|5-4x|=3-x

=>|4x-5|=3-x

=>\(\left\{{}\begin{matrix}3-x>=0\\\left(4x-5\right)^2=\left(3-x\right)^2\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x< =3\\\left(4x-5-x+3\right)\left(4x+5+x-3\right)=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x< =3\\\left(3x-2\right)\left(5x+2\right)=0\end{matrix}\right.\Leftrightarrow x\in\left\{\dfrac{2}{3};-\dfrac{2}{5}\right\}\)

\(\left|6-3x\right|=6+x\)

=>|3x-6|=x+6

=>\(\left\{{}\begin{matrix}x+6>=0\\\left(3x-6\right)^2=\left(x+6\right)^2\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x>=-6\\\left(3x-6-x-6\right)\left(3x-6+x+6\right)=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x>=-6\\4x\left(2x-12\right)=0\end{matrix}\right.\Leftrightarrow x\in\left\{0;6\right\}\)

|4-x|=6

=>|x-4|=6

=>\(\left[{}\begin{matrix}x-4=6\\x-4=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=10\\x=-2\end{matrix}\right.\)

|3-x|=8

=>|x-3|=8

=>\(\left[{}\begin{matrix}x-3=8\\x-3=-8\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=11\\x=-5\end{matrix}\right.\)

|4-x|=2-x

=>|x-4|=2-x

=>\(\left\{{}\begin{matrix}2-x>=0\\\left(x-4\right)^2=\left(2-x\right)^2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x< =2\\\left(x-4-2+x\right)\left(x-4+2-x\right)=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x< =2\\\left(2x-6\right)\cdot\left(-2\right)=0\end{matrix}\right.\)

=>\(x\in\varnothing\)

|3+2x|=2x+5

=>|2x+3|=2x+5

=>\(\left\{{}\begin{matrix}2x+5>=0\\\left(2x+5\right)^2=\left(2x+3\right)^2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>=-\dfrac{5}{3}\\4x^2+20x+25=4x^2+12x+9\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x>=-\dfrac{5}{3}\\20x+25=12x+9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>=-\dfrac{5}{3}\\x=-2\end{matrix}\right.\Leftrightarrow x\in\varnothing\)

Độ dài 1 đường cong là:

$44:4=11$ (đvi độ dài)

Chu vi bốn hình quạt tròn là:

$11\times4=44$ (đvi độ dài)

Từ bốn hình quạt tròn đó ta ghép được 1 hình tròn. Khi đó:

Độ dài cạnh hình vuông là:

$44:\frac{22}{7}=14$ (đvi độ dài)

Diện tích hình vuông là:

$14\times14=196$ (đvi diện tích)

Diện tích bốn hình quạt tròn là:

$\frac{14}{2}\times\frac{14}{2}\times\frac{22}{7}=154$ (đvi diện tích)

Diện tích của phần bên trong đường cong là:

$196-154=42$ (đvi diện tích)

4 curves form 1 circle

The radius of the circle is:

The area of ​​the circle is:

The length of the side of the square is:

7.2 = 14

The area of ​​the square is:

14.14 = 196

The area of the region bounded inside the curves is:

196 - 154 = 42

Bài 1

a: ĐKXĐ: \(n\ne4\)

Để A nguyên thì \(3n+9⋮n-4\)

=>\(3n-12+21⋮n-4\)

=>\(21⋮n-4\)

=>\(n-4\in\left\{1;-1;3;-3;7;-7;21;-21\right\}\)

=>\(n\in\left\{5;3;7;1;11;-3;25;-17\right\}\)

b: ĐKXĐ: n<>1/2

Để B nguyên thì \(6n+5⋮2n-1\)

=>\(6n-3+8⋮2n-1\)

=>\(8⋮2n-1\)

mà 2n-1 lẻ(do n nguyên)

nên \(2n-1\in\left\{1;-1\right\}\)

=>\(n\in\left\{1;0\right\}\)

Bài 2:

a: \(\left|x-\dfrac{1}{2}\right|>=0\forall x\)

=>\(-\dfrac{1}{2}\left|x-2\right|< =0\forall x\)

=>\(A=-\dfrac{1}{2}\left|x-2\right|+\dfrac{3}{2}< =\dfrac{3}{2}\forall x\)

Dấu '=' xảy ra khi x-2=0

=>x=2

b: \(\left|\dfrac{1}{2}-x\right|>=0\forall x\)

=>\(-2,3\left|\dfrac{1}{2}-x\right|< =0\forall x\)

=>\(D=-2,3\left|\dfrac{1}{2}-x\right|+2< =2\forall x\)

Dấu '=' xảy ra khi 1/2-x=0

=>x=1/2

Bài 1: 

\(A=\dfrac{3n+9}{n-4}=\dfrac{3n-12}{n-4}+\dfrac{21}{n-4}=3+\dfrac{21}{n-4}\)

Để A nguyên thì \(\dfrac{21}{n-4}\) phải nguyên hay \(\left(n-4\right)\inƯ\left(21\right)=\left\{1;-1;3;-3;7;-7;21;-21\right\}\)

\(\Rightarrow n\in\left\{5;3;7;1;11;-3;25;-17\right\}\) (thoả mãn điều kiện)

Vậy...

\(B=\dfrac{6n+5}{2n-1}=\dfrac{6n-3}{2n-1}+\dfrac{8}{2n-1}=3+\dfrac{8}{2n-1}\)

Để B nguyên thì \(\dfrac{8}{2n-1}\) phải nguyên hay \(\left(2n-1\right)\inƯ\left(8\right)=\left\{1;-1;2;-2;4;-4;8;-8\right\}\)

Mặt khác: Vì n nguyên nên 2n-1 là số lẻ

Do đó: \(\left(2n-1\right)\in\left\{1;-1\right\}\)

\(\Rightarrow n\in\left\{1;0\right\}\)

Vậy....

|3x+4|=x+2

=>\(\left\{{}\begin{matrix}x+2>=0\\\left(3x+4\right)^2=\left(x+2\right)^2\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x>=-2\\\left(3x+4-x-2\right)\left(3x+4+x+2\right)=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x>=-2\\\left(2x+2\right)\left(4x+6\right)=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x>=-2\\x\in\left\{-1;-\dfrac{3}{2}\right\}\end{matrix}\right.\Leftrightarrow x\in\left\{-1;-\dfrac{3}{2}\right\}\)

|5x-6|=4-x

=>\(\left\{{}\begin{matrix}4-x>=0\\\left(5x-6\right)^2=\left(4-x\right)^2\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x< =4\\\left(5x-6-4+x\right)\left(5x-6+4-x\right)=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x< =4\\\left(6x-10\right)\left(4x-2\right)=0\end{matrix}\right.\Leftrightarrow x\in\left\{\dfrac{5}{3};\dfrac{1}{2}\right\}\)

|5-2x|=x-3

=>|2x-5|=x-3

=>\(\left\{{}\begin{matrix}x-3>=0\\\left(2x-5\right)^2=\left(x-3\right)^2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>=3\\\left(2x-5\right)^2-\left(x-3\right)^2=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x>=3\\\left(2x-5-x+3\right)\left(2x-5+x-3\right)=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x>=3\\\left(x-2\right)\left(3x-8\right)=0\end{matrix}\right.\Leftrightarrow x\in\varnothing\)

|3-2x|=6+4x

=>|2x-3|=4x+6

=>\(\left\{{}\begin{matrix}4x+6>=0\\\left(4x+6\right)^2=\left(2x-3\right)^2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>=-\dfrac{3}{2}\\\left(4x+6-2x+3\right)\left(4x+6+2x-3\right)=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x>=-\dfrac{3}{2}\\\left(2x+9\right)\left(6x+3\right)=0\end{matrix}\right.\Leftrightarrow x=-\dfrac{1}{2}\)

|6-3x|=3x

=>|3x-6|=3x

=>|x-2|=x

=>\(\left\{{}\begin{matrix}x>=0\\\left(x-2\right)^2=x^2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>=0\\-4x+4=0\end{matrix}\right.\Leftrightarrow x=1\)

\(\left|3x+4\right|=x+2\\ \Rightarrow\left[{}\begin{matrix}3x+4=x+2\left(x\ge-\dfrac{4}{3}\right)\\3x+4=-\left(x+2\right)\left(x< -\dfrac{4}{3}\right)\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}3x-x=2-4\\3x+x=-2-4\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}2x=-2\\4x=-6\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-1\left(tm\right)\\x=-\dfrac{6}{4}=-\dfrac{3}{2}\left(tm\right)\end{matrix}\right.\)

______________________

\(\left|5x-6\right|=4-x\\ \Rightarrow\left[{}\begin{matrix}5x-6=4-x\left(x\ge\dfrac{6}{5}\right)\\5x-6=-\left(4-x\right)\left(x< \dfrac{6}{5}\right)\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}5x+x=4+6\\5x-x=-4+6\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}6x=10\\4x=2\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{10}{6}=\dfrac{5}{3}\left(tm\right)\\x=\dfrac{2}{4}=\dfrac{1}{2}\left(tm\right)\end{matrix}\right.\)

________________________

\(\left|5-2x\right|=x-3\\ \Rightarrow\left[{}\begin{matrix}5-2x=x-3\left(x\le\dfrac{5}{2}\right)\\5-2x=-\left(x-3\right)\left(x>\dfrac{5}{2}\right)\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}-2x-x=-3-5\\-2x+x=3-5\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}-3x=-8\\-x=-2\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{-8}{-3}=\dfrac{8}{3}\left(ktm\right)\\x=2\left(ktm\right)\end{matrix}\right.\)

\(\left|3-2x\right|=6+4x\\ \Rightarrow\left[{}\begin{matrix}3-2x=6+4x\left(x\le\dfrac{3}{2}\right)\\3-2x=-\left(6+4x\right)\left(x>\dfrac{3}{2}\right)\end{matrix}\right.\\\Rightarrow\left[{}\begin{matrix}4x+2x=3-6\\-2x+4x=-6-3\end{matrix}\right. \\ \Rightarrow\left[{}\begin{matrix}6x=-3\\2x=-9\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\left(tm\right)\\x=-\dfrac{9}{2}\left(ktm\right)\end{matrix}\right.\)

________________________

\(\left|6-3x\right|=3x\\ \Rightarrow\left[{}\begin{matrix}6-3x=3x\left(x\le2\right)\\6-3x=-3x\left(x>2\right)\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}3x+3x=6\\6=0\left(ktm\right)\end{matrix}\right.\\ \Rightarrow x=\dfrac{6}{6}=1\left(tm\right)\)

1 công nhân xây ngôi nhà đó hết:

$35\times168=5880$ (ngày)

28 công nhân xây ngôi nhà đó hết:

$5880:28=210$ (ngày)

Số công nhân cần có để hoàn thành công việc trong 14 ngày là:

\(56\cdot\dfrac{21}{14}=56\cdot\dfrac{3}{2}=84\left(người\right)\)

Số công nhân cần tăng thêm là:

84-56=28(người)

                         Giải:

Một công nhân hoàn thành công việc đó trong số ngày là:

                  21 x 56 = 1176 (ngày)

Để hoàn thành công việc trong 14 ngày cần số người là:

                     1176 : 14 =  84 (người)

Vậy để hoàn thành công việc trong 14 ngày cần bổ sung thêm số người là:

                      84 - 56 = 28 (người)

Đáp số:.....

a: \(0,5^{1000}=\left(0,5^5\right)^{200}=0,03125^{200}\)

mà \(0,03125< 0,625\)

nên \(0,5^{1000}< 0,625^{200}\)

c: \(A=2+2^2+...+2^{2022}\)

=>\(2A=2^2+2^3+...+2^{2023}\)

=>\(2A-A=2^2+2^3+...+2^{2023}-2-2^2-...-2^{2022}\)

=>\(A=2^{2023}-2\)

=>A<B

e: \(2020A=\dfrac{2020^{2024}-2020}{2020^{2024}-1}=1-\dfrac{2019}{2020^{2024}-1}\)

\(2020B=\dfrac{2020^{2024}+2020}{2020^{2024}+1}=1+\dfrac{2019}{2020^{2024}+1}\)

Vì \(-\dfrac{2019}{2020^{2024}-1}< 0< \dfrac{2019}{2020^{2024}+1}\)

nên \(-\dfrac{2019}{2020^{2024}-1}+1< \dfrac{2019}{2020^{2024}+1}+1\)

=>2020A<2020B

=>A<B

d: \(\left(-\dfrac{3}{2}\right)^{2024}=\left(\dfrac{3}{2}\right)^{2024};\left(-2\right)^{2024}=2^{2024}\)

mà 3/2<2

nên \(\left(-\dfrac{3}{2}\right)^{2024}< 2^{2024}\)

$\left(2\frac13-1,5\right):\left(-6\frac16+5\frac12\right)+2,75$

$=\left(\frac73-\frac32\right):\left(-\frac{37}{6}+\frac{11}{2}\right)+\frac{11}{4}$

$=\frac56:\frac{-2}{3}+\frac{11}{4}$

$=-\frac54+\frac{11}{4}=\frac32$

\(a.\dfrac{3}{7}=\dfrac{2x+1}{3x+5}\\ 3\left(3x+5\right)=7\left(2x+1\right)\\ 9x+15=14x+7\\ 14x-9x=15-7\\ 5x=8\\ x=\dfrac{8}{5}\\ b.\dfrac{x+1}{x-2}=\dfrac{3}{4}\\ 3\left(x-2\right)=4\left(x+1\right)\\ 3x-6=4x+4\\ 4x-3x=-6-4\\ x=-10\\ c.\dfrac{2x+3}{7}=\dfrac{4x-1}{15}\\ 15\left(2x+3\right)=7\left(4x+1\right)\\ 30x+45=28x+7\\ 30x-28x=7-45\\ 2x=-38\\ x=\dfrac{-38}{2}=-19\\ d.\dfrac{6x-5}{-7}=\dfrac{5x-3}{-5}\\ -5\left(6x-5\right)=-7\left(5x-3\right)\\ -30x+25=-35x+21\\ -30x+35x=21-25\\ 5x=-4\\ x=-\dfrac{4}{5}\)

Ở ý c chỗ tích chéo anh bị nhầm dấu rồi ạ.