Chúc bạn học tốt :))

b, Ta có : \(\frac{x-10}{1994}+\frac{x-8}{1996}+\frac{x-6}{1994}+\frac{x-4}{2000}+\frac{x-2}{2002}=\frac{x-2002}{2}+\frac{x-2000}{4}+\frac{x-1998}{6}+\frac{x-1996}{8}+\frac{x-1994}{10}\)

=> \(\frac{x-10}{1994}-1+\frac{x-8}{1996}-1+\frac{x-6}{1994}-1+\frac{x-4}{2000}-1+\frac{x-2}{2002}-1=\frac{x-2002}{2}-1+\frac{x-2000}{4}-1+\frac{x-1998}{6}-1+\frac{x-1996}{8}-1+\frac{x-1994}{10}-1\)

=> \(\frac{x-2004}{1994}+\frac{x-2004}{1996}+\frac{x-2004}{1994}+\frac{x-2004}{2000}+\frac{x-2004}{2002}=\frac{x-2004}{2}+\frac{x-2004}{4}+\frac{x-2004}{6}+\frac{x-2004}{8}+\frac{x-2004}{10}\)

=> \(\frac{x-2004}{1994}+\frac{x-2004}{1996}+\frac{x-2004}{1994}+\frac{x-2004}{2000}+\frac{x-2004}{2002}-\frac{x-2004}{2}-\frac{x-2004}{4}-\frac{x-2004}{6}-\frac{x-2004}{8}-\frac{x-2004}{10}=0\)

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

=> \(x-2004=0\)

=> \(x=2004\)

Vậy phương trình có tập nghiệm là \(S=\left\{2004\right\}\)

a) Sửa đề: \(\frac{x+1}{35}+\frac{x+3}{33}=\frac{x+5}{31}+\frac{x+7}{29}\)

Ta có: \(\frac{x+1}{35}+\frac{x+3}{33}=\frac{x+5}{31}+\frac{x+7}{29}\)

\(\Leftrightarrow\frac{x+1}{35}+1+\frac{x+3}{33}+1=\frac{x+5}{31}+1+\frac{x+7}{29}+1\)

\(\Leftrightarrow\frac{x+36}{35}+\frac{x+36}{33}=\frac{x+36}{31}+\frac{x+36}{29}\)

\(\Leftrightarrow\frac{x+36}{35}+\frac{x+36}{33}-\frac{x+36}{31}-\frac{x+36}{29}=0\)

\(\Leftrightarrow\left(x+36\right)\left(\frac{1}{35}+\frac{1}{33}-\frac{1}{31}-\frac{1}{29}\right)=0\)

Vì \(\frac{1}{35}+\frac{1}{33}-\frac{1}{31}-\frac{1}{29}\ne0\)

nên x+36=0

hay x=-36

Vậy: x=-36

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

\(\Leftrightarrow\frac{x+1}{65}+\frac{x+3}{63}-\frac{x+5}{61}-\frac{x+7}{59}=0\)

\(\Leftrightarrow\frac{x+1}{65}+1+\frac{x+3}{63}+1-\left(\frac{x+5}{61}+1\right)-\left(\frac{x+7}{59}+1\right)=0\)

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

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

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

<=> x=-66

các câu còn lại cũng làm tương tự nhé

\(a.\frac{x+5}{2021}+\frac{x+6}{2020}+\frac{x+7}{2019}=-3\\ \Leftrightarrow\frac{x+5}{2021}+1+\frac{x+6}{2020}+1+\frac{x+7}{2019}+1=0\\ \Leftrightarrow\frac{x+2026}{2021}+\frac{x+2026}{2020}+\frac{x+2026}{2019}=0\\ \Leftrightarrow\left(x+2026\right)\left(\frac{1}{2021}+\frac{1}{2020}+\frac{1}{2019}\right)=0\\\Leftrightarrow x+2026=0\left(Vi\frac{1}{2021}+\frac{1}{2020}+\frac{1}{2019}\ne0\right)\\ \Leftrightarrow x=-2026\)

Vậy tập nghiệm của phương trình trên là \(S=\left\{-2026\right\}\)

\(b.\frac{2-x}{100}-1=\frac{1-x}{101}-\frac{x}{102}\\ \Leftrightarrow\frac{2-x}{100}+1=\frac{1-x}{101}+1+1-\frac{x}{102}\\\Leftrightarrow \frac{102-x}{100}-\frac{102-x}{101}-\frac{102-x}{102}=0\\ \Leftrightarrow\left(102-x\right)\left(\frac{1}{100}-\frac{1}{101}-\frac{1}{102}\right)=0\\ \Leftrightarrow102-x=0\left(Vi\frac{1}{100}-\frac{1}{101}-\frac{1}{102}\ne0\right)\\ \Leftrightarrow x=102\)

Vậy tập nghiệm của phương trình trên là \(S=\left\{102\right\}\)

c/ PT tương đương

\(\frac{x+1}{93}-1+\frac{x-2}{45}-2+\frac{x+4}{32}-3=0\)

\(\Leftrightarrow\frac{x-92}{93}+\frac{x-92}{45}+\frac{x-92}{32}=0\)

\(\Leftrightarrow\left(x-92\right)\left(\frac{1}{93}+\frac{1}{45}+\frac{1}{32}\right)=0\Rightarrow x=92\)

Bài 1:

a) Ta có: \(2,3x-2\left(0,7+2x\right)=3,6-1,7x\)

\(\Leftrightarrow2,3x-1,4-4x-3,6+1,7x=0\)

\(\Leftrightarrow-5=0\)(vl)

Vậy: \(x\in\varnothing\)

b) Ta có: \(\frac{4}{3}x-\frac{5}{6}=\frac{1}{2}\)

\(\Leftrightarrow\frac{4}{3}x=\frac{1}{2}+\frac{5}{6}=\frac{8}{6}=\frac{4}{3}\)

hay x=1

Vậy: x=1

c) Ta có: \(\frac{x}{10}-\left(\frac{x}{30}+\frac{2x}{45}\right)=\frac{4}{5}\)

\(\Leftrightarrow\frac{9x}{90}-\frac{3x}{90}-\frac{4x}{90}-\frac{72}{90}=0\)

\(\Leftrightarrow2x-72=0\)

\(\Leftrightarrow2\left(x-36\right)=0\)

mà 2>0

nên x-36=0

hay x=36

Vậy: x=36

d) Ta có: \(\frac{10x+3}{8}=\frac{7-8x}{12}\)

\(\Leftrightarrow12\left(10x+3\right)=8\left(7-8x\right)\)

\(\Leftrightarrow120x+36=56-64x\)

\(\Leftrightarrow120x+36-56+64x=0\)

\(\Leftrightarrow184x-20=0\)

\(\Leftrightarrow184x=20\)

hay \(x=\frac{5}{46}\)

Vậy: \(x=\frac{5}{46}\)

e) Ta có: \(\frac{10x-5}{18}+\frac{x+3}{12}=\frac{7x+3}{6}-\frac{12-x}{9}\)

\(\Leftrightarrow\frac{2\left(10x-5\right)}{36}+\frac{3\left(x+3\right)}{36}-\frac{6\left(7x+3\right)}{36}+\frac{4\left(12-x\right)}{36}=0\)

\(\Leftrightarrow2\left(10x-5\right)+3\left(x+3\right)-6\left(7x+3\right)+4\left(12-x\right)=0\)

\(\Leftrightarrow20x-10+3x+9-42x-18+48-4x=0\)

\(\Leftrightarrow-23x+29=0\)

\(\Leftrightarrow-23x=-29\)

hay \(x=\frac{29}{23}\)

Vậy: \(x=\frac{29}{23}\)

f) Ta có: \(\frac{x+4}{5}-x-5=\frac{x+3}{2}-\frac{x-2}{2}\)

\(\Leftrightarrow\frac{2\left(x+4\right)}{10}-\frac{10x}{10}-\frac{50}{10}=\frac{25}{10}\)

\(\Leftrightarrow2x+8-10x-50-25=0\)

\(\Leftrightarrow-8x-67=0\)

\(\Leftrightarrow-8x=67\)

hay \(x=\frac{-67}{8}\)

Vậy: \(x=\frac{-67}{8}\)

g) Ta có: \(\frac{2-x}{4}=\frac{2\left(x+1\right)}{5}-\frac{3\left(2x-5\right)}{10}\)

\(\Leftrightarrow5\left(2-x\right)-8\left(x+1\right)+6\left(2x-5\right)=0\)

\(\Leftrightarrow10-5x-8x-8+12x-30=0\)

\(\Leftrightarrow-x-28=0\)

\(\Leftrightarrow-x=28\)

hay x=-28

Vậy: x=-28

h) Ta có: \(\frac{x+2}{3}+\frac{3\left(2x-1\right)}{4}-\frac{5x-3}{6}=x+\frac{5}{12}\)

\(\Leftrightarrow\frac{4\left(x+2\right)}{12}+\frac{9\left(2x-1\right)}{12}-\frac{2\left(5x-3\right)}{12}-\frac{12x}{12}-\frac{5}{12}=0\)

\(\Leftrightarrow4x+8+18x-9-10x+6-12x-5=0\)

\(\Leftrightarrow0x=0\)

Vậy: \(x\in R\)

Bài 2:

a) Ta có: \(5\left(x-1\right)\left(2x-1\right)=3\left(x+8\right)\left(x-1\right)\)

\(\Leftrightarrow5\left(x-1\right)\left(2x-1\right)-3\left(x-1\right)\left(x+8\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left[5\left(2x-1\right)-3\left(x+8\right)\right]=0\)

\(\Leftrightarrow\left(x-1\right)\left(10x-5-3x-24\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(7x-29\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\7x-29=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\7x=29\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\frac{29}{7}\end{matrix}\right.\)

Vậy: Tập nghiệm \(S=\left\{1;\frac{29}{7}\right\}\)

b) Ta có: \(\left(3x-2\right)\left(x+6\right)\left(x^2+5\right)=0\)(1)

Ta có: \(x^2\ge0\forall x\)

\(\Rightarrow x^2+5\ge5\ne0\forall x\)(2)

Từ (1) và (2) suy ra:

\(\left[{}\begin{matrix}3x-2=0\\x+6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=2\\x=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{2}{3}\\x=-6\end{matrix}\right.\)

Vậy: Tập nghiệm \(S=\left\{\frac{2}{3};-6\right\}\)

c) Ta có: \(\left(3x-2\right)\left(9x^2+6x+4\right)-\left(3x-1\right)\left(9x^2-3x+1\right)=x-4\)

\(\Leftrightarrow27x^3-8-\left(27x^3-1\right)-x+4=0\)

\(\Leftrightarrow27x^3-8-27x^3+1-x+4=0\)

\(\Leftrightarrow-x-3=0\)

\(\Leftrightarrow-x=3\)

hay x=-3

Vậy: Tập nghiệm S={-3}

d) Ta có: \(x\left(x-1\right)-\left(x-3\right)\left(x+4\right)=5x\)

\(\Leftrightarrow x^2-x-\left(x^2+x-12\right)-5x=0\)

\(\Leftrightarrow x^2-x-x^2-x+12-5x=0\)

\(\Leftrightarrow12-7x=0\)

\(\Leftrightarrow7x=12\)

hay \(x=\frac{12}{7}\)

Vậy: Tập nghiệm \(S=\left\{\frac{12}{7}\right\}\)

e) Ta có: (2x+1)(2x-1)=4x(x-7)-3x

\(\Leftrightarrow4x^2-1-4x^2+28x+3x=0\)

\(\Leftrightarrow31x-1=0\)

\(\Leftrightarrow31x=1\)

hay \(x=\frac{1}{31}\)

Vậy: Tập nghiệm \(S=\left\{\frac{1}{31}\right\}\)

a, \(\frac{x+1}{35}+\frac{x+3}{33}=\frac{x+5}{31}+\frac{x+7}{29}\)

\(\frac{x+36}{35}+\frac{x+36}{33}-\frac{x+36}{31}-\frac{x+36}{29}=0\)

\(\left(x+36\right)\left(\frac{1}{35}+\frac{1}{33}-\frac{1}{31}-\frac{1}{29}\right)=0\)

\(=>x+36=0\)

\(=>x=36\)

a) \(\left(x-11\right)+\frac{3x}{x-11}=3+\frac{33}{x-11}\)

\(\Leftrightarrow x+\frac{3x}{x-11}-\frac{33}{x-11}=14\)

\(\Leftrightarrow x^2-11x+3x-33=14x-154\)

\(\Leftrightarrow x^2-22x+121=0\)

\(\Leftrightarrow\left(x-11\right)^2=0\Leftrightarrow x=11\)

Vậy .......

b) \(\frac{7-2x}{x-1}=\frac{1-4x}{x+2}\Leftrightarrow\left(7-2x\right)\left(x+2\right)=\left(1-4x\right)\left(x-1\right)\)

\(\Leftrightarrow7x-2x^2+14-4x=x-4x^2-1+4x\)

\(\Leftrightarrow2x^2=-15\)(vô lí)

Vậy pt vô nghiệm

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

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

\(\Leftrightarrow3x-2x^2-6x+4x=2x^2+2x-4x-4+x-4x^2+1-4x\)

\(\Leftrightarrow6x=-3\Leftrightarrow x=-\frac{1}{2}\)

Vậy.........

(gửi trước 3 câu)

d) \(\frac{109x-4}{111x+1}-1=0\Leftrightarrow109x-4=111x+1\Leftrightarrow2x=-5\Leftrightarrow x=-\frac{5}{2}\)

Vậy x=-5/2

e) \(\frac{x^2-7}{x}=x-\frac{1}{2}\Leftrightarrow\frac{x^2-7}{x}-\frac{x^2}{x}=-\frac{1}{2}\Leftrightarrow-\frac{7}{x}=\frac{1}{2}\Leftrightarrow x=-14\)

f) \(\frac{x+1}{x+2}=3\Leftrightarrow x+1=3x+6\Leftrightarrow2x=7\Leftrightarrow x=\frac{7}{2}\)

a/ĐKXĐ: \(y\ne4\)

Đặt \(y-4=x\)

\(1+\frac{45}{x^2}=\frac{14}{x}\Leftrightarrow x^2-14x+45=0\Rightarrow\left[{}\begin{matrix}x=9\\x=5\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}y-4=9\\y-4=5\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}y=13\\y=9\end{matrix}\right.\)

b/ ĐKXĐ: \(x\ne1\)

Đặt \(x-1=y\)

\(\frac{5}{y}-\frac{4}{3y^2}=3\Leftrightarrow9y^2=15y-4\)

\(\Leftrightarrow9y^2-15y+4=0\Rightarrow\left[{}\begin{matrix}y=\frac{4}{3}\\y=\frac{1}{3}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x-1=\frac{4}{3}\\x-1=\frac{1}{3}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\frac{7}{3}\\x=\frac{4}{3}\end{matrix}\right.\)

c/ ĐKXĐ: \(x\ne5\)

\(\Leftrightarrow2x-5=3x-15\)

\(\Leftrightarrow x=10\)

d/ ĐKXĐ: \(x\ne0\)

\(\Leftrightarrow2\left(x^2-12\right)=2x^2+3x\)

\(\Leftrightarrow3x=-24\Rightarrow x=-8\)

e/ ĐKXĐ: \(x\ne2\)

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

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

\(\Rightarrow\left[{}\begin{matrix}x=2\left(l\right)\\x=1\end{matrix}\right.\)

f/ DKXĐ: \(x\ne-\frac{1}{2}\)

\(\Leftrightarrow\left(2x-1\right)\left(2x+1\right)=8\)

\(\Leftrightarrow4x^2-1=8\)

\(\Leftrightarrow x^2=\frac{9}{4}\Rightarrow x=\pm\frac{3}{2}\)