a/ Đặt : \(\left\{{}\begin{matrix}a=4x-3\\b=3x-2\end{matrix}\right.\) \(\Leftrightarrow a+b=7x-5\)
Thay vào pt ta dc :
\(a^3+b^3=\left(a+b\right)^3\)
\(\Leftrightarrow a^3+b^3=a^3+3a^2b+3ab^2+b^3\)
\(\Leftrightarrow3ab\left(a+b\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=0\\b=0\\a+b=0\end{matrix}\right.\)
+) \(a=0\Leftrightarrow4x-3=0\Leftrightarrow x=\dfrac{3}{4}\)
+) \(b=0\Leftrightarrow3x-2=0\Leftrightarrow x=\dfrac{2}{3}\)
+) \(c=0\Leftrightarrow7x-3=0\Leftrightarrow x=\dfrac{3}{7}\)
Vậy...
b/ \(x^3-2x^2-x-6=0\)
\(\Leftrightarrow x^3-3x^2+x^2-3x+2x-6=0\)
\(\Leftrightarrow x^2\left(x-3\right)+x\left(x-3\right)+2\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x^2+x+2\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left[\left(x+\dfrac{1}{2}\right)^2+\dfrac{7}{4}\right]=0\)
Mà \(\left(x+\dfrac{1}{2}\right)^2+\dfrac{7}{4}>0\)
\(\Leftrightarrow x-3=0\Leftrightarrow x=3\)
Vậy..
a) (4x - 3)3 + (3x - 2)3 = (7x - 5)3
\(\Leftrightarrow\) (4x - 3)3 + (3x - 2)3 = (4x - 3)3 + (3x - 2)3 + 3(4x - 3)(3x - 2)(4x - 3 + 3x - 2)
\(\Leftrightarrow\) 3(4x - 3)(3x - 2)(7x - 5) = 0
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=\dfrac{2}{3}\\x=\dfrac{5}{7}\end{matrix}\right.\)
a,(4x - 3)3 + (3x - 2)3 = (7x - 5)3
<=> (4x - 3)3 + (3x - 2)3- (7x - 5)3= 0
<=> (4x - 3 + 3x - 2 - 7x + 5)3 = 0
<=> 4x - 3 + 3x - 2 - 7x + 5 = 0
<=> 4x + 3x - 7x = -5 + 2 + 3
<=> 0x = 0
Vậy pt có vô số nghiệm
b) x3 -2x2 - x - 6 = 0
<=>x3-3x2+x2-3x+2x-6=0
<=>x2(x-3)+x(x-3)+2(x-3)=0
<=>(x-3)(x2+x+2)=0
<=>x-3=0
x2 +x+2=0
<=>x=3
x\(\notin\)R
Vậy pt vó nghiệm =3