2x^ 3 -6x=0
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LL
5
26 tháng 7 2019
\( a)6{x^2} + 7x - 3 < 0\\ \Leftrightarrow 6{x^2} + 9x - 2x - 3 < 0\\ \Leftrightarrow 3x\left( {2x + 3} \right) - \left( {2x + 3} \right) < 0\\ \Leftrightarrow \left( {2x + 3} \right)\left( {3x - 1} \right) < 0\\ \Leftrightarrow \left[ \begin{array}{l} \left\{ \begin{array}{l} 2x + 3 < 0\\ 3x - 1 > 0 \end{array} \right.\\ \left\{ \begin{array}{l} 2x + 3 > 0\\ 3x - 1 < 0 \end{array} \right. \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} \left\{ \begin{array}{l} x < - \dfrac{3}{2}\\ x > \dfrac{1}{3} \end{array} \right.\\ \left\{ \begin{array}{l} x < - \dfrac{3}{2}\\ x < \dfrac{1}{3} \end{array} \right. \end{array} \right. \Leftrightarrow x \in \left( { - \dfrac{3}{2};\dfrac{1}{3}} \right) \)
25 tháng 1 2021
a) Ta có: \(3x^2+2x-1=0\)
\(\Leftrightarrow3x^2+3x-x-1=0\)
\(\Leftrightarrow3x\left(x+1\right)-\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(3x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\3x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\3x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{1}{3}\end{matrix}\right.\)
Vậy: \(S=\left\{-1;\dfrac{1}{3}\right\}\)
b) Ta có: \(x^2-5x+6=0\)
\(\Leftrightarrow x^2-2x-3x+6=0\)
\(\Leftrightarrow x\left(x-2\right)-3\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=3\end{matrix}\right.\)
Vậy: S={2;3}
c) Ta có: \(x^2-3x+2=0\)
\(\Leftrightarrow x^2-x-2x+2=0\)
\(\Leftrightarrow x\left(x-1\right)-2\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
Vậy: S={1;2}
d) Ta có: \(2x^2-6x+1=0\)
\(\Leftrightarrow2\left(x^2-3x+\dfrac{1}{3}\right)=0\)
mà \(2\ne0\)
nên \(x^2-3x+\dfrac{1}{3}=0\)
\(\Leftrightarrow x^2-2\cdot x\cdot\dfrac{3}{2}+\dfrac{9}{4}-\dfrac{23}{12}=0\)
\(\Leftrightarrow\left(x-\dfrac{3}{2}\right)^2=\dfrac{23}{12}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{3}{2}=\dfrac{\sqrt{69}}{6}\\x-\dfrac{3}{2}=\dfrac{-\sqrt{69}}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{9+\sqrt{69}}{6}\\x=\dfrac{9-\sqrt{69}}{6}\end{matrix}\right.\)
Vậy: \(S=\left\{\dfrac{9+\sqrt{69}}{6};\dfrac{9-\sqrt{69}}{6}\right\}\)
e) Ta có: \(4x^2-12x+5=0\)
\(\Leftrightarrow4x^2-10x-2x+5=0\)
\(\Leftrightarrow2x\left(2x-5\right)-\left(2x-5\right)=0\)
\(\Leftrightarrow\left(2x-5\right)\left(2x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-5=0\\2x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=5\\2x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=\dfrac{1}{2}\end{matrix}\right.\)
Vậy: \(S=\left\{\dfrac{5}{2};\dfrac{1}{2}\right\}\)
NX
Tim x,
a,2x^4-6x^3+x^2+6x-3=0
b,x^3-9x^2+26x+24=0
c, P= 2x^4 - 4x^3 + 6x^2 - 4x + 5 biet rang x^2 - x=7
1
28 tháng 11 2016
a)\(2x^4-6x^3+x^2+6x-3=0\)
\(\Leftrightarrow2x^4-6x^3+3x^2-2x^2+6x-3=0\)
\(\Leftrightarrow x^2\left(2x^2-6x+3\right)-\left(2x^2-6x+3\right)=0\)
\(\Leftrightarrow\left(x^2-1\right)\left(2x^2-6x+3\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(2x^2-6x+3\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x-1=0\\x+1=0\\2x^2-6x+3=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=1\\x=-1\\\Delta_{2x^2-6x+3}=\left(-6\right)^2-4\left(2.3\right)=12\end{array}\right.\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=1\\x=-1\\x_{1,2}=\frac{6\pm\sqrt{12}}{4}\end{array}\right.\)
b)\(x^3+9x^2+26x+24=0\)
\(\Leftrightarrow x^3+5x^2+6x+4x^2+20x+24=0\)
\(\Leftrightarrow x\left(x^2+5x+6\right)+4\left(x^2+5x+6\right)=0\)
\(\Leftrightarrow\left(x^2+5x+6\right)\left(x+4\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x+3\right)\left(x+4\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x+2=0\\x+3=0\\x+4=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=-2\\x=-3\\x=-4\end{array}\right.\)
5 tháng 11 2017
Giải như sau.
(1)+(2)⇔x2−2x+1+√x2−2x+5=y2+√y2+4⇔(x2−2x+5)+√x2−2x+5=y2+4+√y2+4⇔√y2+4=√x2−2x+5⇒x=3y(1)+(2)⇔x2−2x+1+x2−2x+5=y2+y2+4⇔(x2−2x+5)+x2−2x+5=y2+4+y2+4⇔y2+4=x2−2x+5⇒x=3y
⇔√y2+4=√x2−2x+5⇔y2+4=x2−2x+5, chỗ này do hàm số f(x)=t2+tf(x)=t2+t đồng biến ∀t≥0∀t≥0
Công việc còn lại là của bạn !
30 tháng 9 2018
\(\left(x+6\right)\left(2x+1\right)=0\)
<=> \(\orbr{\begin{cases}x+6=0\\2x+1=0\end{cases}}\)
<=> \(\orbr{\begin{cases}x=-6\\x=-\frac{1}{2}\end{cases}}\)
Vậy....
hk tốt
^^
CL
3
22 tháng 8 2021
a) \(6x^3-6x=0\Leftrightarrow6x\left(x^2-1\right)=0\Leftrightarrow6x\left(x-1\right)\left(x+1\right)=0\Leftrightarrow\left[{}\begin{matrix}6x=0\\x-1=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=-1\end{matrix}\right.\)b) \(2x\left(3x+7\right)-6x^2=28\Leftrightarrow6x^2+14x-6x^2=28\Leftrightarrow14x=28\Leftrightarrow x=2\)
c) \(2\left(4x+4\right)-5\left(x-3\right)=0\Leftrightarrow8x+8-5x+15=0\Leftrightarrow3x=-23\Leftrightarrow x=-\dfrac{23}{3}\)
QD
1
KT
2 tháng 2 2018
b) \(x^3-6x^2+11x-12=0\)
\(\Leftrightarrow\)\(x^3-4x^2-2x^2+8x+3x-12=0\)
\(\Leftrightarrow\)\(x^2\left(x-4\right)-2x\left(x-4\right)+3\left(x-4\right)\)
\(\Leftrightarrow\)\(\left(x-4\right)\left(x^2-2x+3\right)=0\)
Ta có: \(x^2-2x+3\)
\(=\left(x-1\right)^2+2>0\)
\(\Rightarrow\)\(x-4=0\)
\(\Leftrightarrow\)\(x=4\)
Vậy...
HD
0
15 tháng 8 2017
\(2x^4-6x^3+x^2+6x-3=0\)
\(\Leftrightarrow2x^4-2x^3-4x^3+4x^2-3x^2+3x+3x-3=0\)
\(\Leftrightarrow2x^3\left(x-1\right)-4x^2\left(x-1\right)-3x\left(x-1\right)+3\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(2x^3-4x^2-3x+3\right)=0\)
20 tháng 8 2017
Đã có đáp án:
2x^4-6x^3+x^2+6x-3=0
2x^4-6x^3-3x^2-2x^2-6x-3=0
2x^2(x^2-1)-6x(x^2-1)+3(x^2-1)=0
(x^2-1)(2x^2-6x+3)=0
=> { x^2-1=0 =>x=-1;1
Giả phương trình :(*) 2x^2-6x+3=0
4x^2-12x-6=0
(2x)^2-2.2x.3-3=0
(2x-3)^2- (√3)^2=0
( 2x-3)^2=(√3)^2
=> 2x-3=-√3 => 2x= 3-√3 => x=(3-√3)/2
2x-3=√3 => 2x=√3+3 => x=(√3+3)/2
Vậy x....
\(2x^3-6x=0\)
\(x\left(2x^2-6\right)=0\)
\(\Rightarrow x=0\text{ hoặc }2x^2-6=0\)
\(2x^2=0+6\)
\(2x^2=6\)
\(x^2=6\div2\)
\(x^2=3\)
\(x=\pm\sqrt{3}\)
\(\text{Vậy }x=0;x=\pm\sqrt{3}\)
Ta có 2x3-6x=0
=> x(2x2-6)=0
=>\(\orbr{\begin{cases}x=0\\2x^2-6=0\end{cases}}\)
=>\(\orbr{\begin{cases}x=0\\2\left(x^2-3\right)=0\end{cases}}\)
=>\(\orbr{\begin{cases}x=0\\x^2-3=0\end{cases}}\)
=>\(\orbr{\begin{cases}x=0\\x^2=3\end{cases}}\)
=>\(\orbr{\begin{cases}x=0\\x=\sqrt{3}\end{cases}}\)
Vậy ............
chúc b học tốt nha!!!