\(x^2+9y^2+2x-6y+2=0\)

\(\Leftrightarrow x^2+2x+1+9y^2-6y+1=0\)

\(\Leftrightarrow\left(x+1\right)^2+\left(3y-1\right)^2=0\)

Đẳng thức xảy ra khi \(x=-1;y=\frac{1}{3}\)

tìm giá trị nhỏ nhất: 5x²-4x+7

5x²-4x+7=5(x²-4/5.x+7/5)

\(=\left[\left(x-\frac{2}{5}\right)^2+\frac{31}{25}\right].5\)

\(=5\left(x-\frac{2}{5}\right)^2+\frac{31}{5}\ge\frac{31}{5}\forall x\)

Dấu = xảy ra khi x=2/5

vậy...

Trả lời:

a, \(\left(x-\frac{3}{4}\right)^2=x^2-2.x.\frac{3}{4}+\left(\frac{3}{4}\right)^2=x^2-\frac{3}{2}x+\frac{9}{16}\)

b, \(\left(3a+1\right)^2=\left(3a\right)^2+2.3a.1+1^2=9a^2+6a+1\)

c, \(\left(3a+\frac{1}{3}\right)\left(\frac{1}{3}-3a\right)=\left(\frac{1}{3}\right)^2-\left(3a\right)^2=\frac{1}{9}-9a^2\)

d, \(\left(x^2-2\right)^2=\left(x^2\right)^2-2.x^2.2+2^2=x^4-4x^2+4\)

e, \(\left(9x+2y\right)^2=\left(9x\right)^2+2.9x.2y+\left(2y\right)^2=81x^2+36xy+4y^2\)

g, \(\left(x-5y\right)\left(x+5y\right)=x^2-\left(5y\right)^2=x^2-25y^2\)

h, \(\left(5-x\right)^2=5^2-2.5.x+x^2=25-10x+x\)

x = 3 ; -0,5 ; -2

x=3 , -0,5 , -2

a. ( 2x - 4 ) ( 9 + 3x ) < 0 <=>\(\orbr{\begin{cases}2x-4< 0\\9+3x>0\end{cases}}\)hoặc\(\orbr{\begin{cases}2x-4>0\\9+3x< 0\end{cases}}\)

<=>\(\orbr{\begin{cases}x< 2\\x>-3\end{cases}}\)hoặc\(\orbr{\begin{cases}x>2\\x< -3\end{cases}}\)( mâu thuẫn )

<=> - 3 < x < 2

b. \(\frac{8-4x}{2x+6}>0\)<=>\(\orbr{\begin{cases}8-4x>0\\2x+6>0\end{cases}}\)hoặc\(\orbr{\begin{cases}8-4x< 0\\2x+6< 0\end{cases}}\)

<=>\(\orbr{\begin{cases}x< 2\\x>-3\end{cases}}\)hoặc\(\orbr{\begin{cases}x>2\\x< -3\end{cases}}\)( mâu thuẫn )

<=> - 3 < x < 2

xy²-4³

=xy²-8²

=(xy-8)(xy+8)

a,3³+(2y)^3

=(2y+3)(4y²+6y+9)

b,tương tự câu a)

c,tương tự câu a)

Trả lời:

\(a,27+8y^3=\left(3+2y\right)\left(9-6y+4y^2\right)\)

\(b,\frac{1}{8}x^3-1=\left(\frac{1}{2}x-1\right)\left(\frac{1}{4}x^2+\frac{1}{2}x+1\right)\)

\(c,x^3+27=\left(x+3\right)\left(x^2-3x+9\right)\)

Để x = x ! Thì x \(\le\)2. Nếu x > 2 thì 3 \(\ne\)1 x 2 x3.

- Nếu x = 2 thì 2 = 1 x 2 ( thỏa mãn )

 \(\Leftrightarrow\)x = 2

- Nếu x = 1 thì 1 = 1 ( thỏa mãn )

\(x=\hept{\begin{cases}1\\2\end{cases}}\)

HT

??????? hỏi j dễ zậy