`#\text{nn71}`

`a,`

Ta có: `\text{AB // CD}`

\(\Rightarrow\widehat{\text{A}}+\widehat{\text{D}}=180^0\)

`=>`\(70^0 + \widehat{D} = 180^0\)

`=>`\(\widehat{D} = 180^0 - 70^0 = 110^0\)

`b,`

Vì `\text{AB // CD}`

Mà B `\in` Ax

`=> Ax // CD`

\(\Rightarrow\widehat{\text{C}}=\widehat{\text{xBC}}=70^0\left(\text{2 góc ở vị trí sole trong}\right)\)

Ta có: \(\widehat{\text{A}}=\widehat{\text{xBC}}=70^0\)

Mà `2` góc này ở vị trí đồng vị

`=> \text{AD // BC}`(tính chất).

`c,`

n của \(\text{Ca}\left(\text{OH}\right)_2\) trong phản ứng:

\(\text{n}_{\text{Ca}\left(\text{OH}\right)_2}=\text{C}_{\text{M}}\cdot\text{V}=5\cdot0,4=2\left(\text{mol}\right)\)

PTHH: \(\text{FeCl}_2+\text{Ca}\left(\text{OH}\right)_2\rightarrow\text{Fe}\left(\text{OH}\right)_2+\text{CaCl}_2\)

\(\rightarrow\text{n}_{\text{Ca}\left(\text{OH}\right)_2}=\text{n}_{\text{CaCl}_2}=2\text{ mol}\)

m của CaCl₂ thu được sau phản ứng:

\(\text{m}_{\text{CaCl}_2}=\text{n}_{\text{CaCl}_2}\cdot\text{M}_{\text{CaCl}_2}=2\cdot\left(40+35,5\cdot2\right)=222\left(\text{g}\right)\)

`#\text{dnnv}`

`a)`

n của Fe có trong phản ứng:

\(\text{n}_{\text{Fe}}=\dfrac{\text{m}_{\text{Fe}}}{\text{M}_{\text{Fe}}}=\dfrac{8,4}{56}=0,15\left(\text{mol}\right)\)

PTHH: \(\text{Fe + 2HCl}\rightarrow\text{FeCl}_2+\text{H}_2\)

\(\rightarrow\text{n}_{\text{Fe}}=\dfrac{1}{2}\text{n}_{\text{HCl}}\)

\(\text{n}_{\text{HCl}}=2\cdot0,15=0,3\left(\text{mol}\right)\)

Nồng độ mol HCl đã dùng:

\(\text{C}_{\text{M}}=\dfrac{\text{n}_{\text{HCl}}}{\text{V}_{\text{HCl}}}=\dfrac{0,3}{0,4}=0,75\left(\text{mol/L}\right)\)

`b)`

Từ PT: \(\text{n}_{\text{Fe}}=\text{n}_{\text{H}_2}=0,15\text{ mol}\)

V của H₂ sau phản ứng:

\(\text{V}_{\text{H}_2}=\text{n}_{\text{H}_2}\cdot24,79=0,15\cdot24,79=3,7185\left(\text{l}\right)\)

Bài nào bạn nhỉ? Mình thấy những bài bạn đăng đã được giải rồi.

Ta có: \(a+b+c=2p\Rightarrow p=\dfrac{a+b+c}{2}\)

- Xét VP:

\(4\left(p-b\right)\left(p-c\right)\)

\(=4\left(\dfrac{a+b+c}{2}-b\right)\left(\dfrac{a+b+c}{2}-c\right)\\ =2\left(\dfrac{a+b+c}{2}-b\right)\cdot2\left(\dfrac{a+b+c}{2}-c\right)\\ =\left(2\cdot\dfrac{a+b+c}{2}-2b\right)\left(2\cdot\dfrac{a+b+c}{2}-2c\right)\\ =\left(a+b+c-2b\right)\left(a+b+c-2c\right)\\ =\left(a-b+c\right)\left(a+b-c\right)\\ =\left[a-\left(b-c\right)\right]\left[a+\left(b-c\right)\right]\\ =a^2-\left(b-c\right)^2\\ =a^2-\left(b^2-2bc+c^2\right)\\ =a^2-b^2-c^2+2bc\left(\text{đpcm}\right).\)

Bạn cần giải bài nào?