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HL
0
HN
5 tháng 8 2016
\(x^3-4x^2-8x+8\)
\(\Leftrightarrow\left(x^3-4x^2\right)-\left(8x-8\right)\)
\(\Leftrightarrow x^2\left(x-4\right)-4\left(x-4\right)\)
\(\Leftrightarrow\left(x-4\right)\left(x^2-4\right)\)
14 tháng 9 2020
a) Ta có: \(x^4-16x^2=0\)
\(\Leftrightarrow x^2\left(x^2-16\right)=0\)
\(\Leftrightarrow x^2\left(x-4\right)\left(x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2=0\\x-4=0\\x+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\\x=-4\end{matrix}\right.\)
Vậy: \(x\in\left\{0;4;-4\right\}\)
b) Ta có: \(9x^2+6x+1=0\)
\(\Leftrightarrow\left(3x\right)^2+2\cdot3x\cdot1+1^2=0\)
\(\Leftrightarrow\left(3x+1\right)^2=0\)
\(\Leftrightarrow3x+1=0\)
\(\Leftrightarrow3x=-1\)
hay \(x=-\frac{1}{3}\)
Vậy: \(x=-\frac{1}{3}\)
c) Ta có: \(x^2-6x=16\)
\(\Leftrightarrow x^2-6x-16=0\)
\(\Leftrightarrow x^2-8x+2x-16=0\)
\(\Leftrightarrow x\left(x-8\right)+2\left(x-8\right)=0\)
\(\Leftrightarrow\left(x-8\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-8=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\\x=-2\end{matrix}\right.\)
Vậy: \(x\in\left\{8;-2\right\}\)
d) Ta có: \(9x^2+6x=80\)
\(\Leftrightarrow9x^2+6x-80=0\)
\(\Leftrightarrow9x^2+6x+1-81=0\)
\(\Leftrightarrow\left(3x+1\right)^2-9^2=0\)
\(\Leftrightarrow\left(3x+1-9\right)\left(3x+1+9\right)=0\)
\(\Leftrightarrow\left(3x-8\right)\left(3x+10\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-8=0\\3x+10=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=8\\3x=-10\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{8}{3}\\x=-\frac{10}{3}\end{matrix}\right.\)
Vậy: \(x\in\left\{\frac{8}{3};-\frac{10}{3}\right\}\)
e) Ta có: \(25\left(2x-1\right)^2-9\left(x+1\right)^2=0\)
\(\Leftrightarrow\left(10x-5\right)^2-\left(3x+3\right)^2=0\)
\(\Leftrightarrow\left(10x-5-3x-3\right)\left(10x-5+3x+3\right)=0\)
\(\Leftrightarrow\left(7x-8\right)\left(13x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}7x-8=0\\13x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}7x=8\\13x=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{8}{7}\\x=\frac{2}{13}\end{matrix}\right.\)
Vậy: \(x\in\left\{\frac{8}{7};\frac{2}{13}\right\}\)
TC
5 tháng 1 2018
\(x^3-9x^2+6x+16\)
\(=\left(x^3+x^2\right)-\left(10x^2+10x\right)+\left(16x+16\right)\)
\(=x^2\left(x+1\right)-10x\left(x+1\right)+16\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2-10x+16\right)\)
\(=\left(x+1\right)\left[\left(x^2-2x\right)-\left(8x-16\right)\right]\)
\(=\left(x+1\right)\left(x-2\right)\left(x-8\right)\)
18 tháng 9 2018
\(x^3+9x^2+6x-16=\left(x^3+8x^2\right)+\left(x^2+8x\right)-\left(2x+16\right)\)
\(=x^2\left(x+8\right)+x\left(x+8\right)-2\left(x+8\right)\)
\(=\left(x+8\right)\left(x^2+x-2\right)\)
\(=\left(x+8\right)\left(x^2+2x-x-2\right)\)
\(=\left(x+8\right)\left(x+2\right)\left(x-1\right)\)
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
=> \(\orbr{\begin{cases}7x=8\\13x=2\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{8}{7}\\x=\frac{2}{13}\end{cases}}\)
PQ
4
31 tháng 12 2016
nhẩm x=-1 là nghiệm
\(\left(x+1\right)\left(x^2-10x+16\right)=\left(x+1\right)\left[\left(x-5\right)^2-9\right]=\left(x+1\right)\left(x-8\right)\left(x-2\right)\)
VV
1
HP
7 tháng 8 2018
1) \(x^3+5x^2+9x=-45\)
\(\Rightarrow x^2\left(x+5\right)+9x+45=0\)
\(\Rightarrow x^2\left(x+5\right)+9\left(x+5\right)=0\)
\(\Rightarrow\left(x^2+9\right)\left(x+5\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x^2+9=0\\x+5=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x^2=-9\left(loai\right)\\x=-5\left(nhan\right)\end{cases}}\)
2) \(x^3-6x^2-x+30=0\)
\(\Rightarrow x^3-3x^2-3x^2+9x-10x+30=0\)
\(\Rightarrow x^2\left(x-3\right)-3x\left(x-3\right)-10\left(x-3\right)=0\)
\(\Rightarrow\left(x^2-3x-10\right)\left(x-3\right)=0\)
\(\Rightarrow\left(x^2-5x+2x-10\right)\left(x-3\right)=0\)
\(\Rightarrow\left[x\left(x-5\right)+2\left(x-5\right)\right]\left(x-3\right)=0\)
\(\Rightarrow\left(x+2\right)\left(x-3\right)\left(x-5\right)=0\)
\(\)Từ đây giải x giống câu trên nhé.
3) \(x^2+16=10x\)
\(\Rightarrow x^2-10x+16=0\)
\(\Rightarrow\left(x-8\right)\left(x-2\right)=0\)
Tương tự....
Ta có: \(x^3-9x^2+6x+16\)
\(=x^3-2x^2-7x^2+14x-8x+16\)
\(=x^2\left(x-2\right)-7x\left(x-2\right)-8\left(x-2\right)\)
\(=\left(x-2\right)\left(x^2-7x-8\right)\)
\(=\left(x-2\right)\left(x^2-8x+x-8\right)\)
\(=\left(x-2\right)\left[x\left(x-8\right)+\left(x-8\right)\right]\)
\(=\left(x-2\right)\left(x-8\right)\left(x+1\right)\)
k cho mình nha bn thanks nhìu <3