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\(P=\left(\frac{x-1}{x+3}+\frac{2}{x-3}+\frac{x^2+3}{9-x^2}\right):\left(\frac{2x-1}{2x+1}-1\right)\)\(\left(đkcđ:x\ne\pm3;x\ne-\frac{1}{2}\right)\)
\(=\left(\frac{\left(x-1\right).\left(x-3\right)+2.\left(x+3\right)-\left(x^2+3\right)}{x^2-9}\right):\left(\frac{2x-1-\left(2x+1\right)}{2x+1}\right)\)
\(=\frac{x^2-4x+3+2x+6-x^2-3}{x^2-9}:\frac{-2}{2x+1}\)
\(=\frac{-2x-6}{x^2-9}.\frac{2x+1}{-2}\)
\(=\frac{-2\left(x+3\right)}{\left(x-3\right).\left(x+3\right)}.\frac{2x+1}{-2}\)
\(=\frac{2x+1}{x-3}\)
b)\(\left|x+1\right|=\frac{1}{2}\Leftrightarrow\orbr{\begin{cases}x+1=\frac{1}{2}\\x+1=-\frac{1}{2}\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-\frac{1}{2}\left(koTMđkxđ\right)\\x=-\frac{3}{2}\left(TMđkxđ\right)\end{cases}}}\)
thay \(x=-\frac{3}{2}\) vào P tâ đc: \(P=\frac{2x+1}{x-3}=\frac{2.\left(-\frac{3}{2}\right)+1}{-\frac{3}{2}-3}=\frac{4}{9}\)
c)ta có:\(P=\frac{x}{2}\Leftrightarrow\frac{2x+1}{x-3}=\frac{x}{2}\)
\(\Rightarrow2.\left(2x+1\right)=x.\left(x-3\right)\)
\(\Leftrightarrow4x+2=x^2-3x\)
\(\Leftrightarrow x^2-7x-2=0\)
\(\Leftrightarrow x^2-2.\frac{7}{2}+\frac{49}{4}-\frac{57}{4}=0\)
\(\Leftrightarrow\left(x-\frac{7}{2}\right)^2-\frac{57}{4}=0\)
\(\Leftrightarrow\left(x-\frac{7}{2}-\frac{\sqrt{57}}{2}\right).\left(x-\frac{7}{2}+\frac{\sqrt{57}}{2}\right)\)
bạn tự giải nốt nhé!!
d)\(x\in Z;P\in Z\Leftrightarrow\frac{2x+1}{x-3}\in Z\Leftrightarrow\frac{2x-6+7}{x-3}=2+\frac{7}{x-3}\in Z\)
\(2\in Z\Rightarrow\frac{7}{x-3}\in Z\Leftrightarrow x-3\inƯ\left(7\right)=\left\{\pm1;\pm7\right\}\)
bạn tự làm nốt nhé
a, \(\left(\dfrac{x^2-4x+3+2x+6-x^2-3}{\left(x+3\right)\left(x-3\right)}\right):\left(\dfrac{2x-1-2x-1}{2x+1}\right)\)
\(=\dfrac{-2x+6}{\left(x+3\right)\left(x-3\right)}:\dfrac{-2}{2x+1}=\dfrac{-2\left(x-3\right)\left(2x+1\right)}{-2\left(x+3\right)\left(x-3\right)}=\dfrac{2x+1}{x+3}\)
b, \(\left|x+1\right|=\dfrac{1}{2}\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}-1\\x=-\dfrac{1}{2}-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\left(ktmđk\right)\\x=-\dfrac{3}{2}\end{matrix}\right.\)
Thay x = -3/2 ta được \(\dfrac{2\left(-\dfrac{3}{2}\right)+1}{-\dfrac{3}{2}+3}=\dfrac{-2}{\dfrac{3}{2}}=-\dfrac{4}{3}\)
c) \(\left(3x+5\right)^2-2\left(2x+3\right)\left(3x+5\right)+\left(2x+3\right)^2=\left(x+2\right)^3\)
\(\Leftrightarrow\left[\left(3x+5\right)-\left(2x+3\right)\right]^2=\left(x+2\right)^3\)
\(\Leftrightarrow\left(3x+5-2x-3\right)^2=\left(x+2\right)^3\)
\(\Leftrightarrow\left(x+2\right)^2=\left(x+2\right)^3\)
\(\Leftrightarrow\left(x+2\right)^3-\left(x+2\right)^2=0\)
\(\Leftrightarrow\left(x+2\right)^2.\left(x+2-1\right)=0\)
\(\Leftrightarrow\left(x+2\right)^2.\left(x+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+2=0\\x+1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-2\\x=-1\end{cases}}\)
Vậy tập nghiệm của phương trình là: \(S=\left\{-2;-1\right\}\)
e, (x-1)(x2 + x + 1)-x(x+2)(x-2) = 5
x(x2 +x + 1 ) - (x2 + x +1 )- [ x (x2 - 4)] = 5
x3 +x2 +x - x2 - x - 1 - x3 +4x = 5
4x - 1 = 5
4x = 6
x =\(\dfrac{3}{2}\)
f, (x-1)3 - (x+3)(x2 - 3x +9 ) +3(x2 - 4) = 2
x - 3x2 +3x - 1 - [( x3 - 3x2 + 9x) + (3x2 - 9x +27)] = 2
x3 - 3x2 + 3x - 1 -x3 +3x2 -9x - 3x2 +9x - 27 +3x2 - 12 = 2
3x - 1 - 27 - 12 = 2
3x = 42
x = 14
a) \(\left(x+2\right)^2-9=0\)
\(\Rightarrow\left(x+2\right)^2=9\)
\(\Rightarrow\left(x+2\right)^2=3^2\)
\(\Rightarrow x+2=3\)
\(\Rightarrow x=3-2=1\)
a) ( x + 2 )2 = 9
=> ( x + 2 ) 2 = 9
=> ( x + 2 )2 = 32
=> x + 2 = + 3
=> \(\orbr{\begin{cases}x+2=-3\\x+2=3\end{cases}}\)
=> \(\orbr{\begin{cases}x=-1\\x=5\end{cases}}\)
Vậy x = -1; 5
b) ( x + 2 )2 - x2 + 4 = 0
=> ( x + 2 )2 - ( x2 - 4 ) = 0
=> ( x + 2 )2 - ( x + 2 ) ( x - 2 ) = 0
=> ( x + 2 ) ( x + 2 - x + 2 ) = 0
=> ( x + 2 ) . 4 = 0
=> x + 2 = 0
=> x = - 2
Vậy x = - 2
c) 5 ( 2x - 3 )2 - 5 ( x + 1 )2 - 15( x + 4 ) ( x - 4 ) = - 10
=> 5 ( 4x2 - 12x + 9 ) - 5 ( x2 + 2x + 1 ) - 15 ( x2 - 42 ) = - 10
=> 20x2 - 60x + 45 - 5x2 - 10x - 5 - 15x2 + 240 = -10
=> - 70x + 280 = - 10
=> - 70x = - 290
=> x = \(\frac{29}{7}\)
Vậy x = \(\frac{29}{7}\)
d) x ( x + 5 ) ( x - 5 ) - ( x + 2 ) ( x2 - 2x + 4 ) = 3
=> x ( x2 - 25 ) - ( x3 - 8 ) = 3
=> x3 - 25x - x3 + 8 = 3
=> - 25x + 8 = 3
=> - 25x = -5
=> x = \(\frac{1}{5}\)
Vậy x = \(\frac{1}{5}\)
a. \(\dfrac{x+3}{x-3}-\dfrac{x-3}{x+3}=\dfrac{9}{x^2-9}\) (ĐKXĐ: \(x\ne\pm3\))
\(\Leftrightarrow\left(x+3\right)^2-\left(x-3\right)^2=9\)
\(\Leftrightarrow x^2+6x+9-x^2+6x-9=9\)
\(\Leftrightarrow12x=9\Leftrightarrow x=\dfrac{3}{4}\left(tm\right)\)
\(\Rightarrow S=\left\{\dfrac{3}{4}\right\}\)
b. \(\dfrac{x+2}{4}-x+3=\dfrac{1-x}{8}\)
\(\Leftrightarrow2\left(x+2\right)-8\left(x-3\right)=1-x\)
\(\Leftrightarrow2x+4-8x+24=1-x\)
\(\Leftrightarrow2x-8x+x=1-4-24\)
\(\Leftrightarrow-3x=-27\Leftrightarrow x=9\)
\(\Rightarrow S=\left\{9\right\}\)
-Mệt -.-
\(\frac{x-3}{5}-\frac{2x-1}{10}=\frac{x+1}{2}+\frac{1}{4}\)
\(< =>\frac{\left(x-3\right).4}{20}-\frac{\left(2x-1\right).2}{20}=\frac{\left(x+1\right).10}{20}+\frac{5}{20}\)
\(< =>4x-12-4x+2=10x+10+5\)
\(< =>10x=-10-10-5=-25\)
\(< =>x=-\frac{25}{10}=-\frac{5}{2}\)
\(\frac{x+3}{2}-\frac{2x-1}{3}-1=\frac{x+5}{5}\)
\(< =>\frac{\left(x+3\right).15}{30}-\frac{\left(2x-1\right).10}{30}-\frac{30}{30}=\frac{\left(x+5\right).5}{30}\)\(< =>15x+45-20x+10-30=5x+25\)
\(< =>-5x+25=5x+25< =>10x=0< =>x=0\)
a) Rút gọn VT = 45x + 8. Từ đó tìm được x = 2 15 .
b) Rút gọn VT = -25x – 8. Từ đó tìm được x = − 11 25 .