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Giải:
1) \(9x^2-12xy+4y^2-3\)
\(=\left(3x-2y\right)^2-3\)
\(=\left(3x-2y-\sqrt{3}\right)\left(3x-2y+\sqrt{3}\right)\) (Bước này chắc không cần)
2) \(x^2+4x+1\)
\(=x^2+4x+4-3\)
\(=\left(x+2\right)^2-3\)
\(=\left(x+2-\sqrt{3}\right)\left(x+2+\sqrt{3}\right)\)
(Bước này chắc không cần)
3) \(x^2-4x+7\)
\(=x^2-4x+4+3\)
\(=\left(x-2\right)^2+3\)
4) \(x^2+6x+15\)
\(=x^2+6x+9+6\)
\(=\left(x+3\right)^2+6\)
5) \(x^2-x+\dfrac{1}{3}\)
\(=x^2-x+\dfrac{1}{4}+\dfrac{1}{12}\)
\(=\left(x-\dfrac{1}{2}\right)^2+\dfrac{1}{12}\)
6) \(\dfrac{1}{4}x^2+x\)
\(=\left(\dfrac{1}{2}x\right)^2+x+1-1\)
\(=\left(\dfrac{1}{2}x+1\right)^2-1\)
7) \(3x^2+2x+1\)
\(=x^2+2x+1+2x^2\)
\(=\left(x+1\right)^2+2x^2\)
8) \(2x^2-2x+1\)
\(=x^2-2x+1+x^2\)
\(=\left(x-1\right)^2+x^2\)
9) \(10a^2+5b^2+12ab+4a-6b+15\)
\(=4a^2+6a^2+4b^2+b^2+12ab+4a-6b+15\)
\(=\left(6a^2+b^2+12ab\right)+4a+4a^2-6b+4b^2+15\)
\(=\left(6a+b\right)^2+4a\left(1+a\right)-2b\left(3+2b\right)+15\)
Giải:
1) \(9x^2-12xy+4y^2-3\)
\(=\left(9x^2-12xy+4y^2\right)-3\)
\(=\left(3x-2y\right)^2-3\)
2) \(x^2+4x+1\)
\(=x^2+4x+4-3\)
\(=\left(x+2\right)^2-3\)
3) \(x^2-4x+7\)
\(=x^2-4x+4+3\)
\(=\left(x-2\right)^2+3\)
4) \(x^2+6x+15\)
\(=x^2+6x+9+6\)
\(=\left(x+3\right)^2+6\)
5) \(x^2-x+\dfrac{1}{3}\)
\(=x^2-x+\dfrac{1}{4}+\dfrac{1}{12}\)
\(=\left(x-\dfrac{1}{2}\right)^2+\dfrac{1}{12}\)
6) \(\dfrac{1}{4}x^2+x\)
\(=x\left(\dfrac{1}{4}x+1\right)\)
7) \(3x^2+2x+1\)
\(=x^2+2x+1+2x^2\)
\(=\left(x+1\right)^2+2x^2\)
8) \(2x^2-2x+1\)
\(=x^2-2x+1+x^2\)
\(=\left(x-1\right)^2+x^2\)
9) \(10a^2+5b^2+12ab+4a-6b+15\)
\(=a^2+b^2+9a^2+12ab+4b^2+4a-6b+15\)
\(=9a^2+12ab+4b^2+a^2+4a-6b+b^2+15\)
\(=\left(3a+2b\right)^2+a\left(a+4\right)-b\left(6-b\right)+15\)
Vậy ...
1: \(=\left(3x-2y\right)^2-3\)
2: \(=x^2+4x+4-3=\left(x+2\right)^2-3\)
3: \(=x^2-4x+4+3=\left(x-2\right)^2+3\)
5 \(=x^2-x+\dfrac{1}{4}+\dfrac{1}{12}=\left(x-\dfrac{1}{2}\right)^2+\dfrac{1}{12}\)
6: \(=\dfrac{1}{4}x^2+x+1-1=\left(\dfrac{1}{2}x+1\right)^2-1\)
Giải:
1) \(9x^2-12xy+4y^2-3\)
\(=\left(9x^2-12xy+4y^2\right)-3\)
\(=\left(3x-2y\right)^2-3\)
2) \(x^2+4x+1\)
\(=x^2+4x+4-3\)
\(=\left(x+2\right)^2-3\)
3) \(x^2-4x+7\)
\(=x^2-4x+4+3\)
\(=\left(x-2\right)^2+3\)
4) \(x^2+6x+15\)
\(=x^2+6x+9+6\)
\(=\left(x+3\right)^2+6\)
5) \(x^2-x+\dfrac{1}{3}\)
\(=x^2-x+\dfrac{1}{4}+\dfrac{1}{12}\)
\(=\left(x-\dfrac{1}{2}\right)^2+\dfrac{1}{12}\)
6) \(\dfrac{1}{4}x^2+x\)
\(=x\left(\dfrac{1}{4}x+1\right)\)
7) \(3x^2+2x+1\)
\(=x^2+2x+1+2x^2\)
\(=\left(x+1\right)^2+2x^2\)
8) \(2x^2-2x+1\)
\(=x^2-2x+1+x^2\)
\(=\left(x-1\right)^2+x^2\)
9) \(10a^2+5b^2+12ab+4a-6b+15\)
\(=a^2+b^2+9a^2+12ab+4b^2+4a-6b+15\)
\(=9a^2+12ab+4b^2+a^2+4a-6b+b^2+15\)
\(=\left(3a+2b\right)^2+a\left(a+4\right)-b\left(6-b\right)+15\)
Vậy ...
Giải;
1) \(9x^2-12xy+4y^2-3\)
\(=\left(9x^2-12xy+4y^2\right)-3\)
\(=\left(3x-2y\right)^2-3\)
2) \(x^2+4x+1\)
\(=x^2+4x+4-3\)
\(=\left(x+2\right)^2-3\)
3) \(x^2-4x+7\)
\(=x^2-4x+4+3\)
\(=\left(x-2\right)^2+3\)
4) \(x^2+6x+15\)
\(=x^2+6x+9+6\)
\(=\left(x+3\right)^2+6\)
5) \(x^2-x+\dfrac{1}{3}\)
\(=x^2-x+\dfrac{1}{4}+\dfrac{1}{12}\)
\(=\left(x-\dfrac{1}{2}\right)^2+\dfrac{1}{12}\)
6) \(\dfrac{1}{4}x^2+x\)
\(=x\left(\dfrac{1}{4}x+1\right)\)
7) \(3x^2+2x+1\)
\(=x^2+2x+1+2x^2\)
\(=\left(x+1\right)^2+2x^2\)
8) \(2x^2-2x+1\)
\(=x^2-2x+1+x^2\)
\(=\left(x-1\right)^2+x^2\)
9) \(10a^2+5b^2+12ab+4a-6b+15\)
\(=a^2+b^2+9a^2+12ab+4b^2+4a-6b+15\)
\(=9a^2+12ab+4b^2+a^2+4a-6b+b^2+15\)
\(=\left(3a+2b\right)^2+a\left(a+4\right)-b\left(6-b\right)+15\)
Vậy ...
1: \(=-\left(x^2+2x+2\right)=-\left(x^2+2x+1+1\right)=-\left(x+1\right)^2-1< =-1\)
Dấu '=' xảy ra khi x=-1
2: \(=-\left(4x^2-12x-10\right)\)
\(=-\left(4x^2-12x+9-19\right)\)
\(=-\left(2x-3\right)^2+19< =19\)
Dấu '=' xảy ra khi x=3/2
3: \(=-\left(x^2+4x+4-4\right)=-\left(x+2\right)^2+4< =4\)
Dấu '=' xảy ra khi x=-2
a)\(6x-9-x^2\)
\(=-\left(x^2+6x+9\right)\)
\(=-\left(x+3\right)^2\)
b)\(x^2+4y^2+4xy\)
\(=\left(x+2y\right)^2\)
c)\(x^2+8x+16\)
\(=\left(x+4\right)^2\)
d)\(9x^2-12xy+4y^2\)
\(=\left(3x-2y\right)^2\)
e)\(-25x^2y^2+10xy-1\)
\(=-\left(25x^2y^2-10xy+1\right)\)
\(=-\left(5xy-1\right)^2\)
f)\(4x^2-4x+1\)
\(=\left(2x-1\right)^2\)
j)\(x^2+6x+9\)
\(=\left(x+3\right)^2\)
h)\(9x^2-6x+1\)
\(=\left(3x-1\right)^2\)
#H
a, 6x - 9 - x2 = - x2 + 6x - 9 = - (x2 - 6x + 9) = - (x - 3)2
b, x2 + 4y2 + 4xy = x2 + 2. x . 2y + (2y)2 = (x + 2y)2
c, x2 + 8x + 16 = x2 + 2 . x . 4 + 42 = (x + 4)2
d, 9x2 - 12xy + 4y2 = (3x)2 - 2 . 3x . 2y + (2y)2 = (3x - 2y)2
e, - 25x2y2 + 10xy - 1 = - (25x2y2 - 10xy + 1) = - [(5xy)2 - 2 . 5xy + 1] = - (5xy - 1)2
f, 4x2 - 4x + 1 = (2x)2 - 2 . 2x + 1 = (2x - 1)2
j, x2 + 6x + 9 = x2 + 2 . x . 3 + 32 = (x + 3)2
h, 9x2 - 6x + 1 = (3x)2 - 2 . 3x + 1 = (3x - 1)2
1. \(x^6-2x^3+1=0\Leftrightarrow\left(x^3-1\right)^2=0\Leftrightarrow x=1\)
2. \(x^6+\dfrac{1}{4}x^3+\dfrac{1}{64}=0\Leftrightarrow\left(x^3\right)^2+2.x^3.\dfrac{1}{8}+\left(\dfrac{1}{8}\right)^2=0\Leftrightarrow\left(x+\dfrac{1}{8}\right)^2=0\Leftrightarrow x=-\dfrac{1}{2}\)4. \(x^3-10x^2+25x=0\Leftrightarrow x^3-5x^2-5x^2+25x=0\)
\(\Leftrightarrow x^2\left(x-5\right)-5x\left(x-5\right)=0\)
\(\Leftrightarrow x\left(x-5\right)^2=0\Leftrightarrow x=5\)
5. \(\dfrac{1}{4}x^3-3x^2+9x=0\)
\(\Leftrightarrow x\left(\dfrac{1}{4}x^2-3x+9\right)=0\)
\(\Leftrightarrow x\left[\left(\dfrac{1}{2}x\right)^2-2.\dfrac{1}{2}x.3+3^2\right]=0\)
\(\Leftrightarrow x\left(\dfrac{1}{2}x-3\right)^2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=6\end{matrix}\right.\)
6. \(x^5-16x=0\Leftrightarrow x\left(x^4-16\right)=0\Leftrightarrow x\left(x^2-4\right)\left(x^2+4\right)=0\)
\(\Leftrightarrow x\left(x-2\right)\left(x+2\right)\left(x^2+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-2\\x^2=-4\left(l\right)\end{matrix}\right.\)
7. \(4x^2+4x-3=0\Leftrightarrow4x^2-2x^2-6x-3=0\)
\(\Leftrightarrow2x\left(2x-1\right)-3\left(2x-1\right)=0\)
\(\Leftrightarrow\left(2x-1\right)\left(2x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)
8. \(4x^2+28x+48=0\Leftrightarrow4x^2+12x+14x+48=0\)
\(\Leftrightarrow4x\left(x+3\right)+12\left(x+4\right)=0\)
\(\Leftrightarrow\left(4x+12\right)\left(x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-4\end{matrix}\right.\)
9. \(9x^2-12x+3=0\Leftrightarrow9x^2-9x-3x+3=0\Leftrightarrow9x\left(x-1\right)-3\left(x-1\right)=0\Leftrightarrow\left(x-1\right)\left(9x-3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{3}\end{matrix}\right.\)