bài 1 :

a)\(2,5.16,27.4+7,3=\left(2,5.4\right).16,27+7,3=10.16,27+7,3\)

\(=162,7+7,3=170\)

b) \(\frac{2}{3}+\frac{3}{4}-\frac{5}{6}=\left(\frac{2}{3}-\frac{5}{6}\right)+\frac{3}{4}=\frac{-1}{6}+\frac{3}{4}=\frac{7}{12}\)

c) \(17,6-5,3+16,8-7,6+15,3-6,8\)

\(=\left(17,6-7,6\right)+\left(-5,3+15,3\right)+\left(16,8-6,8\right)\)

\(=10+10+10=30\)

d)\(\frac{38}{11}+\left(13,16+\frac{6}{11}\right)=\frac{38}{11}+13,16+\frac{6}{11}\)

\(=\left(\frac{38}{11}+\frac{6}{11}\right)+13,16=4+13,16=17,16\)

\(2,45.46+8.0,75+54.2,45+0,5.8\\ =\left[2,45.46+54.2,45\right]\)

\(+\left[8.0,75+0,5.8\right]=\left[2,45.\left(46+54\right)\right]+\left[8.\left(0,75+0,5\right)\right]\)

\(=\left(2,45.100\right)+\left(8.1,25\right)=245+10=255\)

 bài 2 :

a) .....

b) \(1-\left(12,5+x-4,25\right):21,75=0\)

\(\Rightarrow1-\left(12,5+x-4,25\right)=21,75\)

\(\Rightarrow12,5+x-4,25=-20,75\\ \Rightarrow12,5+x=-16,5\)

\(\Rightarrow x=-29\)

cậu có thể tham khảo bài làm trên đây ạ, mn ai thấy đúng thì cho mk xin 1 t.i.c.k đúng ạ ,thank nhiều

_{Làm ơn giúp mình với !}

1. 12/18 < 13/17
2. 16/51 < 31/90

\(a)\)

Ta có:

\(\orbr{\begin{cases}\frac{3}{16}< \frac{8}{16}=\frac{1}{2}\\\frac{16}{19}=\frac{32}{38}>\frac{19}{38}\end{cases}}\)\(\Rightarrow\)\(\frac{3}{16}< \frac{1}{2}< \frac{16}{19}\)\(\Rightarrow\)\(\frac{3}{16}< \frac{16}{19}\)

Vậy ...

\(b)\)

Ta có:

\(\orbr{\begin{cases}\frac{4}{17}=\frac{12}{51}\\\frac{6}{25}=\frac{12}{50}\end{cases}}\)\(\Rightarrow\)\(\frac{12}{51}< \frac{12}{50}\)\(\Rightarrow\)\(\frac{4}{17}< \frac{6}{25}\)

Vậy ...

300

465 

a) \(\dfrac{3}{4}=\dfrac{3x4}{4x4}=\dfrac{12}{16},\dfrac{6}{7}=\dfrac{6x2}{7x2}=\dfrac{12}{14}\)

Do 16 > 14 => \(\dfrac{12}{16}< \dfrac{12}{14}hay\dfrac{3}{4}< \dfrac{6}{7}\)

a) $\frac{3}{4}=\frac{3x4}{4x4}=\frac{12}{16},\frac{6}{7}=\frac{6x2}{7x2}=\frac{12}{14}$

Do 16 > 14 => $\frac{12}{16}<\frac{12}{14}hay\frac{3}{4}<\frac{6}{7}$

a: \(4\sqrt{7}=\sqrt{4^2\cdot7}=\sqrt{112}\)

\(3\sqrt{13}=\sqrt{3^2\cdot13}=\sqrt{117}\)

mà 112<117

nên \(4\sqrt{7}< 3\sqrt{13}\)

b: \(3\sqrt{12}=\sqrt{3^2\cdot12}=\sqrt{108}\)

\(2\sqrt{16}=\sqrt{16\cdot2^2}=\sqrt{64}\)

mà 108>64

nên \(3\sqrt{12}>2\sqrt{16}\)

c: \(\dfrac{1}{4}\sqrt{84}=\sqrt{\dfrac{1}{16}\cdot84}=\sqrt{\dfrac{21}{4}}\)

\(6\sqrt{\dfrac{1}{7}}=\sqrt{36\cdot\dfrac{1}{7}}=\sqrt{\dfrac{36}{7}}\)

mà \(\dfrac{21}{4}>\dfrac{36}{7}\)

nên \(\dfrac{1}{4}\sqrt{84}>6\sqrt{\dfrac{1}{7}}\)

d: \(3\sqrt{12}=\sqrt{3^2\cdot12}=\sqrt{108}\)

\(2\sqrt{16}=\sqrt{16\cdot2^2}=\sqrt{64}\)

mà 108>64

nên \(3\sqrt{12}>2\sqrt{16}\)

Tk mình đi mọi người mình bị âm nè!

ai tk mình mình tk lại cho!!!

Cứu tớ với 

a: 

13/17=1-4/17

8/12=1-4/12

mà 4/17<4/12

nên 13/17>8/12=12/18

b: 16/51<17/51=1/3=30/90<31/90

12/18<13/17

16/51>31/90