\(\Leftrightarrow x+66=0\)

hay x=-66

chị ơi, chị giải thích rõ được ko ạ?

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

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

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

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

\(\Leftrightarrow x-66=0\)

\(\Leftrightarrow x=66\)

Vậy x=66.

Thanks bạn nhiều!~

c: \(\Leftrightarrow2x+2x-6=12-2x\)

=>4x-6=12-2x

=>6x=18

hay x=3

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

\(\Leftrightarrow x^2-1+x=2x-1\)

=>x²-x=0

=>x(x-1)=0

=>x=0(loại) hoặc x=1(nhận)

lớp 9 gì như lớp 6 thế

a) đề sai

c) <=>x/3 +x/3 -1 =2-x/3

<=>3.x/3 =3 => x=3

b) x<> 0; -2 <=>

x^2 -1 +x =2x-1

<=>x^2 -x =0 => x =0 (l) x =1 nhận

d ; <=> (x+1)/65+1 +(x+3)/63 +1 =(x+5)/61+1 +(x+7)/59+1

<=>(x+66) [1/65+1/63-1/61-1/59] =0

[...] khác 0

x=-66

bạn làm rõ câu d dc ko bạn

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

\(< =>\left(\dfrac{x+1}{65}+1\right)+\left(\dfrac{x+3}{63}+1\right)=\left(\dfrac{x+5}{61}+1\right)+\left(\dfrac{x+7}{59}+1\right)\)

\(< =>\dfrac{x+66}{65}+\dfrac{x+66}{63}=\dfrac{x+66}{61}+\dfrac{x+66}{59}\)

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

\(< =>x+66=0< =>x=-66\)

Vậy tập nghiệm của phương trình đã cho là: S={-66}

\(\dfrac{x+29}{31}-\dfrac{x+27}{33}=\dfrac{x+17}{43}-\dfrac{x+15}{45}\)

\(< =>\dfrac{x+60}{31}-\dfrac{x+60}{33}=\dfrac{x+60}{43}-\dfrac{x+60}{45}\)

\(< =>\left(x+60\right)\left(\dfrac{1}{31}-\dfrac{1}{33}-\dfrac{1}{43}+\dfrac{1}{45}\right)=0\)

Mà: \(\dfrac{1}{31}-\dfrac{1}{33}-\dfrac{1}{43}+\dfrac{1}{45}\ne0\)

\(=>x+60=0< =>x=-60\)

Vậy tập nghiệm của phương trình đã cho là: s={-60}

a, Theo đề ta có:

\(2.3^x-405=3^{x-1}\)

=> \(2.3^x-405=3^x:3\)

=> \(405=(2.3^x)-(3^x:3)\)

=>\(405=(2.3^x)-(3^x.\dfrac{1}{3})\)

=> \(405=3^x(2-\dfrac{1}{3})\)

=>\(405=3^x(\dfrac{6}{3}-\dfrac{1}{3})\)

=> \(405=3^x.\dfrac{5}{3}\)

=> \(3^x=405:\dfrac{5}{3}\)

=>\(3^x=405.\dfrac{3}{5}\)

=> \(3^x=81.3\)

=> \(3^x=243\)

=> \(3^x=3^5\)

=> x=5

Vậy:..............................

a) \(\dfrac{x+43}{57}+\dfrac{x+46}{54}=\dfrac{x+49}{51}+\dfrac{x+52}{48}\)

\(\left(\dfrac{x+43}{57}+1\right)+\left(\dfrac{x+46}{54}+1\right)=\left(\dfrac{x+49}{51}+1\right)+\left(\dfrac{x+52}{48}\right)\)

\(\dfrac{x+43+57}{57}+\dfrac{x+46+54}{54}-\dfrac{x+49+51}{51}-\dfrac{x+52+48}{48}=0\)

\(\dfrac{x+100}{57}+\dfrac{x+100}{54}-\dfrac{x+100}{51}-\dfrac{x+100}{48}=0\)

\(\left(x+100\right)\left(\dfrac{1}{57}+\dfrac{1}{54}-\dfrac{1}{51}-\dfrac{1}{48}\right)=0\)

Mà \(\dfrac{1}{57}+\dfrac{1}{54}-\dfrac{1}{51}-\dfrac{1}{48}\ne0\)

Nên: \(x+100=0\)

\(x=-100\)