f. (a+1)(a-2) + a +3 / (a+1)(a + 2 ) + a2 - 3a
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\(a\left(a-1\right)+2\left(3-a\right)+3a=a^2-a+6-2a+3a\)
\(=a^2+6-3a+3a\)
\(=a^2+6\)
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2:
a: =>a^2+2ab+b^2-2a^2-2b^2<=0
=>-(a^2-2ab+b^2)<=0
=>(a-b)^2>=0(luôn đúng)
b; =>a^2+b^2+c^2+2ab+2ac+2bc-3a^2-3b^2-3c^2<=0
=>-(2a^2+2b^2+2c^2-2ab-2ac-2bc)<=0
=>(a-b)^2+(b-c)^2+(a-c)^2>=0(luôn đúng)
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a) x 2 - 3 2 x + 9 16 . b) 9 t 2 + 6t + 1.
c) 1 9 − 9 a 2 d) a 4 – 4 a 2 + 4.
![](https://rs.olm.vn/images/avt/0.png?1311)
1) Đúng
2) Đúng
3) Đúng
4)Đúng
5)Đúng
6)Sai
7)Sai
8)Đúng
9)Sai
10)sai
12)sai
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f(3)=g(1)
nên \(1+3\left(3a+1\right)+a^2=1-2a+a^2\)
\(\Leftrightarrow1+9a+3=1-2a\)
=>11a=-3
hay a=-3/11
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VT = (3a + 2b - 1)(a + 5) - 2b(a - 2)
= 3a2 + 2ab - a + 15a + 10b - 5 - 2ab + 4b
= 3a2 + 14a + 14b - 5
= 3a2 + 9a + 5a + 15 + 14b - 20
= 3a(a + 3) + 5(a + 3) + 2(7b - 10)
= (3a + 5)(a + 3) + 2(7b - 10)
= VP (đpcm)
\(\frac{\left(a+1\right)\left(a-2\right)+a+3}{\left(a+1\right)\left(a+2\right)+a^2-3a}=\frac{a^2+a-2a-2+a+3}{a^2+a+2a+2+a^2-3a}=\frac{a^2+1}{2a^2+2}=\frac{a^2+1}{2\left(a^2+1\right)}=\frac{1}{2}\)