Giải pt sau
(9x-1)3 + (2x+3)3 = 27x3 + 8
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T
4 tháng 8 2023
(x-1)3+(2x+3)3=27x3+8
=> (x - 1 + 2x + 3)[(x - 1)2 - (x - 1)(2x + 3) + (2x + 3)2] = (3x)3 + 23
=> (3x + 2)[x2-2x+1-(2x2+x-3)+4x2+12x+9] = (3x + 2)[(3x)2 - 3x.2 + 22]
=> (3x + 2)(3x2 + 9x + 13) = (3x + 2)(9x2 - 6x + 4)
=> (3x + 2)(3x2 + 9x + 13) - (3x + 2)(9x2 - 6x + 4) = 0
=> (3x + 2)(3x2 + 9x + 13 - 9x2 + 6x - 4) = 0
=> (3x + 2)(-6x2 + 15x + 9) = 0
=>\(\left[{}\begin{matrix}3x+2=0\\-6x^2+15x+9=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}3x=-2\\-3\left(2x^2+5x\right)=-9\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{2}{3}\\2x^2+5x=3\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{2}{3}\\2x^2+6x-x-3=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{2}{3}\\2x\left(x+3\right)-\left(x+3\right)=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{2}{3}\\\left(2x-1\right)\left(x+3\right)=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{2}{3}\\x=\dfrac{1}{2}\\x=-3\end{matrix}\right.\)
Vậy phương trình (x-1)3+(2x+3)3=27x3+8 có nghiệm là {-2/3;1/2;-3}
4 tháng 8 2023
=>x^3-3x^2+3x-1+8x^3+36x^2+54x+27=27x^3+8
=>37x^3+51x^2+57x+26-27x^3-8=0
=>10x^3+51x^2+57x+18=0
=>(5x+3)(2x^2+9x+6)=0
=>x=-3/5 hoặc \(x=\dfrac{-9\pm\sqrt{33}}{4}\)
16 tháng 4 2022
a: Đặt x-3=a; x+1=b
Theo đề, ta có: \(a^3+b^3=\left(a+b\right)^3\)
\(\Leftrightarrow3ab\left(a+b\right)=0\)
=>(x-3)(x+1)(2x-2)=0
hay \(x\in\left\{3;-1;1\right\}\)
b: \(\Leftrightarrow\left(2x^2+1\right)^2+2x\left(2x^2+1\right)-15x^2-9x^2=0\)
\(\Leftrightarrow\left(2x^2+1\right)^2+2x\left(2x^2+1\right)-24x^2=0\)
\(\Leftrightarrow\left(2x^2+1\right)^2+6x\left(2x^2+1\right)-4x\left(2x^2+1\right)-24x^2=0\)
\(\Leftrightarrow\left(2x^2+1\right)\left(2x^2+6x+1\right)-4x\left(2x^2+6x+1\right)=0\)
\(\Leftrightarrow\left(2x^2-4x+1\right)\left(2x^2+6x+1\right)=0\)
\(\Leftrightarrow x^2+3x+\dfrac{1}{2}=0\)
\(\Leftrightarrow x^2+3x+\dfrac{9}{4}=\dfrac{7}{4}\)
\(\Leftrightarrow\left(x+\dfrac{3}{2}\right)^2=\dfrac{7}{4}\)
hay \(x\in\left\{\dfrac{\sqrt{7}-3}{2};\dfrac{-\sqrt{7}-3}{2}\right\}\)
NV
Nguyễn Việt Lâm
Giáo viên
2 tháng 4 2019
Nhận thấy \(x=0\) không phải nghiệm, chia cả tử và mẫu của vế trái cho x ta được:
\(\frac{9}{2x+\frac{3}{x}+1}-\frac{1}{2x+\frac{3}{x}-1}=8\)
Đặt \(2x+\frac{3}{x}=a\) pt trở thành:
\(\frac{9}{a+1}-\frac{1}{a-1}=8\)
\(\Leftrightarrow9\left(a-1\right)-\left(a+1\right)=8\left(a^2-1\right)\)
\(\Leftrightarrow8a^2-8a+2=0\Leftrightarrow2\left(2a-1\right)^2=0\Rightarrow a=\frac{1}{2}\)
\(\Rightarrow2x+\frac{3}{x}=\frac{1}{2}\Leftrightarrow2x^2-\frac{1}{2}x+3=0\) \(\Rightarrow\) pt vô nghiệm
HV
2
8 tháng 4 2020
\(\left(x-1\right)^2-\left(x+1\right)^2=2\left(x+3\right)\)
\(\Leftrightarrow\left(x-1+x+1\right)\left(x-1-x-1\right)=2\left(x+3\right)\)
\(\Leftrightarrow2x\left(-2\right)=2\left(x+3\right)\)
\(\Leftrightarrow-4x=2x+6\)
\(\Leftrightarrow-6x=6\)
\(\Leftrightarrow x=-1\)
2) \(\left(2x-1\right)^2-\left(2x+1\right)^2=4\left(x-3\right)\)
\(\Leftrightarrow\left(2x-1+2x+1\right)\left(2x-1-2x-1\right)-4\left(x-3\right)=0\)
\(\Leftrightarrow4x\left(-2\right)-4x+12=0\)
\(\Leftrightarrow-12x=-12\)
\(\Leftrightarrow x=1\)
3)\(\left(2x+3\right)^2-\left(2x+3\right)\left(2x-4\right)+\left(x-2\right)^2=0\)
\(\Leftrightarrow\left(2x+3\right)\left(2x+3-2x+4\right)+\left(x^2-4x+4\right)=0\)
\(\Leftrightarrow7\left(2x+3\right)+x^2-4x+4=0\)
\(\Leftrightarrow x^2+10x+25=0\)
\(\Leftrightarrow\left(x+5\right)^2=0\)
\(\Leftrightarrow x=-5\)
4) \(8x^3-\left(x+1\right)^3=3x-3\)
\(\Leftrightarrow8x^3-\left(x^3+3x+3x^2+1\right)-3x+3=0\)
\(\Leftrightarrow7x^3-3x^2-6x+2=0\)
\(\Leftrightarrow\left(x-1\right)\left(7x^2+4x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\frac{-2+3\sqrt{2}}{7}\\x=\frac{-2-3\sqrt{2}}{7}\end{matrix}\right.\)
5)\(\left(3x-2\right)\left(9x^2+6x+4\right)-\left(3x-1\right)\left(9x^2-3x+1\right)=x-4\)
\(\Leftrightarrow\left(3x\right)^3-2^3-\left(\left(3x\right)^3-1^3\right)=x-4\)
\(\Leftrightarrow27x^3-8-\left(27x^3-1\right)=x-4\)
\(\Leftrightarrow-7=x-4\)
\(\Leftrightarrow x=-3\)
BC
1
24 tháng 9 2021
\(ĐK:\left\{{}\begin{matrix}x\le\dfrac{1}{2};4\le x\\\dfrac{1}{2}\le x\\x\le-11;\dfrac{1}{2}\le x\end{matrix}\right.\Leftrightarrow x\le-11;4\le x\)
\(PT\Leftrightarrow\sqrt{\left(x-4\right)\left(2x-1\right)}+3\sqrt{2x-1}-\sqrt{\left(2x-1\right)\left(x+11\right)}=0\\ \Leftrightarrow\sqrt{2x-1}\left(\sqrt{x-4}-\sqrt{x+11}+3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}2x-1=0\\\sqrt{x-4}-\sqrt{x+11}=-3\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\left(tm\right)\\x-4+x+11-2\sqrt{x^2+7x-44}=9\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow2\sqrt{x^2+7x-44}=2x-2\\ \Leftrightarrow\sqrt{x^2+7x-44}=x-1\\ \Leftrightarrow x^2+7x-44=x^2-2x+1\\ \Leftrightarrow9x=45\Leftrightarrow x=5\left(tm\right)\)
Vậy \(S=\left\{\dfrac{1}{2};5\right\}\)
BC
1
27 tháng 9 2021
https://hoc24.vn/cau-hoi/giai-pt-sqrt2x2-9x43sqrt2x-1sqrt2x221x-11.2005877637936
làm r nha :vv