\(\frac{x+1}{49}+1+\frac{x+2}{48}+1+\frac{x+3}{47}+1+\frac{x+4}{46}+1+\frac{x+5}{45}+1=0\)

\(\Leftrightarrow\frac{x+50}{49}+\frac{x+50}{48}+...+\frac{x+50}{45}=0\)

\(\Leftrightarrow\left(x+50\right)\left(\frac{1}{49}+\frac{1}{48}+...+\frac{1}{45}\right)=0\)

Vì 1/49+1/48+...+1/45 khác 0

Nên x+50=0

do đó x=-50

Ta có : 

\(\frac{x-1}{49}+\frac{x-2}{48}+\frac{x-3}{47}+\frac{x-4}{46}+\frac{x-5}{45}=5\)

\(\Leftrightarrow\)\(\left(\frac{x-1}{49}-1\right)+\left(\frac{x-2}{48}-1\right)+\left(\frac{x-3}{47}-1\right)+\left(\frac{x-4}{46}-1\right)+\left(\frac{x-5}{45}-1\right)=5-5\)

\(\Leftrightarrow\)\(\frac{x-1-49}{49}+\frac{x-2-48}{48}+\frac{x-3-47}{47}+\frac{x-4-46}{46}+\frac{x-5-45}{45}=0\)

\(\Leftrightarrow\)\(\frac{x-50}{49}+\frac{x-50}{48}+\frac{x-50}{47}+\frac{x-50}{46}+\frac{x-50}{45}=0\)

\(\Leftrightarrow\)\(\left(x-50\right)\left(\frac{1}{49}+\frac{1}{48}+\frac{1}{47}+\frac{1}{46}+\frac{1}{45}\right)=0\)

Vì \(\frac{1}{49}+\frac{1}{48}+\frac{1}{47}+\frac{1}{46}+\frac{1}{45}\ne0\) ( vì nó lớn hơn 0 ) 

Nên \(x-50=0\)

\(\Rightarrow\)\(x=50\)

Vậy \(x=50\)

Chúc bạn học tốt ~ 

cảm ơn bạn Phùng Minh Quân

\(x=2\)

tic cho mk nha

\(\frac{x+46}{20}=\frac{5x+2}{5}\)

=> (x+46).5=20.(5x+2)

=> 5x+230=100x+40

=> 5x-100x=40-230

=> -95x=-190

=> x=-190:(-95)

=> x=2

\(\Leftrightarrow\dfrac{46}{7}+\dfrac{81}{35}< =x< =\dfrac{49}{36}\)

\(\Leftrightarrow\dfrac{311}{35}< =x< =\dfrac{49}{36}\)

\(\Leftrightarrow x\in\varnothing\)

\(\frac{1}{\left(x+4\right)\left(x+5\right)}+\frac{1}{\left(x+5\right)\left(x+6\right)}+\frac{1}{\left(x+6\right)\left(x+7\right)}\)

\(\frac{1}{x+4}-\frac{1}{x+5}+\frac{1}{x+5}-\frac{1}{x+6}+\frac{1}{x+6}-\frac{1}{x+7}=\frac{1}{18}\)

\(\frac{1}{x+4}-\frac{1}{x+7}=\frac{1}{18}\)

dit me may

\(\frac{1}{x^2+9x+20}+\frac{1}{x^2+11x+30}+\frac{1}{x^2+13x+42}=\frac{1}{18}\)

\(\Rightarrow\frac{1}{\left(x+4\right)\left(x+5\right)}+\frac{1}{\left(x+5\right)\left(x+6\right)}+\frac{1}{\left(x+6\right)\left(x+7\right)}=\frac{1}{18}\)

\(\Rightarrow\frac{1}{x+4}-\frac{1}{x+5}+\frac{1}{\left(x+5\right)}-\frac{1}{x+6}+\frac{1}{x+6}-\frac{1}{x+7}=\frac{1}{18}\)

\(\Rightarrow\frac{1}{x+4}-\frac{1}{x+7}=\frac{1}{18}\)

\(\Rightarrow\frac{3}{\left(x+4\right)\left(x+7\right)}=\frac{1}{18}\)

\(\Rightarrow x^2+11x+28=54\)

\(\Rightarrow x^2+11x+\frac{121}{4}-\frac{9}{4}=54\)

\(\Rightarrow\left(x+\frac{11}{2}\right)^2=\frac{225}{4}\)

\(\Leftrightarrow\orbr{\begin{cases}x+\frac{11}{2}=\sqrt{\frac{225}{4}}\\x+\frac{11}{2}=-\sqrt{\frac{225}{4}}\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x+\frac{11}{2}=\frac{25}{2}\\x+\frac{11}{2}=-\frac{25}{2}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=7\\x=-18\end{cases}}\)

\(\frac{x+46}{20}=x\frac{2}{5}\)

\(\Rightarrow\frac{x+46}{20}=\frac{5x+2}{5}=\frac{20x+8}{20}\)

=>x+46=20x+8

=>46-8=20x-x

=>38=19x

=>x=2

vậy x=2

a) vì y+z+1/x = x+z+2/y = x+y-3/z = 1/x+y+z

=>

y+z+1/x = x+z+2/y = x+y-3=y+z+1+x+z+2+x+y-3/x+y+z = 2x+2y+2z/x+y+z = 2

=> 2 = 1/ x+y+z => x+y+z=1/2

sau đó áp dụng tính chất dãy tỉ số = hau

bài 1.a)\(A=\frac{9^3.25^3}{18^2.125^2}=\frac{3^6.5^6}{2^2.3^4.5^6}=\frac{9}{4}\)

b) \(B=\frac{18}{37}+\frac{19}{37}+\frac{8}{2017}-\frac{4026}{2017}+\frac{2017}{2018}\)

\(=1-\frac{4014}{2017}+\frac{2017}{2018}=\frac{1997}{2017}+\frac{2017}{2018}\)

Ta có : \(\frac{x-2}{18}=\frac{2}{3}\)nên :

\(\left(x-2\right).3=18.2=36\)

\(\Rightarrow x-2=12\)

\(x=14\)

Vậy \(x=14\).

\(\frac{x+3}{-15}=\frac{1}{3}\)nên :

\(\left(x+3\right).3=-15.1=-15\)

\(\Rightarrow x+3=-5\)

\(x=-8\)

Vậy \(x=-8\).

1) \(\frac{x-2}{18}=\frac{2}{3}\)

\(\Leftrightarrow3\left(x-2\right)=2\cdot18\)

\(\Leftrightarrow3x-6=36\)

\(\Leftrightarrow3x=42\)

<=> x=14

2)\(\frac{x+3}{-15}=\frac{1}{3}\)

\(\Leftrightarrow3\left(x+3\right)=-15\cdot1\)

<=> 3x+9=-15

<=> 3x=-24

<=> x=-8