Tìm x, biết: x(x + 1)(x + 6) - x3 = 5x
Các bác giúp em với.
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HA
1
Những câu hỏi liên quan
9 tháng 10 2021
\(a,\Leftrightarrow\left[{}\begin{matrix}x+5=0\\2x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=-\dfrac{1}{2}\end{matrix}\right.\\ b,\Leftrightarrow\left(x+2\right)\left(x-3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=3\end{matrix}\right.\\ c,\Leftrightarrow2x^2-10x-3x-2x^2=26\\ \Leftrightarrow-13x=26\Leftrightarrow x=-2\\ d,\Leftrightarrow x^2-18x+16=0\\ \Leftrightarrow\left(x^2-18x+81\right)-65=0\\ \Leftrightarrow\left(x-9\right)^2-65=0\\ \Leftrightarrow\left(x-9+\sqrt{65}\right)\left(x-9-\sqrt{65}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=9-\sqrt{65}\\9+\sqrt{65}\end{matrix}\right.\)
\(e,\Leftrightarrow x^2-10x-25=0\\ \Leftrightarrow\left(x-5\right)^2-50=0\\ \Leftrightarrow\left(x-5-5\sqrt{2}\right)\left(x-5+5\sqrt{2}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=5+5\sqrt{2}\\x=5-5\sqrt{2}\end{matrix}\right.\\ f,\Leftrightarrow5x\left(x-1\right)-\left(x-1\right)=0\\ \Leftrightarrow\left(x-1\right)\left(5x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{5}\end{matrix}\right.\\ g,\Leftrightarrow2\left(x+5\right)-x\left(x+5\right)=0\\ \Leftrightarrow\left(2-x\right)\left(x+5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=-5\end{matrix}\right.\\ h,\Leftrightarrow x^2+2x+3x+6=0\\ \Leftrightarrow\left(x+3\right)\left(x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-2\end{matrix}\right.\\ i,\Leftrightarrow4x^2-12x+9-4x^2+4=49\\ \Leftrightarrow-12x=36\Leftrightarrow x=-3\)
\(j,\Leftrightarrow x^2\left(x+1\right)+\left(x+1\right)=0\Leftrightarrow\left(x^2+1\right)\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x^2=-1\left(vô.lí\right)\\x=-1\end{matrix}\right.\Leftrightarrow x=-1\\ k,\Leftrightarrow x^2\left(x-1\right)=4\left(x-1\right)^2\\ \Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\\ \Leftrightarrow\left(x-1\right)\left(x^2-4x+4\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x-2\right)^2=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
NT
2
H
4 tháng 7 2021
\(pt\text{⇔}\left(x^2+3x+2\right)\left(x+5\right)-x^3-8x^2-27=0\text{⇔}x^3+5x^2+3x^2+15x+2x+10-x^3-8x^2-27=0\\ \text{⇔}17x=17\text{⇔}x=1\)
Vậy nghiệm của phương trình : \(S=\left\{1\right\}\)
4 tháng 7 2021
Ta có: \(\left(x+1\right)\left(x+2\right)\left(x+5\right)-x^3-8x^2=27\)
\(\Leftrightarrow\left(x^2+3x+2\right)\left(x+5\right)-x^3-8x^2=27\)
\(\Leftrightarrow x^3+5x^2+3x^2+15x+2x+10-x^3-8x^2=27\)
\(\Leftrightarrow17x=17\)
hay x=1
16 tháng 9 2021
b: Ta có: \(6^{2x-1}=216\)
\(\Leftrightarrow2x-1=3\)
hay x=2
c: Ta có: \(x^3=25x\)
\(\Leftrightarrow x\left(x-5\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=5\\x=-5\end{matrix}\right.\)
NT
2
14 tháng 7 2021
Ta có: x=2
nên x-1=1
Ta có: \(B=\left(x+1\right)\left(x^7-x^6+x^5-x^4+x^3-x^2+x-1\right)\)
\(=\left(x+1\right)\left[x^6\left(x-1\right)+x^4\left(x-1\right)+x^2\left(x-1\right)+\left(x-1\right)\right]\)
\(=\left(x+1\right)\left(x^6+x^4+x^2+1\right)\)
\(=\left(x+1\right)\left(x+1\right)\left(x^4+1\right)\)
\(=\left(2^4+1\right)\left(2+1\right)^2=17\cdot9=153\)
TN
28 tháng 8 2018
\(\frac{2}{3}\left(x-1\right)-x-\frac{3}{4}=1\)
<=> \(\frac{2}{3}x-\frac{2}{3}-x-\frac{3}{4}=1\)
<=> \(-\frac{1}{3}x-\frac{17}{12}=1\)
<=> \(-\frac{1}{3}x=\frac{29}{12}\)
<=> \(x=-\frac{29}{4}\)
\(\frac{5}{6}\left(x+2\right)-x-\frac{1}{2}=\frac{1}{3}\)
<=> \(\frac{5}{6}x+\frac{5}{3}-x-\frac{1}{2}=\frac{1}{3}\)
<=> \(-\frac{1}{6}x+\frac{7}{6}=\frac{1}{3}\)
<=> \(-\frac{1}{6}x=-\frac{5}{6}\)
<=> \(x=5\)
học tốt
NB
1
\(x\left(x+1\right)\left(x+6\right)-x^3=5x \Leftrightarrow x\left(x^2+7x+6\right)-x^3-5x=0\)
\(\Leftrightarrow x^3+7x^2+6x-x^3-5x=0\Leftrightarrow7x^2+x=0\Leftrightarrow=x\left(7x+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=-\frac{1}{7}\end{cases}}\)
VẬY tập nghiệm của PT là \(\left\{0;-\frac{1}{7}\right\}\)