a) b2 + 6b + 10
= b2 + 2.( b ).3 + 33 + 1
= ( b + 3 ) 2 + 1
Vì ( b + 3 ) 2 > hoặc = 0
Nên ( b + 3 ) 2 + 1 > 0
b) B= -a2+ 6a - 15
B= - ( a2 + 2.a.3 + 32 + 8 )
B= - [( a + 3 ) 2 + 8 ]
Vì ( a + 3 )2 > hoặc = 0
Nên ( a + 3 ) 2 + 8 > 0
=> - [( a + 3 ) 2 + 8 ] < 0
Vậy B < 0
a) \(b^2+6b+10\)
=\(b^2+2b.3+3^2-3^2+10\)
=\(\left(b+3\right)^2+1\)
Ta có: \(\left(b+3\right)^2\)\(\ge\)0
Nên: \(\left(b+3\right)^2\)> 0 (với mọi b)
b) \(-a^2+6a-15\)
= \(-\left(a^2-6a+15\right)\)
=\(-\left(a^2-2a.3+3^2-3^2+15\right)\)
=\(-\left[\left(a-3\right)^2+6\right]\)
Ta có: \(\left(a-3\right)^2\ge0\)
Nên: \(\left(a-3\right)^2+6>0\)
Do đó: \(-\left[\left(a-3\right)^2+6\right]< 0\)(với mọi a)
c) Ta có VT=\(\left(a-b\right)^2+\left(ab+1\right)^2\)
\(=a^2-2ab+b^2+a^2b^2+2ab+1\)
\(=a^2+b^2+a^2b^2+1\)
Lại có VP= \(\left(a^2+1\right)\left(b^2+1\right)\)
\(=a^2b^2+a^2+b^2+1=a^2+b^2+a^2b^2+1\)(=VT)
Vậy VT=VP
\(A=b^2+6b+10=b^2+2.b.3+3^2+1=\left(b+3\right)^2+1\)
Vì \(\left(b+3\right)^2\ge0\left(b\in R\right)\)
nên \(\left(b+3\right)^2+1>0\left(b\in R\right)\)
Hay \(b^2+6b+10>0\)với mọi \(b\in R\)
\(B=-a^2+6a-15=-\left(a^2-6a+15\right)=-\left(a^2-2.a.3+3^2+6\right)=-\left[\left(a-3\right)^2+6\right]=-\left(a-3\right)^2-6\)Vì \(\left(a-3\right)^2\ge0\left(a\in R\right)\)
nên \(-\left(a-3\right)^2\le0\left(a\in R\right)\)
do đó \(-\left(a-3\right)^2-6< 0\left(a\in R\right)\)
hay \(-a^2+6a-15< 0\)với mọi \(a\in R\)
C0 \(\left(a-b\right)^2+\left(ab+1\right)^2=\left(a^2+1\right)\left(b^2+1\right)\)
Ta có:
\(\left(a-b\right)^2+\left(ab+1\right)^2=a^2-2ab+b^2+a^2b^2+2ab+1=a^2+a^2b^2+b^2+1=a^2\left(b^2+1\right)+1\left(b^2+1\right)=\left(a^2+1\right)\left(b^2+1\right)\)Hay \(\left(a-b\right)^2+\left(ab+1\right)^2=\left(a^2+1\right)\left(b^2+1\right)\left(đpcm\right)\)
