Tìm x: (x2+x) (x2+x+1)=6
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30 tháng 9 2021
a: Ta có: \(3\left(2x-3\right)+2\left(2-x\right)=-3\)
\(\Leftrightarrow6x-9+4-2x=-3\)
\(\Leftrightarrow4x=2\)
hay \(x=\dfrac{1}{2}\)
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0
NL
1
8 tháng 10 2021
\(a,\Leftrightarrow x^2-x-x^2+4x-4=2\\ \Leftrightarrow3x=6\Leftrightarrow x=2\\ b,\Leftrightarrow\left(x-3\right)\left(x+3\right)-\left(x-3\right)\left(6-x\right)=0\\ \Leftrightarrow\left(x-3\right)\left(x+3-6+x\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{3}{2}\end{matrix}\right.\\ c,\Leftrightarrow x^2+2x-3x-6=0\\ \Leftrightarrow\left(x+2\right)\left(x-3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=3\end{matrix}\right.\)
AH
Akai Haruma
Giáo viên
29 tháng 7 2021
Lời giải:
a.
$|2x-5|=12-3x$
Nếu $x\geq \frac{5}{2}$ thì $2x-5=12-3x$
$\Leftrightarrow x=3,4$ (thỏa mãn)
Nếu $x< \frac{5}{2}$ thì: $5-2x=12-3x$
$\Leftrightarrow x=7$ (loại)
Vậy......
b.
$4x=|x+1|+|x+2|+|x+3|\geq 0$
$\Rightarrow x\geq 0$
Do đó: $|x+1|+|x+2|+|x+3|=(x+1)+(x+2)+(x+3)=3x+6$
Vậy: $3x+6=4x$
$\Leftrightarrow x=6$ (thỏa mãn)
c.
$|x^2+|x+2||=x^2+3$
$\Leftrightarrow x^2+|x+2|=x^2+3$
$\Leftrightarrow |x+2|=3$
$\Leftrightarrow x+2=3$ hoặc $x+2=-3$
$\Leftrightarrow x=1$ hoặc $x=-5$
d.
$|x^2-3|=6$
$\Leftrightarrow x^2-3=6$ hoặc $x^2-3=-6$
$\Leftrightarrow x^2=9$ (chọn) hoặc $x^2=-3< 0$ (loại)
$\Leftrightarrow x=\pm 3$
19 tháng 7 2021
a) Ta có: \(x^2-2x+1=25\)
\(\Leftrightarrow\left(x-1\right)^2=25\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=5\\x-1=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-4\end{matrix}\right.\)
b) Ta có: \(\left(5x+1\right)^2-\left(5x-3\right)\left(5x+3\right)=30\)
\(\Leftrightarrow25x^2+10x+1-25x^2+9=30\)
\(\Leftrightarrow10x=20\)
hay x=2
c) Ta có: \(\left(x-1\right)\left(x^2+x+1\right)-x\left(x+2\right)\left(x-2\right)=5\)
\(\Leftrightarrow x^3-1-x\left(x^2-4\right)=5\)
\(\Leftrightarrow x^3-1-x^3+4x=5\)
\(\Leftrightarrow4x=6\)
hay \(x=\dfrac{3}{2}\)
d) Ta có: \(\left(x-2\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+6\left(x+1\right)^2=15\)
\(\Leftrightarrow x^3-6x^2+12x-8-x^3+27+6\left(x^2+2x+1\right)=15\)
\(\Leftrightarrow-6x^2+12x+19+6x^2+12x+6=15\)
\(\Leftrightarrow24x=-10\)
hay \(x=-\dfrac{5}{12}\)
12 tháng 8 2021
a,\(< =>\left(x-1\right)^2-5^2=0< =>\left(x-1-5\right)\left(x-1+5\right)=0\)
\(< =>\left(x-6\right)\left(x+4\right)=0=>\left[{}\begin{matrix}x=6\\x=-4\end{matrix}\right.\)
b,\(< =>25x^2+10x+1-25x^2+9-30=0\)
\(< =>10x-20=0< =>10\left(x-2\right)=0< =>x=2\)
c,\(< =>x^3-1-x\left(x^2-4\right)-5=0\)
\(< =>x^3-1-x^2+4x-5=0< =>4x-6=0< =>x=\dfrac{6}{4}\)\(d,< =>\left(x-2\right)^3-x^3+3^3+6x^2+12x+6-15=0\)
\(< =>x^3-6x^2+12x-x^3+6x^2+12x+10=0\)
\(< =>24x+10=0< =>x=-\dfrac{5}{12}\)
12 tháng 8 2021
a: Ta có: \(x^2-2x+1=25\)
\(\Leftrightarrow\left(x-4\right)\left(x-6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=6\end{matrix}\right.\)
b: Ta có: \(\left(5x+1\right)^2-\left(5x-3\right)\left(5x+3\right)=30\)
\(\Leftrightarrow25x^2+10x+1-25x^2+9=30\)
\(\Leftrightarrow10x=20\)
hay x=2
c: Ta có: \(\left(x-1\right)\left(x^2+x+1\right)-x\left(x+2\right)\left(x-2\right)=5\)
\(\Leftrightarrow x^3-1-x\left(x^2-4\right)=5\)
\(\Leftrightarrow x^3-1-x^3+4x=5\)
\(\Leftrightarrow4x=6\)
hay \(x=\dfrac{3}{2}\)
13 tháng 8 2021
a: Ta có: \(\left(x-3\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+6\left(x+1\right)^2+3x^2=-33\)
\(\Leftrightarrow x^3-9x^2+27x-27-x^3+27+6x^2+12x+1+3x^2=-33\)
\(\Leftrightarrow39x=-34\)
hay \(x=-\dfrac{34}{39}\)
b: Ta có: \(\left(x-3\right)\left(x^2+3x+9\right)-x\left(x-2\right)\left(x+2\right)=1\)
\(\Leftrightarrow x^3-27-x^3+4x=1\)
\(\Leftrightarrow4x=28\)
hay x=7
c: Ta có: \(\left(x+2\right)\left(x^2-2x+4\right)-x\left(x-3\right)\left(x+3\right)=26\)
\(\Leftrightarrow x^3+8-x^3+9x=26\)
\(\Leftrightarrow x=2\)
21 tháng 7 2021
Bài 10:
a) (x+2)2 -x(x+3) + 5x = -20
=> x2 + 4x + 4 - x2 - 3x + 5x = -20
=> 6x = -20 + (-4)
=> 6x = -24
=> x = -4
b) 5x3-10x2+5x=0
=>5x(x2-2x+1)=0
=>5x(x-1)2 =0
=> 5x=0 hoặc (x-1)2=0
=>x=0 hoặc x=1
c) (x2 - 1)3 - (x4 + x2 + 1)(x2 - 1) = 0
=> (x2 - 1)[(x2 - 1)2 - (x4 + x2 + 1)] = 0
<=> (x2 - 1)(x4 - 2x2 + 1 - x4 - x2 - 1) = 0
<=> (x2 - 1)(-3x2) = 0
<=> (x2 - 1)=0 hoặc (-3x2) =0
<=> x2=1 hoặc x2=0
<=> x=−1;1 hoặc x=0
d)
(x+1)3−(x−1)3−6(x−1)2=-19
⇔x3+3x2+3x+1−(x3−3x2+3x−1)−6(x2−2x+1)+19=0
⇔x3+3x2+3x+1−x3+3x2−3x+1−6x2+12x−6+19=0
⇔12x+13=0⇔12x+13=0
⇔12x=-13
⇔x=-23/12
Học tốt nhé:333
9 tháng 10 2021
\(a,\Leftrightarrow\left[{}\begin{matrix}x+5=0\\2x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=-\dfrac{1}{2}\end{matrix}\right.\\ b,\Leftrightarrow\left(x+2\right)\left(x-3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=3\end{matrix}\right.\\ c,\Leftrightarrow2x^2-10x-3x-2x^2=26\\ \Leftrightarrow-13x=26\Leftrightarrow x=-2\\ d,\Leftrightarrow x^2-18x+16=0\\ \Leftrightarrow\left(x^2-18x+81\right)-65=0\\ \Leftrightarrow\left(x-9\right)^2-65=0\\ \Leftrightarrow\left(x-9+\sqrt{65}\right)\left(x-9-\sqrt{65}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=9-\sqrt{65}\\9+\sqrt{65}\end{matrix}\right.\)
\(e,\Leftrightarrow x^2-10x-25=0\\ \Leftrightarrow\left(x-5\right)^2-50=0\\ \Leftrightarrow\left(x-5-5\sqrt{2}\right)\left(x-5+5\sqrt{2}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=5+5\sqrt{2}\\x=5-5\sqrt{2}\end{matrix}\right.\\ f,\Leftrightarrow5x\left(x-1\right)-\left(x-1\right)=0\\ \Leftrightarrow\left(x-1\right)\left(5x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{5}\end{matrix}\right.\\ g,\Leftrightarrow2\left(x+5\right)-x\left(x+5\right)=0\\ \Leftrightarrow\left(2-x\right)\left(x+5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=-5\end{matrix}\right.\\ h,\Leftrightarrow x^2+2x+3x+6=0\\ \Leftrightarrow\left(x+3\right)\left(x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-2\end{matrix}\right.\\ i,\Leftrightarrow4x^2-12x+9-4x^2+4=49\\ \Leftrightarrow-12x=36\Leftrightarrow x=-3\)
\(j,\Leftrightarrow x^2\left(x+1\right)+\left(x+1\right)=0\Leftrightarrow\left(x^2+1\right)\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x^2=-1\left(vô.lí\right)\\x=-1\end{matrix}\right.\Leftrightarrow x=-1\\ k,\Leftrightarrow x^2\left(x-1\right)=4\left(x-1\right)^2\\ \Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\\ \Leftrightarrow\left(x-1\right)\left(x^2-4x+4\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x-2\right)^2=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
\(\left(x^2+x\right)\left(x^2+x+1\right)=6\)
Đặt x^2 + x = t
\(t\left(t+1\right)=6\Leftrightarrow t^2+t=6\)
\(\Leftrightarrow t^2+t-6=0\Leftrightarrow t^2+3t-2t-6=0\)
\(\Leftrightarrow\left(t-2\right)\left(t+3\right)=0\Leftrightarrow t=2;-3\)
Sr tưởng giải PT
\(\left(x^2+x\right)\left(x^2+x+x\right)=6\)
\(\Leftrightarrow x^4+x^3+x^2+x^3+x^2+x=6\)
\(\Leftrightarrow x^4+2x^3+2x^2+x=6\)
\(\Leftrightarrow x^4+2x^3+2x^2+x-6=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+2\right)\left(x^2+x+3\ne0\right)=0\Leftrightarrow x=1;-2\)