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a: =>5-x+6=12-8x
=>-x+11=12-8x
=>7x=1
hay x=1/7
b: \(\dfrac{3x+2}{2}-\dfrac{3x+1}{6}=2x+\dfrac{5}{3}\)
\(\Leftrightarrow9x+6-3x-1=12x+10\)
=>12x+10=6x+5
=>6x=-5
hay x=-5/6
d: =>(x-2)(x-3)=0
=>x=2 hoặc x=3
3.
a) \(2x+5=20-3x\)
\(\Leftrightarrow2x+3x=20-5\)
\(\Leftrightarrow5x=15\)
\(\Leftrightarrow x=3\)
Vậy \(S=\left\{3\right\}\)
b) \(\left(2x-1\right)^2-\left(x+3\right)^2=0\)
\(\Leftrightarrow\left[\left(2x-1\right)+\left(x+3\right)\right]\left[\left(2x-1\right)-\left(x+3\right)\right]=0\)
\(\Leftrightarrow\left(2x-1+x+3\right)\left(2x-1-x-3\right)=0\)
\(\Leftrightarrow\left(3x+2\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x+2=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{2}{3}\\x=4\end{matrix}\right.\)
Vậy \(S=\left\{-\dfrac{2}{3};4\right\}\)
c) \(\dfrac{5x-4}{2}=\dfrac{16x+1}{7}\)
\(\Leftrightarrow\left(5x-4\right)7=\left(16x+1\right)2\)
\(\Leftrightarrow35x-28=32x+2\)
\(\Leftrightarrow35x-32x=2+28\)
\(\Leftrightarrow2x=30\)
\(\Leftrightarrow x=15\)
Vậy \(S=\left\{15\right\}\)
d) \(\dfrac{2x+1}{6}-\dfrac{x-2}{4}=\dfrac{3-2x}{3}-x\)
\(\Rightarrow\left(2x+1\right)12-\left(x-2\right)18=\left(3-2x\right)24-72x\)
\(\Leftrightarrow24x+12-18x+36=72-48x-72x\)
\(\Leftrightarrow6x+48=72-120x\)
\(\Leftrightarrow6x+120x=72-48\)
\(\Leftrightarrow126x=24\)
\(\Leftrightarrow x=\dfrac{4}{21}\)
Vậy \(S=\left\{\dfrac{4}{21}\right\}\)
a) \(\dfrac{x+1}{2}+\dfrac{3x-2}{3}=\dfrac{x-7}{12}\)
\(\Leftrightarrow\dfrac{6\left(x+1\right)+4\left(3x-2\right)}{12}=\dfrac{x-7}{12}\)
\(\Leftrightarrow6\left(x+1\right)+4\left(3x-2\right)=x-7\)
\(\Leftrightarrow6x+6+12x-8=x-7\)
\(\Leftrightarrow6x+12x-x=-7-6+8\)
\(\Leftrightarrow17x=-5\)
\(\Leftrightarrow x=\dfrac{-5}{17}\)
Vậy .........................
b) \(\dfrac{2x}{x-3}-\dfrac{5}{x+3}=\dfrac{x^2+21}{x^2-9}\left(ĐKXĐ:x\ne\pm3\right)\)
\(\Leftrightarrow\dfrac{2x\left(x+3\right)-5\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{x^2+21}{\left(x-3\right)\left(x+3\right)}\)
\(\Rightarrow2x\left(x+3\right)-5\left(x-3\right)=x^2+21\)
\(\Leftrightarrow2x^2+6x-5x+15=x^2+21\)
\(\Leftrightarrow2x^2-x^2+x+15-21=0\)
\(\Leftrightarrow x^2+x-6=0\)
\(\Leftrightarrow x^2-2x+3x-6=0\)
\(\Leftrightarrow x\left(x-2\right)+3\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\left(n\right)\\x=-3\left(l\right)\end{matrix}\right.\)
Vậy \(S=\left\{2\right\}\)
d) \(\left(x-4\right)\left(7x-3\right)-x^2+16=0\)
\(\Leftrightarrow\left(x-4\right)\left(7x-3\right)-\left(x^2-16\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(7x-3\right)-\left(x-4\right)\left(x+4\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(7x-3-x-4\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(6x-7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\6x-7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=\dfrac{7}{6}\end{matrix}\right.\)
Vậy .........................
P/s: các câu còn lại tương tự, bn tự giải nha
a: \(\left(3x-1\right)^2-\left(x+3\right)^3=\left(2-x\right)\left(x^2+2x+4\right)\)
\(\Leftrightarrow9x^2-6x+1-x^3-9x^2-27x-27=8-x^3\)
\(\Leftrightarrow-x^3-33x-26-8+x^3=0\)
=>-33x=34
hay x=-34/33
b: \(\left(x+1\right)\left(x-1\right)\left(x^2+1\right)-\left(x^2-1\right)^2=2\)
\(\Leftrightarrow\left(x^2+1\right)\left(x^2-1\right)-\left(x^2-1\right)^2=2\)
\(\Leftrightarrow x^4-1-x^4+2x^2-1=2\)
\(\Leftrightarrow2x^2=4\)
hay \(x\in\left\{\sqrt{2};-\sqrt{2}\right\}\)
c: \(x^2-2\sqrt{3}x+3=0\)
\(\Leftrightarrow\left(x-\sqrt{3}\right)^2=0\)
hay \(x=\sqrt{3}\)
d: \(\left(x-\sqrt{2}\right)\left(x+\sqrt{2}\right)-\left(x-\sqrt{2}\right)^2=0\)
\(\Leftrightarrow\left(x-\sqrt{2}\right)\left(x+\sqrt{2}-x+\sqrt{2}\right)=0\)
\(\Leftrightarrow x-\sqrt{2}=0\)
hay \(x=\sqrt{2}\)
a) \(\left(2x+1\right)^2-\left(x+2\right)^2>0\)
\(\Leftrightarrow\left(2x+1-x-2\right)\left(2x+1+x+2\right)>0\)
\(\Leftrightarrow\left(x-1\right)\left(3x+3\right)>0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-1>0\\3x+3>0\end{matrix}\right.\\\left\{{}\begin{matrix}x-1< 0\\3x+3< 0\end{matrix}\right.\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>1\\x>-1\end{matrix}\right.\\\left\{{}\begin{matrix}x< 1\\x< -1\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x>1\\x< -1\end{matrix}\right.\)
Vậy tập nghiệm của bất phương trình là x > 1 hoặc x < -1
b) Sửa lại rồi làm câu b nèk\(\dfrac{5x-3x}{5}+\dfrac{3x+1}{4}>\dfrac{x\left(2x+1\right)}{2}-\dfrac{3}{2}\)
\(\Leftrightarrow4\left(5x-3x\right)+5\left(3x+1\right)>10\left(x+2x\right)-30\)\(\Leftrightarrow20x-12x+15x+5>10x+20x-30\)\(\Leftrightarrow20x-12x+15x-10x-20x>-30-5\)\(\Leftrightarrow-7x>-35\)
\(\Leftrightarrow x< 5\)
c) \(\dfrac{-1}{2x+3}< 0\)
dễ nhé mình học bài hóa mai kt 15 phút nên ko có time để giúp
a) \(\left(2x-1\right)^2-\left(3x+5\right)\left(2x-1\right)=0\)
\(\Leftrightarrow\left(2x-1\right)\left(2x-1-3x-5\right)=0\\ \text{}\Leftrightarrow\left(2x-1\right)\left(-x-6\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}2x-1=0\\-x-6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-6\end{matrix}\right.\)
Vậy \(S=\left\{\dfrac{1}{2};-6\right\}\)
b) \(\dfrac{x+5}{4}-\dfrac{2x-3}{3}=\dfrac{2x-1}{12}\)
\(\Leftrightarrow3\left(x+5\right)-4\left(2x-3\right)=2x-1\\ \Leftrightarrow3x+15-8x+12=2x-1\\ \Leftrightarrow-5x+27=2x-1\\ \Leftrightarrow-5x-2x=-1-27\\ \Leftrightarrow-7x=-28\\ \Leftrightarrow x=4\)
Vậy \(S=\left\{4\right\}\)
\(c)\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}=\dfrac{-4}{1-x^2}\)(ĐKXĐ: \(x\ne\pm1\))
\(\Leftrightarrow\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}-\dfrac{4}{\left(x-1\right)\left(x+1\right)}\\ \dfrac{\left(x+1\right)^2-\left(x-1\right)^2-4}{\left(x-1\right)\left(x+1\right)}=0\\ \Leftrightarrow\dfrac{4x-4}{\left(x-1\right)\left(x+1\right)}=0\\ \Leftrightarrow\dfrac{4\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}=0\\ \Leftrightarrow\dfrac{4}{x+1}=0\)
\(\Leftrightarrow4=0\)(vô lý)
Vậy .....
\(d)\dfrac{1}{x+1}+\dfrac{2x-1}{x^3+1}=\dfrac{2}{x^2-x+1}\)(ĐKXĐ: \(x\ne-1\))
\(\Leftrightarrow\dfrac{1}{x+1}+\dfrac{2x-1}{\left(x+1\right)\left(x^2-x+1\right)}-\dfrac{2}{x^2-x+1}=0\\ \Leftrightarrow\dfrac{x^2-x+1+2x-1-2\left(x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}=0\\ \Leftrightarrow\dfrac{x^2-x+2x-2x-2}{\left(x+1\right)\left(x^2-x+1\right)}=0\\ \Leftrightarrow\dfrac{x^2-x-2}{\left(x+1\right)\left(x^2-x+1\right)}=0\\ \Leftrightarrow x^2-x-2=0\\ \Leftrightarrow x^2+x-2x-2=0\\ \Leftrightarrow x\left(x+1\right)-2\left(x+1\right)=0\\ \Leftrightarrow\left(x+1\right)\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x+1=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\left(KTM\right)\\x=2\left(TM\right)\end{matrix}\right.\)
Vậy ....
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....................
a)\(x+\dfrac{5}{x}>0\left(ĐKXĐ:x\ne0\right)\)
\(\Leftrightarrow\dfrac{x^2+5}{x}>0\)
Mà \(x^2+5>0\)
\(\Rightarrow x>0\)
d)\(\dfrac{x+1}{12}-\dfrac{x-1}{6}>\dfrac{x-2}{8}-\dfrac{x+3}{8}\)
\(\Leftrightarrow\dfrac{x+1}{12}-\dfrac{2x-2}{12}>\dfrac{-5}{8}\)
\(\Leftrightarrow\dfrac{-x+3}{12}>\dfrac{-5}{8}\)
\(\Leftrightarrow-x+3>-\dfrac{15}{2}\)
\(\Leftrightarrow-x>-\dfrac{21}{2}\)
\(\Leftrightarrow x< \dfrac{21}{2}\)
a)(x+1)(x2+2x)=(x+1)x(x+2)=0
\(=>\left\{{}\begin{matrix}x+1=0=>x=-1\\x=0\\x+2=0=>x=-2\end{matrix}\right.\)
b)x(3x-2)-5(2-3x)=x(3x-2)+5(3x-2)=(3x-2)(x+5)=0
\(=>\left\{{}\begin{matrix}3x-2=0=>x=\dfrac{2}{3}\\x+5=0=>x=-5\end{matrix}\right.\)
c)\(\dfrac{4}{9}-25x^2=\left(\dfrac{2}{3}\right)^2-\left(5x\right)^2=\left(\dfrac{2}{3}-5x\right)\left(\dfrac{2}{3}+5x\right)\)
=0
\(=>\left\{{}\begin{matrix}\dfrac{2}{3}-5x=0=>x=\dfrac{2}{15}\\\dfrac{2}{3}+5x=0=>x=\dfrac{-2}{15}\end{matrix}\right.\)
d)\(x^2-x+\dfrac{1}{4}=x^2-2.\dfrac{1}{2}.x+\left(\dfrac{1}{2}\right)^2=\left(x-\dfrac{1}{2}\right)^2=0\)
\(=>x-\dfrac{1}{2}=0=>x=\dfrac{1}{2}\)
a/
\(\left(x+2\right)\left(2-2x\right)-\left(2x+1\right)\left(3-x\right)=0\)
\(\Leftrightarrow-2x-2x^2+4-5x+2x^2-3=0\)
\(\Leftrightarrow-7x+1=0\)
\(\Leftrightarrow\dfrac{1}{7}\)
c/
\(\dfrac{3x+1}{4}-x-1=\dfrac{4-3x}{3}\)
\(\Leftrightarrow\dfrac{3.\left(3x+1\right)}{12}-\dfrac{12x}{12}-\dfrac{12}{12}=\dfrac{4.\left(4-3x\right)}{12}\)
\(\Leftrightarrow9x+3-12x-12-16+12x=0\)
\(\Leftrightarrow9x-25=0\)
\(\Leftrightarrow x=\dfrac{25}{9}\)