125x^3-(2x+1)^3-(3x-1)^3
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a,3x3y3-15x2y2=3x2y2(xy-5)
b,2x(x-5y)+8y(5y-x)=2x(x-5y)-8y(x-5y)=(x-5y).(2x-8y)
c,(3x-1)2-16=(3x-1)2-42=(3x-1+4)(3x-1-4)=(3x+3)(3x-5)
d,x3-3x2+3x-1=x3-1-(3x2+3x)=x3-1-3x(x+1)=(x3-1-3x)(x+1)
e,125x3+1=(5x)3+13=(5x+1)(25x2-5x.1+12)
f,x3+6x2y+12xy2+8y3=x3+3.x2.2y+3.x.(2y)2+(2y)3=(x+2y)3
a: Đặt 2x+1=a; 3x-1=b
Phương trình trở thành \(\left(a+b\right)^3-a^3-b^3=0\)
\(\Leftrightarrow3ab\left(a+b\right)=0\)
=>5x(2x+1)(3x-1)=0
hay \(x\in\left\{0;-\dfrac{1}{2};\dfrac{1}{3}\right\}\)
c: Đặt x-3=a; x+1=b
Theo đề, ta có phương trình \(a^3+b^3=\left(a+b\right)^3\)
=>3ab(a+b)=0
=>(x-3)(x+1)(2x-2)=0
hay \(x\in\left\{3;1;-1\right\}\)
a) \(125x^3=\left(2x+1\right)^3+\left(3x-1\right)^3\)
\(\Leftrightarrow\left(5x\right)^3=\left(2x+1\right)^3+\left(3x-1\right)^3\) (1)
Đặt \(a=2x+1,b=3x-1\)
\(\Rightarrow a+b=5x\)
thay vào pt (1) , ta có : \(\left(a+b\right)^3=a^3+b^3\)
\(\Leftrightarrow a^3+b^3+3a^2b+3ab^2=a^3+b^3\)
\(\Leftrightarrow3a^2b+3ab^2=0\) \(\Leftrightarrow ab\left(a+b\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}ab=0\\a+b=0\end{matrix}\right.\)
Xét \(a+b=0\) \(\Rightarrow5x=0\Leftrightarrow x=0\)
Xét \(ab=0\) \(\Rightarrow\left[{}\begin{matrix}2x+1=0\\3x-1=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{1}{2}\\x=\frac{1}{3}\end{matrix}\right.\)
Vậy tập nghiêm của pt đã cho là : \(S=\left\{0;-\frac{1}{2};\frac{1}{3}\right\}\)
b) tương tự câu a
a) \(x^4-y^4\)
\(=\left(x^2\right)^2-\left(y^2\right)^2\)
\(=\left(x^2-y^2\right)\left(x^2+y^2\right)\)
\(=\left(x+y\right)\left(x-y\right)\left(x^2+y^2\right)\)
b) \(x^2-3y^2\)
\(=x^2-\left(y\sqrt{3}\right)^2\)
\(=\left(x-y\sqrt{3}\right)\left(x+y\sqrt{3}\right)\)
c) \(\left(3x-2y\right)^2-\left(2x-3y\right)^2\)
\(=\left(3x-2y+2x-3y\right)\left(3x-2y-2x+3y\right)\)
\(=\left(5x-5y\right)\left(x+y\right)\)
\(=5\left(x-y\right)\left(x+y\right)\)
d) \(9\left(x-y\right)^2-4\left(x+y\right)^2\)
\(=\left[3\left(x-y\right)+2\left(x+y\right)\right]\left[3\left(x-y\right)-2\left(x+y\right)\right]\)
\(=\left(3x-3y+2x+2y\right)\left(3x-3y-2x-2y\right)\)
\(=\left(5x-y\right)\left(x-5y\right)\)
e) \(\left(4x^2-4x+1\right)-\left(x+1\right)^2\)
\(=\left(2x-1\right)^2-\left(x+1\right)\)
\(=\left(2x-1+x+1\right)\left(2x-1-x-1\right)\)
\(=3x\left(x-2\right)\)
f) \(x^3+27\)
\(=x^3+3^3\)
\(=\left(x+3\right)\left(x^2-3x+9\right)\)
g) \(27x^3-0,001\)
\(=\left(3x\right)^3-\left(0,1\right)^3\)
\(=\left(3x-0,1\right)\left(9x^2+0,3x+0,01\right)\)
h) \(125x^3-1\)
\(=\left(5x\right)^3-1^3\)
\(=\left(5x-1\right)\left(25x^2+5x+1\right)\)
b: Đặt x=a; x-1=b
Theo đề, ta có phương trình: \(a^3+b^3=\left(a+b\right)^3\)
\(\Leftrightarrow3ab\left(a+b\right)=0\)
=>x(x-1)(2x-1)=0
hay \(x\in\left\{0;1;\dfrac{1}{2}\right\}\)
c: Đặt x+1=a; x-2=b
Theo đề, ta có phương trình:
\(a^3+b^3=\left(a+b\right)^3\)
=>3ab(a+b)=0
=>(x+1)(x-2)(2x-1)=0
hay \(x\in\left\{-1;2;\dfrac{1}{2}\right\}\)
bài 1 :
\(\Leftrightarrow-\left(3x-1\right)^3-\left(2x+1\right)^3+125x^3=15x\left(2x+1\right)\left(3x-1\right)\)
\(\Rightarrow x=0\)
\(\Rightarrow2x+1=0\)
\(\Rightarrow2x=-1\)
\(\Rightarrow3x-1=0\)
\(\Rightarrow3x=1\)
vậy x có 3 trường hợp: TH1:x=0
TH2:x=\(\frac{-1}{2}\)
TH3:x=\(\frac{1}{3}\)
bài 2:
\(\Leftrightarrow\left(x-3\right)^3+\left(x+1\right)^3=2\left(x-1\right)\left(x^2-2x+13\right)\)
\(\Rightarrow2\left(x-1\right)\left(x^2-2x+13\right)=8\left(x-1\right)^3\)
=>x=-1;1 hoặc 3