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\(\left[{}\begin{matrix}x+5=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=3\end{matrix}\right.\)
\(a,\Rightarrow\left(x-3\right)\left(x+5\right)=0\Rightarrow\left[{}\begin{matrix}x=3\\x=-5\end{matrix}\right.\\ b,\Rightarrow x^2-x-2x+2=0\\ \Rightarrow\left(x-2\right)\left(x-1\right)=0\Rightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
3x2 + 3x - 5( x + 1 ) = 0
<=> ( 3x2 + 3x ) - 5( x + 1 ) = 0
<=> 3x( x + 1 ) - 5( x + 1 ) = 0
<=> ( x + 1 )( 3x - 5 ) = 0
<=> x + 1 = 0 hoặc 3x - 5 = 0
<=> x = -1 hoặc x = 5/3
(x+2)^2-(x-2)(x+2)=0
=> (x+2)(x+2-x+2)=0
=> (x+2).4=0
=> x+2=0
=> x=-2
mấy câu còn lại tự làm nha
a) (x+2)^2-(x-2)(x+2)=0
(x+2).[x+2-x+2]=0
(x+2).4=0
x+2=0
x=-2
b)(2x - 1)^2 - (2x + 5) (2x - 5 ) = 18
4x2-4x+1-4x2+25=18
26-4x=18
4x=8
x=2
c)( 2x - 1)^2 - 25 = 0
( 2x - 1)^2 - 52 = 0
(2x-1-5)(2x-1+5)=0
(2x-6)(2x+4)=0
\(\Rightarrow\orbr{\begin{cases}2x-6=0\\2x+4=0\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x=3\\x=-2\end{cases}}\)
( 3x - 1 )( x + 3 ) + 9x2 - 1 = 0
<=> 3x2 + 9x - x - 3 + 9x2 - 1 = 0
<=> 12x2 + 8x - 4 = 0
<=> 4( 3x2 + 2x - 1 ) = 0
<=> 3x2 + 2x - 1 = 0
<=> 3x2 + 3x - x - 1 = 0
<=> ( 3x2 + 3x ) - ( x + 1 ) = 0
<=> 3x( x + 1 ) - 1( x + 1 ) = 0
<=> ( 3x - 1 )( x + 1 ) = 0
<=> \(\orbr{\begin{cases}3x-1=0\\x+1=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{1}{3}\\x=-1\end{cases}}\)
Vậy S = { 1/3 ; -1 }
\(\frac{x+1}{3}>\frac{3x-2}{5}\)
\(\Leftrightarrow\frac{5\left(x+1\right)}{15}>\frac{3\left(3x-2\right)}{15}\)
\(\Leftrightarrow5x+5>9x-6\)
\(\Leftrightarrow5x-9x>-6-5\)
\(\Leftrightarrow-4x>-11\)
\(\Leftrightarrow x< \frac{11}{4}\)
Bài làm:
a) \(\left(3x-1\right)\left(x+3\right)+9x^2-1=0\)
\(\Leftrightarrow3x^2+8x-3+9x^2-1=0\)
\(\Leftrightarrow12x^2+8x-4=0\)
\(\Leftrightarrow3x^2+2x-1=0\)
\(\Leftrightarrow\left(3x^2+3x\right)-\left(x+1\right)=0\)
\(\Leftrightarrow3x\left(x+1\right)-\left(x+1\right)=0\)
\(\Leftrightarrow\left(3x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}3x-1=0\\x+1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{1}{3}\\x=-1\end{cases}}\)
Vậy tập nghiệm của PT \(S=\left\{-1;\frac{1}{3}\right\}\)
b) \(\frac{x+1}{3}>\frac{3x-2}{5}\Leftrightarrow\frac{5\left(x+1\right)}{15}>\frac{3\left(3x-2\right)}{15}\)
\(\Rightarrow5x+5>9x-6\)
\(\Leftrightarrow4x< 11\)
\(\Rightarrow x< \frac{11}{4}\)
Lời giải:
$3x^2+4y^2+12x+3y+5=0$
$\Leftrightarrow 3(x^2+4x+4)+4y^2+3y-7=0$
$\Leftrightarrow 3(x+2)^2+(2y+\frac{3}{4})^2-\frac{121}{16}=0$
$\Leftrightarrow 3(x+2)^2=\frac{121}{16}-(2y+\frac{3}{4})^2\leq \frac{121}{16}$
$\Rightarrow (x+2)^2\leq \frac{121}{48}< 4$
$\Rightarrow -2< x+2< 2$
$\Rightarrow -4< x< 0$
$\Rightarrow x\in \left\{-3; -2; -1\right\}$
Đê đây bạn thay giá trị $x$ vào pt ban đầu để tìm $y$ thôi.
\(1.\dfrac{x-1}{3}-x=\dfrac{2x-4}{4}.\Leftrightarrow\dfrac{x-1-3x}{3}=\dfrac{x-2}{2}.\Leftrightarrow\dfrac{-2x-1}{3}-\dfrac{x-2}{2}=0.\)
\(\Leftrightarrow\dfrac{-4x-2-3x+6}{6}=0.\Rightarrow-7x+4=0.\Leftrightarrow x=\dfrac{4}{7}.\)
\(2.\left(x-2\right)\left(2x-1\right)=x^2-2x.\Leftrightarrow\left(x-2\right)\left(2x-1\right)-x\left(x-2\right)=0.\)
\(\Leftrightarrow\left(x-2\right)\left(2x-1-x\right)=0.\Leftrightarrow\left(x-2\right)\left(x-1\right)=0.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2.\\x=1.\end{matrix}\right.\)
\(3.3x^2-4x+1=0.\Leftrightarrow\left(x-1\right)\left(x-\dfrac{1}{3}\right)=0.\Leftrightarrow\left[{}\begin{matrix}x=1.\\x=\dfrac{1}{3}.\end{matrix}\right.\)
\(4.\left|2x-4\right|=0.\Leftrightarrow2x-4=0.\Leftrightarrow x=2.\)
\(5.\left|3x+2\right|=4.\Leftrightarrow\left[{}\begin{matrix}3x+2=4.\\3x+2=-4.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}.\\x=-2.\end{matrix}\right.\)
\(1,\dfrac{x-1}{3}-x=\dfrac{2x-4}{4}\\ \Leftrightarrow\dfrac{x-1}{3}-x=\dfrac{x-2}{2}\\ \Leftrightarrow\dfrac{2\left(x-1\right)-6x}{6}=\dfrac{3\left(x-2\right)}{6}\\ \Leftrightarrow2\left(x-1\right)-6x=3\left(x-2\right)\\ \Leftrightarrow2x-2-6x=3x-6\\ \Leftrightarrow-4x-2=3x-6\)
\(\Leftrightarrow3x-6+4x+2=0\\ \Leftrightarrow7x-4=0\\ \Leftrightarrow x=\dfrac{4}{7}\)
\(2,\left(x-2\right)\left(2x-1\right)=x^2-2x\\ \Leftrightarrow2x^2-4x-x+2=x^2-2x\\ \Leftrightarrow x^2-3x+2=0\\ \Leftrightarrow\left(x^2-2x\right)-\left(x-2\right)=0\\ \Leftrightarrow x\left(x-2\right)-\left(x-2\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
\(3,3x^2-4x+1=0\\ \Leftrightarrow\left(3x^2-3x\right)-\left(x-1\right)=0\\ \Leftrightarrow3x\left(x-1\right)-\left(x-1\right)=0\\ \Leftrightarrow\left(x-1\right)\left(3x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{3}\end{matrix}\right.\)
\(4,\left|2x-4\right|=0\\ \Leftrightarrow2x-4=0\\ \Leftrightarrow2x=4\\ \Leftrightarrow x=2\)
\(5,\left|3x+2\right|=4\\ \Leftrightarrow\left[{}\begin{matrix}3x+2=4\\3x+2=-4\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}3x=2\\3x=-6\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-2\end{matrix}\right.\)
\(6,\left|2x-5\right|=\left|-x+2\right|\\ \Leftrightarrow\left[{}\begin{matrix}2x-5=-x+2\\2x-5=x-2\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}3x=7\\x=3\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{3}\\x=3\end{matrix}\right.\)
\(a,3x-2\left(x-3\right)=0\\ \Leftrightarrow3x-2x+6=0\\ \Leftrightarrow x=-6\\ b,\left(x+1\right)\left(2x-3\right)=\left(2x-1\right)\left(x+5\right)\\ \Leftrightarrow2x^2+2x-3x-3=2x^2-x+10x-5\\ \Leftrightarrow2x^2-x-3=2x^2+9x-5\\ \Leftrightarrow10x-2=0\\ \Leftrightarrow x=\dfrac{1}{5}\\ c,ĐKXĐ:x\ne\pm1\\ \dfrac{2x}{x-1}-\dfrac{x}{x+1}=1\\ \Leftrightarrow\dfrac{2x\left(x+1\right)}{\left(x+1\right)\left(x-1\right)}-\dfrac{x\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}-\dfrac{\left(x+1\right)\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}=0\\ \Leftrightarrow\dfrac{2x^2+2x-x^2+x-x^2+1}{\left(x+1\right)\left(x-1\right)}=0\)
\(\Rightarrow3x+1=0\\ \Leftrightarrow x=-\dfrac{1}{3}\left(tm\right)\)
\(d,\left(2x+3\right)\left(3x-5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}2x+3=0\\3x-5=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{5}{3}\end{matrix}\right.\\ e,ĐKXĐ:x\ne\pm2\\ \dfrac{x-2}{x+2}-\dfrac{3}{x-2}=\dfrac{2\left(x-11\right)}{x^2-4}\\ \Leftrightarrow\dfrac{\left(x-2\right)^2}{\left(x-2\right)\left(x+2\right)}-\dfrac{3\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\dfrac{2x-22}{\left(x-2\right)\left(x+2\right)}=0\)
\(\Leftrightarrow\dfrac{x^2-4x+4-3x-6-2x+22}{\left(x-2\right)\left(x+2\right)}=0\\ \Rightarrow x^2-9x+20=0\\ \Leftrightarrow\left(x^2-5x\right)-\left(4x-20\right)=0\\ \Leftrightarrow x\left(x-5\right)-4\left(x-5\right)=0\\ \Leftrightarrow\left(x-4\right)\left(x-5\right)\\ \Leftrightarrow\left[{}\begin{matrix}x=4\left(tm\right)\\x=5\left(tm\right)\end{matrix}\right.\)
1) -3x2+5x=0
-x(3x-5)=0
suy ra hoặc x=0 hoặc 3x-5=0. giải ra ta có nghiệm phương trình là 0 và 3/5
2) x2+3x-2x-6=0
x(x+3)-2(x+3)=0
(x-2)(x+3)=0
suy ra hoặc x-2=0 hoặc x+3=0. giải ra ta có nghiệm là 2 và -3
3) x2+6x-x-6=0
x(x+6)-(x+6)=0
(x-1)(x+6)=0. vậy nghiệm là 1 và -6
4) x2+2x-3x-6=0
x(x+2)-3(x+2)=0
(x-3)(x+2)=0
vậy nghiệm là -2 và 3
5) x(x-6)-4(x-6)=0
(x-4)(x-6)=0. vậy nghiệm là 4 và 6
6)x(x-8)-3(x-8)=0
(x-3)(x-8)=0
suy ra nghiệm là 3 và 8
7) x2-5x-24=0
x2-8x+3x-24=0
x(x-8)+3(x-8)=0
(x+3)(x-8)=0
vậy nghiệm là -3 và 8
câu 1: -3x2 + 5x = 0
suy ra -x(3x-5)=0
sung ra x = 0 hoặc 3x-5=0 suy ra 3x = 5 suy ra x = 5/3
`(3x+2) (5-x^2)=0`
TH1 :
`->3x+2=0`
`->3x=-2`
`->x=(-2)/3`
TH2 :
`->5-x^2=0`
`->x^2=5`
`->` \(x^2=\left(\sqrt{5}\right)^2\) hoặc \(x=\left(-\sqrt{5}\right)^2\)
`->` \(x=\sqrt{5}\) hoặc \(x=-\sqrt{5}\)
Vậy `x=2,` \(x=\sqrt{5,}x=-\sqrt{5}\)
\(\left(3x+2\right)\left(5-x^2\right)=0\)
TH1 : \(3x+2=0\Leftrightarrow x=-\frac{2}{3}\)
TH2 : \(5-x^2=0\Leftrightarrow x^2=5\Leftrightarrow x=\pm5\)