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

\(\text{a) }\left|2-5x\right|=\left|3x+1\right|\\ \Leftrightarrow\left[{}\begin{matrix}2-5x=3x+1\\2-5x=-3x-1\end{matrix}\right. \Leftrightarrow\left[{}\begin{matrix}-5x-3x=1-2\\-5x+3x=-1-2\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}-8x=-1\\-2x=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{8}\\x=\dfrac{3}{2}\end{matrix}\right.\)

Vậy tập nghiệm phương trình là \(S=\left\{\dfrac{1}{8};\dfrac{3}{2}\right\}\)

\(\text{b) }\dfrac{3}{4x-20}+\dfrac{15}{50-2x^2}+\dfrac{7}{6x+30}=0\)

ĐXKĐ của phương trình \(:x\ne\pm5\)

\(\text{Ta có }:\dfrac{3}{4x-20}+\dfrac{15}{50-2x^2}+\dfrac{7}{6x+30}=0\\ \Rightarrow\dfrac{3}{4\left(x-5\right)}+\dfrac{15}{2\left(25-x^2\right)}+\dfrac{7}{6\left(x+5\right)}=0\\ \Rightarrow\dfrac{3}{4\left(x-5\right)}-\dfrac{15}{2\left(x+5\right)\left(x-5\right)}+\dfrac{7}{6\left(x+5\right)}=0\\ \Rightarrow\dfrac{9\left(x+5\right)}{12\left(x+5\right)\left(x-5\right)}-\dfrac{90}{12\left(x+5\right)\left(x-5\right)}+\dfrac{14\left(x-5\right)}{12\left(x+5\right)\left(x-5\right)}=0\\ \Rightarrow9x+45-90+14x-70=0\\ \Leftrightarrow23x=115\\ \Leftrightarrow x=5\left(KTM\right)\)

Vậy phương trình vô nghiệm

\(\text{c) }\dfrac{x+29}{31}-\dfrac{x+27}{33}=\dfrac{x+17}{43}-\dfrac{x+15}{45}\\ \Leftrightarrow\left(\dfrac{x+29}{31}+1\right)-\left(\dfrac{x+27}{33}+1\right)=\left(\dfrac{x+17}{43}+1\right)-\left(\dfrac{x+15}{45}+1\right)\\ \Leftrightarrow\dfrac{x+60}{31}-\dfrac{x+60}{33}-\dfrac{x+60}{43}+\dfrac{x+60}{45}=0\\ \Leftrightarrow\left(x+60\right)\left(\dfrac{1}{31}-\dfrac{1}{33}-\dfrac{1}{43}+\dfrac{1}{45}\right)=0\\ \Leftrightarrow x+60=0\left(\text{Vì }\dfrac{1}{31}-\dfrac{1}{33}-\dfrac{1}{43}+\dfrac{1}{45}\ne0\right)\\ \Leftrightarrow x=-60\)

Vậy \(x=-60\) là nghiệm của phương trình

a: \(\Leftrightarrow\dfrac{2\left(2x+3\right)}{4x-6}-\dfrac{3}{4x-6}=\dfrac{2}{5}\)

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

=>5(4x+3)=2(4x-6)

=>20x+15=8x-12

=>12x=-27

hay x=-9/4

b: \(\Leftrightarrow\dfrac{x+29}{31}+1-\dfrac{x+27}{33}-1=\dfrac{x+17}{43}+1-\dfrac{x+15}{45}-1\)

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

=>x+60=0

hay x=-60

\(\dfrac{x-1}{2}-\dfrac{x+1}{15}-\dfrac{2x-13}{6}=0\)

\(\Rightarrow\dfrac{15\left(x-1\right)-2\left(x+1\right)-5\left(2x-13\right)}{30}=0\)

\(\Rightarrow15\left(x-1\right)-2\left(x+1\right)-5\left(2x-13\right)=0\)

\(\Rightarrow15x-15-2x-2-10x+65=0\)

\(\Rightarrow3x-48=0\)

\(\Rightarrow3x=48\Rightarrow x=16\)

Chúc bạn học tốt!

Bài Làm:

\(\dfrac{x-1}{2}-\dfrac{x+1}{15}-\dfrac{2x-13}{6}=0\)

\(\Leftrightarrow\)\(15\left(x-1\right)-2\left(x+1\right)-5\left(2x-13\right)=0\)

\(\Leftrightarrow15x-15-2x-2-10x+65=0\)

\(\Leftrightarrow15x-2x-10x=15+2-65\)

\(\Leftrightarrow3x=-48\)

\(\Leftrightarrow x=-16\)

Vậy \(x=-16\).

\(S=\left\{-16\right\}\)

Chúc pạn hok tốt!!!

\(B=\left|157\dfrac{13}{27}-273\dfrac{7}{19}\right|-96\dfrac{14}{27}+15\dfrac{12}{19}\)

\(=273\dfrac{7}{19}-153\dfrac{13}{27}-96\dfrac{14}{27}+15\dfrac{12}{19}\)

\(=\left(273+15+\dfrac{7}{19}+\dfrac{12}{19}\right)-\left(153+96+\dfrac{13}{27}+\dfrac{14}{27}\right)\)

\(=289-250=39\)

1)

\(\dfrac{x-5}{100}+\dfrac{x-4}{101}+\dfrac{x-3}{102}=\dfrac{x-100}{5}+\dfrac{x-101}{4}+\dfrac{x-102}{3}\)

\(\Leftrightarrow\dfrac{x-5}{100}+1+\dfrac{x-4}{101}+1+\dfrac{x-3}{102}+1=\dfrac{x-100}{5}+1+\dfrac{x-101}{4}+1+\dfrac{x-102}{3}+1\)

\(\Leftrightarrow\dfrac{x-105}{100}+\dfrac{x-105}{101}+\dfrac{x-105}{102}=\dfrac{x-105}{5}+\dfrac{x-105}{4}+\dfrac{x-105}{3}+\dfrac{x-105}{2}\)

\(\Leftrightarrow\dfrac{x-105}{100}+\dfrac{x-105}{101}+\dfrac{x-105}{102}-\dfrac{x-105}{5}-\dfrac{x-105}{4}-\dfrac{x-105}{3}-\dfrac{x-105}{2}=0\)

\(\Leftrightarrow\left(x-105\right)\left(\dfrac{1}{100}+\dfrac{1}{101}+\dfrac{1}{102}-\dfrac{1}{5}-\dfrac{1}{4}-\dfrac{1}{3}-\dfrac{1}{2}\right)=0\)\(\Leftrightarrow105-x=0\)

\(\Leftrightarrow x=105\)

b)

\(\dfrac{29-x}{21}+\dfrac{27-x}{23}+\dfrac{25-x}{25}+\dfrac{23-x}{27}+\dfrac{21-x}{29}=0\)

\(\Leftrightarrow\dfrac{29-x}{21}+1+\dfrac{27-x}{23}+1+\dfrac{25-x}{25}+1+\dfrac{23-x}{27}+1+\dfrac{21-x}{29}+1=0\)

\(\Leftrightarrow\dfrac{50-x}{21}+\dfrac{50-x}{23}+\dfrac{50-x}{25}+\dfrac{20-x}{27}+\dfrac{50-x}{29}=0\)

\(\Leftrightarrow\left(50-x\right)\left(\dfrac{1}{21}+\dfrac{1}{23}+\dfrac{1}{25}+\dfrac{1}{27}+\dfrac{1}{29}\right)=0\)

\(\Leftrightarrow50-x=0\)

\(\Leftrightarrow x=50\)

2)

\(\left(5x+1\right)^2=\left(3x-2\right)^2\)

\(\Leftrightarrow\left|5x+1\right|=\left|3x-2\right|\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-3}{2}\\x=\dfrac{1}{8}\end{matrix}\right.\)

b) \(\left(x+2\right)^3=\left(2x+1\right)^3\)

\(\Leftrightarrow x^3+6x^2+12x+8=8x^3+12x^2+6x+1\)

\(\Leftrightarrow-7x^3-6x^2+6x+7=0\)

\(\Leftrightarrow-7x^3+7x^2-13x^2+13x-7x+7=0\)

\(\Leftrightarrow-7x^2\left(x-1\right)-13x\left(x-1\right)-7\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(-7x^2-13x-7\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\-7x^2-13x-7=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\-7\left(x^2+\dfrac{13}{7}x+1\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\-7\left(x+\dfrac{13}{14}\right)^2-\dfrac{169}{196}=0\left(l\right)\end{matrix}\right.\)

\(\Leftrightarrow x=1\)

c. \(\left(x+3\right)^4+\left(x+5\right)^4=2\)

Đặt: \(y=x+4\), ta có:

\(\left(y-1\right)^4+\left(y+1\right)^4=2\)

\(\Leftrightarrow y^4-4y^3+6y^2-4y+1+y^4+4y^3+6y^2+4y+1=2\)

\(\Leftrightarrow2y^4+12y^2=0\)

\(\Leftrightarrow2y^2\left(y^2+6\right)=0\)

\(\Leftrightarrow y=0\)

\(\Leftrightarrow x=-4\)

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

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

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

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

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

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

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

\(\Leftrightarrow x=1\)

Giải phương trình

\(\dfrac{x+2}{13}+\dfrac{2x+45}{15}=\dfrac{3x+8}{37}+\dfrac{4x+69}{9}\)

\(\Leftrightarrow\)\(\dfrac{x+2}{13}+1+\dfrac{2x+45}{15}-1=\dfrac{3x+8}{37}+1+\dfrac{4x+69}{9}-1\)

\(\Leftrightarrow\)\(\dfrac{x+2}{13}+\dfrac{13}{13}+\dfrac{2x+45}{15}-\dfrac{15}{15}=\dfrac{3x+8}{37}+\dfrac{37}{37}+\dfrac{4x+69}{9}-\dfrac{9}{9}\)

\(\Leftrightarrow\dfrac{x+15}{13}+\dfrac{2x+30}{15}=\dfrac{3x+45}{37}+\dfrac{4x+60}{9}\)

\(\Leftrightarrow\dfrac{x+15}{13}+\dfrac{2\left(x+15\right)}{15}=\dfrac{3\left(x+15\right)}{37}+\dfrac{4\left(x+15\right)}{9}\)

\(\Leftrightarrow\left(x+15\right)\left(\dfrac{1}{13}+\dfrac{2}{15}\right)=\left(x+15\right)\left(\dfrac{3}{37}+\dfrac{4}{9}\right)\)

\(\Leftrightarrow\left(x+15\right)\left(\dfrac{1}{13}+\dfrac{2}{15}\right)-\left(x+15\right)\left(\dfrac{3}{37}+\dfrac{4}{9}\right)=0\)

\(\Leftrightarrow\left(x+15\right)\left(\dfrac{1}{13}+\dfrac{2}{15}-\dfrac{3}{37}-\dfrac{4}{9}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+15=0\\\dfrac{1}{13}+\dfrac{2}{15}-\dfrac{3}{37}-\dfrac{4}{9}=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-15\\\dfrac{1}{13}+\dfrac{2}{15}-\dfrac{3}{37}-\dfrac{4}{9}\ne0\end{matrix}\right.\)

Do đó: \(x=-15\)

Vậy \(S=\left\{-15\right\}\)

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