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\(5a^2+10b^2-6ab-4a+2b+3\)
\(=\left(a^2-6ab+9b^2\right)+\left(4a^2-4a+1\right)+\left(b^2+2b+1\right)+1\)
\(=\left(a-3b\right)^2+\left(2a-1\right)^2+\left(b+1\right)^2+1>0\left(đpcm\right)\)
Ta có:
\(\left(3a-2b+c\right)^2=9a^2+4b^2+c^2+2\left(3ac-6ab-2bc\right)\)
\(\Rightarrow b^2=9a^2+4b^2+c^2\)
(vì \(3a-3b+c=0\Leftrightarrow3a-2b+c=-b\), \(6ab+2bc-3ac=0\))
\(\Leftrightarrow9a^2+3b^2+c^2=0\)
\(\Leftrightarrow a=b=c=0\).
Khi đó: \(P=\left(-1\right)^{2019}+\left(-1\right)^{2020}+\left(-1\right)^{2021}=-1\)
Ta có:
(3a−2b+c)2=9a2+4b2+c2+2(3ac−6ab−2bc)
⇒b2=9a2+4b2+c2
(vì 3a−3b+c=0⇔3a−2b+c=−b, 6ab+2bc−3ac=0)
⇔9a2+3b2+c2=0
⇔a=b=c=0.
Khi đó: P=(−1)2019+(−1)2020+(−1)2021=−1
1. = 2(a+b)
2. =2(a-b)
3.=2(a+2b-3c)
4.=3(a-2b-3c)
5.=-4(a+2b+3c)
6.=-5(x+2xy+3y)
7.=-7(a+2ab+3b)
8.=6(xy-2x-3y)
9.=8(xy-3y+2x)
10.=9(ab-2a+1)
Lần sau bn cần ghi đề rõ ràng hơn
11.=x(y-1)
12.=a(x+1)
13.=m(x+y+1)
14.=-a(x+y+1)
15.=-a(x2+x+1)
16.=-2a(x+2y)
17.=2a(x-y+1)
18.=2(2ax-ay-2)
19.=5a(1-2x-3)
20.=-2ab(a+2b+3)
1) 9(a+b)2-4(a-2b)2
=[3(a+b)]2-[2(a-2b)]2=....
2) 8x3+27y3=(2x)3+(3y)3=...
3) (a+b)3-c3=(a+b-c).[(a+b)2+(a+b)c+c2]
4) x3+3x2+3x+1=(x+1)3
Lời giải:
\(a^3+8b^3=1-6b\)
\(\Leftrightarrow a^3+8b^3+6ab-1=0\)
\(\Leftrightarrow (a+2b)^3-3.a.2b(a+2b)+6ab-1=0\)
\(\Leftrightarrow [(a+2b)^3-1]-6ab(a+2b-1)=0\)
\(\Leftrightarrow (a+2b-1)[(a+2b)^2+(a+2b)+1]-6ab(a+2b-1)=0\)
\(\Leftrightarrow (a+2b-1)(a^2+4b^2+1+a+2b-2ab)=0\)
\(\Rightarrow \left[\begin{matrix} a+2b-1=0(1)\\ a^2+4b^2+1+a+2b-2ab=0(2)\end{matrix}\right.\)
Với \((1)\Rightarrow a+2b=1\)
Với \((2)\Rightarrow 2a^2+8b^2+2+2a+4b-4ab=0\)
\(\Leftrightarrow (a^2+4b^2-4ab)+(a^2+2a+1)+(4b^2+4b+1)=0\)
\(\Leftrightarrow (a-2b)^2+(a+1)^2+(2b+1)^2=0\)
\(\Rightarrow (a-2b)^2=(a+1)^2=(2b+1)^2=0\)
\(\Rightarrow \left\{\begin{matrix} a=-1\\ 2b=-1\end{matrix}\right.\Rightarrow a+2b=-2\)
Vậy $a+2b\in \left\{1;-2\right\}
Lời giải:
\(a^3+8b^3=1-6b\)
\(\Leftrightarrow a^3+8b^3+6ab-1=0\)
\(\Leftrightarrow (a+2b)^3-3.a.2b(a+2b)+6ab-1=0\)
\(\Leftrightarrow [(a+2b)^3-1]-6ab(a+2b-1)=0\)
\(\Leftrightarrow (a+2b-1)[(a+2b)^2+(a+2b)+1]-6ab(a+2b-1)=0\)
\(\Leftrightarrow (a+2b-1)(a^2+4b^2+1+a+2b-2ab)=0\)
\(\Rightarrow \left[\begin{matrix} a+2b-1=0(1)\\ a^2+4b^2+1+a+2b-2ab=0(2)\end{matrix}\right.\)
Với \((1)\Rightarrow a+2b=1\)
Với \((2)\Rightarrow 2a^2+8b^2+2+2a+4b-4ab=0\)
\(\Leftrightarrow (a^2+4b^2-4ab)+(a^2+2a+1)+(4b^2+4b+1)=0\)
\(\Leftrightarrow (a-2b)^2+(a+1)^2+(2b+1)^2=0\)
\(\Rightarrow (a-2b)^2=(a+1)^2=(2b+1)^2=0\)
\(\Rightarrow \left\{\begin{matrix} a=-1\\ 2b=-1\end{matrix}\right.\Rightarrow a+2b=-2\)
Vậy $a+2b\in \left\{1;-2\right\}$