\(a^{2020}+b^{2020}+c^{2020}=a^{1010}b^{1010}+b^{1010}c^{1010}+c^{1010}a^{1010}\)

\(\Leftrightarrow a^{2020}+b^{2020}+c^{2020}-a^{1010}b^{1010}-b^{1010}c^{1010}-c^{1010}a^{1010}=0\)

\(\Leftrightarrow2a^{2020}+2b^{2020}+2c^{2020}-2a^{1010}b^{1010}-2b^{1010}c^{1010}-2a^{1010}c^{1010}=0\)

\(\Leftrightarrow\left(a^{2020}-2a^{1010}b^{1010}+b^{2020}\right)+\left(b^{2020}-2b^{1010}c^{1010}+c^{2020}\right)+\left(c^{2020}-2a^{1010}c^{1010}+a^{2020}\right)=0\)

\(\Leftrightarrow\left(a^{1010}-b^{1010}\right)^2+\left(b^{1010}-c^{1010}\right)^2+\left(c^{1010}-a^{1010}\right)^2=0\)

\(\Leftrightarrow\hept{\begin{cases}\left(a^{1010}-b^{1010}\right)=0\\b^{1010}-c^{1010}=0\\c^{1010}-a^{1010}=0\end{cases}}\Leftrightarrow a^{1010}=b^{1010}=c^{1010}\Leftrightarrow\pm a=\pm b=\pm c\)

Rồi thay :> Còn thay kiểu nào thì mình cũng hong biết :">

PTHH : Zn + 2HCl -> ZnCl2 + H2

a, n Zn = m/M = 3,25/65 = 0,05 ( mol ) 

Theo pthh : n Zn/n H2 = 1/1 ; n Zn/n ZnCl2 = 1/1

=> n Zn = n H2 = n ZnCl2 = 0,05 mol

=> V H2 = n.22,4 = 0,05x22,4= 1,12 ( l )

b, m ZnCl2 = n.M = 0,05 . (65+2.35,5) = 6,8(g)

KL

3|x-1| < x+3

3x-3 < x+3(nếu |x+1| >0)

3x-x<3+3

2x < 6

X<3

Chuc bạn hok tốt !!!!!

Nhớ h cho Minh nha

x - 212 - 2x + 13 = 3x + 56 - x

x - 2x - 3x + x = 212 - 13 + 56

-3x = 255

x = - 85

theo đề bài, ta có:

x-212-2x+13=3x+56-x

\(\Rightarrow\)x-2x-3x+x=212+56-13

\(\Rightarrow\)-3x=255

\(\Rightarrow\)x=\(\frac{255}{-3}\)

\(\Rightarrow\)x=-85

Ta có:

\(\left(2x-1\right)^2+\left(x-2\right)\left(x+2\right)=\left(5x+1\right)\left(x-4\right)-12\)

\(\left(4x^2-4x+1\right)+x^2-4=5x^2-19x-4-12\)

\(5x^2-4x-3=5x^2-19x-16\)

\(\left(5x^2-5x^2\right)+\left(19x-4x\right)+\left(16-3\right)=0\)\(15x+13=0\)\(x=-\frac{13}{15}\)