Bài 1:

Ta có: \(-\left|2x+6\right|\le0\)

\(\Rightarrow9-\left|2x+6\right|\le9\)

\(\Rightarrow5-\left(9-\left|2x+6\right|\right)\le5\)

Dấu "=" xảy ra <=> 2x + 6 = 9 <=> x = \(\frac{3}{2}\)

Vậy GTNN của A là 5 khi x = \(\frac{3}{2}\)

Bài 2:

Ta có: \(\left|2x+6\right|\ge0\)

\(\Rightarrow\left|2x+6\right|-3\ge-3\)

\(\Rightarrow-5-\left(\left|2x+6\right|-3\right)\ge-5\)

Dấu "=" xảy ra <=> 2x + 6 = 3 <=> x = \(-\frac{3}{2}\)

Vậy GTLN của A là -5 khi x = \(-\frac{3}{2}\)

B = |\(x\) + 20| + |y - 21| + 2020

 |\(x\) + 20| ≥ 0 ∀ \(x\); |y - 21| ≥ 0 ∀ y

B = |\(x\) + 20| + |y - 21| + 2020 ≥ 2020

B ≥ 2020 dấu bằng xảy ra khi \(\left\{{}\begin{matrix}x+20=0\\y-21=0\end{matrix}\right.\) ⇒ \(\left\{{}\begin{matrix}x=-20\\y=21\end{matrix}\right.\)

B_min = 2020 khi (\(x;y\)) = (-20; 21)

A = |\(x\) + 19| + 1980 

|\(x\) + 19| ≥ 0 \(\forall\) \(x\)

|\(x\) + 19| + 1980 ≥ 1980 ∀ \(x\)

A ≥ 1980 dấu bằng xảy khi \(x\) + 19 = 0 hay \(x\) = -19

Kết luận A đạt giá trị nhỏ nhất là 1980 khi \(x\) = -19

C = 3.|\(x-15\)| + (y + 13)² - 2175

|\(x\) - 15| ≥ 0 ∀ \(x\); (y + 13)² ≥ 0 ∀ y

C = 3.|\(x\) - 15| + (y + 13)² - 2175 ≥ - 2175

C ≥ - 2175 dấu bằng xảy ra khi \(\left\{{}\begin{matrix}x-15=0\\y+13=0\end{matrix}\right.\) ⇒ \(\left\{{}\begin{matrix}x=15\\y=-13\end{matrix}\right.\)

Vậy C_min = -2175 khi  (\(x\); y) = (15; -13)

\(d,\dfrac{5}{7}+\dfrac{9}{23}+-\dfrac{12}{7}+\dfrac{14}{23}\)

\(=\left(\dfrac{5}{7}+\dfrac{-12}{7}\right)+\left(\dfrac{9}{23}+\dfrac{14}{23}\right)\)

\(=-\dfrac{7}{7}+\dfrac{23}{23}\)

\(=\left(-1\right)+1=0\)

\(e,\dfrac{3}{17}+-\dfrac{5}{13}+-\dfrac{18}{35}+\dfrac{14}{17}+\dfrac{17}{-35}+-\dfrac{8}{13}\)

\(=\left(\dfrac{3}{17}+\dfrac{14}{17}\right)+\left(\dfrac{-5}{13}+\dfrac{-8}{13}\right)+\left(\dfrac{-18}{35}+\dfrac{-17}{35}\right)\)

\(=\dfrac{17}{17}+\dfrac{-13}{13}+-\dfrac{35}{35}\)

\(=1+\left(-1\right)+\left(-1\right)=0+\left(-1\right)=-1\)

\(f,\dfrac{-3}{8}. \dfrac{1}{6}+\dfrac{3}{-8}.\dfrac{5}{6}+\dfrac{-10}{16}\)

\(=\dfrac{-3}{8}.\dfrac{1}{6}+\dfrac{-3}{8}.\dfrac{5}{6}+\dfrac{-10}{16}\)

\(=\dfrac{-3}{8}.\left(\dfrac{1}{6}+\dfrac{5}{6}\right)+-\dfrac{10}{16}\)

=\(\dfrac{-3}{8}.1+\dfrac{-10}{16}\)

\(=\dfrac{-3}{8}+\dfrac{-10}{16}\)

\(=\dfrac{-6}{16}+\dfrac{-10}{16}=\dfrac{-16}{16}=-1\)

\(g,\dfrac{-4}{11}.\dfrac{5}{15}.\dfrac{11}{-4}=\dfrac{-4}{11}.\dfrac{5}{15}.\dfrac{-11}{4}\)

\(=\left(\dfrac{-4}{11}.\dfrac{-11}{4}\right).\dfrac{5}{15}\)

\(=1.\dfrac{5}{15}=1.\dfrac{1}{3}=\dfrac{1}{3}\)

\(h,\dfrac{7}{36}-\dfrac{8}{-9}+\dfrac{-2}{3}=\dfrac{7}{36}-\dfrac{-8}{9}+\dfrac{-2}{3}\)

\(=\dfrac{7}{36}-\dfrac{-32}{36}+\dfrac{-24}{36}=\dfrac{7-\left(-32\right)+\left(-24\right)}{36}\)

\(=\dfrac{15}{56}=\dfrac{5}{12}\)

Tick mình nha ^^

d.\(\dfrac{5}{7}\)+\(\dfrac{9}{23}\)+\(\dfrac{12}{7}\)+\(\dfrac{14}{23}\)=(\(\dfrac{5}{7}\)+\(\dfrac{12}{7}\))+(\(\dfrac{9}{23}\)+\(\dfrac{14}{23}\))

                            =\(\dfrac{17}{7}\)+ 1 = \(\dfrac{24}{7}\)

e.\(\dfrac{3}{17}\)+\(\dfrac{-5}{13}\)+\(\dfrac{-18}{35}\)+\(\dfrac{14}{17}\)+\(\dfrac{17}{-35}\)+\(\dfrac{-8}{13}\)

=(\(\dfrac{3}{17}\)+\(\dfrac{14}{17}\))+(\(\dfrac{-5}{13}\)+\(\dfrac{-8}{13}\))+(\(\dfrac{-18}{35}\)+\(\dfrac{17}{-35}\))

= 1+ (-1) + (-1) = -1

f. \(\dfrac{-3}{8}\).\(\dfrac{1}{6}\)+\(\dfrac{3}{-8}\).\(\dfrac{5}{6}\)+\(\dfrac{-10}{16}\)=\(\dfrac{-3}{8}\)(\(\dfrac{1}{6}\)+\(\dfrac{5}{6}\)) + \(\dfrac{-5}{8}\)

                                   =\(\dfrac{-3}{8}\)+\(\dfrac{-5}{8}\)= -1

g. \(\dfrac{-4}{11}\).\(\dfrac{5}{15}\).\(\dfrac{11}{-4}\)=\(\dfrac{5}{15}\)

h.\(\dfrac{7}{36}\)-\(\dfrac{8}{-9}\)+\(\dfrac{-2}{3}\)= \(\dfrac{7}{36}\)-\(\dfrac{-32}{36}\)+\(\dfrac{-24}{36}\)

                       =\(\dfrac{-49}{36}\)

d: \(\dfrac{5}{7}+\dfrac{9}{23}+\dfrac{-12}{7}+\dfrac{14}{23}=1-1=0\)

e: \(\dfrac{3}{17}+\dfrac{-5}{13}+\dfrac{-18}{35}+\dfrac{14}{17}+\dfrac{17}{-35}+\dfrac{-8}{13}=-1-1+1=-1\)

số nước cần phải đổ vào bể để bể đầy là

\(1-\frac{3}{4}=\frac{1}{4}\)(bể)

khi mở vòi nước thì số thời gian bể sẽ đầy là \(\frac{1}{4}:\frac{1}{8}=2\)(giờ)

vậy....

12.5* gọi hai số đó là a và b

ta có:a-b=9 (1)

\(\frac{7}{9}a=\frac{28}{33}b\)

=> \(a=\frac{28}{33}b:\frac{7}{9}\)

\(=>a=\frac{12}{11}b\)(2)

thay (2) vào (1) ta đc

\(\frac{12}{11}b-b=9\)

=> \(\frac{1}{11}b=9\)

=>b=99

=> a=99+9=108

vậy...

A = | x + 5 | + 20

Ta có: | x + 5 | > 0

=> | x + 5 | + 20 > 20

=> GTNN của A là 20

<=> x + 5 = 0

<=> x = -5.

B = | x - 3 | - 10

Ta có: | x - 3 | > 0

=> | x - 3 | - 10 > -10

=> GTNN của B là -10

<=> x - 3 = 0

<=> x = 3.

GTNN A=20

          B=-10

A=20(tại x= -5)

B=0(tại x=13)

sai thì thôi nhé