Ta có : \(\hept{\begin{cases}\left|4x-3\right|\ge0\forall x\\\left|5y+7,5\right|\ge0\forall y\end{cases}}\Rightarrow\left|4x-3\right|+\left|5y+7,5\right|\ge0\)

=> \(F=\left|4x-3\right|+\left|5y+7,5\right|+17,5\ge17,5\)

=> Min F = 17,5

Dấu "=" xảy ra <=> \(\hept{\begin{cases}4x-3=0\\5y+7,5=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=0,75\\y=-1,5\end{cases}}\)

Vậy Min F = 17,5 <=> x = 0,75 ; y = - 1,5

\(x^2\ge0\forall x\in R\Rightarrow|3+x^2|\ge3\forall x\in R.\)

\(\Rightarrow K\ge3+27,7=30,7.\)

Dấu "=" xảy ra <=> x=0

\(x^2+x+8\)

\(=\left(x^2+x+\frac{1}{4}\right)-\frac{1}{4}+8\)

\(=\left(x+\frac{1}{2}\right)^2+\frac{31}{8}>0\) với mọi \(x\)

=> Vô nghiệm

\(a)\)

\(0,333...=3.0,111...=3.\frac{1}{9}=\frac{1}{3}\)

\(b)\)

\(0,454545...=45.0,010101...=45.\frac{1}{99}=\frac{5}{11}\)

\(c)\)

\(0,162162...=162.0,001001...=162.\frac{1}{999}=\frac{6}{37}\)

\(d)\)

\(5,272727...=5+0,272727...=5+27.\frac{1}{99}=5\frac{3}{11}\)

       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

địt ko em

\(\left|3x-4\right|+\left|3y+5\right|=0\)

\(\Leftrightarrow3x-4=0\text{ và }3y+5=0\)

       \(3x=4\)             \(3y=\left(-5\right)\)

       \(x=4\div3\)         \(y=\left(-5\right)\div3\)

        \(x=\frac{4}{3}\)               \(y=\left(-\frac{5}{3}\right)\)

Vậy ...