Câu 1

a, CTHH ở dạng tổng quát :

XH4 ( 4 là chỉ số nha )

b, Theo bài ta có:

X + 4H = 16 ( vì NTK của Oxi = 16)

Suy ra: X = 12

X là Cacbon ( C )

CTHH của hợp chất là CH4 ( 4 là chỉ số nha )

Xin cảm ơn

Hẳn là khai triển...

Ta có: \(\left(a+b+c\right)\left(a-b+c\right)\)

\(=\left[\left(a+c\right)+b\right]\left[\left(a+c\right)-b\right]\)

\(=\left(a+c\right)^2-b^2\)

\(=a^2+2ac+c^2-b^2\)

\(x^{16}-1\)

\(=x^{16}+x^8-x^8-1\)

\(=x^8\left(x^8+1\right)-\left(x^8+1\right)\)

\(=\left(x^8+1\right)\left(x^8-1\right)\)

\(=\left(x^8+1\right)\left(x^8-x^4+x^4-1\right)\)

\(=\left(x^8+1\right)\left[x^4\left(x^4-1\right)+\left(x^4-1\right)\right]\)

\(=\left(x^8+1\right)\left(x^4+1\right)\left(x^4-1\right)\)

\(=\left(x^8+1\right)\left(x^4+1\right)\left(x^4-x^2+x^2-1\right)\)

\(=\left(x^8+1\right)\left(x^4+1\right)\left[x^2\left(x^2-1\right)+\left(x^2-1\right)\right]\)

\(=\left(x^8+1\right)\left(x^4+1\right)\left(x^2+1\right)\left(x^2-1\right)\)

\(=\left(x^8+1\right)\left(x^4+1\right)\left(x^2+1\right)\left(x^2-x+x-1\right)\)

\(=\left(x^8+1\right)\left(x^4+1\right)\left(x^2+1\right)\left[x\left(x-1\right)+\left(x-1\right)\right]\)

\(=\left(x^8+1\right)\left(x^4+1\right)\left(x^2+1\right)\left(x+1\right)\left(x-1\right)\)

Ta có : x² + 7x + 12 = 0

=> x² + 4x + 3x + 12 = 0

=> x(x + 4) + 3(x + 4) = 0

=> (x + 3)(x + 4) = 0

=> \(\orbr{\begin{cases}x+3=0\\x+4=0\end{cases}}\Rightarrow\hept{\begin{cases}x=-3\\x=-4\end{cases}}\)

Vậy \(x\in\left\{-3;-4\right\}\)

x² + 7x + 12 = 0

<=> x² + 3x + 4x + 12 = 0

<=> x( x + 3 ) + 4( x + 3 ) = 0

<=> ( x + 3 )( x + 4 ) = 0

<=> \(\orbr{\begin{cases}x+3=0\\x+4=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-3\\x=-4\end{cases}}\)

Vậy S = { -3 ; -4 }

(x - 1)² - (x + 2)(x - 5) + (4x - 1)² = (-x - 1)(2 - 16x)

=> x² - 2x + 1 - x² + 5x - 2x + 10 + 16x² - 8x + 1 = -2x + 16x² -2 + 16x

=> 16x² - 7x + 12 = 14x - 2  + 16x²

=> -21x = -14

=> 21x = 14

=> x = 2/3

\(\left(x-1\right)^2-\left(x+2\right)\left(x-5\right)+\left(4x-1\right)^2=\left(-x-1\right)\left(2-16x\right)\)

\(\Rightarrow x^2-2x+1-x^2+3x+10+16x^2-8x+1-16x^2-14x+2=0\)

\(\Rightarrow\left(x^2-x^2+16x^2-16x^2\right)+\left(-2x+3x-8x-14x\right)+\left(1+10+1+2\right)=0\)

\(\Rightarrow-21x+14=0\)

\(\Rightarrow-21x=-14\)

\(\Rightarrow x=\frac{2}{3}\)

Bài làm:

Ta có: \(E=5x^2+y^2-4xy+8x-6y+3\)

\(E=\left(4x^2-4xy+y^2\right)+\left(12x-6y\right)+9+\left(x^2-4x+4\right)-10\)

\(E=\left(2x-y\right)^2+6\left(2x-y\right)+9+\left(x-2\right)^2-10\)

\(E=\left(2x-y+3\right)^2+\left(x-2\right)^2-10\ge-10\left(\forall x,y\right)\)

Dấu "=" xảy ra khi: \(\hept{\begin{cases}\left(2x-y+3\right)^2=0\\\left(x-2\right)^2=0\end{cases}}\Rightarrow\hept{\begin{cases}x=2\\y=7\end{cases}}\)

Vậy Min(E) = -10 khi x = 2, y = 7

tìm GTNN/GTLN nha!

\(D=2x^2+9y^2-6xy-6x-12y+2004\)

\(=\left(x^2-6xy+9y^2\right)+\left(4x-12y\right)+4+\left(x^2-10x+25\right)+1975\)

\(=\left(x-3y\right)^2+4\left(x-3y\right)+4+\left(x-5\right)^2+1975\)

\(=\left(x-3y+2\right)^2+\left(x-5\right)^2+1975\)

Vì \(\left(x-3y+2\right)^2\ge0\forall x,y\)

\(\left(x-5\right)^2\ge0\forall x\)

\(\Rightarrow\left(x-3y+2\right)^2+\left(x-5\right)^2\ge0\forall x,y\)

\(\Rightarrow\left(x-3y+2\right)^2+\left(x-5\right)^2+1975\ge1975\forall x,y\)

Dấu " = " xảy ra \(\Leftrightarrow\hept{\begin{cases}x-3y+2=0\\x-5=0\end{cases}}\Leftrightarrow\hept{\begin{cases}7-3y=0\\x=5\end{cases}}\Leftrightarrow\hept{\begin{cases}3y=7\\x=5\end{cases}}\Leftrightarrow\hept{\begin{cases}y=\frac{7}{3}\\x=5\end{cases}}\)

Vậy \(minD=1975\)\(\Leftrightarrow\hept{\begin{cases}x=5\\y=\frac{7}{3}\end{cases}}\)

Sửa đề:64x³ + 48x² + 12x + 1 = 27

64x³ + 48x² + 12x + 1 = 27

(4x+1)³=3³

4x+1=3

4x=2

x=1/2