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Bài làm :
a) x( 2x - 7 ) - 4x + 14 = 0
<=> x( 2x - 7 ) - 2( 2x - 7 ) = 0
<=> ( 2x - 7 )( x - 2 ) = 0
\(\Leftrightarrow\orbr{\begin{cases}2x-7=0\\x-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{7}{2}\\x=2\end{cases}}\)
b) Sửa đề : 5x3 + x2 - 4x + 9 = 0
<=>( 5x3 + 5 ) + (x2 - 4x +4)=0
<=> 5(x3 + 1) + (x-2)2 = 0
<=> 5(x+1)(x2 - x +1) + (x+2)2 =0
\(\Leftrightarrow\orbr{\begin{cases}x=-1\\x=-2\end{cases}}\)
c) 3x3 - 7x2 + 6x - 14 = 0
<=> 3x2( x - 7/3 ) + 6( x - 7/3 ) = 0
<=> ( x - 7/3 )( 3x2 + 6 ) = 0
\(\Leftrightarrow\orbr{\begin{cases}x-\frac{7}{3}=0\\3x^2+6=0\end{cases}}\Leftrightarrow x=\frac{7}{3}\)
d) 5x2 - 5x = 3( x - 1 )
<=> 5x( x - 1 ) - 3( x - 1 ) = 0
<=> ( x - 1 )( 5x - 3 ) = 0
\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\5x-3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\\x=\frac{3}{5}\end{cases}}\)
e) 4x2 - 25 - ( 4x - 10 ) = 0
<=> ( 2x - 5 )( 2x + 5 ) - 2( 2x - 5 ) = 0
<=> ( 2x - 5 )( 2x + 5 - 2 ) = 0
<=> ( 2x - 5 )( 2x + 3 ) = 0
\(\Leftrightarrow\orbr{\begin{cases}2x-5=0\\2x+3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{5}{2}\\x=-\frac{3}{2}\end{cases}}\)
f) x3 + 27 + ( x + 3 )( x - 9 ) = 0
<=> ( x + 3 )( x2 - 3x + 9 ) + ( x + 3 )( x - 9 ) = 0
<=> ( x + 3 )( x2 - 3x + 9 + x - 9 ) = 0
<=> ( x + 3 )( x2 - 2x ) = 0
<=> x( x + 3 )( x - 2 ) = 0
\(\Leftrightarrow\orbr{\begin{cases}\\\end{cases}}\begin{cases}x=0\\x=-3\\x=2\end{cases}\)
a) x( 2x - 7 ) - 4x + 14 = 0
<=> x( 2x - 7 ) - 2( 2x - 7 ) = 0
<=> ( 2x - 7 )( x - 2 ) = 0
<=> \(\orbr{\begin{cases}2x-7=0\\x-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{7}{2}\\x=2\end{cases}}\)
b) 5x3 + x2 - 4x - 9 = 0 ( đề sai )
c) 3x3 - 7x2 + 6x - 14 = 0
<=> 3x2( x - 7/3 ) + 6( x - 7/3 ) = 0
<=> ( x - 7/3 )( 3x2 + 6 ) = 0
<=> \(\orbr{\begin{cases}x-\frac{7}{3}=0\\3x^2+6=0\end{cases}}\Leftrightarrow x=\frac{7}{3}\)( do 3x2 + 6 ≥ 6 > 0 với mọi x )
d) 5x2 - 5x = 3( x - 1 )
<=> 5x( x - 1 ) - 3( x - 1 ) = 0
<=> ( x - 1 )( 5x - 3 ) = 0
<=> \(\orbr{\begin{cases}x-1=0\\5x-3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\\x=\frac{3}{5}\end{cases}}\)
e) 4x2 - 25 - ( 4x - 10 ) = 0
<=> ( 2x - 5 )( 2x + 5 ) - 2( 2x - 5 ) = 0
<=> ( 2x - 5 )( 2x + 5 - 2 ) = 0
<=> ( 2x - 5 )( 2x + 3 ) = 0
<=> \(\orbr{\begin{cases}2x-5=0\\2x+3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{5}{2}\\x=-\frac{3}{2}\end{cases}}\)
f) x3 + 27 + ( x + 3 )( x - 9 ) = 0
<=> ( x + 3 )( x2 - 3x + 9 ) + ( x + 3 )( x - 9 ) = 0
<=> ( x + 3 )( x2 - 3x + 9 + x - 9 ) = 0
<=> ( x + 3 )( x2 - 2x ) = 0
<=> x( x + 3 )( x - 2 ) = 0
<=> x = 0 hoặc x + 3 = 0 hoặc x - 2 = 0
<=> x = 0 hoặc x = -3 hoặc x = 2
Bài 1:
a) \(9\left(4x+3\right)^2=16\left(3x-5\right)^2\)
\(114x^2+216x+81=114x^2-480x+400\)
\(144x^2+216x=144x^2-480x+400-81\)
\(114x^2+216=114x^2-480x+319\)
\(696x=319\)
\(\Rightarrow x=\frac{11}{24}\)
b) \(\left(x^3-x^2\right)^2-4x^2+8x-4=0\)
\(\left(x-1\right)^2\left(x^2+2\right)\left(x+\sqrt{2}\right)\left(x-\sqrt{2}\right)=0\)
\(\Rightarrow x=1\)
c) \(x^5+x^4+x^3+x^2+x+1=0\)
\(\left(x+1\right)\left(x^2+x+1\right)\left(x^2-x+1\right)=0\)
\(\Rightarrow x=-1\)
Bài 2:
a) \(5x^3-7x^2-15x+21=0\)
\(\left(5x-7\right)\left(x+\sqrt{3}\right)\left(x-\sqrt{3}\right)=0\)
\(\Rightarrow x=\frac{7}{5}\)
b) \(\left(x-3\right)^2=4x^2-20x+25\)
\(x^2-6x+9-25=4x^2-20x+25\)
\(x^2-6x+9=4x^2-20x+25-25\)
\(x^2-6x-16=4x^2-20x\)
\(x^2+14x-16=4x^2-4x^2\)
\(-3x^2+14x-16=0\)
\(\Rightarrow\orbr{\begin{cases}x=2\\x=\frac{8}{3}\end{cases}}\)
c) \(\left(x-1\right)^2-5=\left(x+2\right)\left(x-2\right)-x\left(x-1\right)\)
\(x^2-2x=x-4\)
\(x^2-2x=x-4+4\)
\(x^2-2x=x-x\)
\(x^2-3x=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x=3\end{cases}}\)
d) \(\left(2x-3\right)^3-\left(2x+3\right)\left(4x^2-1\right)=-24\)
\(-48x^2+56x-24=-24\)
\(-48x^2+56x=-24+24\)
\(-48x^2+56=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x=\frac{7}{6}\end{cases}}\)
mình ko chắc
\(x\left(2x-7\right)-4x+14=0\Leftrightarrow\left(x-2\right)\left(2x-7\right)=0\Leftrightarrow\left[{}\begin{matrix}x-2=0\\2x-7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\frac{7}{2}\end{matrix}\right.\)
\(x^2\left(x-1\right)-4\left(x-1\right)=\left(x^2-4\right)\left(x-1\right)=\left(x-2\right)\left(x+2\right)\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+2=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\\x=1\end{matrix}\right.\)
\(x^4-x^3-x^2+x=x\left(x^3+1\right)-x^2\left(x+1\right)=x\left(x+1\right)\left(x^2-x+1-x^2\right)=x\left(x+1\right)\left(1-x\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x+1=0\\1-x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\pm1\end{matrix}\right.\)
a) \(x\left(2x-7\right)-4x+14-0\Leftrightarrow2x^2-11x+14=0\Leftrightarrow2x^2-4x-7x+14=0\Leftrightarrow2x\left(x-2\right)-7\left(x-2\right)=0\Leftrightarrow\left(2x-7\right)\left(x-2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=3,5\\x=2\end{matrix}\right.\)
b) \(x^2\left(x-1\right)-4x+4=0\Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)=0\Leftrightarrow\left(x-1\right)\left(x-2\right)\left(x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\\x=-2\end{matrix}\right.\)
c) \(x+x^2-x^3-x^4=0\Leftrightarrow x\left(x^3+x^2-x-1\right)=0\Leftrightarrow x\left[x\left(x^2-1\right)+\left(x^2-1\right)\right]=0\Leftrightarrow x\left(x+1\right)\left(x^2-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=-1\end{matrix}\right.\)
d) \(2x^3+3x^2+2x+3=0\Leftrightarrow x^2\left(2x+3\right)+2x+3=0\Leftrightarrow\left(x^2+1\right)\left(2x+3\right)=0\Leftrightarrow x=-1,5\left(x^2+1>0\forall x\right)\)
e) \(4x^2-25-\left(2x-5\right)\left(2x+7\right)=0\Leftrightarrow\left(2x-5\right)\left(2x+5\right)-\left(2x-5\right)\left(2x+7\right)=0\Leftrightarrow\left(2x-5\right)\left(2x+5-2x-7\right)=0\Leftrightarrow2x-5=0\Leftrightarrow x=2,5\)
g) \(x^3+27+\left(x+3\right)\left(x-9\right)=0\Leftrightarrow\left(x+3\right)\left(x^2-3x+9\right)+\left(x+3\right)\left(x-9\right)=0\Leftrightarrow\left(x+3\right)\left(x^2-3x+9+x-9\right)=0\Leftrightarrow x\left(x+3\right)\left(x-2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\\x=2\end{matrix}\right.\)
(2x-3)2-(x+5)2=0
<=>(2x-3-x-5)(2x-3+x+5)=0
<=>(x-8)(3x+2)=0
<=>x-8=0 hoặc 3x+2=0
<=>x=8 hoặc x=-2/3
(2x-3)2
-(x+5)2=0
<=>(2x-3-x-5)(2x-3+x+5)=0
<=>(x-8)(3x+2)=0
<=>x-8=0 hoặc 3x+2=0
<=>x=8 hoặc x=-2/3
chcú cậu hok tốt @_@
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) \(\left(2x+5\right)\left(x-4\right)=\left(x-4\right)\left(5-x\right)\)
\(\Leftrightarrow\left(2x+5\right)\left(x-4\right)+\left(x-5\right)\left(x-4\right)=0\)
\(\Leftrightarrow3x\left(x-4\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=4\end{cases}}\)
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) \(9\left(x+2\right)^2=\left(x+3\right)^2\)
\(\Leftrightarrow\orbr{\begin{cases}3x+6=x+3\\3x+6=-x-3\end{cases}}\Leftrightarrow\orbr{\begin{cases}2x=-3\\4x=-9\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-\frac{3}{2}\\x=-\frac{9}{4}\end{cases}}\)
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
a) \(\left(2x-1\right)^2-25=0\)
\(\Leftrightarrow\left(2x-6\right)\left(2x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-6=0\\2x+4=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
Vậy \(x\in\left\{-2;3\right\}\)
b) \(\left(x+8\right)^2=121\)
\(\Leftrightarrow\left(x+8\right)^2-121=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+19\right)=0\) \(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-19\end{matrix}\right.\)
Vậy \(x\in\left\{-19;3\right\}\)
c) \(x^3-4x^2+4x=0\)
\(\Leftrightarrow x\left(x^2-4x+4\right)=0\)
\(\Leftrightarrow x\left(x-2\right)^2=0\) \(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)
Vậy \(x\in\left\{0;2\right\}\)
d) \(4x^2-4x=-1\Leftrightarrow4x^2-4x+1=0\)
\(\Leftrightarrow\left(2x-1\right)^2=0\Leftrightarrow x=\frac{1}{2}\)
Vậy \(x=\frac{1}{2}\)
tìm x, biết
a) (2x-1)2 -25 =0
(2x-1)2 =25
(2x-1)2 =52
(2x-1) =5
2x =6
x =3
b) (x+8)2 =121
(x+8)2 =112
(x+8) =11
x =3