\(x=2\)

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

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

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

=>x+46=20x+8

=>46-8=20x-x

=>38=19x

=>x=2

vậy x=2

a) \(\frac{-1}{5}\le\frac{x}{8}\le\frac{1}{4}\)

\(\frac{-8}{40}\le\frac{5x}{40}\le\frac{10}{40}\)

\(\Rightarrow-8\le5x\le10\)

\(\Rightarrow x\in\left\{-5;0;5;10\right\}\)

5.x = - 5

x = -5 ÷ 5 = -1

5.x = 0

=> x = 0

5.x = 5

=> x = 1

5.x = 10

=> x = 2

Vậy, \(x\in\left\{-1;0;1;2\right\}\)

Câu b) mình không biết làm, bạn thông cảm nha!!!

Cbht

a. \(-\frac{1}{5}\le\frac{x}{8}\le\frac{1}{4}\)

<=> \(-\frac{8}{40}\le\frac{5x}{40}\le\frac{10}{40}\)

<=> -8 \(\le\)5x \(\le\)10

<=> x + 5 \(\in\){\(\pm\)8;\(\pm\)7; \(\pm\)6;\(\pm\)5;\(\pm\)4;\(\pm\)3;\(\pm\)2; \(\pm\)1; 0; 9;10}

<=> x \(\in\left\{3;-13;2;-12;1;-11;0;-10;-1;-9;-2;-8;-3;-7;-4;-6;-5;4;5\right\}\)

b. \(\frac{x+46}{20}=x\frac{2}{5}\)

<=> 5x + 230 = 40x

<=> 35x = 230

<=> x = \(\frac{46}{7}\)

Câu a sai nha !!! Mình sửa lại sau chỗ 

-8 \(\le\)5x\(\le\)10

<=> x \(\in\){0; \(\pm\)1;2}

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

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

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

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