x-5 = x-10+5= (x-10)+5 > x-10

\(\frac{3x-3}{7-x}=\frac{3}{5}\)

\(\frac{3\left(x-1\right)}{3}=\frac{7-x}{5}\)

\(x-1=\frac{7-x}{5}\)

\(5\left(x-1\right)=7-x\)

\(5x-5=7-x\)

\(5x+x=7+5\)

\(6x=12\)

\(x=2\)

\(\frac{3x-3}{7-x}=\frac{3}{5}\)

\(\Rightarrow\left(3x-3\right).5=\left(7-x\right).3\)

\(\Rightarrow15x-15=21-3x\)

\(\Rightarrow15x+3x=21+15\)

\(\Rightarrow18x=36\)

\(\Rightarrow x=2\)

\(x-\frac{3}{4}=\frac{2}{-6}\)

\(x-\frac{3}{4}=\frac{-1}{3}\)

\(x=\frac{-1}{3}+\frac{3}{4}\)

\(x=\frac{5}{12}\)

mk giải lun ak

x=2/-6+3/4

x=5/12

/ là dấu phân số nha bạn

k mk

A = \(\frac{1}{3}+\frac{13}{35}+\frac{33}{35}+\frac{61}{63}+\frac{97}{99}+\frac{141}{143}\)

\(=\left(1-\frac{2}{3}\right)+\left(1-\frac{2}{15}\right)+\left(1-\frac{2}{35}\right)+\left(1-\frac{2}{63}\right)+\left(1-\frac{2}{99}\right)+\left(1-\frac{2}{143}\right)\)

\(=\left(1+1+1+1+1+1\right)-\left(\frac{2}{3}+\frac{2}{15}+\frac{2}{35}+\frac{2}{63}+\frac{2}{99}+\frac{2}{143}\right)\)

\(=6-\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+\frac{2}{11.13}\right)\)

\(=6-\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}\right)\)

\(=6-\left(1-\frac{1}{13}\right)\)

\(=6-1+\frac{1}{13}\)

\(=5+\frac{1}{13}\)

\(=\frac{66}{13}\)

\(\text{Vậy }A=\frac{66}{13}\)

vler bài dễ tke mak cx đăn

bài nào z Linh

Ta có:

a) A = |x - 2| + |x - 4| + 2017|

=> A = |x - 2| + |4 - x| + 2017 \(\ge\)|x - 2 + 4 - x| + 2017 = |2| + 2017=2019

Dấu "=" xảy ra <=> (x - 2)(4 - x) \(\ge\)0

<=> 2 \(\le\)x \(\le\)4

Vậy MinA = 2019 <=> 2 \(\le\)x \(\)4

b) Ta có: B = |2019 - x| + |2020 - x|

=> B = |x - 2019| + |2020 - x| \(\ge\)|x - 2019 + 2020 - x| = |1| =  1

Dấu "=" xảy ra <=> (x - 2019)(2020 - x) \(\ge\)0

<=> 2019 \(\le\)x \(\le\)2020

Vậy MinB = 1 <=> 2019 \(\le\)x \(\le\)2020

ta có 

        /x-2/> hoặc= x-2

       /x-4/= /4-x/> hoặc=4-x      

=> /x-2/+/x-4/+2017> hoặc= (x-2)+(4-x)+2017=2019

           hay A> hoặc= 2019

           => GTNN của A là 2019

b,

       Vì /2019-x/ > hoặc= 2019-x

            /2020-x/=/x-2020/> hoặc=x-2020

      =>/2019-x/+/2020-x/>hoặc=(2019-x)+(x-2020)=-1

         Hay B> hoặc=-1

               =>B=1

\(\frac{2^{19}\cdot27^4}{6^7\cdot16^3}\)

\(=\frac{2^{19}\cdot3^{12}}{2^7\cdot3^7\cdot2^{12}}\)

\(=\frac{2^{19}\cdot3^{12}}{2^{19}\cdot3^7}\)

\(=3^5\)

\(=243\)

\(\frac{2^{19}.27^4}{6^7.16^3}\)

\(=\frac{2^{19}.\left(3^3\right)^4}{2^7.3^7.\left(2^4\right)^3}\)

\(=\frac{2^{19}.3^{12}}{2^7.3^7.2^{12}}\)

=\(3^5=243\)

sai đề 

Ví dụ : a = 3 , b = -2 . Ta có : \(\left|a+b\right|=\left|3+(-2)\right|=\left|1\right|=1\)

Còn \(\left|a\right|+\left|b\right|=3+\left|-2\right|=3+2=5\). Do đó \(\left|a+b\right|=\left|a\right|+\left|b\right|\)là sai vì \(1\ne5\).

Chúng ta còn nhận ra rằng :

\(\left|a-b\right|=\left|3-(-2)\right|=\left|5\right|=5\)

\(\left|a\right|-\left|b\right|=\left|3\right|-\left|-2\right|=3-2=1\)

Do vậy \(\left|a-b\right|=\left|a\right|-\left|b\right|\)là sai vì \(1\ne5\).

Cho ta bài toán tương tự.Với mọi x,y \(\inℚ\)thì \(\left|a-b\right|\ge\left|a\right|-\left|b\right|\).

Nhưng nếu cho x = 3 , y = 2 hoặc x = -3 , y = -2 ta lại thấy :

\(\left|a+b\right|=\left|a\right|+\left|b\right|\)là đúng . Vậy với điều kiện nào thì \(\left|a+b\right|=\left|a\right|+\left|b\right|\)?

Không khó khắn lắm chúng ta thấy rằng \(\left|a+b\right|=\left|a\right|+\left|b\right|\)khi \(ab\ge0\)