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DT
4
NN
8 tháng 12 2019
\(2x\left(x^2-25\right)=0\)
\(\Rightarrow\orbr{\begin{cases}2x=0\\x^2-25=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x=\pm5\end{cases}}\)
\(2x\left(3x-5\right)+\left(3x-5\right)=0\)
\(\left(2x+1\right)\left(3x-5\right)=0\)
\(\Rightarrow\orbr{\begin{cases}2x+1=0\\3x-5=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=-\frac{1}{2}\\x=\frac{5}{3}\end{cases}}\)
NN
8 tháng 12 2019
\(9\left(3x-2\right)-x\left(2-3x\right)=0\)
\(9\left(3x-2\right)+x\left(3x-2\right)=0\)
\(\left(9+x\right)\left(3x-2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}9+x=0\\3x-2=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=-9\\x=\frac{2}{3}\end{cases}}\)
\(\left(2x-1\right)^2=25\)
\(\Rightarrow\orbr{\begin{cases}2x-1=5\\2x-1=-5\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=3\\x=-2\end{cases}}\)
30 tháng 5 2022
a: \(6x^4+25x^3+12x^2-25x+6=0\)
\(\Leftrightarrow6x^4+12x^3+13x^3+26x^2-14x^2-28x+3x+6=0\)
\(\Leftrightarrow\left(x+2\right)\left(6x^3+13x^2-14x+3\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(6x^3+18x^2-5x^2-15x+x+3\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x+3\right)\left(6x^2-5x+1\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x+3\right)\left(3x-1\right)\left(2x-1\right)=0\)
hay \(x\in\left\{-2;-3;\dfrac{1}{3};\dfrac{1}{2}\right\}\)
b: \(x^5+2x^4+3x^3+3x^2+2x+1=0\)
\(\Leftrightarrow x^5+x^4+x^4+x^3+2x^3+2x^2+x^2+x+x+1=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^4+x^3+2x^2+x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^4+x^2+x^3+x+x^2+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^2+x+1\right)\left(x^2+1\right)=0\)
=>x+1=0
hay x=-1
c: \(x^2\left(x^2+2\right)-x^2-2=0\)
\(\Leftrightarrow\left(x^2+2\right)\left(x^2-1\right)=0\)
=>x=1 hoặc x=-1
6 tháng 7 2018
\(1.6x\left(x-10\right)-2x+20=0\)
⇔\(6x\left(x-10\right)-2\left(x-10\right)=0\)
⇔ \(2\left(x-10\right)\left(3x-1\right)=0\)
⇔ x = 10 hoặc x = \(\dfrac{1}{3}\)
KL....
\(2.3x^2\left(x-3\right)+3\left(3-x\right)=0\)
⇔ \(3\left(x-3\right)\left(x^2-1\right)=0\)
⇔ \(x=+-1\) hoặc \(x=3\)
KL....
\(3.x^2-8x+16=2\left(x-4\right)\)
⇔ \(\left(x-4\right)^2-2\left(x-4\right)=0\)
⇔ \(\left(x-4\right)\left(x-6\right)=0\)
⇔ \(x=4\) hoặc \(x=6\)
KL.....
\(4.x^2-16+7x\left(x+4\right)=0\)
\(\text{⇔}4\left(x+4\right)\left(2x-1\right)=0\)
⇔ \(x=-4hoacx=\dfrac{1}{2}\)
KL.....
\(5.x^2-13x-14=0\)
⇔ \(x^2+x-14x-14=0\)
\(\text{⇔}\left(x+1\right)\left(x-14\right)=0\)
\(\text{⇔}x=14hoacx=-1\)
KL......
Còn lại tương tự ( dài quá ~ )
QT
3
19 tháng 7 2018
A) x3-6x2+12x-8=0
<=>(x-2)3=0
<=>x-2=0
<=>x=2
B)4(x-3)2 -(2x-1)(2x+1)=13
<=>4(x2-6x+9)-4x2+1=13
<=>4x2-24x+36-4x2+1=13
<=>-24x+37=13
<=>24x=37-13
<=>24x=24
<=>x=1
C)25x2-6(x+1)2=0
<=>(5x-\(\sqrt{6}\left(x+1\right)\))(5x+\(\sqrt{6}\left(x+1\right)\))=0
<=>5x-\(\sqrt{6}\left(x+1\right)\)=0 hoặc 5x+\(\sqrt{6}\left(x+1\right)\))=0
<=>5x-\(\sqrt{6}x-\sqrt{6}\)=0 <=>5x+\(\sqrt{6}x+\sqrt{6}\)=0
<=>x(5-\(\sqrt{6}\))=\(\sqrt{6}\) <=>x(5+\(\sqrt{6}\))=\(-\sqrt{6}\)
<=>x=\(\frac{\sqrt{6}}{5-\sqrt{6}}\) <=>x=\(\frac{-\sqrt{6}}{5+\sqrt{6}}\)
19 tháng 7 2018
Rút gọn C=(4+2A+A^2).(4-A^2).(4-2a+a^2) GIẢI GIÚP MIK ĐI
17 tháng 9 2018
Bài 1 :
\(a)\)\(\left(x-1\right)\left(x^2+x+1\right)-x\left(x+3\right)\left(x-3\right)=15\)
\(\Leftrightarrow\)\(x^3-1-x\left(x^2-3^2\right)=15\)
\(\Leftrightarrow\)\(x^3-1-x^3+9x=15\)
\(\Leftrightarrow\)\(9x=16\)
\(\Leftrightarrow\)\(x=\frac{16}{9}\)
Vậy \(x=\frac{16}{9}\)
Chúc bạn học tốt ~
HH
1 tháng 7 2018
a/ \(25x^2-9=0\)
<=> \(\left(5x-3\right)\left(5x+3\right)=0\)
<=> \(\orbr{\begin{cases}5x-3=0\\5x+3=0\end{cases}}\)
<=> \(\orbr{\begin{cases}5x=3\\5x=-3\end{cases}}\)
<=> \(\orbr{\begin{cases}x=\frac{3}{5}\\x=-\frac{3}{5}\end{cases}}\)
b/ \(\left(x+4\right)^2-\left(x+9\right)\left(x-1\right)=16\)
<=> \(x^2+8x+16-x^2+8x-9=16\)
<=> \(16x+7=16\)
<=> \(16x=9\)
<=> \(x=\frac{9}{16}\)
1 tháng 7 2018
a) \(25x^2-9=0\)
\(\Leftrightarrow\left(5x-3\right)\left(5x+3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}5x-3=0\\5x+3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}5x=3\\5x=-3\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{3}{5}\\x=-\frac{3}{5}\end{cases}}}\)
Vậy S = {3/5 ; -3/5}
b) \(\left(x+4\right)^2-\left(x+9\right)\left(x-1\right)=16\)
\(\Leftrightarrow\left(x+4\right)^2-4^2-\left(x+9\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left(x+4-4\right)\left(x+4+4\right)-\left(x+9\right)\left(x-1\right)=0\)
\(\Leftrightarrow x\left(x+8\right)-\left(x+9\right)\left(x-1\right)=0\)
\(\Leftrightarrow x^2+8x-x^2-8x+9=0\)
\(\Leftrightarrow9=0\left(vl\right)\)
Vậy S = \(\varnothing\)
NN
Tìm x
a) 9(3x-2)=x(2-3x)
b) 25x2-2=0
c) x2-25=6x-9
d) (x+2)2-(x-2)(x+2)=0
e) x3-8=(x-2)3
f) x3+5x2-4x-20=0
1
9 tháng 6 2017
a) 9(3x - 2) = x(2 - 3x)
\(\Leftrightarrow\)-9(2 - 3x) = x(2 - 3x)
\(\Leftrightarrow\)-9(2 - 3x) - x(2 - 3x) = 0
\(\Leftrightarrow\)(2 - 3x)(- 9 - x) = 0
\(\Leftrightarrow\)2 - 3x = 0 hay - 9 - x = 0
\(\Leftrightarrow\) 3x = 2 \(\Leftrightarrow\) x = - 9
\(\Leftrightarrow\) x = 2/3
b) 25x2 - 2 = 0
\(\Leftrightarrow\)(5x)2 - (\(\sqrt{2}\))2 = 0
\(\Leftrightarrow\)(5x - \(\sqrt{2}\))(5x + \(\sqrt{2}\)) = 0
\(\Leftrightarrow\)5x - \(\sqrt{2}\)= 0 hay 5x + \(\sqrt{2}\)= 0
\(\Leftrightarrow\)5x = \(\sqrt{2}\) \(\Leftrightarrow\)5x = -\(\sqrt{2}\)
\(\Leftrightarrow\) x = \(\sqrt{2}\)/5 \(\Leftrightarrow\) x = -\(\sqrt{2}\)/5
c) x2 - 25 = 6x - 9
\(\Leftrightarrow\)(x2 - 6x + 9) - 25 = 0
\(\Leftrightarrow\)(x - 3)2 - 52 = 0
\(\Leftrightarrow\)(x - 3 - 5)(x - 3 + 5) = 0
\(\Leftrightarrow\)(x - 7)(x + 2) = 0
\(\Leftrightarrow\)x - 7 = 0 hay x + 2 = 0
\(\Leftrightarrow\)x = 7 \(\Leftrightarrow\)x = -2
d) (x + 2)2 - (x - 2)(x + 2) = 0
\(\Leftrightarrow\)(x + 2)(x + 2) - (x - 2)(x + 2) = 0
\(\Leftrightarrow\)(x + 2)(x + 2 - x + 2) = 0
\(\Leftrightarrow\)(x + 2)4 = 0 (hay 4(x + 2) = 0)
\(\Leftrightarrow\)x + 2 = 0 (vì 4 \(\ne\)0)
\(\Leftrightarrow\)x = -2
e) x3 - 8 = (x - 2)3
\(\Leftrightarrow\)x3 - 23 = (x - 2)3
\(\Leftrightarrow\)(x - 2)(x2 + 2x + 4) = (x - 2)3
\(\Leftrightarrow\)(x - 2)(x2 + 2x + 4) - (x - 2)3 = 0
\(\Leftrightarrow\)(x - 2)(x2 + 2x + 4) - (x - 2)(x - 2)2 = 0
\(\Leftrightarrow\)(x - 2)[x2 + 2x + 4 - (x - 2)2] = 0
\(\Leftrightarrow\)(x - 2)[x2 + 2x + 4 - (x2 - 4x + 4)] = 0
\(\Leftrightarrow\)(x - 2)(x2 + 2x + 4 - x2 + 4x - 4) = 0
\(\Leftrightarrow\)(x - 2)6x = 0 (hay 6x(x - 2) = 0)
\(\Leftrightarrow\)x - 2 = 0 hay x = 0 (vì 6\(\ne\)0)
\(\Leftrightarrow\)x = 2
f) x3 + 5x2 - 4x - 20 = 0
\(\Leftrightarrow\)x2(x + 5) - 4(x + 5) = 0
\(\Leftrightarrow\)(x + 5)(x2 - 4) = 0
\(\Leftrightarrow\)(x + 5)(x - 2)(x + 2) = 0
\(\Leftrightarrow\)x + 5 = 0 hay x - 2 = 0 hay x + 2 = 0
\(\Leftrightarrow\)x = -5 \(\Leftrightarrow\)x = 2 \(\Leftrightarrow\)x = -2