Question

[Nguyễn Trần Phương Uyên]

Answer 1

[Nguyễn Lê Phước Thịnh]

Bài 4:

\(\left(x-\dfrac{2}{5}\right)^2>=0\forall x\)

\(\left(y+20\right)^{10}>=0\forall y\)

Do đó: \(\left(x-\dfrac{2}{5}\right)^2+\left(y+20\right)^{10}>=0\forall x,y\)

=>\(A=\left(x-\dfrac{2}{5}\right)^2+\left(y+20\right)^{10}+2010>=2010\forall x,y\)

Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}x-\dfrac{2}{5}=0\\y+20=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=\dfrac{2}{5}\\y=-20\end{matrix}\right.\)

Bài 3:

\(\left(ad+bc\right)^2=4bacd\)

=>\(a^2d^2+b^2c^2+2adbc-4adbc=0\)

=>\(\left(ad\right)^2+\left(bc\right)^2-2adbc=0\)

=>(ad-bc)²=0

=>ad-bc=0

=>ad=bc

=>\(\dfrac{a}{b}=\dfrac{c}{d}\)

=>ĐPCM

Bài 2:

a: |2x-1|+3=15

=>|2x-1|=15-3=12

=>\(\left[{}\begin{matrix}2x-1=12\\2x-1=-12\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{13}{2}\\x=-\dfrac{11}{2}\end{matrix}\right.\)

b: \(\left|x-3,2\right|+\left|2x-\dfrac{1}{5}\right|=x+3\)(1)

TH1: x<1/10

(1) sẽ trở thành \(\dfrac{1}{5}-2x+3,2-x=x+3\)

=>-3x+3,4=x+3

=>-4x=3-3,4=-0,4

=>x=0,1(loại)

TH2: 1/10<=x<3,2

(1) sẽ trở thành \(2x-\dfrac{1}{5}+3,2-x=x+3\)

=>x+3=x+3(luôn đúng)

TH3: x>=3,2

(1) sẽ trở thành \(x-3,2+2x-\dfrac{1}{5}=x+3\)

=>3x-3,4=x+3

=>2x=6,4

=>x=3,2(nhận)

 Vậy: 1/10<=x<=3,2