*câu 5
`A=sqrt{x^2+6x+9}+sqrt{x^2-2x+1}`
`=sqrt{(x+3)^2}+\sqrt{(x-1)^2}`
`=|x+3|+|x-1|`
`=|x+3|+|1-x|`
Áp dụng BĐT `|A|+|B|>=|A+B|`
`=>A>=|x+3+1-x|=4`
Dấu "=" `<=>AB>=0`
`<=>(x+3)(1-x)>=0`
`<=>(x+3)(x-1)<=0`
`<=>-3<=x<=1`
Vậy `min_A=4<=>-3<=x<=1`
câu 5:
\(A=\sqrt{x^2+6x+9}+\sqrt{x^2-2x+1}\)
\(A=\sqrt[]{\left(x+3\right)^2}+\sqrt{\left(x-1\right)^2}\)
\(A=\left|x+3\right|+|x-1|\)\(=\left|x+3\right|+\left|1-x\right|\ge\left|x+3+1-x\right|=4\)
=>\(A\ge4\)
4.2) Ta có: \(sinMFE=\dfrac{ME}{EF}\Rightarrow\dfrac{ME}{EF}=\dfrac{3}{4}\Rightarrow\dfrac{4}{EF}=\dfrac{3}{4}\Rightarrow EF=\dfrac{16}{3}\)
Ta có: \(ME^2=EI.EF\Rightarrow EI=\dfrac{ME^2}{EF}=\dfrac{4^2}{\dfrac{16}{3}}=3\)
\(\Rightarrow FI=EF-EI=\dfrac{16}{3}-3=\dfrac{7}{3}\)
Ta có: \(MI^2=EI.FI=3.\dfrac{7}{3}=7\Rightarrow MI=\sqrt{7}\)
b) Áp dụng hệ thức lượng \(\Rightarrow\left\{{}\begin{matrix}PM.PE=PI^2\\MQ.QF=QI^2\end{matrix}\right.\)
\(\Rightarrow PM.PE+QM.QF=PI^2+QI^2=PQ^2\)
Ta có: \(\angle IPM=\angle IQM=\angle PMQ=90\)
\(\Rightarrow IPMQ\) là hình chữ nhật \(\Rightarrow PQ=MI\)
\(\Rightarrow PM.PE+QM.QF=MI^2\)
