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\(a,\left(2x-3\right)\left(x^2-4\right)=0\\ \Leftrightarrow\left(2x-3\right)\left(x-2\right)\left(x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=2\\x=-2\end{matrix}\right.\\ b,2x-\left(3-5x\right)=4\left(x+3\right)\\ \Leftrightarrow2x-3+5x=4x+12\\ \Leftrightarrow7x-3-4x-12=0\\ \Leftrightarrow3x-15=0\\ \Leftrightarrow x=5\)
\(c,ĐKXĐ:\left\{{}\begin{matrix}x\ne-1\\x\ne2\end{matrix}\right.\)
\(\dfrac{1}{x-2}-\dfrac{2}{x+1}=\dfrac{11-3x}{\left(x+1\right)\left(x-2\right)}\\ \Leftrightarrow\dfrac{x+1}{\left(x-2\right)\left(x+1\right)}-\dfrac{x-2}{\left(x+1\right)\left(x-2\right)}-\dfrac{11-3x}{\left(x+1\right)\left(x-2\right)}=0\\ \Leftrightarrow\dfrac{x+1-x+2-11+3x}{\left(x+1\right)\left(x-2\right)}=0\\ \Rightarrow3x-8=0\\ \Leftrightarrow x=\dfrac{8}{3}\left(tm\right)\)
1/ ( x-3) 2=16
\(\Rightarrow\left[{}\begin{matrix}x-3=4\\x-3=-4\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=7\\x=-1\end{matrix}\right.\)
2/ (3x-1)3=8
\(\Rightarrow3x-1=2\\ \Rightarrow3x=3\\ \Rightarrow x=1\)
3/ (x-11)3=-27
\(\Rightarrow x-11=-3\\ \Rightarrow x=8\)
phần 4 mình ko rõ đề
\(a,\dfrac{x-3}{x}=\dfrac{x-3}{x+3}\)\(\left(đk:x\ne0,-3\right)\)
\(\Leftrightarrow\dfrac{x-3}{x}-\dfrac{x-3}{x+3}=0\)
\(\Leftrightarrow\dfrac{\left(x-3\right)\left(x+3\right)-x\left(x-3\right)}{x\left(x+3\right)}=0\)
\(\Leftrightarrow x^2-9-x^2+3x=0\)
\(\Leftrightarrow3x-9=0\)
\(\Leftrightarrow3x=9\)
\(\Leftrightarrow x=3\left(n\right)\)
Vậy \(S=\left\{3\right\}\)
\(b,\dfrac{4x-3}{4}>\dfrac{3x-5}{3}-\dfrac{2x-7}{12}\)
\(\Leftrightarrow\dfrac{4x-3}{4}-\dfrac{3x-5}{3}+\dfrac{2x-7}{12}>0\)
\(\Leftrightarrow\dfrac{3\left(4x-3\right)-4\left(3x-5\right)+2x-7}{12}>0\)
\(\Leftrightarrow12x-9-12x+20+2x-7>0\)
\(\Leftrightarrow2x+4>0\)
\(\Leftrightarrow2x>-4\)
\(\Leftrightarrow x>-2\)
a, làm tương tự với phần b bài nãy bạn đăng
b, \(\left(x+1\right)^2-5=x^2+11\)
\(\Leftrightarrow x^2+2x+1-5=x^2+11\)
\(\Leftrightarrow2x-10=0\Leftrightarrow x=5\)
Vậy tập nghiệm của phương trình là S = { 5 } ( kết luận như thế với các phần sau nhé ! )
c, \(3\left(3x-1\right)=3x+5\Leftrightarrow9x-3-3x-5=0\)
\(\Leftrightarrow6x-8=0\Leftrightarrow x=\frac{4}{3}\)
d, \(3x\left(2x-3\right)-3\left(3+2x^2\right)=0\)
\(\Leftrightarrow6x^2-9x-9-6x^2=0\Leftrightarrow-9x=9\Leftrightarrow x=-1\)
e, khai triển nó ra rút gọn rồi giải thôi nhé! ( tự làm )
f, \(\left(x-1\right)^2-x\left(x+1\right)+3\left(x-2\right)+5=0\)
\(\Leftrightarrow x^2-2x+1-x^2+x+3x-6+5=0\)
\(\Leftrightarrow2x=0\Leftrightarrow x=\frac{0}{2}\)vô lí
Vậy phương trình vô nghiệm
a: 11x+4=-3/2
=>\(11x=-\dfrac{3}{2}-4=-\dfrac{11}{2}\)
=>\(x=-\dfrac{1}{2}\)
b: \(x^2-9+2\left(x-3\right)=0\)
=>\(\left(x-3\right)\left(x+3\right)+2\left(x-3\right)=0\)
=>\(\left(x-3\right)\left(x+3+2\right)=0\)
=>(x-3)(x+5)=0
=>\(\left[{}\begin{matrix}x-3=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-5\end{matrix}\right.\)
c: \(\dfrac{x-3}{5}+\dfrac{1+2x}{3}=6\)
=>\(\dfrac{3\left(x-3\right)+5\left(2x+1\right)}{15}=6\)
=>\(3x-9+10x+5=90\)
=>13x-4=90
=>13x=94
=>\(x=\dfrac{94}{13}\)
d: \(\dfrac{2}{x+1}-\dfrac{1}{x-2}=\dfrac{3x-11}{\left(x+1\right)\left(x-2\right)}\)(ĐKXĐ: \(x\notin\left\{-1;2\right\}\))
=>\(\dfrac{2\left(x-2\right)-\left(x+1\right)}{\left(x+1\right)\left(x-2\right)}=\dfrac{3x-11}{\left(x-2\right)\left(x+1\right)}\)
=>3x-11=2x-4-x-1
=>3x-11=x-5
=>2x=6
=>x=3(nhận)
a)3x-4=5=>3x=5+4=9=>x=3
b)(x+2)(x-3)=0<=>x+2=0 hoặc x-3=0<=>x=-2 hoặc x=3
c)\(\frac{2}{x+1}-\frac{1}{x-2}=\frac{3x-11}{\left(x+1\right)\left(x-2\right)}\)
\(\Leftrightarrow\frac{2.\left(x-2\right)}{\left(x+1\right).\left(x-2\right)}-\frac{x+1}{\left(x+1\right).\left(x-2\right)}=\frac{3x-11}{\left(x+1\right).\left(x-2\right)}\)
\(\Leftrightarrow\frac{2x-4-\left(x+1\right)}{\left(x+1\right).\left(x-2\right)}=\frac{3x-11}{\left(x+1\right).\left(x-2\right)}\)
\(\Leftrightarrow\frac{2x-4-x-1}{\left(x+1\right).\left(x-2\right)}=\frac{3x-11}{\left(x+1\right).\left(x-2\right)}\) \(\Leftrightarrow2x-4-x-1=3x-11\Leftrightarrow x-5=3x-11\)
<=>3x-x=-5+11<=>2x=6<=>x=3
a/3x-4=5
<=>3x=9
<=>x=3
b/(x+2)(x-3)=0
<=>x+2=0 hoac x-3=0
<=>x=-2 hoac x=3
vay
cau c de khi nao co thoi gian mk giai not cho
1:
a: =>3x=6
=>x=2
b: =>4x=16
=>x=4
c: =>4x-6=9-x
=>5x=15
=>x=3
d: =>7x-12=x+6
=>6x=18
=>x=3
2:
a: =>2x<=-8
=>x<=-4
b: =>x+5<0
=>x<-5
c: =>2x>8
=>x>4
(2 – 3x)(x + 11) = (3x – 2)(2 – 5x)
⇔ (2 – 3x)(x + 11) – (3x – 2)(2 – 5x) = 0
⇔ (2 – 3x)(x + 11) + (2 – 3x)(2 – 5x) = 0
⇔ (2 – 3x)[(x + 11) + (2 – 5x)] = 0
⇔ (2 – 3x)(x + 11 + 2 – 5x) = 0
⇔ (2 – 3x)(13 – 4x) = 0 ⇔ 2 – 3x = 0 hoặc 13 – 4x = 0
2 – 3x = 0 ⇔ x = 2/3
13 – 4x = 0 ⇔ x = 13/4
Vậy phương trình có nghiệm x = 2/3 hoặc x = 13/4
\(\dfrac{3x}{2}+\dfrac{11}{x}=\dfrac{4}{3}\)\(\left(dkxd:x\ne0\right)\)
\(\Leftrightarrow\dfrac{3x}{2}+\dfrac{11}{x}-\dfrac{4}{3}=0\)
\(\Leftrightarrow\dfrac{3x.3x+11.6-4.2x}{6x}=0\)
\(\Leftrightarrow9x^2+66-8x=0\)
\(\Leftrightarrow9x^2-8x+66=0\)
\(\Leftrightarrow\) Pt vô nghiệm