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14 tháng 8 2020
a) 16x^2 - (4x - 5)^2 = 15
<=> 16x^2 - 16x^2 + 40x - 25 = 15
<=> 40x = 40
<=> x = 1
b) (2x + 3)^2 - 4(x - 1)(x + 1) = 49
<=> 4x^2 + 12x + 9 - 4x^2 - 4x + 4x + 4 = 49
<=> 12x + 13 = 49
<=> 12x = 36
<=> x = 3
c) (2x + 1)(1 - 2x) + (1 - 2x)^2 = 18
<=> 1 - 4x^2 + 1 - 4x + 4x^2 = 18
<=> 2 - 4x = 18
<=> -4x = 16
<=> x = -4
d)2(x + 1)^2 - (x - 3)(x + 3) - (x - 4)^2 = 0
<=> 2x^2 + 4x + 2 - x^2 + 3^2 - x^2 + 8x - 16 = 0
<=> 12x - 5 = 0
<=> 12x = 5
<=> x = 5/12
e) (x - 5)^2 - x(x - 4) = 9
<=> x^2 - 10x + 25 - x^2 + 4x = 9
<=> -6x + 25 = 9
<=> -6x = 9 - 25
<=> -6x = -16
<=> x = -16/-6 = 8/3
f) (x - 5)^2 + (x - 4)(1 - x) = 0
<=> x^2 - 10x + 25 + x - x^2 - x - 4 + 4x = 0
<=> -5x + 21 = 0
<=> -5x = -21
<=> x = 21/5
NM
2
17 tháng 8 2015
a) x^4 - 2x^2 + 1 = 0
=> ( x^2 - 1 )^2 = 0
=> x^2 - 1 = 0
=> x^2 = 1
=> x = 1 hoặc x = -1
CW
4 tháng 12 2016
a) x4-2x2+1=0
(thang Tran giải rồi nhé)
b) x4-2x2-8=0
<=> x^4 - 2x^2 +1 -9 =0
<=> (x^2 -1)^2 -9 =0
\(\Leftrightarrow\orbr{\begin{cases}x^2-1=-3\\x^2-1=3\end{cases}\Leftrightarrow\orbr{\begin{cases}x^2=-2\left(VN\right)\\x=+_-\sqrt{2}\end{cases}}}\)
Vậy x=+- căn 2
c) x4-4x2-60=0
\(\Leftrightarrow x^4-4x^2+4-64=0\)
\(\Leftrightarrow\left(x^2-2\right)-64=0\)
\(\Leftrightarrow\left(x^2+62\right)\left(x^2-66\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x^2+62=0\\x^2-66=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x^2=-62\left(VN\right)\\x^2=+_-\sqrt{66}\end{cases}}}\)
Vậy x=+- căn 66
d) x6-16x2+64=0
6 tháng 10 2020
a) \(4x^3-9x=0\)
\(\Leftrightarrow x\left(4x^2-9\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\4x^2=9\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\pm\frac{3}{2}\end{cases}}\)
b) \(3x\left(x-2\right)-5x+10=0\)
\(\Leftrightarrow\left(3x-5\right)\left(x-2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{5}{3}\\x=2\end{cases}}\)
c) \(4x\left(x+3\right)-x^2+9=0\)
\(\Leftrightarrow4x\left(x+3\right)-\left(x-3\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left(3x+3\right)\left(x+3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=-1\\x=-3\end{cases}}\)
d) \(\left(2x+5\right)\left(x-4\right)=\left(x-4\right)\left(5-x\right)\)
\(\Leftrightarrow\left(2x+5\right)\left(x-4\right)+\left(x-5\right)\left(x-4\right)=0\)
\(\Leftrightarrow3x\left(x-4\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=4\end{cases}}\)
6 tháng 10 2020
e) \(16x^2-25=\left(4x-5\right)\left(2x+1\right)\)
\(\Leftrightarrow\left(4x-5\right)\left(4x+5\right)-\left(4x-5\right)\left(2x+1\right)=0\)
\(\Leftrightarrow\left(4x-5\right)\left(2x+4\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{5}{4}\\x=-2\end{cases}}\)
f) \(\left(x+\frac{1}{5}\right)^2=\frac{64}{9}\)
\(\Leftrightarrow\orbr{\begin{cases}x+\frac{1}{5}=\frac{8}{3}\\x+\frac{1}{5}=-\frac{8}{3}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{37}{15}\\x=-\frac{43}{15}\end{cases}}\)
g) \(9\left(x+2\right)^2=\left(x+3\right)^2\)
\(\Leftrightarrow\orbr{\begin{cases}3x+6=x+3\\3x+6=-x-3\end{cases}}\Leftrightarrow\orbr{\begin{cases}2x=-3\\4x=-9\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-\frac{3}{2}\\x=-\frac{9}{4}\end{cases}}\)
6 tháng 10 2020
a) 4x3 - 9x = 0
<=> x( 4x2 - 9 ) = 0
<=> x( 2x - 3 )( 2x + 3 ) = 0
<=> x = 0 hoặc 2x - 3 = 0 hoặc 2x + 3 = 0
<=> x = 0 hoặc x = ±3/2
b) 3x( x - 2 ) - 5x + 10 = 0
<=> 3x( x - 2 ) - 5( x - 2 ) = 0
<=> ( x - 2 )( 3x - 5 ) = 0
<=> x - 2 = 0 hoặc 3x - 5 = 0
<=> x = 2 hoặc x = 5/3
c) 4x( x + 3 ) - x2 + 9 = 0
<=> 4x( x + 3 ) - ( x2 - 9 ) = 0
<=> 4x( x + 3 ) - ( x - 3 )( x + 3 ) = 0
<=> ( x + 3 )[ 4x - ( x - 3 ) ] = 0
<=> ( x + 3 )( 4x - x + 3 ) = 0
<=> ( x + 3 )( 3x + 3 ) = 0
<=> x + 3 = 0 hoặc 3x + 3 = 0
<=> x = -3 hoặc x= -1
d) ( 2x + 5 )( x - 4 ) = ( x - 4 )( 5 - x )
<=> ( 2x + 5 )( x - 4 ) - ( x - 4 )( 5 - x ) = 0
<=> ( x - 4 )[ ( 2x + 5 ) - ( 5 - x ) ] = 0
<=> ( x - 4 )( 2x + 5 - 5 + x ) = 0
<=> ( x - 4 ).3x = 0
<=> x - 4 = 0 hoặc 3x = 0
<=> x = 4 hoặc x = 0
e) 16x2 - 25 = ( 4x - 5 )( 2x + 1 )
<=> ( 4x - 5 )( 4x + 5 ) - ( 4x - 5 )( 2x + 1 ) = 0
<=> ( 4x - 5 )[ ( 4x + 5 ) - ( 2x + 1 ) ] = 0
<=> ( 4x - 5 )( 4x + 5 - 2x - 1 ) = 0
<=> ( 4x - 5 )( 2x + 4 ) = 0
<=> 4x - 5 = 0 hoặc 2x + 4 = 0
<=> x = 5/4 hoặc x = -2
f) ( x + 1/5 )2 = 64/9
<=> ( x + 1/5 )2 = ( ±8/3 )2
<=> x + 1/5 = 8/3 hoặc x + 1/5 = -8/3
<=> x = 37/15 hoặc x = -43/15
g) 9( x + 2 )2 = ( x + 3 )2
<=> 32( x + 2 )2 - ( x + 3 )2 = 0
<=> [ 3( x + 2 ) ]2 - ( x + 3 )2 = 0
<=> ( 3x + 6 )2 - ( x + 3 )2 = 0
<=> [ ( 3x + 6 ) - ( x + 3 ) ][ ( 3x + 6 ) + ( x + 3 ) ] = 0
<=> ( 3x + 6 - x - 3 )( 3x + 6 + x + 3 ) = 0
<=> ( 2x + 3 )( 4x + 9 ) = 0
<=> 2x + 3 = 0 hoặc 4x + 9 = 0
<=> x = -3/2 hoặc x = -9/4
13 tháng 9 2020
a) x4 - 16x2 = 0
<=> ( x2 )2 - ( 4x )2 = 0
<=> ( x2 - 4x )( x2 + 4x ) = 0
<=> [ x( x - 4 ) ][ x( x + 4 ) ] = 0
<=> x( x - 4 )x( x + 4 ) = 0
<=> x2( x - 4 )( x + 4 ) = 0
<=> \(\hept{\begin{cases}x^2=0\\x-4=0\\x+4=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\pm4\end{cases}}\)( thay bằng dấu hoặc hộ mình nhé )
b) 9x2 + 6x + 1 = 0
<=> ( 3x )2 + 2.3x.1 + 12 = 0
<=> ( 3x + 1 )2 = 0
<=> 3x + 1 = 0
<=> 3x = -1
<=> x = -1/3
c) x2 - 6x = 16
<=> x2 - 6x - 16 = 0
<=> x2 + 2x - 8x - 16 = 0
<=> x( x + 2 ) - 8( x + 2 ) = 0
<=> ( x + 2 )( x - 8 ) = 0
<=> \(\orbr{\begin{cases}x+2=0\\x-8=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-2\\x=8\end{cases}}\)
d) 9x2 + 6x = 80
<=> 9x2 + 6x - 80 = 0
<=> 9x2 + 30x - 24x - 80 = 0
<=> 9x( x + 10/3 ) - 24( x + 10/3 ) = 0
<=> ( x + 10/3 )( 9x - 24 ) = 0
<=> \(\orbr{\begin{cases}x+\frac{10}{3}=0\\9x-24=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-\frac{10}{3}\\x=\frac{8}{3}\end{cases}}\)
e) Áp dụng công thức an.bn = ( ab )n ta có :
25( 2x - 1 )2 - 9( x + 1 )2 = 0
<=> 52( 2x - 1 )2 - 32( x + 1 )2 = 0
<=> [ 5( 2x - 1 ) ]2 - [ 3( x + 1 ) ]2 = 0
<=> ( 10x - 5 )2 - ( 3x + 3 )2 = 0
<=> [ ( 10x - 5 ) - ( 3x + 3 ) ][ ( 10x - 5 ) + ( 3x + 3 ) ] = 0
<=> ( 10x - 5 - 3x - 3 )( 10x - 5 + 3x + 3 ) = 0
<=> ( 7x - 8 )( 13x - 2 ) = 0
<=> \(\orbr{\begin{cases}7x-8=0\\13x-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{8}{7}\\x=\frac{2}{13}\end{cases}}\)
B
16 tháng 9 2020
Bài làm :
a) x4 - 16x2 = 0
<=> ( x2 )2 - ( 4x )2 = 0
<=> ( x2 - 4x )( x2 + 4x ) = 0
<=> [ x( x - 4 ) ][ x( x + 4 ) ] = 0
<=> x( x - 4 )x( x + 4 ) = 0
<=> x2( x - 4 )( x + 4 ) = 0
Vậy x=0 hoặc x=±4
b) 9x2 + 6x + 1 = 0
<=> ( 3x )2 + 2.3x.1 + 12 = 0
<=> ( 3x + 1 )2 = 0
<=> 3x + 1 = 0
<=> 3x = -1
<=> x = -1/3
c) x2 - 6x = 16
<=> x2 - 6x - 16 = 0
<=> x2 + 2x - 8x - 16 = 0
<=> x( x + 2 ) - 8( x + 2 ) = 0
<=> ( x + 2 )( x - 8 ) = 0
\(\Leftrightarrow\orbr{\begin{cases}x+2=0\\x-8=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-2\\x=8\end{cases}}\)
d) 9x2 + 6x = 80
<=> 9x2 + 6x - 80 = 0
<=> 9x2 + 30x - 24x - 80 = 0
<=> 9x( x + 10/3 ) - 24( x + 10/3 ) = 0
<=> ( x + 10/3 )( 9x - 24 ) = 0
\(\Leftrightarrow\orbr{\begin{cases}x+\frac{10}{3}=0\\9x-24=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-\frac{10}{3}\\x=\frac{8}{3}\end{cases}}\)
e) 25( 2x - 1 )2 - 9( x + 1 )2 = 0
<=> 52( 2x - 1 )2 - 32( x + 1 )2 = 0
<=> [ 5( 2x - 1 ) ]2 - [ 3( x + 1 ) ]2 = 0
<=> ( 10x - 5 )2 - ( 3x + 3 )2 = 0
<=> [ ( 10x - 5 ) - ( 3x + 3 ) ][ ( 10x - 5 ) + ( 3x + 3 ) ] = 0
<=> ( 10x - 5 - 3x - 3 )( 10x - 5 + 3x + 3 ) = 0
<=> ( 7x - 8 )( 13x - 2 ) = 0
\(\Leftrightarrow\orbr{\begin{cases}7x-8=0\\13x-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{8}{7}\\x=\frac{2}{13}\end{cases}}\)
XO
13 tháng 9 2020
a) Ta có : x4 - 16x2 = 0
=> x4 - 8x2 - 8x2 + 64 = 64
=> x2(x2 - 8) - 8(x2 - 8) = 64
=> (x2 - 8)2 = 64
=> \(\orbr{\begin{cases}x^2-8=8\\x^2-8=-8\end{cases}}\Rightarrow\orbr{\begin{cases}x^2=16\\x^2=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\pm4\\x=0\end{cases}}\Rightarrow x\in\left\{4;-4;0\right\}\)
b) Ta có 9x2 + 6x + 1 = 0
=> 9x2 + 3x + 3x + 1 = 0
=> 3x(3x + 1) + (3x + 1) = 0
=> (3x + 1)2 = 0
=> 3x + 1 = 0
=> x = -1/3
c) Ta có x2 - 6x = 16
=> x2 - 6x + 9 = 25
=> (x - 3)2 = 25
=> \(\orbr{\begin{cases}x-3=5\\x-3=-5\end{cases}}\Rightarrow\orbr{\begin{cases}x=8\\x=-2\end{cases}}\Rightarrow x\in\left\{8;-2\right\}\)
d) 9x2 + 6x = 80
=> 9x2 + 6x + 1 = 81
=> (3x + 1)2 = 81
=> \(\orbr{\begin{cases}3x+1=9\\3x+1=-9\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{8}{3}\\x=-\frac{10}{3}\end{cases}\Rightarrow x\in}\left\{\frac{8}{3};\frac{-10}{3}\right\}\)
e) 25(2x - 1)2 - 9(x + 1)2 = 0
=> [5(2x - 1)]2 - [3(x + 1)]2 = 0
=> (10x - 5)2 - (3x + 3)2 = 0
=> (10x - 5 - 3x - 3)(10x - 5 + 3x + 3) = 0
=> (7x - 8)(13x - 2) = 0
=> \(\orbr{\begin{cases}7x=8\\13x=2\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{8}{7}\\x=\frac{2}{13}\end{cases}}\)
12 tháng 10 2020
a) 2x (x-5) -(x2-10x +25)=0
\(\Leftrightarrow\)2x(x-5)-(x-5)2=0
\(\Leftrightarrow\)(x-5)(2x-x+5)=0
\(\Leftrightarrow\)(x-5)(x+5)=0
\(\Leftrightarrow\)\(\left[{}\begin{matrix}x-5=0\\x+5=0\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left[{}\begin{matrix}x=5\\x=-5\end{matrix}\right.\)
b) x2 - 9 +3x(x+3) = 0
\(\Leftrightarrow\)(x2 - 9) +3x(x+3) =0
\(\Leftrightarrow\)(x-3)(x+3)+3x(x+3)=0
\(\Leftrightarrow\)(x+3)(x-3+3x)=0
\(\Leftrightarrow\)(x+3)(4x-3)=0
\(\Leftrightarrow\)\(\left[{}\begin{matrix}x+3=0\\4x-3=0\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left[{}\begin{matrix}x=-3\\4x=3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=\frac{3}{4}\end{matrix}\right.\)
c) x3 - 16x = 0
\(\Leftrightarrow\)x(x2-16)=0
\(\Leftrightarrow\)x(x-4)(x+4)=0
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-4=0\\x+4=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\\x=-4\end{matrix}\right.\)
d) (2x+3)(x-2) - (x2 -4x+4) = 0
\(\Leftrightarrow\)(2x+3)(x-2) -(x-2)2=0
\(\Leftrightarrow\)(x-2)(2x+3-x+2)=0
\(\Leftrightarrow\)(x-2)(x+5)=0
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+5=0\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left[{}\begin{matrix}x=2\\x=-5\end{matrix}\right.\)
e) 9x2 -(x2 -2x +1)=0
\(\Leftrightarrow\)(3x)2-(x-1)2=0
\(\Leftrightarrow\)(3x-x+1)(3x+x-1)=0
\(\Leftrightarrow\)(2x+1)(4x-1)=0
\(\Leftrightarrow\)\(\left[{}\begin{matrix}2x+1=0\\4x-1=0\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left[{}\begin{matrix}2x=-1\\4x=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{-1}{2}\\x=\frac{1}{4}\end{matrix}\right.\)
f)x3-4x2 -9x +36 = 0
\(\Leftrightarrow\)(x3-9x)-(4x2-36)=0
\(\Leftrightarrow\)x(x2-9)-4(x2-9)=0
\(\Leftrightarrow\)(x-4)(x2-9)=0
\(\Leftrightarrow\)(x-4)(x-3)(x+3)=0
\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\x-3=0\\x+3=0\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left[{}\begin{matrix}x=4\\x=3\\x=-3\end{matrix}\right.\)
g) 3x - 6 = (x-1).(x-2)
\(\Leftrightarrow\)3(x-2)=(x-1)(x-2)
\(\Leftrightarrow\)x-1=3
\(\Leftrightarrow\)x=4
i) (x-2).(x+2) +(2x+1)2 =-5x.(x-3) =5 (?? đề sao vậy ??)
k) x2 -1 = (x-1).(2x+3)
\(\Leftrightarrow\)(x-1)(x+1)=(x-1)(2x+3)
\(\Leftrightarrow\)x+1=2x+3
\(\Leftrightarrow\)x-2x=3-1
\(\Leftrightarrow\)-x=2
\(\Leftrightarrow\)x=-2
l) (2x-1)2 +(x+3).(x-3) -5x(x-2)=6
\(\Leftrightarrow\)4x2-4x+1+x2-9-5x2+10x=6
\(\Leftrightarrow\)6x-8=6
\(\Leftrightarrow\)6x=14
\(\Leftrightarrow\)x=\(\frac{7}{3}\)
TT
2
TN
23 tháng 7 2017
\(a,x^3-16x=0\)
\(\Leftrightarrow x\left(x^2-16\right)=0\)
\(\Leftrightarrow x\left(x-4\right)\left(x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-4=0\\x+4=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=4\\x=-4\end{matrix}\right.\)
\(b,x^4-2x^3+10x^2-20x=0\)
\(\Leftrightarrow x^3\left(x-2\right)+10x\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^3+10x\right)=0\)
\(\Leftrightarrow\left(x-2\right)x\left(x^2+10\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x=0\\x^2+10=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\\x=0\\\left[{}\begin{matrix}x^2=10\\x^2=-10\end{matrix}\right.\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\\x=0\\x=\sqrt{10}\\x=-\sqrt{10}\end{matrix}\right.\)\(c,\left(2x-1\right)^2=\left(x+3\right)^2\)
\(\Leftrightarrow4x^2-4x+1=x^2+6x+9\)
\(\Leftrightarrow4x^2-4x+1-x^2-6x-9=0\)
\(\Leftrightarrow3x^2-10x-8=0\)
\(\Leftrightarrow3x^2-12x+2x-8=0\)
\(\Leftrightarrow3x\left(x-4\right)+2\left(x-4\right)=0\)
\(\Leftrightarrow\left(3x-2\right)\left(x-4\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}3x-2=0\\x-4=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-\dfrac{2}{3}\\x=4\end{matrix}\right.\)
Phần d tương tự
23 tháng 7 2017
Câu a :
\(x^3-16x=0\)
\(\Leftrightarrow x\left(x^2-4^2\right)=0\)
\(\Leftrightarrow x\left[\left(x+4\right)\left(x-4\right)\right]=0\)
\(\Rightarrow\) \(x=0\)
\(\Rightarrow\) \(x+4=0\Rightarrow x=-4\)
\(\Rightarrow x-4=0\Rightarrow x=4\)
Câu b :
\(x^4-2x^3+10x^2-20x=0\)
\(\Leftrightarrow x^3\left(x-2\right)+10x\left(x-2\right)\) \(=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^3+10x\right)=0\)
\(\Leftrightarrow x\left(x-2\right)\left(x^2+10\right)=0\)
\(\Rightarrow x=0\)
\(\left(x-2\right)=0\Rightarrow x=2\)
\(x^2+10=0\) \(\Rightarrow\) x ( loại )
16x2 - 9(x + 1)2 = 0
[4x - 3(x + 1)][4x + 3(x + 1)] = 0
(4x - 3x - 3)(4x + 3x + 3) = 0
(x - 3)(7x + 3) = 0
\(\left[\begin{array}{nghiempt}x-3=0\\7x+3=0\end{array}\right.\)
\(\left[\begin{array}{nghiempt}x=3\\x=-\frac{3}{7}\end{array}\right.\)