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b) \(25-x^2+14xy-49y^2\)
\(=25-\left(x^2-14xy+49y^2\right)\)
\(=25-\left[x^2-2\cdot7y\cdot x+\left(7y\right)^2\right]\)
\(=25-\left(x-7y\right)^2\)
\(=5^2-\left(x-7y\right)^2\)
\(=\left[5-\left(x-7y\right)\right]\left[5+\left(x-7y\right)\right]\)
\(=\left(5-x+7y\right)\left(5+x-7y\right)\)
c) \(x^5+x^4+1\)
\(=x^5+x^4+1+x^3-x^3\)
\(=\left(x^5+x^4+x^3\right)+\left(1-x^3\right)\)
\(=x^3\left(x^2+x+1\right)+\left(1-x\right)\left(x^2+x+1\right)\)
\(=\left(x^2+x+1\right)\left[x^3+\left(1-x\right)\right]\)
\(=\left(x^2+x+1\right)\left(x^3+1-x\right)\)
b: 25-x^2+14xy-49y^2
=25-(x-7y)^2
=(5-x+7y)(5+x-7y)
c: =x^5+x^4+x^3+1-x^3
=x^3(x^2+x+1)+(1-x)(x^2+x+1)
=(x^2+x+1)(x^3+1-x)
\(A=x^2-4x+9=\left(x-2\right)^2+5\ge5\forall x\)
Dấu '=' xảy ra khi x=2
\(B=x^2-x+1=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\forall x\)
Dấu '=' xảy ra khi \(x=\dfrac{1}{2}\)
1)\(6x-x^2=x\left(6-x\right)\)
2)\(5x^2z-15xyz+30xz^2=5x\left(xz-3y+6z\right)\)
3)\(x^3-6x^2+9x=x\left(x^2-6x+9\right)=x\left(x-3\right)^2\)
ĐKXĐ: \(x\ge\dfrac{2}{3}\)
\(x^2+5x-6+\left(\sqrt{2x+7}-3\right)+\left(\sqrt{3x-2}-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+6\right)+\dfrac{2\left(x-1\right)}{\sqrt{2x+7}+3}+\dfrac{3\left(x-1\right)}{\sqrt{3x-2}+1}=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+6+\dfrac{2}{\sqrt{2x+7}+3}+\dfrac{3}{\sqrt{3x-2}+2}\right)=0\)
\(\Leftrightarrow x-1=0\)
\(\Leftrightarrow x=1\)
\(\dfrac{1+x}{1-x}+3=\dfrac{x-3}{x-1}\)
\(ĐK:x\ne1\)
\(\Leftrightarrow\dfrac{1+x}{1-x}+3=\dfrac{3-x}{1-x}\)
\(\Leftrightarrow\dfrac{\left(1+x\right)+3\left(1-x\right)}{1-x}=\dfrac{3-x}{1-x}\)
\(\Leftrightarrow\left(1+x\right)+3\left(1-x\right)=3-x\)
\(\Leftrightarrow1+x+3-3x=3-x\)
\(\Leftrightarrow-x=-1\)
\(\Leftrightarrow x=1\left(ktm\right)\)
Vậy pt vô nghiệm
\(\dfrac{1+x}{1-x}+3=\dfrac{x-3}{x-1}\) đề như thế này phải ko?
A = -x² - 6x + 1
= -(x² + 6x - 1)
= -(x² + 6x + 9 - 10)
= -[(x + 3)² - 10]
= -(x + 3)² + 10
Do (x + 3)² ≥ 0 với mọi x ∈ R
⇒ -(x + 3)² ≤ 0 với mọi x ∈ R
⇒ -(x + 3)² + 10 ≤ 10 với mọi x ∈ R
Vậy GTLN của A là 10 khi x = -3
\(A=-x^2-6x+1\)
\(A=-\left(x^2+6x-1\right)\)
\(A=-\left(x^2+6x+9-10\right)\)
\(A=-\left(x^2+2\cdot x\cdot3+3^2\right)+10\)
\(A=-\left(x+3\right)^2+10\)
Có: \(\left(x+3\right)^2\ge0\forall x\Rightarrow-\left(x+3\right)^2\le0\)
\(\Rightarrow-\left(x+3\right)^2+10\le10\)
\(\Rightarrow A\le10\)
Dấu "=" xảy ra khi \(\left(x+3\right)^2=0\Leftrightarrow x+3=0\)
\(\Leftrightarrow x=-3\)
Vậy: \(A_{min}=10\Leftrightarrow x=-3\)
(x -1)x2 - 4x(x - 1) + 4(x - 1)
= (x - 1)x - 4(x - 1)2
= (x - 1)[(x - 4(x - 1)]
= (x - 1)(-3x + 4)
Thay x = 3 vào biểu thức :
(3 - 1)(-3.3 + 4) = 2.(-5) = -10
\(\Leftrightarrow6x^2-14x+4-6x^2-12x+18-7x+3=0\)
\(\Leftrightarrow-33x=-25\Rightarrow x=\frac{25}{33}\)
2( 3x - 1 )( x - 2 ) - 6( x - 1 )( x + 3 ) = 7x - 3
<=> 2( 3x2 - 7x + 2 ) - 6( x2 + 2x - 3 ) = 7x - 3
<=> 6x2 - 14x + 4 - 6x2 - 12x + 18 = 7x - 3
<=> -26x + 22 = 7x - 3
<=> -26x - 7x = -3 - 22
<=> -33x = -25
<=> x = 25/33
<=> -36x =