+1}\)ta có: \(A=\frac{1}{2+1}=\frac{1}{3}\)

a) ĐKXĐ : \(\hept{\begin{cases}x\ne0\\x\ne1\end{cases}}\)

b)

\(A=\left(\frac{x}{x-1}-\frac{1}{x^2-x}\right):\frac{x^2+2x+1}{x}\)

\(A=\left(\frac{x}{x-1}-\frac{1}{x\left(x-1\right)}\right)\cdot\frac{x}{x^2+2x+1}\)

\(A=\left(\frac{x\cdot x}{x\left(x-1\right)}-\frac{1}{x\left(x-1\right)}\right)\cdot\frac{x}{\left(x+1\right)^2}\)

\(A=\frac{x^2-1}{x\left(x-1\right)}\cdot\frac{x}{\left(x+1\right)^2}=\frac{\left(x^2-1\right)\cdot x}{x\left(x-1\right)\left(x+1\right)^2}=\frac{\left(x+1\right)\left(x-1\right)\cdot x}{x\left(x-1\right)\left(x+1\right)^2}=\frac{1}{x+1}\)

c) \(A=\frac{1}{x+1}=\frac{1}{2+1}=\frac{1}{3}\)

Vậy \(A=\frac{1}{3}\)

Đặt f(x) = x³ + ax + b

      g(x) = x² + x - 2 = x² - x + 2x - 2 = x( x - 1 ) + 2( x - 1 ) = ( x - 1 )( x + 2 )

f(x) ⋮ g(x) <=> ( x³ + ax + b ) ⋮ ( x - 1 )( x + 2 )

<=> \(\hept{\begin{cases}\left(x^3+ax+b\right)\text{⋮}\left(x-1\right)\left[1\right]\\\left(x^3+ax+b\right)\text{⋮}\left(x+2\right)\left[2\right]\end{cases}}\)

Áp dụng định lí Bézout vào [1] :

f(x) ⋮ ( x - 1 ) <=> f(1) = 0

<=> 1 + a + b = 0

<=> a + b = -1 (1)

Áp dụng định lí Bézout vào [2] :

f(x) ⋮ ( x + 2 ) <=> f(-2) = 0

<=> -8 - 2a + b = 0

<=> -2a + b = 8 (2)

Từ (1) và (2) => \(\hept{\begin{cases}a+b=-1\\-2a+b=8\end{cases}}\Leftrightarrow\hept{\begin{cases}a=-3\\b=2\end{cases}}\)( hpt lớp 9 mới học nên làm sơ sơ :33 )

Vậy a = -3 ; b = 2

P/s: Dùng hệ số bất định cũng được

:33 bt làm r nhwung vẫn k bruh