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

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

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

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

\(x^{16}-1\)

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

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

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

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

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

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

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

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

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

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

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

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

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