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Ta có: 5x + 3x2 = 0
<=> x(3x + 5) = 0
<=> \(\orbr{\begin{cases}x=0\\3x+5=0\end{cases}}\)
<=> \(\orbr{\begin{cases}x=0\\x=-\frac{5}{3}\end{cases}}\) Vậy S = {0; -5/3)
5(x2 - 2x) = (3 + 5x)(x - 1)
<=> 5x2 - 10x = 5x2 - 2x - 3
<=> 5x2 - 10x - 5x2 + 2x = -3
<=> -8x = -3
<=> x = 3/8 Vậy S = {3/8}
(4x + 3)2 = 4(x - 1)2
<=> (4x + 3)2 - (2x - 2)2 = 0
<=> (4x + 3 - 2x + 2)(4x +3 + 2x - 2) = 0
<=> (2x + 5)(6x + 1) = 0
<=> \(\orbr{\begin{cases}2x+5=0\\6x+1=0\end{cases}}\)
<=> \(\orbr{\begin{cases}x=-\frac{5}{2}\\x=-\frac{1}{6}\end{cases}}\) Vậy S = {-5/3; -1/6}
a) 5x + 3.x2 = 0
<=>x . ( 5 + 3x ) = 0
<=> \(\orbr{\begin{cases}x=0\\5+3.x=0\end{cases}}\)
<=>\(\orbr{\begin{cases}x=0\\z=-\frac{5}{3}\end{cases}}\)
Nghiệm cuối cùng là :{ 0;\(-\frac{5}{3}\)}
b) 5.( x2 - 2.x ) = ( 3 + 5.x ) . ( x- 1 )
<=>5.x2 - 10.x = 3.x -3 + 5.x2 - 5.x
<=> -10.x = 3.x - 3-5.x
<=> -10.x = -2.x - 3
<=> -8.x = -3
<=> x = \(\frac{3}{8}\)
Vậy x = \(\frac{3}{8}\)
c) ( 4x + 3 )2 = 4. ( x - 1 )2
<=> 16.x2 + 24.x + 9 = 4.( x2 -2.x + 1 )
<=> 16.x2+24.x + 9 = 4.x2 -8.x + 4
<=> 16.x2 +24.x + 9 -4.x2 + 8.x - 4= 0
<=> 12.x2 + 32.x + 5 = 0
<=> 12.x2 + 30.x + 2.x + 5 = 0
<=> 6.x . ( 2.x + 5 ) + 2.x + 5 =0
<=> ( 2.x + 5 ) . ( 6.x + 1 ) =0
<=> \(\orbr{\begin{cases}2.x+5=0\\6.x+1=0\end{cases}}\)
<=> \(\orbr{\begin{cases}x=-\frac{5}{2}\\x=-\frac{1}{6}\end{cases}}\)
Nghiệm cuối cùng là : { \(-\frac{5}{2};-\frac{1}{6}\)}
1: Ta có: \(\dfrac{5x^2-12}{x^2-1}+\dfrac{3}{x-1}=\dfrac{5x}{x+1}\)
\(\Leftrightarrow\dfrac{5x^2-12}{\left(x-1\right)\left(x+1\right)}+\dfrac{3x+3}{\left(x-1\right)\left(x+1\right)}=\dfrac{5x^2-5x}{\left(x+1\right)\left(x-1\right)}\)
Suy ra: \(5x^2+3x-9=5x^2-5x\)
\(\Leftrightarrow8x=9\)
hay \(x=\dfrac{9}{8}\left(tm\right)\)
2: Ta có: \(\dfrac{3}{x-5}-\dfrac{15-3x}{x^2-25}=\dfrac{3}{x+5}\)
\(\Leftrightarrow\dfrac{3x+15}{\left(x-5\right)\left(x+5\right)}+\dfrac{3x-15}{\left(x-5\right)\left(x+5\right)}=\dfrac{3x-15}{\left(x+5\right)\left(x-5\right)}\)
Suy ra: \(6x=3x-15\)
\(\Leftrightarrow3x=-15\)
hay \(x=-5\left(loại\right)\)
2. ĐKXĐ: $x\neq \pm 5$
PT \(\Leftrightarrow \frac{3}{x-5}+\frac{3x-15}{x^2-25}=\frac{3}{x+5}\)
\(\Leftrightarrow \frac{3}{x-5}+\frac{3(x-5)}{(x-5)(x+5)}=\frac{3}{x+5}\)
\(\Leftrightarrow \frac{3}{x-5}+\frac{3}{x+5}=\frac{3}{x+5}\Leftrightarrow \frac{3}{x-5}=0\) (vô lý)
Vậy pt vô nghiệm.
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,\(\frac{4}{x+2}-\frac{4x-6}{4x-x^3}=\frac{x-3}{x\left(x-2\right)}\left(đk:x\ne-2;0;2\right)\)
\(< =>\frac{4x\left(x-2\right)}{x\left(x-2\right)\left(x+2\right)}+\frac{4x-6}{x\left(x-2\right)\left(2+x\right)}=\frac{\left(x-3\right)\left(x+2\right)}{x\left(x-2\right)\left(x+2\right)}\)
\(< =>4x^2-8x+4x-6=x^2-x-6\)
\(< =>4x^2-x^2-4x+x-6+6=0\)
\(< =>3x^2-3x=0< =>3x\left(x-1\right)=0< =>\orbr{\begin{cases}x=0\left(loai\right)\\x=1\left(tm\right)\end{cases}}\)
a: 3x-5>15-x
=>4x>20
hay x>5
b: \(3\left(x-2\right)\left(x+2\right)< 3x^2+x\)
=>3x2+x>3x2-12
=>x>-12
a, \(12-2\left(1-x\right)^2=\left(3x-2\right)\left(2x-3\right)\)
\(< =>12-2\left(1-2x+x^2\right)=6x^2-9x-4x+6\)
\(< =>12-2+4x-2x^2=6x^2-13x+6\)
\(< =>10+4x-2x^2-6x^2+13x-6=0\)
\(< =>-8x^2+17x+4=0< =>\orbr{\begin{cases}x=\frac{17-\sqrt{417}}{16}\\x=\frac{17+\sqrt{417}}{16}\end{cases}}\)
b, \(10x+3-5x=4x+12< =>5x+3-4x-12=0\)
\(< =>x-9=0< =>x=9\)
c, \(11x+42-2x=100-9x-22< =>9x+42-100+9x+22=0\)
\(< =>18x+64-100=0< =>18x-36=0< =>x=\frac{36}{18}=2\)
d, \(2x-\left(3-5x\right)=4\left(x+3\right)< =>2x-3+5x=4x+12\)
\(< =>7x-3-4x-12=0< =>3x-15=0< =>x=\frac{15}{3}=5\)
e, \(2\left(x-3\right)+5x\left(x-1\right)=5x^2< =>2x-6+5x^2-5=5x^2\)
\(< =>2x-11+5x^2-5x^2=0< =>2x-11=0< =>x=\frac{11}{2}\)
f, \(-6\left(1,5-2x\right)=3\left(-15+2x\right)< =>-6\left(\frac{3}{2}-2x\right)=3\left(2x-15\right)\)
\(< =>-9+12x-6x+45=0< =>6x+36=0< =>x=-6\)
g, \(14x-\left(2x+7\right)=3x+12x-13< =>14x-2x-7=15x-13\)
\(< =>12x-7-15x+13=0< =>-3x+6=0< =>x=-2\)
h, \(\left(x-4\right)\left(x+4\right)-2\left(3x-2\right)=\left(x-4\right)^2\)
\(< =>x^2-16-6x+4=x^2-8x+16\)
\(< =>x^2-6x-12-x^2+8x-16=0\)
\(< =>2x-28=0< =>x=\frac{28}{2}=14\)
q, \(4\left(x-2\right)-\left(x-3\right)\left(2x-5\right)=?\)thiếu đề
1) -2(x - 3) + 5x (x - 1) = 5x (x + 1)
<=> -2x + 6 + 5x2 - 5x = 5x2 + 5x
<=> 6 = 5x2 + 5x + 2x - 5x2 + 5x
<=> 6 = 12x
<=> \(\dfrac{6}{12}\) = x = 0,5
vậy tập nghiệm S ={0,5}
2) 7 - (2x + 4) = -(x + 4)
<=> 7 - 2x - 4 = -x - 4
<=> 7 - 4 + 4 = -x + 2x
<=> 7 = x
vậy tập nghiệm S ={7}