\(\frac{3388}{8228}\),\(ƯCLN\left[3388,8228\right]=484\)

\(\Rightarrow\frac{3388}{8228}=\frac{3388:484}{8228:484}=\frac{7}{17}\)

\(\frac{6930}{9450},ƯCLN\left[6930,8228\right]=630\)

\(\Rightarrow\frac{6930}{9450}=\frac{6930:630}{9450:630}=\frac{11}{15}\)

Lời giải:

a. $ƯC(a,b)\in Ư(36)=\left\{\pm 1; \pm 2; \pm 3; \pm 4; \pm 6; \pm 9; \pm 12; \pm 18; \pm 36\right\}$

b. $Ư(a,b)\in Ư(50)=\left\{\pm 1; \pm 2; \pm 5; \pm 10; \pm 25; \pm 50\right\}$
Suy ra ước có 2 chữ số của $a,b$ là:
$\left\{\pm 10; \pm 25; \pm 50\right\}$

bn bấm máy tính, là nó ra ngay

a: UCLN(16;42)=2

UC(16;42)={1;2}

b: UCLN(168;120;144)=24

UC(168;120;144)={1;2;3;4;6;8;12;24}

a) \(\left|x\right|+x\)

Vì \(\left|x\right|\ge0\) nên ta có 3TH:

TH1: \(x>0\)

\(\Rightarrow\left|x\right|+x=2x\)

TH2: \(x=0\)

\(\Rightarrow\left|x\right|+x=0\)

TH3: \(x< 0\)

\(\Rightarrow\left|x\right|+x=0\)

b) \(N=\left|x\right|:x\)

Vì \(\left|x\right|\ge0\) và \(x\ne0\) nên ta có 2TH:

TH1: \(x>0\)

\(\Rightarrow\left|x\right|:x=1\)

TH2: \(x< 0\)

\(\Rightarrow\left|x\right|:x=-1\)

a: \(M=\left|x\right|+x=\left[{}\begin{matrix}x+x=2x\left(x\ge0\right)\\-x+x=0\left(x< 0\right)\end{matrix}\right.\)

b: \(N=\left|x\right|:x=\pm1\)

27/36=3/4                                       75/100=3/4

26/91=2/7                                         1515/1919=15/19

27/36=3/4                                       75/100= 3/4

26/91=   2/7                                      1515/1919= 15/19

a: \(A=\dfrac{2\cdot8^4\cdot27^2+44\cdot6^9}{2^7\cdot6^7+2^7\cdot40\cdot9^4}\)

\(=\dfrac{2\cdot2^{12}\cdot3^6+2^2\cdot11\cdot2^9\cdot3^9}{2^7\cdot3^7\cdot2^7+2^7\cdot2^3\cdot5\cdot3^8}\)

\(=\dfrac{2^{13}\cdot3^6+2^{11}\cdot3^9\cdot11}{2^{14}\cdot3^7+2^{10}\cdot5\cdot3^8}\)

\(=\dfrac{2^{11}\cdot3^6\left(2^2+3^3\cdot11\right)}{2^{10}\cdot3^7\left(2^4+5\cdot3\right)}\)

\(=\dfrac{2\cdot301}{3\cdot31}=\dfrac{602}{93}\)