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BA
25 tháng 10 2019
\(2x^3+5x^2-12x=0\)
\(\Rightarrow x\cdot\left(2x^2+5x-12\right)=0\)
\(\Rightarrow x\cdot\left(2x^2-3x+8x-12\right)=0\)
\(\Rightarrow x\cdot\left[x\cdot\left(2x-3\right)+4\cdot\left(2x-3\right)\right]=0\)
\(\Rightarrow x\cdot\left(2x-3\right)\cdot\left(x+4\right)=0\)
\(\Rightarrow\hept{\begin{cases}x=0\\2x-3=0\\x+4=0\end{cases}}\Rightarrow\hept{\begin{cases}x=0\\x=\frac{3}{2}\\x=-4\end{cases}}\)
\(x^2-5x-24=0\)
\(\Rightarrow x^2+3x-8x-24=0\)
\(\Rightarrow x\cdot\left(x+3\right)-8\cdot\left(x+3\right)=0\)
\(\Rightarrow\left(x+3\right)\cdot\left(x-8\right)=0\)
\(\Rightarrow\hept{\begin{cases}x+3=0\\x-8=0\end{cases}\Rightarrow\hept{\begin{cases}x=-3\\x=8\end{cases}}}\)
\(x^2-6x+8=0\)
\(\Rightarrow x^2-2x-4x+8=0\)
\(\Rightarrow x\cdot\left(x-2\right)-4\cdot\left(x-2\right)=0\)
\(\Rightarrow\left(x-2\right)\cdot\left(x-4\right)=0\)
\(\Rightarrow\hept{\begin{cases}x-2=0\\x-4=0\end{cases}\Rightarrow\hept{\begin{cases}x=2\\x=4\end{cases}}}\)
BA
26 tháng 10 2019
\(x^2-5x-24=0\)
\(x^2+3x-8x-24=0\)
\(x\cdot\left(x+3\right)-8\cdot\left(x+3\right)=0\)
\(\left(x+3\right)\cdot\left(x-8\right)=0\)
\(\hept{\begin{cases}x+3=0\\x-8=0\end{cases}\Rightarrow\hept{\begin{cases}x=-3\\x=8\end{cases}}}\)
\(x^2-6x+8=0\)
\(x^2-2x-4x+8=0\)
\(x\cdot\left(x-2\right)-4\cdot\left(x-2\right)=0\)
\(\left(x-2\right)\cdot\left(x-4\right)=0\)
\(\hept{\begin{cases}x-2=0\\x-4=0\end{cases}\Rightarrow\hept{\begin{cases}x=2\\x=4\end{cases}}}\)
KN
26 tháng 10 2019
\(2x^2+5x^2-12x=0\)
\(\Leftrightarrow7x^2-12x=0\)
\(\Leftrightarrow x\left(7x-12\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\7x-12=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{12}{7}\end{cases}}\)
A
1 tháng 10 2020
a, \(x^2-12x-2x+24=0\Leftrightarrow x^2-14x+24=0\Leftrightarrow\left(x-12\right)\left(x-2\right)=0\)
TH1 : x = 12 ; TH2 : x = 2
b, \(x^2-5x-24=0\Leftrightarrow\left(x-8\right)\left(x+3\right)=0\)
TH1 : x = 8 ; TH2 : x = -3
c, \(4x^2-12x-7=0\Leftrightarrow\left(2x+1\right)\left(2x-7\right)=0\)
TH1 : x = -1/2 ; TH2 : x = 7/2
d, \(x^3+6x^2+12x+8=0\Leftrightarrow\left(x+2\right)^3=0\Leftrightarrow x=-2\)
Tương tự HĐT thôi :)
1 tháng 10 2020
a) x2 - 12x - 2x + 24 = 0
<=> x( x - 12 ) - 2( x - 12 ) = 0
<=> ( x - 12 )( x - 2 ) = 0
<=> \(\orbr{\begin{cases}x-12=0\\x-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=12\\x=2\end{cases}}\)
b) x2 - 5x - 24 = 0
<=> x2 + 3x - 8x - 24 = 0
<=> x( x + 3 ) - 8( x + 3 ) = 0
<=> ( x + 3 )( x - 8 ) = 0
<=> \(\orbr{\begin{cases}x+3=0\\x-8=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-3\\x=8\end{cases}}\)
c) 4x2 - 12x - 7 = 0
<=> 4x2 + 2x - 14x - 7 = 0
<=> 2x( 2x + 1 ) - 7( 2x + 1 ) = 0
<=> ( 2x + 1 )( 2x - 7 ) = 0
<=> \(\orbr{\begin{cases}2x+1=0\\2x-7=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-\frac{1}{2}\\x=\frac{7}{2}\end{cases}}\)
d) x3 + 6x2 + 12x + 8 = 0
<=> ( x + 2 )3 = 0
<=> x + 2 = 0
<=> x = -2
e) ( x + 2 )2 - x2 + 4 = 0
<=> x2 + 4x + 4 - x2 + 4 = 0
<=> 4x + 8 = 0
<=> 4x = -8
<=> x = -2
f) 2( x + 5 ) = x2 + 5x
<=> x2 + 5x - 2x - 10 = 0
<=> x( x + 5 ) - 2( x + 5 ) = 0
<=> ( x + 5 )( x - 2 ) = 0
<=> \(\orbr{\begin{cases}x+5=0\\x-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-5\\x=2\end{cases}}\)
m) 16( 2x - 3 )2 - 25( x - 5 )2 = 0
<=> 42( 2x - 3 )2 - 52( x - 5 )2 = 0
<=> [ 4( 2x - 3 ) ]2 - [ 5( x - 5 ) ]2 = 0
<=> ( 8x - 12 )2 - ( 5x - 25 )2 = 0
<=> [ 8x - 12 - ( 5x - 25 ) ][ 8x - 12 + ( 5x - 25 ) ] = 0
<=> ( 8x - 12 - 5x + 25 )( 8x - 12 + 5x - 25 ) = 0
<=> ( 3x + 13 )( 13x - 37 ) = 0
<=> \(\orbr{\begin{cases}3x+13=0\\13x-37=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-\frac{13}{3}\\x=\frac{37}{13}\end{cases}}\)
n) x2 - 6x + 4 = 0
<=> ( x2 - 6x + 9 ) - 5 = 0
<=> ( x - 3 )2 - ( √5 )2 = 0
<=> ( x - 3 - √5 )( x - 3 + √5 ) = 0
<=> \(\orbr{\begin{cases}x-3-\sqrt{5}=0\\x-3+\sqrt{5}=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3+\sqrt{5}\\x=3-\sqrt{5}\end{cases}}\)
NC
1 tháng 10 2020
a) \(x^2-12x-2x+24=0\)
\(\Leftrightarrow x\left(x-12\right)-2\left(x-12\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-12\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=12\\x=2\end{cases}}\)
b) \(x^2-5x-24=0\)
\(\Leftrightarrow\left(x^2+3x\right)-\left(8x+24\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x-8\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=-3\\x=8\end{cases}}\)
c) \(4x^2-12x-7=0\)
\(\Leftrightarrow\left(4x^2-14x\right)+\left(2x-7\right)=0\)
\(\Leftrightarrow\left(2x-7\right)\left(2x+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{7}{2}\\x=-\frac{1}{2}\end{cases}}\)
d) \(x^3+6x^2+12x+8=0\)
\(\Leftrightarrow\left(x+2\right)^3=0\)
\(\Rightarrow x=-2\)
NC
1 tháng 10 2020
e) \(\left(x+2\right)^2-x^2+4=0\)
\(\Leftrightarrow4x+8=0\)
\(\Rightarrow x=-2\)
f) \(2\left(x+5\right)=x^2+5x\)
\(\Leftrightarrow2\left(x+5\right)-x\left(x+5\right)=0\)
\(\Leftrightarrow\left(x+5\right)\left(2-x\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=-5\\x=2\end{cases}}\)
m) \(16\left(2x-3\right)^2-25\left(x-5\right)^2=0\)
\(\Leftrightarrow\orbr{\begin{cases}8x-12=5x-25\\8x-12=25-5x\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}3x=-13\\13x=37\end{cases}}\Rightarrow\orbr{\begin{cases}x=-\frac{13}{3}\\x=\frac{37}{13}\end{cases}}\)
n) \(x^2-6x+4=0\)
\(\Leftrightarrow\left(x-3\right)^2-5=0\)
\(\Leftrightarrow\left(x-3+\sqrt{5}\right)\left(x-3-\sqrt{5}\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=3+\sqrt{5}\\x=3-\sqrt{5}\end{cases}}\)
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: \(S=\left\{\dfrac{5}{2};\dfrac{1}{2}\right\}\)
4 tháng 3 2020
a,\(2x^2-6x+1=0\)
\(=>x.\left(2x-6\right)=1\)
\(th1:\orbr{\begin{cases}x=1\\2x-6=1\end{cases}=>\orbr{\begin{cases}x=1\\x=\frac{7}{2}\end{cases}}}\)
\(th2:\orbr{\begin{cases}x=-1\\2x-6=-1\end{cases}=>\orbr{\begin{cases}x=-1\\x=\frac{5}{2}\end{cases}}}\)
b,\(4x^2-12x+5=0\)
\(=>x.\left(4x-12\right)=-5\)
\(th1:\orbr{\begin{cases}x=1\\4x-12=-5\end{cases}=>\orbr{\begin{cases}x=1\\x=\frac{7}{4}\end{cases}}}\)
\(th2:\orbr{\begin{cases}x=-1\\4x-12=5\end{cases}=>\orbr{\begin{cases}x=-1\\x=\frac{17}{4}\end{cases}}}\)
\(th3:\orbr{\begin{cases}x=5\\4x-12=-1\end{cases}=>\orbr{\begin{cases}x=5\\x=\frac{11}{4}\end{cases}}}\)
\(th4:\orbr{\begin{cases}x=-5\\4x-12=1\end{cases}=>\orbr{\begin{cases}x=-5\\x=\frac{13}{4}\end{cases}}}\)
KN
4 tháng 3 2020
\(x^2+6x-16=0\)
Ta có \(\Delta=6^2+4.16=100,\sqrt{\Delta}=10\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{-6+10}{2}=2\\x=\frac{-6-10}{2}=-8\end{cases}}\)
\(5x^3+6x^2+12x+8=0\)
\(\Leftrightarrow4x^3+\left(x^2+6x+12x+8\right)=0\)
\(\Leftrightarrow4x^3+\left(x+2\right)^3=0\)
\(\Leftrightarrow\left(x+2\right)^3=-4x^3\)
\(\Leftrightarrow x+2-\sqrt[3]{4x}=0\)
\(\Leftrightarrow x\left(1-\sqrt[3]{4}\right)=0\)
\(\Leftrightarrow x=-\frac{2}{1-\sqrt[3]{4}}=\frac{2}{\sqrt[3]{4}-1}\)
bài này chắc là xét mấy TH ta ??? nhưng tìm 1 KQ dễ lắm bn
\(5x^3+6x^2+12x+8=0\)
\(5.x.x.x+6x.x+12x=-8\)
\(23x=-8\)
\(x=-\frac{8}{23}\)