a) Ta có: \(\left(a^2+b^2\right)^2-4a^2b^2=\left(a^2+b^2\right)^2-\left(2ab\right)^2\)

\(=\left(a^2+b^2-2ab\right)\left(a^2+b^2+2ab\right)=\left(a-b\right)^2.\left(a+b\right)^2\)( đpcm )

b) Ta có: \(\left(a-b\right)^3+\left(b-c\right)^3+\left(c-a\right)^3-3\left(a-b\right)\left(b-c\right)\left(c-a\right)\)

\(=\left(a-b+b-c\right)^3-3\left(a-b\right)\left(b-c\right)\left(a-b+b-c\right)+\left(c-a\right)^3\)

\(-3\left(a-b\right)\left(b-c\right)\left(c-a\right)\)

\(=\left(a-c\right)^3-3\left(a-b\right)\left(b-c\right)\left(a-c\right)+\left(c-a\right)^3-3\left(a-b\right)\left(b-c\right)\left(c-a\right)\)

\(=\left(a-c\right)^3+\left(c-a\right)^3-3\left(a-b\right)\left(b-c\right)\left(a-c\right)-3\left(a-b\right)\left(b-c\right)\left(c-a\right)\)

\(=\left(a-c\right)^3-\left(a-c\right)^3+3\left(a-b\right)\left(b-c\right)\left(c-a\right)-3\left(a-b\right)\left(b-c\right)\left(c-a\right)=0\)

\(\Rightarrow\left(a-b\right)^3+\left(b-c\right)^3+\left(c-a\right)^3=3\left(a-b\right)\left(b-c\right)\left(c-a\right)\)( đpcm )

1) Ta có: \(\left(a^2+b^2\right)^2-4a^2b^2\)

\(=a^4+2a^2b^2+b^4-4a^2b^2\)

\(=a^4-2a^2b^2+b^4\)

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

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

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

2) Ta có: \(\left(a-b\right)^3+\left(b-c\right)^3+\left(c-a\right)^3\)

\(=\left(a-b+b-c\right)\left[\left(a-b\right)^2-\left(a-b\right)\left(b-c\right)+\left(b-c\right)^2\right]+\left(c-a\right)^3\)

\(=\left(a-c\right)\left(a^2-2ab+b^2-ab+ac+b^2-bc+b^2-2bc+c^2\right)+\left(c-a\right)^3\)

\(=-\left(c-a\right)\left(a^2+3b^2+c^2-3ab+ac-3bc\right)+\left(c-a\right)\left(c^2-2ca+a^2\right)\)

\(=\left(c-a\right)\left(c^2-2ca+a^2-a^2-3b^2-c^2+3ab-ac+3bc\right)\)

\(=\left(c-a\right)\left(3ab+3bc-3b^2-3ac\right)\)

\(=3\left(c-a\right)\left(ab-b^2-ac+bc\right)\)

\(=3\left(c-a\right)\left[b\left(a-b\right)-c\left(a-b\right)\right]\)

\(=3\left(a-b\right)\left(b-c\right)\left(c-a\right)\)

a) \(\frac{2x\left(3x-5\right)}{x^2+1}< 0\)

Ta có \(x^2+1\ge1>0\forall x\)

Để bpt < 0 => 2x( 3x - 5 ) < 0

Xét hai trường hợp :

1/ \(\hept{\begin{cases}2x>0\\3x-5< 0\end{cases}}\Leftrightarrow\hept{\begin{cases}x>0\\x< \frac{5}{3}\end{cases}\Rightarrow}0< x< \frac{5}{3}\)

2. \(\hept{\begin{cases}2x< 0\\3x-5>0\end{cases}}\Leftrightarrow\hept{\begin{cases}x< 0\\x>\frac{5}{3}\end{cases}}\)( loại )

Vậy nghiệm của bất phương trình là 0 < x < 5/3

b) \(\frac{x}{x-2}+\frac{x+2}{x}>2\)( ĐKXĐ : \(x\ne0,x\ne2\))

<=> \(\frac{x}{x-2}+\frac{x+2}{x}-2>0\)

<=> \(\frac{x^2}{x\left(x-2\right)}+\frac{\left(x+2\right)\left(x-2\right)}{x\left(x-2\right)}-\frac{2x\left(x-2\right)}{x\left(x-2\right)}>0\)

<=> \(\frac{x^2+x^2-4-2x^2+4x}{x\left(x-2\right)}>0\)

<=> \(\frac{4x-4}{x\left(x-2\right)}>0\)

\(x\left(x-2\right)>0\Leftrightarrow\orbr{\begin{cases}x>2\\x< 0\end{cases}}\)

\(x\left(x-2\right)< 0\Leftrightarrow0< x< 2\)

Xét các trường hợp

1/ \(\hept{\begin{cases}4x-4>0\\x\left(x-2\right)>0\end{cases}}\)

+) \(\hept{\begin{cases}4x-4>0\\x>2\end{cases}}\Leftrightarrow\hept{\begin{cases}x>1\\x>2\end{cases}}\Leftrightarrow x>2\)

+) \(\hept{\begin{cases}4x-4>0\\x< 0\end{cases}}\Leftrightarrow\hept{\begin{cases}x>1\\x< 0\end{cases}}\)( loại )

2/ \(\hept{\begin{cases}4x-4< 0\\x\left(x-2\right)< 0\end{cases}}\Rightarrow\hept{\begin{cases}x< 1\\0< x< 2\end{cases}}\Rightarrow0< x< 1\)

Vậy nghiệm của bất phương trình là x > 2 hoặc 0 < x < 1

c) \(\frac{2x-3}{x+5}\ge3\)( ĐKXĐ : \(x\ne-5\))

\(\Leftrightarrow\frac{2x-3}{x+5}-3\ge0\)

\(\Leftrightarrow\frac{2x-3}{x+5}-\frac{3\left(x+5\right)}{\left(x+5\right)}\ge0\)

\(\Leftrightarrow\frac{2x-3-3x-15}{x+5}\ge0\)

\(\Leftrightarrow\frac{-x-18}{x+5}\ge0\)

Xét hai trường hợp

1/ \(\hept{\begin{cases}-x-18\ge0\\x+5>0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\le-18\\x>-5\end{cases}}\)( loại )

2/ \(\hept{\begin{cases}-x-18\le0\\x+5< 0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ge-18\\x< -5\end{cases}}\Leftrightarrow-18\le x< -5\)

Vậy nghiệm của bất phương trình là \(-18\le x< -5\)

d) \(\frac{x-1}{x-3}>1\)( ĐKXĐ : \(x\ne3\))

\(\Leftrightarrow\frac{x-1}{x-3}-1>0\)

\(\Leftrightarrow\frac{x-1}{x-3}-\frac{x-3}{x-3}>0\)

\(\Leftrightarrow\frac{x-1-x+3}{x-3}>0\)

\(\Leftrightarrow\frac{2}{x-3}>0\)

\(\Leftrightarrow x-3>0\)

\(\Leftrightarrow x>3\)

Vậy nghiệm của bất phương trình là x > 3

a) Ta có: \(\left(2x+1\right)^2+\left(1-x\right)3x\le\left(x+2\right)^2\)

\(\Leftrightarrow x^2+4x+4\ge4x^2+4x+1+3x-3x^2\)

\(\Leftrightarrow x^2+4x+4\ge x^2+7x+1\)

\(\Leftrightarrow3\ge3x\)

\(\Rightarrow x\le1\)

b) Ta có: \(\left(x-4\right)\left(x+4\right)\ge\left(x+3\right)^2+5\)

\(\Leftrightarrow x^2-16\ge x^2+6x+9+5\)

\(\Leftrightarrow6x\le-30\)

\(\Leftrightarrow x\le-5\)

a) ( 2x + 1 )² + ( 1 - x )3x ≤ ( x + 2 )²

<=> 4x² + 4x + 1 + 3x - 3x² ≤ x² + 4x + 4

<=> 4x² + 4x + 3x - 3x² - x² - 4x ≤ 4 - 1

<=> 3x ≤ 3

<=> x ≤ 1

b) ( x - 4 )( x + 4 ) ≥ ( x + 3 )² + 5

<=> x² - 16 ≥ x² + 6x + 9 + 5

<=> x² - x² - 6x ≥ 9 + 5 + 16

<=> -6x ≥ 30

<=> x ≤ -5

Bài 1.

a) ( 7x - 3 )² - 5x( 9x + 2 ) - 4x² = 18

<=> 49x² - 42x + 9 - 45x² - 10x - 4x² = 18

<=> -52x + 9 = 18

<=> -52x = 9

<=> x = -9/52 

b) ( x - 7 )² - 9( x + 4 )² = 0

<=> x² - 14x + 49 - 9( x² + 8x + 16 ) = 0

<=> x² - 14x + 49 - 9x² - 72x - 144 = 0

<=> -8x² - 86x - 95 = 0 

<=> -8x² - 10x - 76x - 95 = 0

<=> -8x( x + 5/4 ) - 76( x + 5/4 ) = 0

<=> ( x + 5/4 )( -8x - 76 ) = 0

<=> \(\orbr{\begin{cases}x+\frac{5}{4}=0\\-8x-76=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-\frac{5}{4}\\x=-\frac{19}{2}\end{cases}}\)

c) ( 2x + 1 )² + ( 4x - 1 )( x + 5 ) = 36

<=> 4x² + 4x + 1 + 4x² + 19x - 5 = 36

<=> 8x² + 23x - 4 - 36 = 0

<=> 8x² + 23x - 40 = 0

=> Vô nghiệm ( lớp 8 chưa học nghiệm vô tỉ nghen ) :))

Bài 2.

a) x² - 12x + 39 = ( x² - 12x + 36 ) + 3 = ( x - 6 )² + 3 ≥ 3 > 0 ∀ x ( đpcm )

b) 17 - 8x + x² = ( x² - 8x + 16 ) + 1 = ( x - 4 )² + 1 ≥ 1 > 0 ∀ x ( đpcm )

c) -x² + 6x - 11 = -( x² - 6x + 9 ) - 2 = -( x - 3 )² - 2 ≤ -2 < 0 ∀ x ( đpcm )

d) -x² + 18x - 83 = -( x² - 18x + 81 ) - 2 = -( x - 9 )² - 2 ≤ -2 < 0 ∀ x ( đpcm )

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

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

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

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

\(x^{16}-1\)

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

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

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

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

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

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

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

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

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

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

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

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