a) Cacbonnic + canxi hidroxit \(\Rightarrow\)canxi cacbonat.

b) Hidro peoxit \(\Rightarrow\)nước + oxi

c) Canxi cacbonat \(\Rightarrow\)Canxi oxit + cacbobic

mik trả lời hơi chậm, nhưng cũng làm bài cho bn rồi, nên bn nhớ giữ lời hứa k cho mik 3 k đấy nhé bn.

học tốt

mik chưa học, bạn nhek, lên google đi bn

O_O Ể, lp 7 có môn này à?

AB // BC ???

Đúng đề rồi bạn

\(\left|x-\frac{1}{2}\right|+\left|y+\frac{2}{3}\right|+\left|x^2+xz\right|=0\)

\(\Leftrightarrow\)\(\left|x-\frac{1}{2}\right|+\left|y+\frac{2}{3}\right|+\left|x\left(x+z\right)\right|=0\)

Ta có : 

\(\left|x-\frac{1}{2}\right|\ge0\)

\(\left|y+\frac{2}{3}\right|\ge0\)

\(\left|x\left(x+z\right)\right|\ge0\)

Mà \(\left|x-\frac{1}{2}\right|+\left|y+\frac{2}{3}\right|+\left|x\left(x+z\right)\right|=0\)

\(\Rightarrow\)\(\hept{\begin{cases}\left|x-\frac{1}{2}\right|=0\\\left|y+\frac{2}{3}\right|=0\\\left|x\left(x+z\right)\right|=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=\frac{1}{2}\\y=\frac{-2}{3}\\z=\frac{-1}{2}\end{cases}}}\)

Vậy \(x=\frac{1}{2}\)\(;\)\(y=\frac{-2}{3}\) và \(z=\frac{-1}{2}\)

Chúc bạn học tốt ~ 

Ta có: (2x-1)²⁰¹⁸≥0 ; (y-2/5)²⁰¹⁸≥0 ; |x+y-z|≥0

=>\(\hept{\begin{cases}\left(2x-1\right)^{2018}=0\\\left(y-\frac{2}{5}\right)^{2018}=0\\\left|x+y-z\right|=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=\frac{1}{2}\\y=\frac{2}{5}\\z=\frac{9}{10}\end{cases}}}\)

Chúc bạn học tốt!

Ta có : 

\(\left(2x-1\right)^{2018}\ge0\)

\(\left(y-\frac{2}{5}\right)^{2018}\ge0\)

\(\left|x+y-z\right|\ge0\)

Mà \(\left(2x-1\right)^{2018}+\left(y-\frac{2}{5}\right)^{2018}+\left|x+y-z\right|=0\) ( Giả thiết ) 

\(\Rightarrow\)\(\hept{\begin{cases}\left(2x-1\right)^{2018}=0\\\left(y-\frac{2}{5}\right)^{2018}=0\\\left|x+y-z\right|=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=\frac{1}{2}\\y=\frac{2}{5}\\z=\frac{9}{10}\end{cases}}}\)

Vậy \(x=\frac{1}{2}\)\(;\)\(y=\frac{2}{5}\) và \(z=\frac{9}{10}\)

Chúc bạn học tốt ~ 

\(\frac{x+2}{327}+\frac{x+3}{326}+\frac{x+4}{325}+\frac{x+5}{324}+\frac{x+329}{5}=0\)

\(\Leftrightarrow\)\(\left(\frac{x+2}{327}+1\right)+\left(\frac{x+3}{326}+1\right)+\left(\frac{x+4}{325}+1\right)+\left(\frac{x+5}{324}+1\right)+\frac{x+329}{5}=0\)

\(\Leftrightarrow\)\(\frac{x+2+327}{327}+\frac{x+3+326}{326}+\frac{x+4+325}{325}+\frac{x+5+324}{324}+\frac{x+329}{5}=0\)

\(\Leftrightarrow\)\(\frac{x+329}{327}+\frac{x+329}{326}+\frac{x+329}{325}+\frac{x+329}{324}+\frac{x+329}{5}=0\)

\(\Leftrightarrow\)\(\left(x+329\right)\left(\frac{1}{327}+\frac{1}{326}+\frac{1}{325}+\frac{1}{324}+\frac{1}{5}\right)=0\)

Vì \(\frac{1}{327}+\frac{1}{326}+\frac{1}{325}+\frac{1}{324}+\frac{1}{5}\ne0\)

Nên \(x+329=0\)

\(\Leftrightarrow\)\(x=-329\)

Vậy \(x=-329\)

Chúc bạn học tốt ~ 

\(\frac{x+2}{327}+\frac{x+3}{326}+\frac{x+4}{325}+\frac{x+5}{324}+\frac{x+329}{5}=0\)

\(\Rightarrow\frac{x+2}{327}+\frac{x+3}{326}+\frac{x+4}{325}+\frac{x+5}{324}+\frac{x+329+20}{5}=0\)

\(\Rightarrow\frac{x+2}{327}+\frac{x+3}{326}+\frac{x+4}{325}+\frac{x+5}{324}+\frac{x+329}{5}+4=0\)

\(\Rightarrow\frac{x+2}{327}+1+\frac{x+3}{326}+1+\frac{x+4}{325}+1+\frac{x+5}{324}+1+\frac{x+329}{5}=0\)

\(\Rightarrow\frac{x+329}{327}+\frac{x+329}{326}+\frac{x+329}{325}+\frac{x+329}{324}+\frac{x+329}{5}=0\)

\(\Rightarrow\left(x+329\right)\left(\frac{1}{327}+\frac{1}{326}+\frac{1}{325}+\frac{1}{324}+\frac{1}{5}\right)=0\)

\(\Rightarrow x+329=0\Leftrightarrow\left(\frac{1}{327}+\frac{1}{326}+\frac{1}{325}+\frac{1}{324}+\frac{1}{5}\right)\ne0\)

\(\Rightarrow x=-329\)

Vậy \(\Rightarrow x=-329\)

a, 10⁸ . 2⁸ = ( 2.5 )⁸ . 2⁸

                = 2⁸ . 5⁸ .2⁸

               = 2¹⁶ . 5⁸

b, 27² : 25³ = ( 3³ )² : ( 5² )³

                  = 3⁶ : 5⁶

Bài làm

a) 27² : 25³ =