Đề bài thiếu 1 vế của pt

a)MTC 15

\(\dfrac{\left(x-3\right)\times3}{15}=\dfrac{6.15-\left(1-2x\right)\times5}{15}=\dfrac{3x-9}{15}=\dfrac{90-5-10x}{15}=3x-9=90-5-10x\Leftrightarrow3x+10x=90-5+9\)

Chưa nghỉ tết à :))

\(a,\dfrac{x-3}{5}=6-\dfrac{1-2x}{3}\)

\(\Rightarrow3\left(x-3\right)=6.15-5\left(1-2x\right)\)

\(\Leftrightarrow3x-9=90-5+10x\)

\(\Leftrightarrow3x-10x=90-5+9\)

\(\Leftrightarrow-7x=94\)

\(\Leftrightarrow x=-\dfrac{94}{7}\)

Vậy.....

\(b,\dfrac{3x-2}{6}-5=\dfrac{3-2\left(x+7\right)}{4}\)

\(\Rightarrow2\left(3x-2\right)-5.12=3\left[3-2\left(x+7\right)\right]\)

\(\Leftrightarrow6x-4-60=-6x-33\)

\(\Leftrightarrow6x+6x=-33+60+4\)

\(\Leftrightarrow12x=31\)

\(\Leftrightarrow x=\dfrac{31}{12}\)

Vậy.....

\(c,2\left(x+\dfrac{3}{5}\right)=5-\left(\dfrac{13}{5}+x\right)\)

\(\Leftrightarrow2x+\dfrac{6}{5}=5-\dfrac{13}{5}-x\)

\(\Leftrightarrow2x+x=5-\dfrac{13}{5}-\dfrac{6}{5}\)

\(\Leftrightarrow3x=\dfrac{6}{5}\)

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

Vậy.....

\(d,\dfrac{5\left(x-1\right)+2}{6}-\dfrac{7x-1}{4}=\dfrac{2\left(2x+1\right)}{7}-5\)

\(\Rightarrow28\left[5\left(x-1\right)+2\right]-42\left(7x-1\right)=24\left[2\left(2x+1\right)\right]-5.168\)

\(\Leftrightarrow140x-84-294x+42=96x+48-840\)

\(\Leftrightarrow140x-294x-96x=48-840-42+84\)

\(\Leftrightarrow-250x=-750\)

\(\Leftrightarrow x=3\)

Vậy.....

\(e,\dfrac{x-1}{2}+\dfrac{x-1}{4}=1-\dfrac{2\left(x-1\right)}{3}\)

\(\Rightarrow6\left(x-1\right)+3\left(x-1\right)=12-4\left[2\left(x-1\right)\right]\)

\(\Leftrightarrow6x-6+3x-3=12-8x+8\)

\(\Leftrightarrow6x+3x+8x=12+8+3+6\)

\(\Leftrightarrow17x=29\)

\(\Leftrightarrow x=\dfrac{29}{17}\)

Vậy.....

\(g,\dfrac{2-x}{2001}-1=\dfrac{1-x}{2002}-\dfrac{x}{2003}\)

\(\Leftrightarrow\dfrac{2}{2001}-\dfrac{x}{2001}-1=\dfrac{1}{2002}-\dfrac{x}{2002}-\dfrac{x}{2003}\)

\(\Leftrightarrow-\dfrac{x}{2001}+\dfrac{x}{2002}+\dfrac{x}{2003}=\dfrac{1}{2002}+1-\dfrac{2}{2001}\)

\(\Leftrightarrow x\left(-\dfrac{1}{2001}+\dfrac{1}{2002}+\dfrac{1}{2003}\right)=1+\dfrac{1}{2002}-\dfrac{2}{2001}\)

\(\Leftrightarrow x=\dfrac{\left(1+\dfrac{1}{2002}-\dfrac{2}{2001}\right)}{\left(-\dfrac{1}{2001}+\dfrac{1}{2002}+\dfrac{1}{2003}\right)}=2003\)

Vậy.....

Câu 1:

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

=>\(\left[{}\begin{matrix}x+\dfrac{2}{3}=0\\x-\dfrac{1}{2}=0\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=-\dfrac{2}{3}\\x=\dfrac{1}{2}\end{matrix}\right.\)

Câu 2:

x+1=2x+3

=>x-2x=3-1

=>-x=2

=>x=-2

=>-2 là nghiệm

Câu 3:

ĐKXĐ: x<>-5

\(\dfrac{\left(-x+2\right)\left(2x+10\right)}{x^2+10x+25}=0\)

=>\(\dfrac{\left(-x+2\right)\cdot2\cdot\left(x+5\right)}{\left(x+5\right)^2}=0\)

=>\(\dfrac{2\left(-x+2\right)}{\left(x+5\right)}=0\)

=>-x+2=0

=>x=2(nhận)

Câu 4:

ĐKXĐ: \(\left\{{}\begin{matrix}x-1\ne0\\x-2\ne0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x\ne1\\x\ne2\end{matrix}\right.\)

Câu 10: ĐKXĐ: x<>1

\(x^2+\dfrac{1}{x-1}=1+\dfrac{1}{1-x}\)

=>\(x^2-1+\dfrac{1}{x-1}+\dfrac{1}{x-1}=0\)

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

=>\(\dfrac{\left(x^2-1\right)\cdot\left(x-1\right)+2}{x-1}=0\)

=>\(x^3-x^2-x+1+2=0\)

=>\(x^3-x^2-x+3=0\)

=>\(x\simeq-1,36\)

\(1,\dfrac{4x-4}{3}=\dfrac{7-x}{5}\\ \Leftrightarrow5\left(4x-4\right)=3\left(7-x\right)\\ \Leftrightarrow20x-20=21-3x\\ \Leftrightarrow17x=41\Leftrightarrow x=\dfrac{41}{17}\)

\(2,\dfrac{3x-9}{5}=\dfrac{3-x}{2}\\ \Leftrightarrow6x-18=15-5x\\ \Leftrightarrow11x=33\\ \Leftrightarrow x=3\)

\(3,\dfrac{2x-1}{5}-\dfrac{3-x}{3}=1\\ \Leftrightarrow\dfrac{6x-3-15+5x}{15}=1\\ \Leftrightarrow11x-18=1\\ \Leftrightarrow x=\dfrac{19}{11}\)

\(4,\dfrac{x-5}{3}+\dfrac{3x+4}{2}=\dfrac{5x+2}{6}\\ \Leftrightarrow2x-10+9x+12=5x+2\\ \Leftrightarrow6x=0\Leftrightarrow x=0\)

\(5,\dfrac{x-3}{2}+\dfrac{2x+3}{5}=\dfrac{2x+5}{10}\\ \Leftrightarrow5x-15+4x+6=2x+5\\ \Leftrightarrow7x=14\\ \Leftrightarrow x=2\)

Tick nha

2: Ta có: \(\dfrac{3x-9}{5}=\dfrac{3-x}{2}\)

\(\Leftrightarrow6x-18=15-5x\)

\(\Leftrightarrow11x=33\)

hay x=3

Đề bài tào lao thật sự

Vừa độ vừa radian trong 1 phương trình là không chính xác. Đã độ thì độ hết, đã radian thì radian hết

a:

\(\dfrac{3\left(2x+1\right)}{4}-5-\dfrac{3x+2}{10}=\dfrac{2\left(3x-1\right)}{5}\)

\(\Leftrightarrow\dfrac{15\left(2x+1\right)-100-2\left(3x+2\right)}{20}=\dfrac{8\left(3x-1\right)}{20}\)

\(\Leftrightarrow15\left(2x+1\right)-100-2\left(3x+2\right)=8\left(3x-1\right)\)

\(\Leftrightarrow30x+15-100-6x+4=24x-8\)\(\Leftrightarrow30x-6x-24x=100-4-8\)

\(\Leftrightarrow0x=88\)

Vậy pt vô nghiệm

b:

\(\dfrac{x-15}{23}+\dfrac{x-23}{15}-2=0\)

\(\Leftrightarrow\dfrac{x-15}{23}+\dfrac{x-23}{15}=2\)

\(\Leftrightarrow\dfrac{x-15}{23}-1+\dfrac{x-23}{15}-1=2-2\)

\(\Leftrightarrow\dfrac{x-15-23}{23}+\dfrac{x-23-15}{15}=0\)

\(\Leftrightarrow\dfrac{x-38}{23}+\dfrac{x-23}{15}=0\)

\(\Leftrightarrow\left(x+38\right)\left(\dfrac{1}{23}+\dfrac{1}{15}\right)=0\)

Vì \(\dfrac{1}{23}+\dfrac{1}{15}\ne0\) nên x + 38 =0 \(\Leftrightarrow x=-38\)

Vậy tập nghiện của pt S= {-38}

c:

\(\dfrac{3\left(2x+1\right)}{4}-\dfrac{5x+3}{6}+\dfrac{x+1}{3}=x+\dfrac{7}{12}\)

\(\Leftrightarrow\dfrac{9\left(2x+1\right)-2\left(5x+3\right)+4\left(x+1\right)}{12}=\dfrac{12x+7}{12}\)

\(\Leftrightarrow9\left(2x+1\right)-2\left(5x+3\right)+4\left(x+1\right)=12x+7\)

\(\Leftrightarrow18x+9-10x-6+4x+4=12x+7\)

\(\Leftrightarrow18x-10x+4x-12x=7-9+6-4\)

\(\Leftrightarrow0x=0\)

Vậy pt vô số nghiệm

a) Ta có: \(\left(2x-3\right)^2=\left(2x-3\right)\left(x+1\right)\)

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

\(\Leftrightarrow\left(2x-3\right)\left(2x-3-x-1\right)=0\)

\(\Leftrightarrow\left(2x-3\right)\left(x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-3=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=3\\x=4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=4\end{matrix}\right.\)

Vậy: \(S=\left\{\dfrac{3}{2};4\right\}\)

b) Ta có: \(x\left(2x-9\right)=3x\left(x-5\right)\)

\(\Leftrightarrow x\left(2x-9\right)-3x\left(x-5\right)=0\)

\(\Leftrightarrow x\left(2x-9\right)-x\left(3x-15\right)=0\)

\(\Leftrightarrow x\left(2x-9-3x+15\right)=0\)

\(\Leftrightarrow x\left(6-x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\6-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=6\end{matrix}\right.\)

Vậy: S={0;6}

c) Ta có: \(3x-15=2x\left(x-5\right)\)

\(\Leftrightarrow3\left(x-5\right)-2x\left(x-5\right)=0\)

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

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

Vậy: \(S=\left\{5;\dfrac{3}{2}\right\}\)

d) Ta có: \(\dfrac{5-x}{2}=\dfrac{3x-4}{6}\)

\(\Leftrightarrow6\left(5-x\right)=2\left(3x-4\right)\)

\(\Leftrightarrow30-6x=6x-8\)

\(\Leftrightarrow30-6x-6x+8=0\)

\(\Leftrightarrow-12x+38=0\)

\(\Leftrightarrow-12x=-38\)

\(\Leftrightarrow x=\dfrac{19}{6}\)

Vậy: \(S=\left\{\dfrac{19}{6}\right\}\)

e) Ta có: \(\dfrac{3x+2}{2}-\dfrac{3x+1}{6}=2x+\dfrac{5}{3}\)

\(\Leftrightarrow\dfrac{3\left(3x+2\right)}{6}-\dfrac{3x+1}{6}=\dfrac{12x}{6}+\dfrac{10}{6}\)

\(\Leftrightarrow6x+4-3x-1=12x+10\)

\(\Leftrightarrow3x+3-12x-10=0\)

\(\Leftrightarrow-9x-7=0\)

\(\Leftrightarrow-9x=7\)

\(\Leftrightarrow x=-\dfrac{7}{9}\)

Vậy: \(S=\left\{-\dfrac{7}{9}\right\}\)

1) PT \(\Leftrightarrow\dfrac{x+3}{15}=\dfrac{4}{15}\) \(\Rightarrow x+3=4\) \(\Rightarrow x=1\)

  Vậy ...

2) Mạnh dạn đoán đề là \(\left(2x-5\right)\left(x-3\right)=0\)

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

  Vậy ...

3) PT \(\Rightarrow3x-4-2x+5=3\)

          \(\Rightarrow x=2\)

 Vậy ...

4) PT \(\Rightarrow\left[{}\begin{matrix}2x+1=0\\\dfrac{1}{2}x-1=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=2\end{matrix}\right.\)

  Vậy ...

3) Ta có: \(\left(3x-4\right)-\left(2x-5\right)=3\)

\(\Leftrightarrow3x-4-2x+5=3\)

\(\Leftrightarrow x+1=3\)

hay x=2

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

\(\Leftrightarrow\left[{}\begin{matrix}2x-3=\dfrac{1}{3}\\2x-3=-\dfrac{1}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\\x=\dfrac{4}{3}\end{matrix}\right.\)

b, tương tự 

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

TH1 : \(2x-1=x+\dfrac{1}{3}\Leftrightarrow x=\dfrac{4}{3}\)

TH2 : \(2x-1=-x-\dfrac{1}{3}\Leftrightarrow3x=\dfrac{2}{3}\Leftrightarrow x=\dfrac{2}{9}\)

d, \(3x-\left|x+15\right|=\dfrac{5}{4}\Leftrightarrow\left|x+15\right|=3x-\dfrac{5}{4}\)ĐK : x >= 5/12

TH1 : \(x+15=3x-\dfrac{5}{4}\Leftrightarrow-2x=-\dfrac{65}{4}\Leftrightarrow x=\dfrac{65}{8}\)( tm )

TH2 : \(x+15=\dfrac{5}{3}-3x\Leftrightarrow4x=-\dfrac{40}{3}\Leftrightarrow x=-\dfrac{10}{3}\)

b: ta có: \(\dfrac{5}{6}-\left|x+\dfrac{1}{4}\right|=\dfrac{1}{4}\)

\(\Leftrightarrow\left|x+\dfrac{1}{4}\right|=\dfrac{7}{12}\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{4}=\dfrac{7}{12}\\x+\dfrac{1}{4}=-\dfrac{7}{12}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=-\dfrac{5}{6}\end{matrix}\right.\)