Bài 4:

\(P=\dfrac{x^2-2x+2022}{x^2}=\dfrac{2022x^2-2.2022x+2022^2}{2022x^2}=\dfrac{\left(x^2-2.2022x+2022^2\right)+2021x^2}{2022x^2}=\dfrac{\left(x-2022\right)^2}{2022x^2}+\dfrac{2021}{2022}\ge\dfrac{2021}{2022}\)\(P_{min}=\dfrac{2021}{2022}\Leftrightarrow x=2022\)

- 1999 . 19981998 + 19991999 . 1998

= - 1999 . 1998 . 10001 + 1999 . 10001 . 1998

= 10001 . 1998 ( - 1999 + 1999 )

= 10001 . 1998 . 0

= 0

-1999.19981998+19991999.1998=0

hok tốt

=-1998(19981998-19991999)

=-1998.-1

=1998

kb vs **** cho mik nha

Đề bị lỗi hiển thị rồi bạn.

\(\frac{x}{4}=\frac{y}{2}=\frac{z}{-3}\)= \(\frac{x+y+z}{4+2-3}\)= \(\frac{240}{3}=80\)

ta có 

\(\frac{x}{4}=80=>x=320\)

\(\frac{y}{2}=80=>y=160\)

\(\frac{z}{-3}=80=>z=-240\)

quỳnh vũ ơi bạn nhầm đề rồi ý mình xyz la x.y.z

giup minh voi may ban oi 

a: =-3/4+1/2-1/13+3/13=-1/4+2/13=-13/52+8/52=-5/52

b: =10/11+1/11-1/8=1-1/8=7/8

c: =4(2,86+3,14)-30,05+9x0,75

=24-30,05+6,75

=0,7

Xét x=1,x=2,1<x<2,x<1,x>2 

cam on ban nha. ban hoc truong nao vay?

a: \(=\dfrac{1}{3}+\dfrac{1}{8}-\dfrac{3}{16}\cdot4=\dfrac{11}{24}-\dfrac{3}{64}=\dfrac{88}{192}-\dfrac{9}{192}=\dfrac{79}{192}\)

b: \(=\dfrac{5}{8}\cdot\left(-16\right)=-10\)

c: \(=\left(\dfrac{8}{5}+\dfrac{1}{3}\right):\dfrac{29}{5}-\dfrac{4}{81}\cdot9\)

\(=\dfrac{29}{15}\cdot\dfrac{5}{29}-\dfrac{4}{9}=\dfrac{1}{3}-\dfrac{4}{9}=-\dfrac{1}{9}\)

\(c,=4\cdot\left(-\dfrac{1}{8}\right)-2\cdot\dfrac{1}{4}-\dfrac{3}{2}+1=-\dfrac{1}{2}-\dfrac{1}{2}-\dfrac{3}{2}+1=-\dfrac{3}{2}\\ \left(5x+\dfrac{2}{3}\right)^2=\dfrac{1}{4}\Rightarrow\left[{}\begin{matrix}5x+\dfrac{2}{3}=\dfrac{1}{2}\\5x+\dfrac{2}{3}=-\dfrac{1}{2}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}5x=-\dfrac{1}{6}\\5x=-\dfrac{7}{6}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{30}\\x=-\dfrac{7}{30}\end{matrix}\right.\\ c,x:2=\left(-4\right):5\Rightarrow x=-\dfrac{4}{5}\cdot2=-\dfrac{8}{5}\)