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

\(\Leftrightarrow\dfrac{\left(x+1\right).33}{35.33}+\dfrac{\left(x+3\right).35}{33.35}=\dfrac{\left(x+5\right).29}{31.29}+\dfrac{\left(x+7\right).31}{29.31}\)

\(\Leftrightarrow\dfrac{33\left(x+1\right)+35\left(x+3\right)}{1155}=\dfrac{29\left(x+5\right)+31\left(x+7\right)}{899}\)

\(\Leftrightarrow\dfrac{33x+33+35x+35.3}{1155}=\dfrac{29x+29.5+31x+31.7}{899}\)

\(\Leftrightarrow\dfrac{68x+138}{1155}=\dfrac{60x+362}{899}\)

\(\Leftrightarrow\dfrac{\left(68x+138\right).899}{1155.899}=\dfrac{\left(60x+362\right).1155}{899.115}\)

\(\Rightarrow\left(68x+138\right).899=\left(60x+362\right).1155\)

<=> 61132x+124062=69300x+418110

<=> 61132x-69300x=418110-124062

<=> -8168x=294048

<=> x=-36

Vậy phương trình có nghiệm x=-36

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

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

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

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

\(\Leftrightarrow x=-36\)

1) PT \(\Leftrightarrow\left(\dfrac{x+1}{35}+1\right)+\left(\dfrac{x+3}{33}+1\right)=\left(\dfrac{x+5}{31}+1\right)+\left(\dfrac{x+7}{29}+1\right)\)

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

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

\(\Leftrightarrow x+36=0\) (Do \(\dfrac{1}{29}+\dfrac{1}{31}-\dfrac{1}{33}-\dfrac{1}{35}>0\))

\(\Leftrightarrow x=-36\).

Vậy nghiệm của pt là x = -36.

2) x(x+1)(x+2)(x+3)= 24

⇔ x.(x+3)  .   (x+2).(x+1)  = 24

⇔(\(x^2\) + 3x) . (\(x^2\) + 3x + 2) = 24

Đặt \(x^2\)+ 3x = b

⇒ b . (b+2)= 24

Hay: \(b^2\) +2b = 24

⇔\(b^2\) + 2b + 1 = 25

⇔\(\left(b+1\right)^2\)= 25

+ Xét b+1 = 5 ⇒ b=4 ⇒  \(x^2\)+ 3x = 4 ⇒ \(x^2\)+4x-x-4=0 ⇒x(x+4)-(x+4)=0

⇒(x-1)(x+4)=0⇒x=1 và x=-4

+ Xét b+1 = -5 ⇒ b=-6 ⇒ \(x^2\)+3x=-6 ⇒\(x^2\) + 3x + 6=0

⇒\(x^2\) + 2.x.\(\dfrac{3}{2}\) + (\(\dfrac{3}{2}\))² = - \(\dfrac{15}{4}\)  Hay ( \(x^2\) +\(\dfrac{3}{2}\) )²= -\(\dfrac{15}{4}\) (vô lí)

⇒x= 1 và x= 4

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

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

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

\(=>x+36=0\)

\(=>x=36\)

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

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

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

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

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

\(\Leftrightarrow x+36=0\).Do \(\frac{1}{35}+\frac{1}{33}-\frac{1}{31}-\frac{1}{29}\ne0\)

\(\Leftrightarrow x=-36\)

A, \(\frac{x+3}{2}\)-\(\frac{x-1}{3}\)=\(\frac{x+5}{6}\)+1

⇔ \(\frac{3\left(x+3\right)}{6}\)-\(\frac{2\left(x-1\right)}{6}\)=\(\frac{x+5}{6}\)+\(\frac{6}{6}\)

⇔ 3x+9-2x+2=x+5+6

⇔ 3x-2x-x=5+6-9-2

⇔0x=0 (luôn đúng với mọi x)

Vậy phương trình có vô số nghiêm:S=R

a) \(x+\frac{3}{2}-x-\frac{1}{3}=x+\frac{5}{6}+1\)

⇔ \(\frac{3}{2}-x-\frac{1}{3}=\frac{5}{6}+1\)

⇔ \(\frac{7}{6}-x=\frac{5}{6}+1\)

⇔ \(\frac{7}{6}-x=\frac{11}{6}\)

⇔ \(-x=\frac{11}{6}-\frac{7}{6}\)

⇔ \(-x=\frac{2}{3}\)

⇔ \(x=\frac{-2}{3}\)

Vậy tập nghiệm của pt là S = \(\left\{\frac{-2}{3}\right\}\)

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

đề chính xác là như này đúng k cậu=)đề k rõ ràng k aii giúp đc nhé!

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

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

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

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

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

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

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

\(\Leftrightarrow x+36=0\)

\(\Leftrightarrow x=0-36\)

\(\Leftrightarrow x=-36.\)

Vậy phương trình có tập hợp nghiệm là: \(S=\left\{-36\right\}.\)

Chúc bạn học tốt!

a: =>3x=-25

=>x=-25/3

b: =>(x-5)(4x+3-31)=0

=>(4x-28)(x-5)=0

=>x=5 hoặc x=7

c: =>3x-3-x-3=5x-33

=>2x-6=5x-33

=>-3x=-27

=>x=9(nhận)

a: \(\Rightarrow\left(\dfrac{x+1}{35}+1\right)+\left(\dfrac{x+3}{33}+1\right)=\left(\dfrac{x+5}{31}+1\right)+\left(\dfrac{x+7}{29}+1\right)\)

=>x+36=0

=>x=-36

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

=>x-2004=0

=>x=2004