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\(P=2x^2+x+1\)
\(P=\left(\sqrt{2}.x\right)^2+2.\sqrt{2}.\frac{1}{2}x+\frac{1}{2}-\frac{1}{2}+1\)
\(P=\left(\sqrt{2}.x+\sqrt{\frac{1}{2}}\right)^2-\left(\frac{1}{2}-1\right)\)
\(P=\left(x\sqrt{2}+\sqrt{\frac{1}{2}}\right)^2-\left(-\frac{1}{2}\right)\)
\(P=\left(x\sqrt{2}+\sqrt{\frac{1}{2}}\right)^2-\left(-\sqrt{\frac{1}{2}}\right)^2\)
\(P=\left(x\sqrt{2}+\sqrt{\frac{1}{2}}-\sqrt{\frac{1}{2}}\right)\left(x\sqrt{2}+\sqrt{\frac{1}{2}}-\sqrt{\frac{1}{2}}\right)\)
\(P=\left(x\sqrt{2}\right)\left(x\sqrt{2}\right)\)
\(P=\left(x\sqrt{2}\right)^2\)
\(P=2x^2\)
1. \(\left(A+B\right)^2=A^2+2AB+B^2\)
2. \(\left(A-B\right)^2=A^2-2AB+B^2\)
3. \(A^2-B^2=\left(A-B\right)\left(A+B\right)\)
4. \(A^3-B^3=\left(A-B\right)\left(A^2+AB+B^2\right)\)
5. \(A^2+B^2=\left(A+B\right)\left(A^2-AB+B^2\right)\)
6. \(\left(A+B\right)^3=A^3+3A^2B+3AB^2+B^3\)
7. \(\left(A-B\right)^2=A^3-3A^2B+3AB^2-B^3\)
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Lời giải:
$2x^2+y^2+2xy-8x-6y+30$
$=(x^2+y^2+2xy)+x^2-8x-6y+30$
$=(x+y)^2-6(x+y)+(x^2-2x)+30$
$=(x+y)^2-6(x+y)+9+(x^2-2x+1)+20$
$=(x+y-3)^2+(x-1)^2+20\geq 20$
Vậy GTNN của biểu thức là $20$ khi $x+y-3=x-1=0$
$\Leftrightarrow x=1; y=2$
7 hằng đẳng thức cơ bản:
1, (a + b)2 = a2 + 2ab + b2
2, (a _ b)2 = a2 _ 2ab + b2
3, a2 - b2 = ( a - b ). (a + b )
4. (A+B)3= A3+3A2B +3AB2+B3
5. (A – B)3 = A3- 3A2B+ 3AB2- B3
6. A3 + B3= (A+B)(A2- AB +B2)
7. A3- B3= (A- B)(A2+ AB+ B2)
Mở rộng :
8. (A+B+C)2= A2+ B2+C2+2 AB+ 2AC+ 2BC
9. (a+b−c)2=a2+b2+c2+2ab−2bc−2ac(a+b−c)2=a2+b2+c2+2ab−2bc−2ac
10. (a−b−c)2=a2+b2+c2−2ab−2ac+2bc(a−b−c)2=a2+b2+c2−2ab−2ac+2bc
11. a3+b3=(a+b)3−3ab(a+b)a3+b3=(a+b)3−3ab(a+b)
12. a3−b3=(a−b)3+3ab(a−b)a3−b3=(a−b)3+3ab(a−b)
13. (a+b+c)3=a3+b3+c3+3(a+b)(b+c)(c+a)(a+b+c)3=a3+b3+c3+3(a+b)(b+c)(c+a)
14. a3+b3+c3−3abc=(a+b+c)(a2+b2+c2−ab−bc−ac)a3+b3+c3−3abc=(a+b+c)(a2+b2+c2−ab−bc−ac)
15. (a−b)3+(b−c)3+(c−a)3=3(a−b)(b−c)(c−a)(a−b)3+(b−c)3+(c−a)3=3(a−b)(b−c)(c−a)
16. (a+b)(b+c)(c+a)−8abc=a(b−c)2+b(c−a)2+c(a−b)2(a+b)(b+c)(c+a)−8abc=a(b−c)2+b(c−a)2+c(a−b)2
17. (a+b)(b+c)(c+a)=(a+b+c)(ab+bc+ca)−abc(a+b)(b+c)(c+a)=(a+b+c)(ab+bc+ca)−abc
19. ab2+bc2+ca2−a2b−b2c−c2a=(a−b)3+(b−c)3+(c−a)33ab2+bc2+ca2−a2b−b2c−c2a=(a−b)3+(b−c)3+(c−a)33
20.ab3+bc3+ca3−a3b−b3c−c3a=(a+b+c)[(a−b)3+(b−c)3+(c−a)3]3ab3+bc3+ca3−a3b−b3c−c3a=(a+b+c)[(a−b)3+(b−c)3+(c−a)3]3
