1,(3x-2)(4x+5)=0

\(\Leftrightarrow\left\{{}\begin{matrix}3x-2=0\\4x+5=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x=2\\4x=-5\end{matrix}\right.\)

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

Vậy phương trình có tập nghiệm là ...

2,\(5\left(2x-3\right)-4\left(5x-7\right)=19-2\left(x+11\right)\)

\(\Leftrightarrow10x-15-20x+28=19-2x-22\)

\(\Leftrightarrow10x-20x+2x=15-28+19-22\)

\(\Leftrightarrow-8x=-16\)

=> x= 2

vậy..

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

\(\Leftrightarrow\left(x^2-2.x.\frac{1}{2}+\frac{1}{4}-\frac{1}{4}+1\right)-4=0\)

\(\Leftrightarrow\left(x^2-2.x.\frac{1}{2}+\frac{1}{4}\right)+\frac{3}{4}-4=0\)

\(\Leftrightarrow\left(x-\frac{1}{2}\right)^2-\frac{13}{4}=0\) ( vô nghiệm )

(vì \(\left(x-\frac{1}{2}\right)^2\ge0\Rightarrow\left(x-\frac{1}{2}\right)^2-\frac{13}{4}\ge0\) )

từ đó suy ra phương trình vô nghiệm

5,\(\frac{4x+3}{2}-2+3x=\frac{2x-1}{10}+\frac{19x+2}{5}-1\)

\(\Leftrightarrow\frac{5\left(4x+3\right)}{10}-\frac{10\left(2-3x\right)}{10}=\frac{2x-1}{10}+\frac{2\left(19x+2\right)}{10}-\frac{10}{10}\)

\(\Leftrightarrow\frac{20x+15}{10}-\frac{20-30x}{10}=\frac{2x-1}{10}+\frac{38x+4}{10}-\frac{10}{10}\)

\(\Rightarrow20x+15-20+30x=2x-1+38x+4-10\)

\(\Leftrightarrow20x+30x-2x-38x=-15+20-1+4-10\)

\(\Leftrightarrow10x=-2\)

\(\Leftrightarrow x=-5\)

Vậy ....

p/s : thực ra mk cx chỉ ms học th nên giải bài tập về phương trình vẫn còn nhiều chỗ sai nữa,có gì mong mn giúp đỡ :)

\(5X\left(X-2020\right)+X=2020\)

\(\Leftrightarrow5X^2-10100X+X=2020\)

\(\Leftrightarrow5X^2-10099X=2020\)

\(\Leftrightarrow5X^2-10099X-2020=0\)

\(\Leftrightarrow5X^2-10100X+x-2020=0\)

\(\Leftrightarrow5X\left(X-2020\right)+X-2020=0\)

\(\Leftrightarrow\left(X-2020\right)\left(5X+1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=2020\\x=-\frac{1}{5}\end{cases}}\)

\(4\left(x-5\right)^2-\left(2x+1\right)^2=0\)

\(\Leftrightarrow\left[2\left(x-5\right)\right]^2-\left(2x+1\right)^2=0\)

\(\Leftrightarrow\left[2\left(x-5\right)-2x-1\right]\left[2\left(x-5\right)+2x+1\right]=0\)

\(\Leftrightarrow\left(2x-10-2x-1\right)\left(2x-10+2x+1\right)=0\)

\(\Leftrightarrow-11\left(4x-9\right)=0\)

\(\Leftrightarrow x=\frac{9}{4}\)

\(a,5x\left(x-2020\right)+x=2020\)

\(< =>5x\left(x-2020\right)+x-2020=0\)

\(< =>\left(5x+1\right)\left(x-2020\right)=0\)

\(< =>\orbr{\begin{cases}5x+1=0\\x-2020=0\end{cases}}\)

\(< =>\orbr{\begin{cases}5x=-1\\x=2020\end{cases}< =>\orbr{\begin{cases}x=-\frac{1}{5}\\x=2020\end{cases}}}\)

\(b,4\left(x-5\right)^2-\left(2x+1\right)^2=0\)

\(< =>4\left(x^2-20x+25\right)-\left(4x^2+4x+1\right)=0\)

\(< =>4x^2-80x+100-4x^2-4x-1=0\)

\(< =>-84x+99=0< =>84x=99< =>x=\frac{99}{84}\)

1) Ta có : \(4x+20=0\)

=> \(x=-\frac{20}{4}=-5\)

Vậy phương trình có tập nghiệm là \(S=\left\{-5\right\}\)

2) Ta có : \(3x+15=30\)

=> \(3x=15\)

=> \(x=5\)

Vậy phương trình có tập nghiệm là \(S=\left\{5\right\}\)

3) Ta có : \(8x-7=2x+11\)

=> \(8x-2x=11+7=18\)

=> \(6x=18\)

=> \(x=3\)

Vậy phương trình có tập nghiệm là \(S=\left\{3\right\}\)

4) Ta có : \(2x+4\left(36-x\right)=100\)

=> \(2x+144-4x=100\)

=> \(-2x=-44\)

=> \(x=22\)

Vậy phương trình có tập nghiệm là \(S=\left\{22\right\}\)

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

=> \(2x-3+5=4x+12\)

=> \(-2x=10\)

=> \(x=-5\)

Vậy phương trình có tập nghiệm là \(S=\left\{-5\right\}\)

1) 4x+20=0

\(\Leftrightarrow\) 4x=-20

\(\Leftrightarrow\) x=-5

Vậy pt trên có tập nghiệm là S={-5}

2) 3x+15=30

\(\Leftrightarrow\) 3x=15

\(\Leftrightarrow\) x=5

Vậy pt trên có tập nghiệm là S={5}

3) 8x-7=2x+11

\(\Leftrightarrow\) 8x-2x=11+7

\(\Leftrightarrow\) 6x=18

\(\Leftrightarrow\) x=3

Vậy pt trên có tập nghiệm là S={3}

4) 2x+4(36-x)=100

\(\Leftrightarrow\) 2x+144-4x=100

\(\Leftrightarrow\) -2x+144=100

\(\Leftrightarrow\) -2x=-44

\(\Leftrightarrow\) x=22

Vậy pt trên có tập nghiệm là S={22}

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

\(\Leftrightarrow\) 2x-3+5x=4x+12

\(\Leftrightarrow\) 2x+5x-4x=12+3

\(\Leftrightarrow\) 3x=15

\(\Leftrightarrow\) x=5

Vậy pt trên có tập nghiệm là S={5}

6) 3x(x+2)=3(x-2)²

\(\Leftrightarrow\) 3x²+6x=3(x²-2x.2+2²)

\(\Leftrightarrow\) 3x²+6x=3x²-12x+12

\(\Leftrightarrow\) 3x²-3x²+6x+12x=12

\(\Leftrightarrow\) 18x=12

\(\Leftrightarrow\) x=\(\frac{2}{3}\)

7) \(\frac{5x-2}{3}\)=3

\(\Leftrightarrow\) 5x-2=9

\(\Leftrightarrow\) 5x=11

\(\Leftrightarrow\) x=\(\frac{11}{5}\)

8) (6x+3)(5x-20)=0

\(\Rightarrow\) 6x+3=0 hoặc 5x-20=0

\(\Rightarrow\) 6x=-3 hoặc 5x=20

\(\Rightarrow\) x=\(\frac{-1}{2}\) hoặc x=4

Vậy pt trên có tập nghiệm là S={\(\frac{-1}{2}\);4}

10) \(\frac{2x-5}{x+5}=3\)

\(\Leftrightarrow\) \(\frac{2x-5}{x+5}\)= \(\frac{3\left(x+5\right)}{x+5}\)

\(\Leftrightarrow\) 2x-5=3x+15

\(\Leftrightarrow\) 2x-3x=15+5

\(\Leftrightarrow\) -x=20

\(\Leftrightarrow\) x=-20

Vậy pt trên có tập nghiệm là S={-20}

11) \(\frac{1}{x-2}+4=\frac{x-3}{2-x}\)

\(\Leftrightarrow\) \(\frac{1}{x-2}+\frac{4\left(x-2\right)}{x-2}=\frac{3-x}{x-2}\)

\(\Leftrightarrow\) 1+4x-8=3-x

\(\Leftrightarrow\) 4x+x=3+8-1

\(\Leftrightarrow\) 5x=10

\(\Leftrightarrow\) x=2

Vậy pt trên có tập nghiệm là S={2}

12) \(\frac{x+2}{x-2}-1=\frac{2x}{x\left(x-2\right)}\)

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

\(\Leftrightarrow\) x²+2x-x²+2x=2x

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

\(\Leftrightarrow\) 2x=0

\(\Leftrightarrow\) x=0

Vậy pt trên có tập nghiệm là S={0}

a) 3x - 2(5 + 2x) =45 - 2x

=> 3x - 10 - 4x = 45 - 2x

=> 3x - 4x + 2x = 45 + 10

=> x = 55

b) \(\frac{x-3}{5}=6-\frac{1-2x}{3}\)

=> \(\frac{x-3}{5}=\frac{2x+17}{3}\)

=> 5(2x + 17) = 3(x - 3)

=> 10x + 85 = 3x - 9

=> 7x = -94

=> x = -94/7

c) \(\frac{5\left(x-1\right)+2}{6}-\frac{7x-1}{4}=\frac{2\left(2x+1\right)}{7}-5\)

=> \(\frac{5x-3}{6}-\frac{7x-1}{4}=\frac{4x-33}{7}\)

=> \(\frac{10x-6}{12}-\frac{21x-3}{12}=\frac{4x-33}{7}\)

=> \(\frac{-11x-3}{12}=\frac{4x-33}{7}\)

=> (-11x - 3).7 = (4x - 33).12

= -77x - 21 = 48x - 396

=> x = 3

d) (x - 1)(5x + 3) = (3x - 8)(x - 1)

=> (x - 1)(5x + 3) - (3x - 8)(x -1) = 0

=> (x - 1)(2x + 11) = 0

=> \(\orbr{\begin{cases}x-1=0\\2x+11=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=1\\x=-5,5\end{cases}}\) 

e) (x - 1)(x² + 5x - 2) - (x³ - 1) = 0

=> (x - 1)(x² + 5x - 2) - (x - 1)(x² + x + 1) = 0

=> (x - 1)(4x - 3) = 0

=> \(\orbr{\begin{cases}x-1=0\\4x-3=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=1\\x=0,75\end{cases}}\)

f) \(\frac{x-17}{33}+\frac{x-21}{29}+\frac{x}{25}=4\) 

=> \(\left(\frac{x-17}{33}-1\right)+\left(\frac{x-21}{29}-1\right)+\left(\frac{x}{25}-2\right)=0\)

=> \(\frac{x-50}{33}+\frac{x-50}{29}+\frac{x-50}{25}=0\)

=> \(\left(x-50\right)\left(\frac{1}{33}+\frac{1}{29}+\frac{1}{25}\right)=0\)

=> x - 50 = 0 (Vì \(\frac{1}{33}+\frac{1}{29}+\frac{1}{25}\ne0\))

=> x = 50

b, \(\frac{x-3}{5}=6-\frac{1-2x}{3}\)

\(\Leftrightarrow\frac{x-3}{5}=\frac{17+2x}{3}\Leftrightarrow3x-9=85+10x\)

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

f, sửa : \(\frac{x+1}{65}+\frac{x+3}{63}=\frac{x+5}{61}+\frac{x+7}{59}\)

\(\Leftrightarrow\frac{x+1}{65}+1+\frac{x+3}{63}+1=\frac{x+5}{61}+1+\frac{x+7}{59}+1\)

\(\Leftrightarrow\frac{x+66}{65}+\frac{x+66}{63}=\frac{x+66}{61}+\frac{x+66}{59}\)

\(\Leftrightarrow\frac{x+66}{65}+\frac{x+66}{63}-\frac{x+66}{61}-\frac{x+66}{59}=0\)

\(\Leftrightarrow\left(x+66\right)\left(\frac{1}{65}+\frac{1}{63}-\frac{1}{61}-\frac{1}{59}\ne0\right)=0\)

\(\Leftrightarrow x=-66\)

\( a)5\left( {x - 3} \right) - 4 = 2\left( {x - 1} \right) + 7\\ \Leftrightarrow 5x - 15 - 4 = 2x - 2 + 7\\ \Leftrightarrow 5x - 19 = 2x + 5\\ \Leftrightarrow 5x - 2x = 5 + 19\\ \Leftrightarrow 3x = 24\\ \Leftrightarrow x = 8\\ b)\dfrac{{8x - 3}}{4} - \dfrac{{3x - 2}}{2} = \dfrac{{2x - 1}}{2} + \dfrac{{x + 3}}{4}\\ \Leftrightarrow 8x - 3 - \left( {3x - 2} \right).2 = \left( {2x - 1} \right).2 + x + 3\\ \Leftrightarrow 8x - 3 - 6x + 4 = 4x - 2 + x + 3\\ \Leftrightarrow 2x + 1 = 5x + 1\\ \Leftrightarrow 2x - 5x = 0\\ \Leftrightarrow - 3x = 0\\ \Leftrightarrow x = 0 \)

\( c)\dfrac{{2\left( {x + 5} \right)}}{3} + \dfrac{{x + 12}}{2} - \dfrac{{5\left( {x - 2} \right)}}{6} = \dfrac{x}{3} + 11\\ \Leftrightarrow 4\left( {x + 5} \right) + 3\left( {x + 12} \right) - \left[ {5\left( {x - 2} \right)} \right] = 2x + 66\\ \Leftrightarrow 4x + 20 + 3x + 36 - 5x + 10 = 2x + 66\\ \Leftrightarrow 2x + 66 = 2x + 66\\ \Leftrightarrow 0x = 0\left( {VSN} \right)\\ \Leftrightarrow x = 0 \)

\(d)\dfrac{x-10}{1994}+\dfrac{x-8}{1996}+\dfrac{x-6}{1998}+\dfrac{x-4}{2000}+\dfrac{x-2}{2002}=\dfrac{x-2002}{2}+\dfrac{x-2000}{4}+\dfrac{x-1998}{6}+\dfrac{x-1996}{8}+\dfrac{x-1994}{10}\\ \Leftrightarrow \dfrac{x-10}{1994}-1+\dfrac{x-8}{1996}-1+\dfrac{x-6}{1998}-1+\dfrac{x-4}{2000}-1+\dfrac{x-2}{2002}-1=\dfrac{x-2002}{2}-1+\dfrac{x-2000}{4}-1+\dfrac{x-1998}{6}-1+\dfrac{x-1996}{8}-1+\dfrac{x-1994}{10}-1\\ \Leftrightarrow \dfrac{x-2004}{1994}+\dfrac{x-2004}{1996}+\dfrac{x-2004}{1998}+\dfrac{x-2004}{2000}\dfrac{x-2004}{2002}=\dfrac{x-2004}{2}+\dfrac{x-2004}{4}+\dfrac{x-2004}{6}+\dfrac{x-2004}{8}+\dfrac{x-2004}{10}\\ \Leftrightarrow \dfrac{x-2004}{1994}+\dfrac{x-2004}{1996}+\dfrac{x-2004}{1998}+\dfrac{x-2004}{2000}\dfrac{x-2004}{2002}-\dfrac{x-2004}{2}-\dfrac{x-2004}{4}-\dfrac{x-2004}{6}-\dfrac{x-2004}{8}-\dfrac{x-2004}{10}=0\\ \Leftrightarrow \left(x-2004\right)\left(\dfrac{1}{1994}+\dfrac{1}{1996}+\dfrac{1}{1998}+\dfrac{1}{2000}+\dfrac{1}{2002}-\dfrac{1}{2}-\dfrac{1}{4}-\dfrac{1}{6}-\dfrac{1}{8}-\dfrac{1}{10}=0\right)\\ \Leftrightarrow x-2004=0\\ \Leftrightarrow x=2004\)

a, 5(x-3)-4=2(x-1)+7

<=>\(5x-15-4=2x-2+7\)

\(\Leftrightarrow5x-2x=15+4-2+7\)

\(\Leftrightarrow3x=24\)

\(\Leftrightarrow x=8\)

b, \(\frac{8x-3}{4}-\frac{3x-2}{2}=\frac{2x-1}{2}+\frac{x+3}{4}\)

\(\Leftrightarrow\frac{8x-3}{4}-\frac{2\left(3x-2\right)}{4}=\frac{2\left(2x-1\right)}{4}+\frac{x+3}{4}\)

\(\Rightarrow8x-3-6x+4=4x-2+x+3\)

\(\Leftrightarrow8x-6x-4x-x=3+4-2+3\)

\(\Leftrightarrow-3x=8\)

\(\Leftrightarrow x=\frac{-8}{3}\)

c,\(\frac{2\left(x+5\right)}{3}+\frac{x+12}{2}-\frac{5\left(x-2\right)}{6}=\frac{x}{3}+11\)

<=>\(\frac{4\left(x+5\right)}{6}+\frac{3\left(x+12\right)}{6}-\frac{5\left(x-2\right)}{6}=\frac{2x}{6}+\frac{66}{6}\)

\(\Leftrightarrow\frac{4x+20}{6}+\frac{3x+36}{6}-\frac{5x-10}{6}=\frac{2x}{6}+\frac{66}{6}\)

\(\Rightarrow4x+20+3x+36-5x+10=2x+66\)

\(\Leftrightarrow4x+3x-5x-2x=66-20-36-10\)

\(\Leftrightarrow0=0\)