ko bít

\(B=|2x-4036|-|4039-2x|+2020.\)

\(=|2x-4036|-|2x-4039|+2020\)

\(\Rightarrow B\le|2x-4036-2x+4039|+2020=|3|+2020=2023\)

\(\Rightarrow B_{max}=3\Leftrightarrow\left(2x-4036\right)\left(2x-4039\right)\ge0\)

Chia 2 trường hợp cùng dương và cùng âm rồi tìm ra x nốt nhé

#)Giải :

Bài 1 :

\(A=\frac{15}{6.16}+\frac{15}{16.26}+\frac{15}{26.36}\)

\(A=3\left(\frac{5}{6.16}+\frac{5}{16.26}+\frac{5}{26.36}\right)\)

\(A=3\left(\frac{1}{6}-\frac{1}{16}+\frac{1}{16}-\frac{1}{26}+\frac{1}{26}-\frac{1}{36}\right)\)

\(A=3\left(\frac{1}{6}-\frac{1}{36}\right)\)

\(A=3.\frac{5}{36}\)

\(A=\frac{15}{36}\)

\(B=\frac{33}{6.16}-\frac{63}{16.26}+\frac{93}{26.36}\)

\(B=\frac{3.11}{6.16}-\frac{3.21}{16.26}+\frac{3.31}{26.36}\)

\(B=\frac{1}{2.16}.11+\frac{1}{2.16}.\frac{3.21}{13}+\frac{3.31}{26.36}\)

\(B=\frac{1}{32}.\frac{143-63}{13}+\frac{3.31}{26.12.3}\)

\(B=\frac{1}{32}.\frac{80}{13}+\frac{31}{36.12}\)

\(B=\frac{5}{26}+\frac{31}{26.12}=\frac{60+31}{26.12}=\frac{91}{26.12}=\frac{13.7}{13.2.12}=\frac{7}{24}\)

Bài 2 :

a)\(28.157-28.267+28.10\)

\(=28\left(157-267+10\right)\)

\(=28.-100\)

\(=-2800\)

b)\(\left(-5\right).4.\left(-2\right).23.\left(-25\right)\)

\(=\left[4.\left(-25\right)\right].\left[\left(-5\right).2\right].23\)

\(=\left(-100\right).\left(-10\right).23\)

\(=23000\)

2)

28 . 157  -  28 .267 + 28.10 

=28.( 157 - 267 + 10 )

= 28 . -100

= -2800

3)

[(-2).(-5)]. 4 .-25.23

= 10 . -100 .23

=-23000

còn câu 1 mik suy nghĩ đã

Số phần tử của M = \(2012-1985+1=28\)

số phần tử là

\(\left(2012-1985\right):1+1=28\)

\(\Rightarrow28\)là phần tử tập hợp M

\(5x^2-4x+2xy+y^2-5=\left(2x-1\right)^2+\left(x+y\right)^2-6\ge-6\)

Dấu "=" xra khi \(\hept{\begin{cases}2x-1=0\\x+y=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=\frac{1}{2}\\y=-x=-\frac{1}{2}\end{cases}}}\)

TL:

\(=4x^2-4x+1+x^2+2xy+y^2-6\) 

 \(=\left(2x-1\right)^2+\left(x+y\right)^2-6\) \(\ge-6\forall x;y\in R\) 

Dấu "=" xảy ra <=>  \(\left(2x-1\right)^2=0\) và  \(\left(x+y\right)^2=0\) 

                       <=>\(x=\frac{1}{2}\)   và   \(y=\frac{-1}{2}\) (thay x=1/2) 

Vậy........

hc tốt

#)Giải :

\(123-5\left(x+4\right)=38\)

\(\Leftrightarrow5\left(x+4\right)=85\)

\(\Leftrightarrow x+4=17\)

\(\Leftrightarrow x=13\)

\(\left(3x-16\right)343=2.2401\)

\(\Leftrightarrow\left(3x-16\right)343=4802\)

\(\Leftrightarrow3x-16=14\)

\(\Leftrightarrow3x=30\)

\(\Leftrightarrow x=10\)

\(\left(2600+6400\right)-3x=1200\)

\(\Leftrightarrow9000-3x=1200\)

\(\Leftrightarrow3x=7800\)

\(\Leftrightarrow x=2600\)

P/s : Câu cuối cùng lỗi chỗ b4 ph k ? hay là tìm cả b và x thế ???

\(123-5\left(x+4\right)=38\)

\(=>123-5x-20=38\)

\(=>-5x=-123+20+38\)

\(=>-5x=-65\)

\(=>x=13\)

b, \(\left(3x-16\right)343=2.2401\)

=>1029x -5488= 4802

=> 1029x=5488+4802

=> 1029x= 10290

=> x=10

c, (2600+6400)-3x=1200

=> 9000-3x=1200

=> -3x= -9000+1200

=> -3x= -7800

=> x=2600

d,

k hiểu đề

\(\frac{x+2}{327}+\frac{x+3}{326}+\frac{x+4}{325}+\frac{x+5}{324}+\frac{x+349}{5}=0\)

\(\Leftrightarrow\frac{x+2}{327}+1+\frac{x+3}{326}+1+\frac{x+4}{325}+1+\frac{x+5}{324}+1+\frac{x+349}{5}-4=0\)

\(\Leftrightarrow\frac{x+329}{327}+\frac{x+329}{326}+\frac{x+329}{325}+\frac{x+329}{324}+\frac{x+329}{5}=0\)

\(\Leftrightarrow\left[x+329\right]\cdot\left[\frac{1}{327}+\frac{1}{326}+\frac{1}{325}+\frac{1}{324}+\frac{1}{5}\right]=0\)

\(\Leftrightarrow x+329=0\) vì \(\frac{1}{327}+\frac{1}{326}+\frac{1}{325}+\frac{1}{324}+\frac{1}{5}>0\)

\(\Leftrightarrow x=-329\)

Vậy x = -329

Trả lời:

\(\frac{x+2}{327}+\frac{x+3}{326}+\frac{x+4}{325}+\frac{x+5}{324}+\frac{x+349}{5}=0\)

\(\frac{x+2}{327}+1+\frac{x+3}{326}+1+\frac{x+4}{325}+1+\frac{x+5}{324}+1+\frac{x+349}{5}-4=0\)

\(\frac{x+329}{327}+\frac{x+329}{326}+\frac{x+329}{325}+\frac{x+329}{324}+\frac{x+329}{5}=0\)

\(\left(x+329\right)\left(\frac{1}{327}+\frac{1}{326}+\frac{1}{325}+\frac{1}{324}+\frac{1}{5}\right)=0\)

\(\Leftrightarrow x+329=0\Leftrightarrow x=-329\)

Vậy \(x=-329\)

     ~ Học tốt ~

đề là \(x^2-\frac{1}{x^2}\)hay là \(x^2+\frac{1}{x^2}\)vậy? Xem lại đề thử xem!

\(x^2+\frac{1}{x^2}+y^2+\frac{1}{y^2}=4\)

\(\Leftrightarrow\left(x^2-2+\frac{1}{x^2}\right)+\left(y^2-2+\frac{1}{y^2}\right)=0\)

\(\Leftrightarrow\left(x-\frac{1}{x}\right)^2+\left(y-\frac{1}{y}\right)^2=0\)

\(\Leftrightarrow\left(x;y\right)=\left(1;1\right);\left(1;-1\right);\left(-1;1\right);\left(-1;-1\right)\) 

#)Giải :

a)\(\left(x-149\right)\div35=270\)

\(\Leftrightarrow x-149=9450\)

\(\Leftrightarrow x=9599\)

\(a,(x-149):35=270\)

\(\Leftrightarrow x-149=9450\)

\(\Leftrightarrow x=9450+149=9599\)

Vậy x = 9599

\(b,(2009-x):29+32=100\)

\(\Leftrightarrow(2009-x):29=68\)

\(\Leftrightarrow2009-x=1972\)

\(\Leftrightarrow x=2009-1972=37\)

Vậy x = 37

\(c,x+472=918\)

\(\Leftrightarrow x=918-472=446\)

Vậy x = 446