1) (3x² + \(\dfrac{y}{4}\) )³ = (3x²)³ + 3.(3x²)². \(\dfrac{y}{4}\) + 3.3x² . \(\dfrac{y^2}{16}\) + \(\dfrac{y^3}{64}\)

2) (4 - 26)³ = 4³ - 3.4².26 + 3.4.26² - 26³

3) (2a + 3b)² = (2a)² + 2.2a.3b + (3b)²

4) (\(\dfrac{b}{2}\) - 5).(\(\dfrac{b}{2}\) + 5) = \(\left(\dfrac{b}{2}\right)^2\) - 5²

a.

\(\left(x^2+x\right)^2-6x^2-6x+8=0\)

\(\Leftrightarrow\left(x^2+x\right)^2-6\left(x^2+x\right)+8=0\)

\(\Leftrightarrow\left(x^2+x\right)^2-2\left(x^2+x\right)-4\left(x^2+x\right)+8=0\)

\(\Leftrightarrow\left(x^2+x\right)\left(x^2+x-2\right)-4\left(x^2+x-2\right)=0\)

\(\Leftrightarrow\left(x^2+x-2\right)\left(x^2+x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2+x-2=0\\x^2+x-4=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\\x=\dfrac{-1\pm\sqrt{17}}{2}\end{matrix}\right.\)

b.

\(\left(x^2+10x+16\right)\left(x^2+10x+24\right)-20=0\)

\(\Leftrightarrow\left(x^2+10x+16\right)\left(x^2+10x+16+8\right)-20=0\)

\(\Leftrightarrow\left(x^2+10x+16\right)^2+8\left(x^2+10x+16\right)-20=0\)

\(\Leftrightarrow\left(x^2+10x+16\right)^2-2\left(x^2+10x+16\right)+10\left(x^2+10x+16\right)-20=0\)

\(\Leftrightarrow\left(x^2+10x+16\right)\left(x^2+10x+14\right)+10\left(x^2+10x+14\right)=0\)

\(\Leftrightarrow\left(x^2+10x+14\right)\left(x^2+10x+26\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2+10x+14=0\\x^2+10x+26=0\left(vn\right)\end{matrix}\right.\)

\(\Leftrightarrow x=-5\pm\sqrt{11}\)

\(\Leftrightarrow x^4+2x^3+2x^2-2x^2-4x-4=0\)

\(\Leftrightarrow x^2\left(x^2+2x+2\right)-2\left(x^2+2x+2\right)=0\)

\(\Leftrightarrow\left(x^2-2\right)\left(x^2+2x+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2-2=0\\x^2+2x+2=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2=2\\\left(x+1\right)^2+1=0\left(\text{vô nghiệm}\right)\end{matrix}\right.\) 

\(\Leftrightarrow x=\pm\sqrt{2}\)

Gọi CTHH của hợp chất M là $N_xO_y$
Ta có : 

$\dfrac{m_N}{m_O} = \dfrac{14x}{16y} = \dfrac{7}{16}$

và $PTK = 14x + 16y = 46$

Suy ra : x = 1 ; y = 2

Vậy CTHH của M là $NO_2$

\(CaSiO_3:20+14+8.3=58\\ H_3PO_4:3+15+8.4=50\\ Al_2\left(SO_4\right)_3:13.2+\left(16+8.4\right).3=170\\ KMnO_4:19+25+8.4=76\\ Na_2S:11.2+16=38\\ BaCl_2:56+17.2=90\)

bạn viết tiếng việt rồi đổi thành tiếng anh đi =))

a b c là dấu cộng ph k bạn?

1/a 1/b nữa là dấu gì vậy bạn?

\(7x^2-x+1=7\left(x^2-\dfrac{x}{7}+\dfrac{1}{196}\right)+\dfrac{27}{28}\)

\(=7\left(x-\dfrac{1}{14}\right)^2+\dfrac{27}{28}\ge\dfrac{27}{28}\forall x\)

\(Min=\dfrac{27}{28}\Leftrightarrow x=\dfrac{1}{14}\)

A=   7x² - x + 1 

A=  7( x² - 2.1/14x + 1/196) + 27/28 

A= 7(x - 1/14)² + 27/28

A = 7(x - 1/14)² ≥ 0  ⇔ 7(x-1/14)² +27/28 ≥ 27/28 

A(min)= 27/28 ⇔ x = 1/14