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\(\frac{x-2}{x+2}-\frac{3}{x-2}=\frac{2\left(x-11\right)}{x^2-4}\left(x\ne\pm2\right)\)
\(\Leftrightarrow\frac{\left(x-2\right)^2}{\left(x-2\right)\left(x+2\right)}-\frac{3\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}=\frac{2x-22}{\left(x-2\right)\left(x+2\right)}\)
=> x2 - 4x + 4 - 3x - 6 = 2x - 22
<=> x2 - 9x + 20 = 0
<=> x2 - 4x - 5x + 20 = 0
<=> x( x - 4 ) - 5( x - 4 ) = 0
<=> ( x - 4 )( x - 5 ) = 0
<=> x - 4 = 0 hoặc x - 5 = 0
<=> x = 4 (tm) hoặc x = 5 (tm)
Vậy ...
\(\frac{x-2}{x+2}-\frac{3}{x-2}=\frac{2\left(x-11\right)}{x^2-4}\)ĐKXĐ : \(x\ne\pm2\)
\(\Leftrightarrow\frac{\left(x-2\right)^2-3\left(x+2\right)}{\left(x+2\right)\left(x-2\right)}=\frac{2x-22}{\left(x-2\right)\left(x+2\right)}\)
\(\Rightarrow x^2-4x+4-3x-6=2x-22\)
\(\Leftrightarrow x^2-7x-2=2x-22\Leftrightarrow x^2-9x+20=0\)
\(\Leftrightarrow\left(x-4\right)\left(x-5\right)=0\Leftrightarrow x=4;x=5\)( tmđk )
Vậy tập nghiệm phương trình là S = { 4 ; 5 }
ĐKXĐ : \(\hept{\begin{cases}x^2+x-6\ne0\\x^2+4x+3\ne0\\2x-1\ne0\end{cases}\Leftrightarrow\hept{\begin{cases}\left(x+3\right)\left(x-2\right)\ne0\\\left(x+1\right)\left(x+3\right)\ne0\\x\ne\frac{1}{2}\end{cases}\Rightarrow\hept{\begin{cases}x\ne2;-3\\x\ne-1;-3\\x\ne\frac{1}{2}\end{cases}}}}\)
TXĐ : \(x\ne\left\{-3;-1;\frac{1}{2};2\right\}\)
\(pt\Leftrightarrow\frac{5}{\left(x+3\right)\left(x-2\right)}-\frac{2}{\left(x+1\right)\left(x+3\right)}=\frac{-3}{2x-1}\)
\(\Leftrightarrow\frac{5\left(x+1\right)-2\left(x-2\right)}{\left(x-2\right)\left(x+1\right)\left(x+3\right)}=\frac{-3}{2x-1}\)
\(\Leftrightarrow\frac{3x+9}{\left(x-2\right)\left(x+1\right)\left(x+3\right)}=\frac{-3}{2x-1}\)
\(\Leftrightarrow\frac{3}{\left(x-2\right)\left(x+1\right)}=\frac{-3}{2x-1}\)
\(\Leftrightarrow\frac{1}{x^2-x-2}=\frac{1}{1-2x}\)
\(\Leftrightarrow x^2-x-2-1+2x=0\)
\(\Leftrightarrow x^2+x-3=0\)
\(\Leftrightarrow\left(x^2+2.\frac{1}{2}.x+\frac{1}{4}\right)-\frac{13}{4}=0\)
\(\Leftrightarrow\left(x+\frac{1}{2}\right)^2-\left(\frac{\sqrt{13}}{2}\right)^2=0\)
\(\Leftrightarrow\left(x+\frac{1-\sqrt{13}}{2}\right)\left(x+\frac{1+\sqrt{13}}{2}\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{\sqrt{13}-1}{2}\\x=\frac{-\sqrt{13}-1}{2}\end{cases}}\)
\(\frac{5}{x^2+x-6}-\frac{2}{x^2+4+3}=-\frac{3}{2x-1}\)
<=> \(\frac{5}{\left(x-2\right)\left(x+3\right)}-\frac{2}{\left(x+1\right)\left(x+3\right)}=-\frac{3}{2x-1}\)
<=> \(\frac{5\left(x+1\right)-2\left(x-2\right)}{\left(x-2\right)\left(x+3\right)}=-\frac{3}{2x-1}\)
<=> \(\frac{5x+5-2x+4}{\left(x-2\right)\left(x+3\right)}=-\frac{3}{2x-1}\)
<=> \(\frac{3x+9}{\left(x-2\right)\left(x+3\right)}=-\frac{3}{2x-1}\)
<=> \(\frac{3\left(x+3\right)}{\left(x-2\right)\left(x+3\right)}=-\frac{3}{2x-1}\)
<=> \(\frac{1}{x-2}=-\frac{1}{2x-1}\)
<=> x-2=1-2x <=> 3x=3
=> x=1
Đáp số: x=1
a: \(\Leftrightarrow\dfrac{y+5}{y\left(y-5\right)}-\dfrac{y-5}{2y\left(y+5\right)}=\dfrac{y+25}{2\left(y-5\right)\left(y+5\right)}\)
\(\Leftrightarrow2\left(y+5\right)^2-\left(y-5\right)^2=y^2+25y\)
=>\(2y^2+20y+50-y^2+10y-25=y^2+25y\)
=>30y+25=25y
=>5y=-25
=>y=-5(loại)
b: \(\Leftrightarrow x\left(x+1\right)+x\left(x-3\right)=4x\)
=>x^2+x+x^2-3x-4x=0
=>2x^2-6x=0
=>2x(x-3)=0
=>x=0(nhận) hoặc x=3(loại)
c: =>x^2-9-6(2x+7)=-13(x+3)
=>x^2-9-12x-42+13x+39=0
=>x^2+x-6=0
=>(x+3)(x-2)=0
=>x=2(nhận) hoặc x=-3(loại)
\(a.\frac{x-5}{4}-2x+1=\frac{x}{3}-\frac{2-x}{6}\\\Leftrightarrow \frac{3\left(x-5\right)}{12}-\frac{24}{12}x+\frac{12}{12}=\frac{4x}{12}-\frac{2\left(2-x\right)}{12}\\\Leftrightarrow 3\left(x-5\right)-24x+12=4x-2\left(2-x\right)\\\Leftrightarrow 3x-15-24x+12=4x-4+2x\\ \Leftrightarrow3x-15-24x+12-4x+4-2x=0\\ \Leftrightarrow-27x+1=0\\ \Leftrightarrow-27x=-1\\ \Leftrightarrow x=\frac{1}{27}\)
\(b.\left(2x-1\right)^2=\left(x-2\right)\left(2x-1\right)\\ \Leftrightarrow\left(2x-1\right)^2-\left(x-2\right)\left(2x-1\right)=0\\ \Leftrightarrow\left(2x-1\right)\left[\left(2x-1\right)-\left(x-2\right)\right]=0\\ \Leftrightarrow\left(2x-1\right)\left(2x-1-x+2\right)=0\\ \Leftrightarrow\left(2x-1\right)\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}2x-1=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{2}\\x=-1\end{matrix}\right.\)
\(c.\frac{x+5}{x-5}-\frac{x-5}{x+5}=\frac{-3}{25-x^2}\\\Leftrightarrow \frac{x+5}{x-5}-\frac{x-5}{x+5}=\frac{3}{x^2-25}\\\Leftrightarrow \frac{x+5}{x-5}-\frac{x-5}{x+5}=\frac{3}{\left(x-5\right)\left(x+5\right)}\\ \Leftrightarrow\frac{\left(x+5\right)\left(x+5\right)}{\left(x-5\right)\left(x+5\right)}-\frac{\left(x-5\right)\left(x-5\right)}{\left(x-5\right)\left(x+5\right)}=\frac{3}{\left(x-5\right)\left(x+5\right)}\\ \Leftrightarrow\left(x+5\right)\left(x+5\right)-\left(x-5\right)\left(x-5\right)=3\\\Leftrightarrow x^2+5x+5x+25-\left(x^2-5x-5x+25\right)=3\\\Leftrightarrow x^2+5x+5x+25-x^2+5x+5x-25=3\\ \Leftrightarrow20x=3\\ \Leftrightarrow x=\frac{3}{20}\)
\(d.x^2-x-12=0\\\Leftrightarrow x^2-4x+3x-12=0\\\Leftrightarrow \left(x^2-4x\right)+\left(3x-12\right)=0\\ \Leftrightarrow x\left(x-4\right)+3\left(x-4\right)=0\\ \Leftrightarrow\left(x-4\right)\left(x+3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-4=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-3\end{matrix}\right.\)
( 2x - 1 ) - x = 0
=> 2x - 1 = x
=> 2x - x = 1
=> x = 1
( x - 1 )( 2x - 3) = 0
=> \(\orbr{\begin{cases}x-1=0\\2x-3=0\end{cases}}\)=> \(\orbr{\begin{cases}x=1\\x=\frac{3}{2}\end{cases}}\)
Vậy tập nghiệm của phương trình là S = { 1 ; 3/2 }
\(\frac{x}{x+1}=\frac{x+2}{x-1}\)( đkxđ : \(x\ne\pm1\))
( Chỗ này chưa học kĩ nên chưa hiểu lắm :]
`b,(x+5)(2x-3)=0`
`<=>` $\left[ \begin{array}{l}x+5=0\\2x-3=0\end{array} \right.$
`<=>` $\left[ \begin{array}{l}x=\dfrac{3}{2}\\x=-5\end{array} \right.$
Vậy `S={-5,3/2}`
a: =>3(3x-7)+2(x+1)=-96
=>9x-21+2x+2=-96
=>11x-19=-96
=>11x=-96+19=-75
=>x=-75/11
b: \(x-\dfrac{x+1}{3}=\dfrac{2x+1}{5}\)
=>15x-5(x+1)=3(2x+1)
=>15x-5x-5=6x+3
=>10x-5=6x+3
=>4x=8
=>x=2
a)
\(\dfrac{3x-7}{2}+\dfrac{x+1}{3}=-16\)
\(< =>9x-21+2x+2=-96\)
\(< =>9x+2x=-96+21-2\\ < =>11x=-77\\ < =>x=-7\)
b)
\(\dfrac{x-x+1}{3}=\dfrac{2x+1}{5}\\ < =>5=6x+3\\ < =>6x=5-3\\ < =>6x=2\\ < =>x=\dfrac{1}{3}\)
\(2x^2+4x+5=2\left(x^2+2x+\frac{5}{2}\right)=2\left[\left(x^2+2.x.1+1\right)+\frac{3}{2}\right]=2\left(x+1\right)^2+3\ge3\)
Min=3 khi x=-1
Còn phần cô giáo thì zầy nè
\(\frac{1}{2x^2+4x+5}=\frac{1}{2\left(x^2+2x+\frac{5}{2}\right)}=\frac{1}{2\left[\left(x^2+2.x.1+1\right)+\frac{3}{2}\right]}=\frac{1}{2\left(x+1\right)^2+3}\)
muốn \(\frac{1}{2x^2+4x+5}\) lớn nhất thì \(2x^2+4x+5\)nhỏ nhất
\(2x^2+4x+5=2\left(x^2+2x+\frac{5}{2}\right)=2\left[\left(x^2+2.x.1+1\right)+\frac{3}{2}\right]=2\left(x+1\right)^2+3\ge3\)
Min=3 khi x=-1
Nếu cảm thấy đúng thì
\(\frac{x+5}{2x-1}-\frac{1-2x}{x+5}-2=0\left(x\ne\frac{1}{2};x\ne-5\right)\)
<=> \(\frac{\left(x+5\right)^2}{\left(2x-1\right)\left(x+5\right)}+\frac{\left(2x-1\right)^2}{\left(2x-1\right)\left(x+5\right)}-\frac{2\left(2x-1\right)\left(x+5\right)}{\left(2x-1\right)\left(x+5\right)}=0\)
=> x2 + 10x + 25 + 4x2 - 4x + 1 - 2( 2x2 + 9x - 5 ) = 0
<=> 5x2 + 6x + 26 - 4x2 - 18x + 10 = 0
<=> x2 - 12x + 36 = 0
<=> ( x - 6 )2 = 0
<=> x - 6 = 0 <=> x = 6 (tm)
Vậy ...
\(\frac{x+5}{2x-1}-\frac{1-2x}{x+5}-2=0\)ĐKXĐ : \(x\ne-5;\frac{1}{2}\)
\(\Leftrightarrow\frac{\left(x+5\right)^2-\left(1-2x\right)\left(2x-1\right)}{\left(2x-1\right)\left(x+5\right)}-\frac{2\left(x+5\right)\left(2x-1\right)}{\left(x+5\right)\left(2x-1\right)}=0\)
\(\Leftrightarrow\frac{\left(x+5\right)^2+\left(2x-1\right)^2-2\left(x+5\right)\left(2x-1\right)}{\left(x+5\right)\left(2x-1\right)}=0\)
\(\Rightarrow x^2+10x+25+\left(4x^2-4x+1\right)-2\left(2x^2-x+10x-5\right)=0\)
\(\Leftrightarrow x^2+10x+25+4x^2-4x+1-4x^2-18x+10=0\)
\(\Leftrightarrow x^2-12x+36=0\Leftrightarrow\left(x-6\right)^2=0\Leftrightarrow x=6\)
Vậy tập nghiệm của phương trình là S = { 6 }