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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) \(9x^2-12xy+4y^2-3\)
\(=\left(3x-2y\right)^2-3\)
\(=\left(3x-2y-\sqrt{3}\right)\left(3x-2y+\sqrt{3}\right)\) (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. \(x^6-2x^3+1=0\Leftrightarrow\left(x^3-1\right)^2=0\Leftrightarrow x=1\)
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. \(9x^2-12x+3=0\Leftrightarrow9x^2-9x-3x+3=0\Leftrightarrow9x\left(x-1\right)-3\left(x-1\right)=0\Leftrightarrow\left(x-1\right)\left(9x-3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{3}\end{matrix}\right.\)
Tìm x:
1. \(25x^2-20x+4=0\)
⇔ \(\left(5x-2\right)^2=0\)
⇔ \(5x-2=0\)
⇔ \(5x=2\)
⇔ \(x=\dfrac{2}{5}\)
⇒ S = \(\left\{\dfrac{2}{5}\right\}\)
2. \(\left(2x-3\right)^2-\left(2x+1\right).\left(2x-1\right)=0\)
⇔ \(4x^2-12x+9-\left(4x^2-1\right)=0\)
⇔ \(4x^2-12x+9-4x^2+1=0\)
⇔ \(-12x+10=0\)
⇔ \(-12x=-10\)
⇔ \(x=\dfrac{5}{6}\)
⇒ S \(=\left\{\dfrac{5}{6}\right\}\)
3. \(\left(\dfrac{1}{2}x-1\right)\left(\dfrac{1}{2}x+1\right)-\left(\dfrac{1}{2}x-1\right)^2=0\)
⇔ \(\dfrac{1}{4}x^2-1-\left(\dfrac{1}{4}x^2-x+1\right)=0\)
⇔ \(\dfrac{1}{4}x^2-1-\dfrac{1}{4}x^2+x-1=0\)
⇔ \(-2+x=0\)
⇔ \(x=2\)
⇒ S \(=\left\{2\right\}\)
4. \(\left(2x-3\right)^2+\left(2x+5\right)^2=8\left(x+1\right)^2\)
⇔ \(4x^2-12x+9+4x^2+20x+25=8\left(x^2+2x+1\right)\)
⇔ \(8x^2+8x+34=8x^2+16x+8\)
⇔ \(8x+34=16x+8\)
⇔ \(8x-16x=8-34\)
⇔ \(-8x=-26\)
⇔ \(x=\dfrac{13}{4}\)
⇒ S \(=\left\{\dfrac{13}{4}\right\}\)
5.\(4x^2+12x-7=0\)
⇔ \(4x^2+14x-2x-7=0\)
⇔ \(2x\left(2x+7\right)-\left(2x+7\right)=0\)
⇔ \(\left(2x+7\right)\left(2x-1\right)=0\)
⇔ \(\left[{}\begin{matrix}2x+7=0\\2x-1=0\end{matrix}\right.\)
⇔ \(\left[{}\begin{matrix}x=\dfrac{-7}{2}\\x=\dfrac{1}{2}\end{matrix}\right.\)
⇒ S \(=\left\{\dfrac{-7}{2};\dfrac{1}{2}\right\}\)
6. \(\dfrac{1}{4}x^2+\dfrac{2}{3}x-\dfrac{5}{9}=0\)
⇔ \(9x^2+24x-20=0\)
⇔ \(9x^2+30x-6x-20=0\)
⇔ \(3x\left(3x+10\right)-2\left(3x+10\right)=0\)
⇔ \(\left(3x+10\right)\left(3x-2\right)=0\)
⇔ \(\left[{}\begin{matrix}3x+10=0\\3x-2=0\end{matrix}\right.\)
⇔ \(\left[{}\begin{matrix}x=\dfrac{-10}{3}\\x=\dfrac{2}{3}\end{matrix}\right.\)
⇒ S \(=\left\{\dfrac{-10}{3};\dfrac{2}{3}\right\}\)
7. \(24\dfrac{8}{9}-\dfrac{1}{4}x^2-\dfrac{1}{3}x=0\)
⇔ \(\dfrac{224}{9}-\dfrac{1}{4}x^2-\dfrac{1}{3}x=0\)
⇔ \(896-9x^2-12x=0\)
⇔ \(-896+9x^2+12x=0\)
⇔ \(9x^2+12x-896=0\)
⇔ \(9x^2-84x+96x-896=0\)
⇔ \(3x\left(3x-28\right)+32\left(3x-28\right)=0\)
⇔ \(\left(3x-28\right)\left(3x+32\right)=0\)
⇔ \(\left[{}\begin{matrix}3x-28=0\\3x+32=0\end{matrix}\right.\)
⇔ \(\left[{}\begin{matrix}x=\dfrac{28}{3}\\x=\dfrac{-32}{3}\end{matrix}\right.\)
⇒ S \(=\left\{\dfrac{-32}{3};\dfrac{28}{3}\right\}\)
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 ...