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29 tháng 10 2018

a) x(x+1)+3(x+1)=0

⇌ (x+1)(x+3)=0

\(\Rightarrow\left[{}\begin{matrix}x+1=0\\x+3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-1\\x=-3\end{matrix}\right.\)

b)3x(12x-4)-2x(18x+3)=0

⇒36x2-12x-36x2+6x=0

⇒ -6x = 0

⇒ x=0

a: \(\Leftrightarrow x^4-24x^2+144=0\)

\(\Leftrightarrow\left(x^2-12\right)^2=0\)

hay \(x=\pm2\sqrt{3}\)

b: \(\Leftrightarrow x^4-2x^2+12x-8=0\)

\(\Leftrightarrow\left(x^2-2x+4\right)\left(x^2+2x-2\right)=0\)

\(\Leftrightarrow x^2+2x-2=0\)

hay \(x\in\left\{-1+\sqrt{3};-1-\sqrt{3}\right\}\)

2 tháng 10 2017

t.i.c.k mik mik t.i.c.k lại

19 tháng 6 2019

Giải pt :

a) \(2x\left(x+5\right)-\left(x-3\right)^2=x^2+6\)

\(\Leftrightarrow2x^2+10x-x^2+6x-9-x^2-6=0\)

\(\Leftrightarrow16x-15=0\)

\(\Leftrightarrow x=\frac{15}{16}\)

b) \(6\left(x-3\right)+\left(x-1\right)^2-\left(x+1\right)^2=2x\)

\(\Leftrightarrow2x-18=2x\)

\(\Leftrightarrow-18=0\)( vô lí )

=> x thuộc rỗng

c)d) tương tự

e) \(\frac{5x-2}{6}+\frac{3-4x}{2}=2-\frac{x+7}{3}\)

\(\Leftrightarrow\frac{5x-2}{6}+\frac{9-12x}{6}=\frac{12}{6}-\frac{2x+14}{6}\)

\(\Leftrightarrow5x-2+9-12x=12-2x-14\)

\(\Leftrightarrow-5x+9=0\)

\(\Leftrightarrow x=\frac{9}{5}\)

f) \(\frac{2x-1}{2}=\frac{2x+1}{4}-\frac{1-2x}{8}\)

\(\Leftrightarrow\frac{4\left(2x-1\right)}{8}=\frac{2\left(2x+1\right)}{8}-\frac{1-2x}{8}\)

\(\Leftrightarrow8x-4=4x+2-1+2x\)

\(\Leftrightarrow2x-5=0\)

\(\Leftrightarrow x=\frac{5}{2}\)

19 tháng 6 2019

Tìm x :

a) \(3x^3-27x=0\)

\(\Leftrightarrow3x\left(x^2-9\right)=0\)

\(\Leftrightarrow3x\left(x-3\right)\left(x+3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}3x=0\\x-3=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\\x=-3\end{matrix}\right.\)

b) \(2x^3-12x^2+18x=0\)

\(\Leftrightarrow2x\left(x^2-6x+9\right)=0\)

\(\Leftrightarrow2x\left(x-3\right)^2=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)

\(a,2x^2-18x+28=0\)

\(\Leftrightarrow2\left(x^2-9x+14\right)=0\)

\(\Leftrightarrow x^2-9x+14=0\)

\(\Leftrightarrow\left(x-7\right)\left(x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-7=0\\x-2=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=7\\x=2\end{matrix}\right.\)

\(b,\dfrac{x-2}{x^2-9}+\dfrac{3x-1}{x+3}=\dfrac{2x+1}{x-3}+1\left(ĐKXĐ:x\ne\pm3\right)\)

\(\Leftrightarrow\dfrac{x-2}{\left(x-3\right)\left(x+3\right)}+\dfrac{\left(3x-1\right)\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}-\dfrac{\left(2x+1\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-1=0\)

\(\Leftrightarrow\dfrac{x-2}{\left(x-3\right)\left(x+3\right)}+\dfrac{3x^2-10x+3}{\left(x-3\right)\left(x+3\right)}-\dfrac{2x^2+7x+3}{\left(x-3\right)\left(x+3\right)}-\dfrac{\left(x-3\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=0\)\(\Rightarrow x-2+3x^2-10x+3-2x^2-7x-3-x^2+9=0\)

\(\Leftrightarrow-16x+7=0\)

\(\Leftrightarrow-16x=-7\)

\(\Leftrightarrow x=\dfrac{7}{16}\left(tm\right)\)

\(VậyS=\left\{\dfrac{7}{16}\right\}\)

a: =>x^2-9x+14=0

=>(x-2)(x-7)=0

=>x=2 hoặc x=7

b: =>x-2+(3x-1)(x-3)=(2x+1)(x+3)+x^2-9

=>x-2+3x^2-9x-x+3=2x^2+7x+3+x^2-9

=>3x^2-9x+1=3x^2+7x-6

=>-16x=-7

=>x=7/16

21 tháng 12 2018

1) \(2x\left(x-3\right)+5x-15=0\)

\(2x\left(x-3\right)+5\left(x-3\right)=0\)

\(\left(x-3\right)\left(2x+5\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x-3=0\\2x+5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{-5}{2}\end{matrix}\right.\)

2) \(x\left(2x-7\right)-4x+14=0\)

\(x\left(2x-7\right)-2\left(2x-7\right)=0\)

\(\left(2x-7\right)\left(x-2\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}2x-7=0\\x-2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=2\end{matrix}\right.\)

3) \(x^2-12x+36=0\)

\(\left(x-6\right)^2=0\)

\(x-6=0\)

\(x=6\)

4) \(\left(x+3\right)\left(x^2-3x+9\right)-x\left(x-1\right)\left(x+1\right)-27=0\)

\(\left(x^3+3^3\right)-x\left(x^2-1\right)-27=0\)

\(x^3+27-x^3+x-27=0\)

\(x=0\)

25 tháng 3 2018

a) ĐKXĐ: x khác 0

\(x+\dfrac{5}{x}>0\)

\(\Leftrightarrow x^2+5>0\) ( luôn đúng)

Vậy bất pt vô số nghiệm ( loại x = 0)

d)

\(\dfrac{x+1}{12}-\dfrac{x-1}{6}>\dfrac{x-2}{8}-\dfrac{x+3}{8}\)

\(\Leftrightarrow\dfrac{x+1}{12}-\dfrac{x-1}{6}>\dfrac{x-2-x-3}{8}\)

\(\Leftrightarrow\dfrac{x+1}{12}-\dfrac{x-1}{6}>\dfrac{-5}{8}\)

\(\Leftrightarrow2x+2-4x+4>-15\)

\(\Leftrightarrow-2x>-21\)

\(\Leftrightarrow x< \dfrac{21}{2}\)

Vậy....................

25 tháng 3 2018

a)\(x+\dfrac{5}{x}>0\left(ĐKXĐ:x\ne0\right)\)

\(\Leftrightarrow\dfrac{x^2+5}{x}>0\)

\(x^2+5>0\)

\(\Rightarrow x>0\)

d)\(\dfrac{x+1}{12}-\dfrac{x-1}{6}>\dfrac{x-2}{8}-\dfrac{x+3}{8}\)

\(\Leftrightarrow\dfrac{x+1}{12}-\dfrac{2x-2}{12}>\dfrac{-5}{8}\)

\(\Leftrightarrow\dfrac{-x+3}{12}>\dfrac{-5}{8}\)

\(\Leftrightarrow-x+3>-\dfrac{15}{2}\)

\(\Leftrightarrow-x>-\dfrac{21}{2}\)

\(\Leftrightarrow x< \dfrac{21}{2}\)