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HM
0
DD
1
27 tháng 9 2019
x^3 - 8 - (x - 2)(x - 12) = 0
x^3 - x^2 + 14x - 32 = 0
(x - 2)(x^2 + x + 16) = 0
vì x^2 + x + 16 # 0
=> x - 2 = 0
=> x = 2
14 tháng 12 2017
a) \(x\left(5+3x\right)-\left(x+1\right)\left(3x-2\right)=12\)
\(5x+3x^2-3x^2+2x-3x+2=12\)
\(4x=10\)
\(x=\frac{5}{2}\)
vậy \(x=\frac{5}{2}\)
\(13x\left(x-8\right)-x+8=0\)
\(13x\left(x-8\right)-\left(x-8\right)=0\)
\(\left(13x-1\right)\left(x-8\right)=0\)
\(\Rightarrow\orbr{\begin{cases}13x-1=0\\x-8=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{1}{13}\\x=8\end{cases}}\)
vậy \(\orbr{\begin{cases}x=\frac{1}{13}\\x=8\end{cases}}\)
3N
5
14 tháng 10 2020
a) x3 - 8 = ( x - 2 )( x - 12 )
<=> ( x - 2 )( x2 + 2x + 4 ) - ( x - 2 )( x - 12 ) = 0
<=> ( x - 2 )( x2 + 2x + 4 - x + 12 ) = 0
<=> ( x - 2 )( x2 + x + 16 ) = 0
<=> \(\orbr{\begin{cases}x-2=0\\x^2+x+16=0\end{cases}}\Leftrightarrow x=2\)( vì x2 + x + 16 = ( x2 + x + 1/4 ) + 63/4 = ( x + 1/2 )2 + 63/4 ≥ 63/4 > 0 ∀ x )
b) x2( x2 + 4 ) - x2 = 4
<=> x2( x2 + 4 ) - x2 - 4 = 0
<=> x2( x2 + 4 ) - ( x2 + 4 ) = 0
<=> ( x2 + 4 )( x2 - 1 ) = 0
<=> \(\orbr{\begin{cases}x^2+4=0\\x^2-1=0\end{cases}}\Leftrightarrow x=\pm1\)( vì x2 + 4 ≥ 4 > 0 ∀ x )
A
14 tháng 10 2020
Ko khó nè :3, đừng tách ra nhé !
a, \(x^3-8=\left(x-2\right)\left(x-12\right)\)
\(\Leftrightarrow\left(x-2\right)\left(x^2+2x+4\right)=\left(x-2\right)\left(x-12\right)\)
\(\Leftrightarrow\left(x-2\right)\left(x^2+2x+4-x+12\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^2+x+16\right)=0\Leftrightarrow x=2\)
b, \(x^2\left(x^2+4\right)-x^2=4\Leftrightarrow-x^2\left(-x^2-4\right)-x^2=4\)
\(\Leftrightarrow-x^2\left(4-x^2\right)-x^2=4\Leftrightarrow-x^2\left(2-x\right)\left(2+x\right)-x^2-4=0\)
\(\Leftrightarrow-x^2\left(2-x\right)\left(2+x\right)+\left(-x^2-4\right)=0\)
\(\Leftrightarrow-x^2\left(2-x\right)\left(2+x\right)+\left(2-x\right)\left(2+x\right)=0\)
\(\Leftrightarrow\left(-x^2+1\right)\left(2-x\right)\left(2+x\right)=0\Leftrightarrow x=\pm1;\pm2\)
Check hộ dáp án nhá :), ko chắc lắm nếu khai triển sẽ dễ nhìn hơn đấy.
NH
11 tháng 11 2020
a)(x+2).(x+3)-(x-2).(x+5)=10
( x^2 +3x+2x+6)-(x^2 +5x-2x-10)=10
x^2 +3x+2x+6-x^2 -5x+2x+10-10=0
2x+6=0
2x=-6
x=-3
HD
"1
P=1/12*(x^2+y^2)^2-2x^2*y^2 biết x^2-y^2=4
2 tìm x
X^9+x^8+x-1=0
3phân tích
X^2+2x-4y^2+4y
3xy-z-3x+yz
0
NT
2
MT
27 tháng 7 2015
(x-3)^2-(x-2)(x-8)=12
<=>x2-6x+9-(x2-10x+16)=12
<=>x2-6x+9-x2+10x-16=12
<=>4x-7=12
<=>4x=19
<=>x=\(\frac{19}{4}\)
(2x+5)^2=(3x-8)^2
<=>(2x+5)2-(3x-8)2=0
<=>(2x-5-3x+8)(2x-5+3x-8)=0
<=>(3-x)(5x-13)=0
<=>3-x=0 hoặc 5x-13=0
<=>x=3 hoặc x=\(\frac{13}{5}\)
MT
27 tháng 7 2015
sửa lại :
(x-3)^2-(x-2)(x-8)=12
<=>x2-6x+9-x2+10x-16=12
<=>4x-7=12
<=>4x=19
<=>x=19/4
(2x+5)^2=(3x-8)^2
<=>(2x+5)2-(3x-8)2=0
<=>(2x+5-3x+8)(2x+5+3x-8)=0
<=>(11-x)(5x-3)=0
<=>11-x=0 hoặc 5x-3=0
<=>x=11 hoặc x=3/5
13 tháng 7 2016
\(a,5\left(3x+5\right)-4\left(2x-3\right)=5x+8\left(2x+12\right)+1\)
\(\Rightarrow5\left(3x+5\right)-4\left(2x-3\right)-5x-8\left(2x+12\right)-1=0\)
\(\Rightarrow15x+25-8x+12-5x-16x-96-1=0\)
\(\Rightarrow-14x-60=0\)
\(\Rightarrow-14x=60\) \(\Rightarrow x=-\frac{60}{14}=\frac{-30}{7}\)
\(b,\left(2x+3\right)\left(x-4\right)-\left(3x-5\right)\left(x-4\right)=\left(5-x\right)\left(x-2\right)\)
\(\Rightarrow2x^2+3x-8x-12-3x^2+5x+12x-20=5x-x^2-10+2x\)
\(\Rightarrow-x^2+12x-32=7x-x^2-10\)
\(\Rightarrow-x^2+12x-32-7x+x^2+10=0\)
\(\Rightarrow5x-22=0\)
\(\Rightarrow5x=22\Rightarrow x=\frac{22}{5}\)
PA
13 tháng 7 2016
a) 5(3x+5)-4(2x-3) = 5x+8(2x+12)+1
15x + 25 - 8x + 12 = 5x + 16x + 96 + 1
15x - 8x - 5x - 16x = 96 + 1 - 25 - 12
-14x = 60
x = \(\frac{60}{-14}\)
x = \(-\frac{30}{7}\)
b) (2x+3)(x-4)-(3x-5)(x-4) = (5-x).(x-2)
(x - 4)(2x + 3 - 3x +5) = 5x - 10 - x2 + 2x
(x - 4)[(2x - 3x) + (3 + 5)] = 5x - 10 - x2 + 2x
(x - 4)(-x + 8) = 5x - 10 - x2 + 2x
-x2 + 8x + 4x - 32 = 5x - 10 - x2 + 2x
(-x2 + x2) + (8x + 4x - 5x - 2x) = -10 + 32
5x = 22
x = \(\frac{22}{5}\)
11 tháng 10 2020
a) \(x\left(x-2\right)-7x+14=0\)
\(\Leftrightarrow x\left(x-2\right)-7\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-7\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=2\\x=7\end{cases}}\)
b) \(x^2\left(x-3\right)+12-4x=0\)
\(\Leftrightarrow x^2\left(x-3\right)-4\left(x-3\right)=0\)
\(\Leftrightarrow\left(x^2-4\right)\left(x-3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=3\\x^2=4\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\\x=\pm2\end{cases}}\)
c) \(x^2+12x-13=0\)
\(\Leftrightarrow\left(x^2-x\right)+\left(13x-13\right)=0\)
\(\Leftrightarrow x\left(x-1\right)+13\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+13\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=1\\x=-13\end{cases}}\)
d) \(4x^2-4x=8\)
\(\Leftrightarrow x^2-x-2=0\)
\(\Leftrightarrow\left(x+1\right)\left(x-2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=-1\\x=2\end{cases}}\)
e) \(x^2-6x=1\)
\(\Leftrightarrow\left(x-3\right)^2=10\)
\(\Leftrightarrow\orbr{\begin{cases}x-3=\sqrt{10}\\x-3=-\sqrt{10}\end{cases}}\Rightarrow\orbr{\begin{cases}x=3+\sqrt{10}\\x=3-\sqrt{10}\end{cases}}\)
11 tháng 10 2020
a) x( x - 2 ) - 7x + 14 = 0
<=> x( x - 2 ) - 7( x - 2 ) = 0
<=> ( x - 2 )( x - 7 ) = 0
<=> \(\orbr{\begin{cases}x-2=0\\x-7=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=2\\x=7\end{cases}}\)
b) x2( x - 3 ) + 12 - 4x = 0
<=> x2( x - 3 ) - 4( x - 3 ) = 0
<=> ( x - 3 )( x2 - 4 ) = 0
<=> \(\orbr{\begin{cases}x-3=0\\x^2-4=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\\x=\pm2\end{cases}}\)
c) x2 + 12x - 13 = 0
<=> x2 - x + 13x - 13 = 0
<=> x( x - 1 ) + 13( x - 1 ) = 0
<=> ( x - 1 )( x + 13 ) = 0
<=> \(\orbr{\begin{cases}x-1=0\\x+13=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\\x=-13\end{cases}}\)
d) 4x2 - 4x = 8
<=> 4( x2 - x ) = 8
<=> x2 - x = 2
<=> x2 - x - 2 = 0
<=> x2 + x - 2x - 2 = 0
<=> x( x + 1 ) - 2( x + 1 ) = 0
<=> ( x + 1 )( x - 2 ) = 0
<=> \(\orbr{\begin{cases}x+1=0\\x-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-1\\x=2\end{cases}}\)
e) x2 - 6x = 1
<=> x2 - 6x + 9 = 1 + 9
<=> ( x - 3 )2 = 10
<=> ( x - 3 )2 = ( ±√10 )2
<=> \(\orbr{\begin{cases}x-3=\sqrt{10}\\x-3=-\sqrt{10}\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=3+\sqrt{10}\\x=3-\sqrt{10}\end{cases}}\)
\(8-12:\left(x-2\right)=2\)
=>\(\dfrac{12}{x-2}=8-2=6\)
=>\(x-2=\dfrac{12}{6}=2\)
=>x=2+2=4