Q = x^2 + 8x + 20

= (x^2 + 8x + 16) + 4

= (x+4)^2 + 4 ≥ 4 với mọi x

Dấu = xảy ra khi :

x+4=0 hay x = -4

VẬY MIN Q = 4 tại x = -4

Q = x²+ 8x + 20 

    = x²+ 2.4.x + 16+ 4

    = (x+4)²+4

Vì (x+4)² \(\ge\) 0 với mọi x \(\Rightarrow\) (x+4)²+ 4\(\ge\) 0+4 

                                        hay Q\(\ge\) 4

Dấu "=" xảy ra \(\Leftrightarrow\) (x+4)²=0 \(\Leftrightarrow\) x+4=0 \(\Leftrightarrow\) x= -4

 Vậy Q đạt giá trị nhỏ nhất khi x= -4

\(x^2-y^2+2x+1\\=(x^2+2x+1)-y^2\\=(x+1)^2-y^2\\=(x+1-y)(x+1+y)\)

a, Ta có: \(n_{Na}=\dfrac{2,3}{23}=0,1\left(mol\right)\)

\(m_{HCl}=60.7,3\%=4,38\left(g\right)\Rightarrow n_{HCl}=\dfrac{4,38}{36,5}=0,12\left(mol\right)\)

PT: \(2Na+2HCl\rightarrow2NaCl+H_2\)

Xét tỉ lệ: \(\dfrac{0,1}{2}< \dfrac{0,12}{2}\), ta được HCl dư.

Theo PT: \(n_{H_2}=\dfrac{1}{2}n_{Na}=0,05\left(mol\right)\Rightarrow V_{H_2}=0,05.24,79=1,2395\left(l\right)\)

b, \(n_{HCl\left(pư\right)}=n_{NaCl}=n_{Na}=0,1\left(mol\right)\)

\(\Rightarrow n_{HCl\left(dư\right)}=0,12-0,1=0,02\left(mol\right)\)

Ta có: m _{dd sau pư} = 2,3 + 60 - 0,05.2 = 62,2 (g)

\(\Rightarrow\left\{{}\begin{matrix}C\%_{NaCl}=\dfrac{0,1.58,5}{62,2}.100\%\approx9,41\%\\C\%_{HCl}=\dfrac{0,02.36,5}{62,2}.100\%\approx1,17\%\end{matrix}\right.\)

5.66,46

bn diễn ra đc ko vì nó cần tìm x

\(4(x-1)^2-25(2-3x)\\=4(x^2-2x+1)-50+75x\\=4x^2-8x+4-50+75x\\=4x^2+67x-46\)

cảm ơn bn nhìu

a) \(\left(2x+5\right)^2-8^2\)

\(=\left(2x+5-8\right)\left(2x+5+8\right)\)

\(=\left(2x-3\right)\left(2x+13\right)\)

b) \(\left(2x-1\right)^2-\left(3x-1\right)^2\)

\(=\left[\left(2x-1\right)-\left(3x-1\right)\right]\left[\left(2x-1\right)+\left(3x-1\right)\right]\)

\(=\left(2x-1-3x+1\right)\left(2x-1+3x-1\right)\)

\(=-x\left(5x-2\right)\)

1 June is not as hot as July

2 Although it was a comedy, most audience fell asleep

3 When they got divorced, they were married for five years

4 John was very sad because he failed the final exam

1 Cấu trúc so sánh ngang bằng: S1 + be as + adj + as S2

3 When sẽ đưa vào cùng với mệnh đề có hành động xen vào