2 . 3/4 - ( x + 1/2 ) = 4/5

   3/2 - ( x + 1/2 ) =4/5

             x + 1/2 = 4/5 - 3/2 

             x + 1/2 = 7/10

             x          = 7/10 -1/2

             x          = 1/5

cho k lấy ko

|2x - 2| - 5 = 2x

<=> |2x - 2| = 2x + 5

Đk : 2x + 5 \(\ge\)0 

<=> 2x \(\ge\)-5

<=> x\(\ge\)\(\frac{-5}{2}\)

Khi đó : 2x - 2 = 2x + 5   or   2x - 2 = -2x - 5

<=>  2x - 2 - 2x = 5         or   2x - 2 + 2x = -5

<=>   -2 = 5 (vô lý)          or    4x - 2  = -5

<=> x \(\in\varnothing\)              or     4x = -3

<=>                                            x = \(\frac{-3}{4}\)(t/m ĐK)

Vậy x = -3/4

ADTCCDTSBN Ta có

\(\frac{b+c+d}{a}+\frac{a+c+d}{b}+\frac{a+b+d}{c}+\frac{a+b+c}{d}\)

\(=\frac{b+c+d+a+c+d+a+b+d+a+b+c}{a+b+c+d}=3\)

\(=>k=3\)

Thay vào M Ta có:

\(M=\left(k-3\right)^{2019}=\left(3-3\right)^{2019}=0\)

\(=>M=0\)

P/S:Ko chắc~!!

Áp dụng TC của dãy tỉ số bằng nhau , ta có :

\(\frac{b+c+d}{a}=\frac{a+c+d}{b}=\frac{a+b+d}{c}=\frac{a+b+c}{d}\left(1\right)\)

\(=\frac{b+c+d+a+c+d+a+b+d+a+b+c}{a+b+c+d}\)

\(=\frac{3a+3b+3c+3d}{a+b+c+d}=\frac{3\left(a+b+c+d\right)}{a+b+c+d}=3\)

Mà \(\left(1\right)=k\Rightarrow k=3\)

Ta có : \(M=\left(k-3\right)^{2019}\)

\(\Leftrightarrow M=\left(3-3\right)^{2019}\)

\(\Leftrightarrow M=0\)

đề sai

\(\frac{1}{2005.2004}-\frac{1}{2004.2003}-.....-\frac{1}{2}\)

a)A=2+x-3x+1

A=3-2x

b)0,4x-4x+12

=12-3,6x

|x+5/3 | -1/2=3/4

|x-5/3| = 3/4 +1/2

|x-5/3|=3/4 + 2/4

|x-5/3|=5/4

=> x- 5/3=5/4 hay x-5/3=-5/4

x= 5/4+5/3            x=-5/4+5/3

=> x=35/12          x=5/12

Áp dụng t/c của dãy tỉ số bằng nhau, ta có:

 \(\frac{b+c+d}{a}=\frac{a+c+d}{b}=\frac{a+b+d}{c}=\frac{a+b+c}{d}=k\)

\(=\frac{b+c+d+a+c+d+a+b+d+a+b+c}{a+b+c+d}\)

= \(\frac{3b+3c+3a+3d}{a+b+c+d}=\frac{3\left(a+b+c+d\right)}{a+b+c+d}=3\)(Do a + b + c + d \(\ne\)0)

=> k = 3

Với k = 3 => M = (3 - 3)²⁰¹⁹ = 0

I love 3000

gọi 3 cạnh tam giác lần lượt là a b c

theo gt ta có a / 3 = b / 5 = c / 7

áp dụng dãy tỉ số bằng nhau , ta có a / 3 = b / 5 = c / 7 = a + b + c / 3 + 5 + 7 = 45 / 15 = 3

=> a / 3 = 3    => a = 3 * 3 = 9

     b / 5 = 3          b = 3 * 5 = 15

     c / 7 = 3          c = 3 * 7 = 21

Ta có:

\(x^2+y^2=1\Rightarrow\left(x^2+y^2\right)^2=1\)(1)

Thay (1) vào \(\frac{x^4}{a}+\frac{y^4}{b}=\frac{1}{a+b}\)ta có:

\(\frac{x^4}{a}+\frac{y^4}{b}=\frac{\left(x^2+y^2\right)^2}{a+b}\Leftrightarrow\frac{x^4b+y^4a}{ab}=\frac{x^4+2x^2y^2+y^4}{a+b}\)

\(\Leftrightarrow\left(x^4b+y^4a\right)\left(a+b\right)=\left(x^4+2x^2y^2+y^4\right).ab\)

\(\Leftrightarrow x^4ab+x^4b^2+y^4a^2+y^4ab=x^4ab+2x^2y^2ab+y^4ab\)

\(\Leftrightarrow x^4b^2+y^4a^2=2x^2y^2ab\)

\(\Leftrightarrow\left(x^2b\right)^2-2x^2y^2ab+\left(y^2a\right)^2=0\)

\(\Leftrightarrow\left(x^2b-y^2a\right)^2=0\)

\(\Leftrightarrow x^2b-y^2a=0\)

\(\Leftrightarrow x^2b=y^2a\)

\(\Rightarrow\frac{x^2}{a}=\frac{y^2}{b}=\frac{x^2+y^2}{a+b}=\frac{1}{a+b}\)

\(\Rightarrow\left(\frac{x^2}{a}\right)^{1002}=\left(\frac{y^2}{b}\right)^{1002}=\left(\frac{1}{a+b}\right)^{1002}\)

\(\Rightarrow\frac{x^{2004}}{a^{1002}}=\frac{y^{2004}}{b^{1002}}=\frac{1}{\left(a+b\right)^{1002}}\)

\(\Rightarrow\frac{x^{2004}}{a^{1002}}+\frac{y^{2004}}{b^{1002}}=\frac{1}{\left(a+b\right)^{1002}}+\frac{1}{\left(a+b\right)^{1002}}=\frac{2}{\left(a+b\right)^{1002}}\left(đpcm\right)\)

Chúc bạn học tốt!