Question

1. Tìm GTNN:

a)A = |x| + |1+x|

b) B = |2-x| + 3

2. Chứng minh

a) a⁴ +1 \(\ge\) a(a² +1)

b) \(\dfrac{a^2}{a^4+1}\le\dfrac{1}{2}\)

3. Giải pt:

a) |-4x| + 3x = 1

b) |x-2|=|3x|

c) |x-2|=3x

Answer 1

2.

a)Xét hiệu:

\(a^4+1-a\left(a^2+1\right)\)

\(=a^4+1-a^3-a\)

\(=a^3\left(a-1\right)-\left(a-1\right)\)

\(=\left(a^3-1\right)\left(a-1\right)\)

\(=\left(a-1\right)\left(a^2+a+1\right)\left(a-1\right)\)

\(=\left(a-1\right)^2\left(a^2+a+1\right)\)

Ta có:

\(a^2+a+1=a^2+2.a.\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{2}{4}=\left(a+\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\)

Suy ra:

\(\left(a-1\right)^2\left(a^2+a+1\right)\ge0\)

=> \(a^4+1\ge a\left(a^2+1\right)\)

Answer 2

1)a)\(A=\left|x\right|+\left|1+x\right|=\left|x\right|+\left|-1-x\right|\ge\left|x-1-x\right|=1\)

\(\Rightarrow MINA=1\)

b)MINB=3

2b)Ta có:\(\left(a^2-1\right)^2\ge0\)

\(\Leftrightarrow a^4-2a^2+1\ge0\)

\(\Leftrightarrow a^4+1\ge2a^2\)

\(\Rightarrow\dfrac{a^2}{a^4+1}\le\dfrac{a^2}{2a^2}=\dfrac{1}{2}\left(đpcm\right)\)