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c) \(\dfrac{x}{x-2}+\dfrac{x}{x+2}=\dfrac{4x}{x^2-4}.ĐKXĐ:x\ne2;-2\)
<=>\(\dfrac{x\left(x+2\right)}{x^2-4}+\dfrac{x\left(x-2\right)}{x^2-4}=\dfrac{4x}{x^2-4}\)
<=>x2+2x+x2-2x=4x
<=>2x2-4x=0
<=>2x(x-2)=0
<=>\(\left[{}\begin{matrix}2x=0< =>x=0\\x-2=0< =>x=2\left(loại\right)\end{matrix}\right.\)
Vậy pt trên có nghiệm là S={0}
d) 11x-9=5x+3
<=>11x-5x=9+3
<=>6x=12
<=>x=2
Vậy pt trên có nghiệm là S={2}
e) (2x+3)(3x-4) =0
<=> \(\left[{}\begin{matrix}2x+3=0< =>x=\dfrac{-3}{2}\\3x-4=0< =>x=\dfrac{4}{3}\end{matrix}\right.\)
Vậy pt trên có tập nghiệm là S={\(\dfrac{-3}{2};\dfrac{4}{3}\)}
a) 5x+9 =2x
<=> 5x-2x=9
<=> 3x=9
<=> x=3
Vậy pt trên có nghiệm là S={3}
b) (x+1)(4x-3)=(2x+5)(x+1)
<=> (x+1)(4x-3)-(2x+5)(x+1)=0
<=>(x+1)(2x-8)=0
<=>\(\left[{}\begin{matrix}x+1=0< =>x=-1\\2x-8=0< =>2x=8< =>x=4\end{matrix}\right.\)
Vậy pt trên có tập nghiệm là S={-1;4}
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Bài 1.
a) ( x - 3 )( x + 7 ) = 0
<=> x - 3 = 0 hoặc x + 7 = 0
<=> x = 3 hoặc x = -7
Vậy S = { 3 ; -7 }
b) ( x - 2 )2 + ( x - 2 )( x - 3 ) = 0
<=> ( x - 2 )( x - 2 + x - 3 ) = 0
<=> ( x - 2 )( 2x - 5 ) = 0
<=> x - 2 = 0 hoặc 2x - 5 = 0
<=> x = 2 hoặc x = 5/2
Vậy S = { 2 ; 5/2 }
c) x2 - 5x + 6 = 0
<=> x2 - 2x - 3x + 6 = 0
<=> x( x - 2 ) - 3( x - 2 ) = 0
<=> ( x - 2 )( x - 3 ) = 0
<=> x - 2 = 0 hoặc x - 3 = 0
<=> x = 2 hoặc x = 3
\(a,2x-5=-x+4\\ \Leftrightarrow3x=9\\ \Leftrightarrow x=3\\ b,\left(4x-10\right)\left(25+5x\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}4x-10=0\\25+5x=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-5\end{matrix}\right.\\ c,\dfrac{x}{3}-\dfrac{2x+1}{2}=\dfrac{x}{6}-x\\ \Leftrightarrow\dfrac{2x}{6}-\dfrac{3\left(2x+1\right)}{6}-\dfrac{x}{6}+\dfrac{6x}{6}=0\\ \Leftrightarrow2x-6x-3-x+6x=0\\ \Leftrightarrow x-3=0\\ \Leftrightarrow x=3\)
d, ĐKXĐ:\(x\ne-2,x\ne3\)
\(1+\dfrac{x}{3-x}=\dfrac{5x}{\left(x+2\right)\left(3-x\right)}+\dfrac{2}{x+2}\\ \Leftrightarrow\dfrac{\left(x+2\right)\left(3-x\right)}{\left(x+2\right)\left(3-x\right)}+\dfrac{x\left(x+2\right)}{\left(x+2\right)\left(3-x\right)}-\dfrac{5x}{\left(x+2\right)\left(3-x\right)}-\dfrac{2\left(3-x\right)}{\left(x+2\right)\left(3-x\right)}=0\\ \Leftrightarrow\dfrac{-x^2+x+6}{\left(x+2\right)\left(3-x\right)}+\dfrac{x^2+2x}{\left(x+2\right)\left(3-x\right)}-\dfrac{5x}{\left(x+2\right)\left(3-x\right)}-\dfrac{6-2x}{\left(x+2\right)\left(3-x\right)}=0\)
\(\Leftrightarrow\dfrac{-x^2+x+6+x^2+2x-5x-6+2x}{\left(x+2\right)\left(3-x\right)}=0\\ \Rightarrow0=0\left(luôn.đúng\right)\)
`a,(2x-5)(12+5x)=0`
\(\Leftrightarrow\left[{}\begin{matrix}2x-5=0\\12+5x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=5\\5x=-12\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-\dfrac{12}{5}\end{matrix}\right.\)
`b, (x-3)(x-4)-2(x-3)=0`
`<=>(x-3)(x-4-2)=0`
`<=>(x-3)(x-6)=0`
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x-6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=6\end{matrix}\right.\)
`c, x(x-1)(x+1)=0`
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-1=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=-1\end{matrix}\right.\)
`d, (2x)/3 +(2x-1)/6=0`
`<=> (4x)/6 +(2x-1)/6=0`
`<=> (4x+2x-1)/6=0`
`<=> (6x-1)/6=0`
`<=> 6x-1=0`
`<=> 6x=1`
`<=>x=1/6` ( đề là vậy à bạn )
a) \(\left(2x-5\right)\left(12+5x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-5=0\\12+5x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=5\\5x=-12\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2,5\\x=-2,4\end{matrix}\right.\)
b) \(\left(x-3\right)\left(x-4\right)-2\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left[\left(x-4\right)-2\right]=0\)
\(\Leftrightarrow\left(x-3\right)\left(x-6\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x-6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=6\end{matrix}\right.\)
c) \(x\left(x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\x-1=0\\x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=1\\x=0\end{matrix}\right.\)
d) \(\dfrac{2x}{3}+\dfrac{2x-1}{6}=0\)
\(\Leftrightarrow\dfrac{4x+2x-1}{6}=0\)
\(\Leftrightarrow6x-1=0\)
\(\Leftrightarrow6x=1\Leftrightarrow x=\dfrac{1}{6}\)
a,\(2x+5=2-x\)
\(< =>2x+x+5-2=0\)
\(< =>3x+3=0\)
\(< =>x=-1\)
b, \(/x-7/=2x+3\)
Với \(x\ge7\)thì \(PT< =>x-7=2x+3\)
\(< =>2x-x+3+7=0\)
\(< =>x+10=0< =>x=-10\)( lọai )
Với \(x< 7\)thì \(PT< =>7-x=2x+3\)
\(< =>2x+x+3-7=0\)
\(< =>3x-4=0< =>x=\frac{4}{3}\) ( loại )
c,\(\frac{4}{x+2}-\frac{4x-6}{4x-x^3}=\frac{x-3}{x\left(x-2\right)}\left(đk:x\ne-2;0;2\right)\)
\(< =>\frac{4x\left(x-2\right)}{x\left(x-2\right)\left(x+2\right)}+\frac{4x-6}{x\left(x-2\right)\left(2+x\right)}=\frac{\left(x-3\right)\left(x+2\right)}{x\left(x-2\right)\left(x+2\right)}\)
\(< =>4x^2-8x+4x-6=x^2-x-6\)
\(< =>4x^2-x^2-4x+x-6+6=0\)
\(< =>3x^2-3x=0< =>3x\left(x-1\right)=0< =>\orbr{\begin{cases}x=0\left(loai\right)\\x=1\left(tm\right)\end{cases}}\)
1/
<=> 3x + 12 = 0
<=> 3x = -12
<=> x = -4
2/
<=> x + x = 3 + 5
<=> 2x = 8
<=> x = 4
3/
<=> 2x - 3 + 5x = 4x + 12
<=> 2x + 5x - 4x = 12 + 3
<=> 3x = 15
<=> x = 5
4/ Câu này bạn vào kí hiệu toán học của Olm viết lại đề cho rõ xíu nhé!
a: \(\Leftrightarrow7\left(7-3x\right)+12\left(5x+2\right)=84\left(x+13\right)\)
\(\Leftrightarrow49-21x+60x+24=84x+1092\)
\(\Leftrightarrow39x-84x=1092-73\)
=>-45x=1019
hay x=-1019/45
b: \(\Leftrightarrow21\left(x+3\right)-14=4\left(5x+9\right)-7\left(7x-9\right)\)
=>21x+63-14=20x+36-49x+63
=>21x+49=-29x+99
=>50x=50
hay x=1
c: \(\Leftrightarrow7\left(2x+1\right)-3\left(5x+2\right)=21x+63\)
=>14x+7-15x-6-21x-63=0
=>-22x-64=0
hay x=-32/11
d: \(\Leftrightarrow35\left(2x-3\right)-15\left(2x+3\right)=21\left(4x+3\right)-17\cdot105\)
=>70x-105-30x-45=84x+63-1785
=>40x-150-84x+1722=0
=>-44x+1572=0
hay x=393/11
a: \(\Leftrightarrow x^2-2x+1-x^2-2x-1=2x-6\)
=>2x-6=-4x
=>6x=6
hay x=1
b: \(\Leftrightarrow\left(x-3\right)\left(x+3\right)-\left(x-3\right)\left(5x+2\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+3-5x-2\right)=0\)
=>(x-3)(-4x+1)=0
=>x=3 hoặc x=1/4
c: \(\Leftrightarrow4x^2+12x+9-3\left(x^2-16\right)-x^2+4x-4=0\)
\(\Leftrightarrow3x^2+16x+5-3x^2+48=0\)
=>16x+53=0
hay x=-53/16
d: \(\Leftrightarrow x^3+4x^2-9x-36=0\)
\(\Leftrightarrow\left(x+4\right)\left(x^2-9\right)=0\)
hay \(x\in\left\{-4;3;-3\right\}\)
b)x^2-9=(x-3)(5x+2)
\(\Leftrightarrow\left(x-3\right)\left(x+3\right)-\left(x-3\right)\left(5x+2\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+3-5x-2\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(1-4x\right)=0\)
\(\Rightarrow\left\{{}\begin{matrix}x-3=0\\1-4x=0\end{matrix}\right.\left\{{}\begin{matrix}x=0+3\\x=1:4\end{matrix}\right.\left\{{}\begin{matrix}x=3\\x=\dfrac{1}{4}\end{matrix}\right.\)
a) x4 + 2x3 + 5x2 + 4x - 12 = 0
=> x4 - x3+ 3x3- 3x2 + 8x2 -8x + 12x - 12 = 0
=> x3( x - 1) + 3x2( x - 1) + 8x( x - 1) + 12 ( x - 1 ) = 0
=> ( x - 1)( x3 + 3x2 + 8x + 12 ) = 0
=> ( x - 1)( x3 + 2x2 + x2 + 2x + 6x + 12 ) = 0
=> ( x - 1)[ x2( x + 2) + x( x + 2) + 6( x + 2) ] = 0
=> ( x - 1)( x + 2)( x2 + x + 6 ) = 0
Ta thấy : x2 + x + 6
= x2 + 2.\(\dfrac{1}{2}x+\dfrac{1}{4}-\dfrac{1}{4}+6=\left(x+\dfrac{1}{2}\right)^2+\dfrac{23}{4}\text{≥}\dfrac{23}{4}>0\text{∀}x\)
=> ( x - 1)( x + 2 ) = 0
=> x = 1 hoặc x = -2
Vậy,....
b) ( x + 1)3 + ( x - 2)3 = ( 2x - 1)3
=>x3+ 3x2 + 3x + 1 + x3 - 6x2 + 12x - 8 - ( 8x3 - 12x2 + 6x - 1)=0
=> 2x3 - 3x2 + 15x - 7 - 8x3 + 12x2 - 6x + 1 = 0
=> 9x2 - 6x3 + 9x - 6 = 0
=> 9x( x + 1) -6( x3 + 1 ) = 0
=> 9x( x + 1) - 6( x + 1)( x2 - x + 1) = 0
=> 3( x + 1)( 3x - 2x2 + 2x - 2) = 0
=> 3( x + 1)( - 2x2 + 5x - 2) = 0
=> 3( x + 1)( - 2x2 + x + 4x - 2) = 0
=> 3( x + 1)[ x( 1 - 2x ) - 2( 1 - 2x ) ] = 0
=> 3( x + 1)( 1 - 2x )( x - 2) = 0
Suy ra :
* x + 1 = 0 => x = -1
* 1 - 2x = 0 => x = \(\dfrac{1}{2}\)
* x - 2 = 0 => x = 2
Vậy,.....
Ok , No broblem Trang Mi