Giải phương trình
\(125x^3-\left(2x+1\right)^3-\left(3x-1\right)^3=0 \)
\(\left(x-3\right)^3+\left(x+1\right)^3=8\left(x-1\right)^3\)
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Giải phương trìnha. frac{1}{27}cdotleft(x-3right)^3-frac{1}{125}cdotleft(x-5right)^30b.125x^3-left(2x+1right)^3-left(3x-1right)^30c.left(x-3right)^3+left(x+1right)^38cdotleft(x-1right)^3d.left(x^2-3x+2right)cdotleft(x^2+15x+56right)+80e.left(2x^2-3x+1right)cdotleft(2x^2+5x+1right)-9x^20f.left(x+6right)^4+left(x+8right)^4272
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Giải phương trình
a. \(\frac{1}{27}\cdot\left(x-3\right)^3-\frac{1}{125}\cdot\left(x-5\right)^3=0\)
b.\(125x^3-\left(2x+1\right)^3-\left(3x-1\right)^3=0\)
c.\(\left(x-3\right)^3+\left(x+1\right)^3=8\cdot\left(x-1\right)^3\)
d.\(\left(x^2-3x+2\right)\cdot\left(x^2+15x+56\right)+8=0\)
e.\(\left(2x^2-3x+1\right)\cdot\left(2x^2+5x+1\right)-9x^2=0\)
f.\(\left(x+6\right)^4+\left(x+8\right)^4=272\)
Giải phương trình sau;
a) \(125x^3-\left(2x+1\right)^3-\left(3x-1\right)^3=0\)
b) \(\left(x-1\right)^3+\left(x+1\right)^3=8\left(x-1\right)^3\)
c) \(\frac{x+19}{27}+\frac{x+17}{29}=\frac{x+15}{31}-\frac{x+13}{33}\)
Giải các phương trình sau:
a) \(125x^3-\left(2x+1\right)^3-\left(3x-1\right)^3=0\)
b) \(\left(x-1\right)^3+\left(x+1\right)^3=8\left(x-1\right)^3\)
c)\(\frac{x+19}{27}+\frac{x+17}{29}=\frac{x+15}{31}-\frac{x+13}{33}\)
Giải pt:
a, \(\dfrac{1}{27}.\left(x-3\right)^2-\dfrac{1}{125}.\left(x-5\right)^3=0\)
b, \(125x^3-\left(2x+1\right)^3-\left(3x-1\right)^3=0\)
c, \(\left(x-3\right)^3+\left(x+1\right)^3=8.\left(x-1\right)^3\)
a: \(\Leftrightarrow\left(\dfrac{1}{3}x-1\right)^3=\left(\dfrac{1}{5}x-1\right)^3\)
=>1/3x-1=1/5x-1
=>2/15x=0
hay x=0
b: Đặt 2x+1=a; 3x-1=b
Theo đề, ta có \(\left(a+b\right)^3-a^3-b^3=0\)
=>3ab(a+b)=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: \(\left(a+b\right)^3=a^3+b^3\)
=>3ab(a+b)=0
=>(x-3)(x+1)(2x-2)=0
hay \(x\in\left\{3;-1;1\right\}\)
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Giải phương trình sau:
a) \(125x^3-\left(2x+1\right)^3-\left(3x-1\right)^3=0\) \(\)
b) \(\left(x-1\right)^3+\left(x+1\right)^3=8\left(x-1\right)^3\)
c) \(\frac{x+19}{27}+\frac{x+17}{29}=\frac{x+15}{31}-\frac{x+13}{33}\)
Bài tập. Giải các phương trình sau:
a) \(\left|7-x\right|+2x=3\)
b) \(\left|2x-3\right|-4x-9=0\)
c) \(\left|3x+5\right|=\left|2-5x\right|\)
d) \(x\left|x-3\right|-\left|x^2+x+1\right|=1\)
a: =>|x-7|=3-2x
\(\Leftrightarrow\left\{{}\begin{matrix}x< =\dfrac{3}{2}\\\left(-2x+3\right)^2-\left(x-7\right)^2=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x< =\dfrac{3}{2}\\\left(2x-3-x+7\right)\left(2x-3+x-7\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x< =\dfrac{3}{2}\\\left(x+4\right)\left(3x-10\right)=0\end{matrix}\right.\Leftrightarrow x=-4\)
b: =>|2x-3|=4x+9
\(\Leftrightarrow\left\{{}\begin{matrix}x>=-\dfrac{9}{4}\\\left(4x+9-2x+3\right)\left(4x+9+2x-3\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x>=-\dfrac{9}{4}\\\left(2x+12\right)\left(6x+6\right)=0\end{matrix}\right.\Leftrightarrow x=-1\)
c: =>3x+5=2-5x hoặc 3x+5=5x-2
=>8x=-3 hoặc -2x=-7
=>x=-3/8 hoặc x=7/2
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Giải các phương trình sau:
a) \(\frac{x}{2016}+\frac{x-1}{2015}+\frac{x-2}{2014}+\frac{x-3}{2013}=4\)
b) \(125x^3=\left(2x+1\right)^3+\left(3x-1\right)^3\)
c) \(\left(2x-5\right)^3+27\left(x-1\right)^3+\left(8-5x\right)^3=0\)
P/s: câu c giải theo cách (cho a + b + c = 0; CM: \(a^3+b^3+c^3\))
\(\frac{x}{2016}+\frac{x-1}{2015}+\frac{x-2}{2014}+\frac{x-3}{2013}=4\)
\(\Leftrightarrow\left(\frac{x}{2016}-1\right)+\left(\frac{x-1}{2015}-1\right)+\left(\frac{x-2}{2014}-1\right)+\left(\frac{x-3}{2013}-1\right)=0\)
\(\Leftrightarrow\frac{x-2016}{2016}+\frac{x-2016}{2015}+\frac{x-2016}{2014}+\frac{x-2016}{2013}=0\)
\(\Leftrightarrow\left(x-2016\right)\left(\frac{1}{2016}+\frac{1}{2015}+\frac{1}{2014}+\frac{1}{2013}\right)=0\)
Dễ thấy cái vế sau > 0 nên x=2016
Câu b có cách nào hay hơn bằng cách phá ko ta,hóng quá:)
\(125x^3=\left(2x+1\right)^3+\left(3x-1\right)^3\)
\(\Leftrightarrow8x^3+12x^2+6x+1+27x^3-27x^2+9x-1=125x^3\)
\(\Leftrightarrow35x^3-15x^2+15x=125x^3\)
\(\Leftrightarrow90x^3+15x^2-15x=0\)
\(\Leftrightarrow x\left(90x^2+15x-15\right)=0\)
\(\Leftrightarrow x\left(3x-1\right)\left(2x+1\right)=0\)
\(\Leftrightarrow x=0;x=-\frac{1}{2};x=\frac{1}{3}\)
Câu c có cách giải rất hay đó nha :)
\(\left(2x-5\right)^3+27\left(x-1\right)^3+\left(8-5x\right)^3=0\)
\(\Leftrightarrow\left(2x-5\right)^3+\left(3x-3\right)^3+\left(8-5x\right)^3=0\)
Đặt \(2x-5=a;3x-3=b;8-5x=c\Rightarrow a+b+c=0\)
\(\Rightarrow a+b=-c\)
\(\Rightarrow\left(a+b\right)^3=\left(-c\right)^3\)
\(\Leftrightarrow a^3+3ab\left(a+b\right)+b^3=-c^3\)
\(\Leftrightarrow a^3+b^3+c^3=-3ab\left(a+b\right)=-3ab\left(-c\right)=3abc\)
Khi đó:
\(\left(2x-5\right)^3+27\left(x-1\right)^3+\left(8-5x\right)^3=0\)
\(\Leftrightarrow3\left(2x-5\right)\left(3x-3\right)\left(8-5x\right)=0\)
\(\Leftrightarrow x=\frac{5}{2};x=1;x=\frac{8}{5}\)
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Giải phương trình:
1, left(x^2+x+1right)left(x^4+2x^3+7x^2+26x+37right)5left(x+3right)^3
2, left(x+1right)^3+left(x+3right)^3+6left(x+1right)left(x+7right)left(x+3right)8left(x+2right)^3
3, x^3+left(x-1right)^3+3xleft(x-1right)left(x^4+xright)left(2x-1right)^3
4, dfrac{left(x+1right)^3}{3x+1}+dfrac{x^3+5x+2}{x^3+2x+1}x+3
5, dfrac{5x^3+x^2+x+1}{4x^2+1}+dfrac{6left(4x^2+1right)}{x^3+x^2+1}x+7
6, left(x^2-4x+1right)^3+left(8x-x^2+4right)^3+left(x-5right)^3125x^3
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Giải phương trình:
1, \(\left(x^2+x+1\right)\left(x^4+2x^3+7x^2+26x+37\right)=5\left(x+3\right)^3\)
2, \(\left(x+1\right)^3+\left(x+3\right)^3+6\left(x+1\right)\left(x+7\right)\left(x+3\right)=8\left(x+2\right)^3\)
3, \(x^3+\left(x-1\right)^3+3x\left(x-1\right)\left(x^4+x\right)=\left(2x-1\right)^3\)
4, \(\dfrac{\left(x+1\right)^3}{3x+1}+\dfrac{x^3+5x+2}{x^3+2x+1}=x+3\)
5, \(\dfrac{5x^3+x^2+x+1}{4x^2+1}+\dfrac{6\left(4x^2+1\right)}{x^3+x^2+1}=x+7\)
6, \(\left(x^2-4x+1\right)^3+\left(8x-x^2+4\right)^3+\left(x-5\right)^3=125x^3\)
Giải phương trình sau
1. \(5x^2-16x+7+\left(x+1\right)\sqrt{x^2+3x-1}=0\)
2. \(3\left(\sqrt{2x^2+1}-1\right)=x\left(1+3x+8\sqrt{2x^2+1}\right)\)
\(\left(\frac{2x-1}{2-x}+2\sqrt{2-x}\right)^3=27\left(2x-1\right)\)
Giải phương trình nghiệm nguyên sau:
\(3x^3-13x^2+30x-4=\sqrt{\left(6x+2\right)\left(3x-4\right)^3}\)