\(\dfrac{a}{b}=\dfrac{b}{d}=>b^2=ad\)

Ta có: 

\(VT=\dfrac{a^2+b^2}{b^2+d^2}=\dfrac{a^2+ad}{ad+d^2}=\dfrac{a\left(a+d\right)}{d\left(a+d\right)}=\dfrac{a}{d}=VP\)

Đặt: \(\dfrac{a}{b}=\dfrac{c}{d}=k=>\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)

Ta có:

\(VT=\dfrac{5a+3b}{5a-3b}=\dfrac{5bk+3b}{5bk-3b}=\dfrac{b\left(5k+3\right)}{b\left(5k-3\right)}=\dfrac{5k+3}{5k-3}\\ =\dfrac{d\left(5k+3\right)}{d\left(5k-3\right)}=\dfrac{5dk+3d}{5dk-3d}=\dfrac{5c+3d}{5c-3d}=VP\)

hết cứu

\(\dfrac{2}{15}-\dfrac{7}{10}\\ =\dfrac{4}{30}-\dfrac{21}{30}\\ =\dfrac{4-21}{30}\\ =\dfrac{-17}{30}\)

Đặt: \(\dfrac{a}{b}=\dfrac{c}{d}=k=>\left\{{}\begin{matrix}a=bk\\b=dk\end{matrix}\right.\)

Ta có: 

\(VT=\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{\left(bk\right)^2+\left(dk\right)^2}{b^2+d^2}=\dfrac{b^2k^2+d^2k^2}{b^2+d^2}\\ =\dfrac{k^2\left(b^2+d^2\right)}{b^2+d^2}=k^2\left(1\right)\)

\(VP=\dfrac{ac}{bd}=\dfrac{bk\cdot dk}{bd}=\dfrac{bd\cdot k^2}{bd}=k^2\left(2\right)\)

Từ (1) và (2) => \(\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{ac}{bd}\)

\(\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}+\dfrac{1}{100}\\ =\dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}+\dfrac{1}{6\cdot7}+\dfrac{1}{7\cdot8}+\dfrac{1}{8\cdot9}+\dfrac{1}{9\cdot10}+\dfrac{1}{100}\\ =\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}+\dfrac{1}{100}\\ =\dfrac{1}{4}-\dfrac{1}{10}+\dfrac{1}{100}\\ =\dfrac{25}{100}-\dfrac{10}{100}+\dfrac{1}{100}\\ =\dfrac{16}{100} =\dfrac{4}{25}\)

\(\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}+\dfrac{1}{100}\\ =\dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}+\dfrac{1}{6\cdot7}+\dfrac{1}{7\cdot8}+\dfrac{1}{8\cdot9}+\dfrac{1}{9\cdot10}+\dfrac{1}{100}\\ =\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}+\dfrac{1}{100}\\ =\dfrac{1}{4}-\dfrac{1}{10}+\dfrac{1}{100}\\ =\dfrac{4}{25}\)

\(a.\dfrac{2}{3}-\left(-\dfrac{1}{2}-x\right)=-\dfrac{4}{5}\\ \dfrac{2}{3}+\dfrac{1}{2}+x=-\dfrac{4}{5}\\ x=-\dfrac{4}{5}-\dfrac{2}{3}-\dfrac{1}{2}\\ x=-\dfrac{59}{30}\\ b.\left(-x-3\dfrac{1}{4}\right)-\left(1\dfrac{2}{3}-2\dfrac{3}{4}\right)=\dfrac{-5}{6}\\ \left(-x-\dfrac{13}{4}\right)-\left(\dfrac{5}{3}-\dfrac{11}{4}\right)=\dfrac{-5}{6}\\ -x-\dfrac{13}{4}-\dfrac{5}{3}+\dfrac{11}{4}=-\dfrac{5}{6}\\ -x-\dfrac{5}{3}-\dfrac{1}{2}=-\dfrac{5}{6}\\ x=\dfrac{5}{6}-\dfrac{5}{3}-\dfrac{1}{2}\\ x=-\dfrac{4}{3}\\ c.\dfrac{8}{23}\cdot\dfrac{46}{24}-\dfrac{1}{2}x=\dfrac{1}{3}\\ \dfrac{2}{3}-\dfrac{1}{2}x=\dfrac{1}{3}\\ \dfrac{1}{2}x=\dfrac{2}{3}-\dfrac{1}{3}=\dfrac{1}{3}\\ x=\dfrac{1}{3}:\dfrac{1}{2}=\dfrac{2}{3}\\ d.\dfrac{x-1}{16}=\dfrac{3}{x+1}\\ \left(x-1\right)\left(x+1\right)=3\cdot16=48\\ x^2-1=48\\ x^2=49\\ x^2=7^2\\ x=\pm7\)

\(e.\left(1,2\right)^3x^2=\left(1,2\right)^5\\ x^2=\dfrac{\left(1,2\right)^5}{\left(1,2\right)^3}\\ x^2=\left(1,2\right)^2\\ x=\pm1,2\\ f.\left(\dfrac{2}{3}x-\dfrac{1}{4}\right)^2=4\\ \left(\dfrac{2}{3}x-\dfrac{1}{4}\right)^2=2^2\\TH1:\dfrac{2}{3}x-\dfrac{1}{4}=2\\ \dfrac{2}{3}x=2+\dfrac{1}{4}=\dfrac{9}{4}\\ x=\dfrac{9}{4}:\dfrac{2}{3}=\dfrac{27}{8}\\ TH2:\dfrac{2}{3}x-\dfrac{1}{4}=-2\\ \dfrac{2}{3}x=-2+\dfrac{1}{4}=-\dfrac{7}{4}\\ x=\dfrac{-7}{4}:\dfrac{2}{3}=-\dfrac{21}{8}\\ g.\left(\dfrac{1}{6}x-3\right)^2=\dfrac{4}{9}\\ \left(\dfrac{1}{6}x-3\right)^2=\left(\dfrac{2}{3}\right)^2\\ TH1:\dfrac{1}{6}x-3=\dfrac{2}{3}\\ \dfrac{1}{6}x=\dfrac{2}{3}+3=\dfrac{11}{3}\\ x=\dfrac{11}{3}:\dfrac{1}{6}=22\\ TH2:\dfrac{1}{6}x-3=-\dfrac{2}{3}\\ \dfrac{1}{6}x=-\dfrac{2}{3}+3=\dfrac{7}{3}\\ x=\dfrac{7}{3}:\dfrac{1}{6}=14\)

|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

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