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a) Ta có: \(2x^3+5x^2-3x=0\)
\(\Leftrightarrow x\left(2x^2+5x-3\right)=0\)
\(\Leftrightarrow x\left(2x^2+6x-x-3\right)=0\)
\(\Leftrightarrow x\left[2x\left(x+3\right)-\left(x+3\right)\right]=0\)
\(\Leftrightarrow x\left(x+3\right)\left(2x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+3=0\\2x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\\2x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\\x=\dfrac{1}{2}\end{matrix}\right.\)
Vậy: \(S=\left\{0;-3;\dfrac{1}{2}\right\}\)
b) Ta có: \(2x^3+6x^2=x^2+3x\)
\(\Leftrightarrow2x^2\left(x+3\right)=x\left(x+3\right)\)
\(\Leftrightarrow2x^2\left(x+3\right)-x\left(x+3\right)=0\)
\(\Leftrightarrow x\left(x+3\right)\left(2x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+3=0\\2x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\\2x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\\x=\dfrac{1}{2}\end{matrix}\right.\)
Vậy: \(S=\left\{0;-3;\dfrac{1}{2}\right\}\)
c) Ta có: \(x^2+\left(x+2\right)\left(11x-7\right)=4\)
\(\Leftrightarrow x^2+11x^2-7x+22x-14-4=0\)
\(\Leftrightarrow12x^2+15x-18=0\)
\(\Leftrightarrow12x^2+24x-9x-18=0\)
\(\Leftrightarrow12x\left(x+2\right)-9\left(x+2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(12x-9\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\12x-9=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\12x=9\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{3}{4}\end{matrix}\right.\)
Vậy: \(S=\left\{-2;\dfrac{3}{4}\right\}\)
a. (x - 22) - 1 = 0
<=> x - 4 - 1 = 0
<=> x = 5
b. 4 - (x - 2)2 = 0
<=> 22 - (x - 2)2 = 0
<=> (2 - x + 2)(2 + x - 2) = 0
<=> x(4 - x) = 0
<=> \(\left[{}\begin{matrix}x=0\\4-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)
d. (3x - 2)2 - (2x + 3)2 = 5(x + 4)(x - 4)
<=> (3x - 2 - 2x - 3)(3x - 2 + 2x + 3) = 5(x2 - 16)
<=> (x - 5)(5x + 1) = 5x2 - 80
<=> 5x2 + x - 25x - 5 = 5x2 - 80
<=> 5x2 - 5x2 + x - 25x = -80 + 5
<=> -24x = -75
<=> x = \(\dfrac{25}{8}\)
a/ \(x=\dfrac{-5}{12}\)
b/ \(x\approx-1,9526\)
c/ \(x=\dfrac{21-i\sqrt{199}}{10}\)
d/ \(x=\dfrac{-20}{13}\)
1) Ta có: \(\left(3-x^2\right)+6-2x=0\)
\(\Leftrightarrow3-x^2+6-2x=0\)
\(\Leftrightarrow-x^2-2x+9=0\)
\(\Leftrightarrow x^2+2x-9=0\)
\(\Leftrightarrow x^2+2x+1=10\)
\(\Leftrightarrow\left(x+1\right)^2=10\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=\sqrt{10}\\x+1=-\sqrt{10}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{10}-1\\x=-\sqrt{10}-1\end{matrix}\right.\)
Vậy: \(S=\left\{\sqrt{10}-1;-\sqrt{10}-1\right\}\)
2) Ta có: \(5\left(2x-1\right)+7=4\left(2-x\right)+2\)
\(\Leftrightarrow10x-5+7=8-4x+2\)
\(\Leftrightarrow10x+4x=8+2+5-7\)
\(\Leftrightarrow14x=8\)
\(\Leftrightarrow x=\dfrac{4}{7}\)
Vậy: \(S=\left\{\dfrac{4}{7}\right\}\)
1: \(\Leftrightarrow2x^2-10x-3x-2x^2=0\)
=>-13x=0
=>x=0
2: \(\Leftrightarrow5x-2x^2+2x^2-2x=13\)
=>3x=13
=>x=13/3
3: \(\Leftrightarrow4x^4-6x^3-4x^3+6x^3-2x^2=0\)
=>-2x^2=0
=>x=0
4: \(\Leftrightarrow5x^2-5x-5x^2+7x-10x+14=6\)
=>-8x=6-14=-8
=>x=1
`1)2x(x-5)-(3x+2x^2)=0`
`<=>2x^2-10x-3x-2x^2=0`
`<=>-13x=0`
`<=>x=0`
___________________________________________________
`2)x(5-2x)+2x(x-1)=13`
`<=>5x-2x^2+2x^2-2x=13`
`<=>3x=13<=>x=13/3`
___________________________________________________
`3)2x^3(2x-3)-x^2(4x^2-6x+2)=0`
`<=>4x^4-6x^3-4x^4+6x^3-2x^2=0`
`<=>x=0`
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`4)5x(x-1)-(x+2)(5x-7)=0`
`<=>5x^2-5x-5x^2+7x-10x+14=0`
`<=>-8x=-14`
`<=>x=7/4`
___________________________________________________
`5)6x^2-(2x-3)(3x+2)=1`
`<=>6x^2-6x^2-4x+9x+6=1`
`<=>5x=-5<=>x=-1`
___________________________________________________
`6)2x(1-x)+5=9-2x^2`
`<=>2x-2x^2+5=9-2x^2`
`<=>2x=4<=>x=2`
a) đặt \(\left(x^2+x\right)\)là \(y\)
ta có: \(3y^2-7y+4\)\(=0\)
<=>\(\left(3y-4\right)\left(y-1\right)=0\)
còn lại bạn tự xử nhé
a)(x - 4)2 - 25= 0
<--> ( x - 4)2 - 52 = 0
<--> ( x - 4 - 5 )( x - 4 + 5 ) = 0
<--> ( x - 4 - 5 ) = 0 <--> x - 9 = 0 <--> x = 9
hoặc
<--> ( x - 4 + 5 ) = 0 <--> x + 1 = 0 <--> x = -1
b)bài này tương tự bài a
\(a,\left(x-4\right)^2-25=0\)
\(\Rightarrow\left(x-4-5\right)\left(x-4+5\right)=0\)
\(\Rightarrow\left(x-9\right)\left(x+1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-9=0\\x+1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=9\\x=-1\end{matrix}\right.\)
\(b,\left(x-3\right)^2-\left(x+1\right)^2=0\)
\(\Rightarrow\left(x-3-x-1\right)\left(x-3+x+1\right)=0\)
\(\Rightarrow-4\left(2x-2\right)=0\)
\(\Rightarrow2\left(x-1\right)=0\)
\(\Rightarrow x-1=0\)
\(\Rightarrow x=1\)
\(c,\left(x^2-4\right)\left(2x+3\right)=\left(x^2-4\right)\left(x-1\right)\)
\(\Rightarrow2x+3=x-1\)
\(\Rightarrow2x-x=-1-3\)
\(\Rightarrow x=-4\)
\(d,\left(3x-7\right)^2-4\left(x+1\right)^2=0\)
\(\Rightarrow\left(3x-7\right)-\left[2\left(x+1\right)\right]^2=0\)
\(\Rightarrow\left(3x-7\right)^2-\left(2x+2\right)^2=0\)
\(\Rightarrow\left(3x-7-2x-2\right)\left(3x-7+2x+2\right)=0\)
\(\Rightarrow\left(x-9\right)\left(5x-5\right)=0\)
\(\Rightarrow5\left(x-9\right)\left(x-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-9=0\\x-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=9\\x=1\end{matrix}\right.\)
\(8,1-\left(x-6\right)=4\left(2-2x\right)\)
\(\Leftrightarrow1-x+6=8-8x\)
\(\Leftrightarrow-x+8x=8-1-6\)
\(\Leftrightarrow7x=1\)
\(\Leftrightarrow x=\dfrac{1}{7}\)
\(9,\left(3x-2\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-2=0\\x+5=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-5\end{matrix}\right.\)
\(10,\left(x+3\right)\left(x^2+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x^2+2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=\varnothing\end{matrix}\right.\)
`8)1-(x-5)=4(2-2x)`
`<=>1-x+5=8-6x`
`<=>5x=2<=>x=2/5`
`9)(3x-2)(x+5)=0`
`<=>[(x=2/3),(x=-5):}`
`10)(x+3)(x^2+2)=0`
Mà `x^2+2 > 0 AA x`
`=>x+3=0`
`<=>x=-3`
`11)(5x-1)(x^2-9)=0`
`<=>(5x-1)(x-3)(x+3)=0`
`<=>[(x=1/5),(x=3),(x=-3):}`
`12)x(x-3)+3(x-3)=0`
`<=>(x-3)(x+3)=0`
`<=>[(x=3),(x=-3):}`
`13)x(x-5)-4x+20=0`
`<=>x(x-5)-4(x-5)=0`
`<=>(x-5)(x-4)=0`
`<=>[(x=5),(x=4):}`
`14)x^2+4x-5=0`
`<=>x^2+5x-x-5=0`
`<=>(x+5)(x-1)=0`
`<=>[(x=-5),(x=1):}`
Bài làm
a) ( x - 4 )2 - 25 = 0
<=> ( x - 4 - 5 )( x - 4 + 5 ) = 0
<=> ( x - 9 )( x + 1 ) = 0
<=> \(\orbr{\begin{cases}x-9=0\\x+1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=9\\x=-1\end{cases}}}\)
Vậy tập nghiệm phương trình S = { -2; 9 }
b) ( x - 3 )2 - ( x + 1 )2 = 0
<=> ( x - 3 - x - 1 )( x - 3 + x + 1 ) = 0
<=> -4( 3x - 2 ) = 0
<=> 3x - 2 = 0
<=> \(x=\frac{2}{3}\)
Vậy \(x=\frac{2}{3}\)là nghiệm phương trình.
c) ( x2 - 4 )( 2x + 3 ) = ( x2 - 4 )( x - 1 )
<=> ( x2 - 4 )( 2x + 3 ) - ( x2 - 4 )( x - 1 ) = 0
<=> ( x2 - 4 )( 2x + 3 - x - 1 ) = 0
<=> ( x2 - 4 )( x + 2 ) = 0
<=> \(\orbr{\begin{cases}x^2-4=0\\x+2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\pm2\\x=-2\end{cases}}}\)
Vậy tập nghiệm phương trình là S = { 2; -2 }
d) ( 3x - 7 )2 - 4( x + 1 )2 = 0
<=> ( 3x - 7 )2 - [ 2( x + 1 ) ] 2 = 0
<=> [ ( 3x - 7 ) - 2( x + 1 ) ][ ( 3x - 7 ) + 2( x + 1 )] = 0
<=> ( 3x - 7 - 2x - 2 )( 3x - 7 + 2x + 1 ) = 0
<=> ( x - 9 )( 5x - 6 ) = 0
<=> \(\orbr{\begin{cases}x-9=0\\5x-6=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=9\\x=\frac{6}{5}\end{cases}}}\)
Vậy tập nghiệm phương trình S = { 9; 6/5 }
# Học tốt #