Bài giải

'THAM KHẢO

a,

Điều kiện: x+2≥0⇔x≥−2x+2≥0⇔x≥-2

|2x+3|=x+2|2x+3|=x+2

⇔[2x+3=x+22x+3=−x−2⇔[2x+3=x+22x+3=−x−2

⇔[x=−13x=−5⇔[x=−13x=−5

⇔⎡⎣x=−1(t/m)x=−53(t/m)⇔[x=−1(t/m)x=−53(t/m)

Vậy x∈{−1;−53}x∈{-1;-53}

b,

A=|x−2006|+|2007−x|≥|x−2006+2007−x|=|1|=1A=|x−2006|+|2007−x|≥|x−2006+2007−x|=|1|=1

Đẳng thức xảy ra ⇔(x−2006)(2007−x)≥0⇔(x−2006)(2007−x)≥0

⇔(x−2006)(x−2007)≤0⇔(x−2006)(x−2007)≤0

Vì x−2006>x−2007x−2006>x−2007

⇒{x−2006≥0x−2007≤0⇒{x−2006≥0x−2007≤0

⇔{x≥2006x≤2007⇔{x≥2006x≤2007

⇔2006≤x≤2007⇔2006≤x≤2007

Vậy Amin=1⇔2006≤x≤2007

Bạn dfsfsdfsd bớt bớt lại nhé

Ta có : A = |x - 2006| + |2007 - x| ≥ |x - 2006 + 2007 - x|

                                                 = |(x - x) - 2006 + 2007| = |1| = 1

Dấu "=" xảy ra khi (x - 2006)(2007 - x) ≥ 0 => 2006 ≤ x ≤ 2007

Vậy gtnn của A là 1 tại 2006 ≤ x ≤ 2007

mọi người ơi giúp mình với mình sắp phải nộp rồi

ap dung bdt \(|a|+|b|\ge|a+b|\) voi \(a.b\ge0\)

thi \(A\ge|x-2016+2007-x|=|1|=1\)

vay GTNN cua A = 1 . Dat duoc khi \(\left(x-2016\right)\left(2017-x\right)\ge0\)

                                                           <=> \(2016\le x\le2017\)

chuc ban hoc tot

\(A=|x-2006|+|2007-x|\ge|x-2006+2007-x|\)

\(\Rightarrow A\ge1\)

Dấu"="xảy ra \(\Leftrightarrow\left(x-2006\right)\left(2007-x\right)\ge0\)

\(\Leftrightarrow\hept{\begin{cases}x-2006\ge0\\2007-x\ge0\end{cases}}\)hoặc \(\hept{\begin{cases}x-2006< 0\\2007-x< 0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x\ge2006\\x\le2007\end{cases}}\)hoặc \(\hept{\begin{cases}x< 2006\\x>2007\end{cases}}\)(loại )

\(\Leftrightarrow2006\le x\le2007\)

Vậy \(A_{min}=1\)\(\Leftrightarrow2006\le x\le2007\)

a) \(A=x^{15}+3x^{14}+5\)

\(=x^{14}\left(x+3\right)+5\)

\(=x^{14}.0+5\)

= 5

b) x = -3 => x + 3 = 0

\(B=\left(x^{2007}+3x^{2006}+1\right)^{2007}\)

\(=\left[x^{2006}\left(x+3\right)+1\right]^{2007}\)

\(=\left(x^{2006}.0+1\right)^{2007}\)

\(=1^{2007}=1\)

\(A=x^{15}+3.x^{14}+5\text{ biết x+3=0}\)

\(A=x^{14}.\left(x+3\right)+5\)

\(\text{Do x+3=0}\Rightarrow A=x^{14}.0+5\)

\(A=0+5\)

\(A=5\)        \(\text{Vậy }A=5\text{ với x+3=0}\)

\(B=\left(x^{2007}+3.x^{2006}+1\right)^{2007}\text{ biết x=-3}\)

\(B=\left[x^{2006}.\left(x+3\right)+1\right]^{2007}\)

\(\text{Do x=-3}\Rightarrow B=\left[x^{2006}.\left(-3+3\right)+1\right]^{2007}\)

\(B=\left(x^{2006}.0+1\right)^{2007}\)

\(B=\left(0+1\right)^{2007}\)

\(B=1^{2007}\)

\(B=1\)           \(\text{Vậy }B=1\text{ với x=-3}\)

Ngu thì câm mồm đi

\(\frac{x-1}{2009}+\frac{x-2}{2008}=\frac{x-3}{2007}+\frac{x-4}{2006}\)

\(\Leftrightarrow\frac{x-1}{2009}-1+\frac{x-2}{2008}-1=\frac{x-3}{2007}-1+\frac{x-4}{2006}-1\)

\(\Leftrightarrow\frac{x-2010}{2009}+\frac{x-2010}{2008}-\frac{x-2010}{2007}-\frac{x-2010}{2006}=0\)

\(\Leftrightarrow\left(x-2010\right)\left(\frac{1}{2009}+\frac{1}{2008}-\frac{1}{2007}-\frac{1}{2006}\ne0\right)=0\)

\(\Leftrightarrow x=2010\)