\(a)\) Chưa biết -_- 

\(b)\) \(12^2-8^2=4^2.3^2-4^2.2^2=4^2\left(3^2-2^2\right)>4^2\)

Vậy \(12^2-8^2>4^2\)

\(c)\) \(7^8+7^9=7^8\left(1+7\right)=7^8.8< 7^8.49=7^8.7^2=7^{10}\)

Vậy \(7^8+7^9< 7^{10}\)

Chúc bạn học tốt ~ 

a)  13^2 = (10+3)^2 = 10^2 + 3^2 + 2.10.3 > 10^2 + 3^2

b) 12^2 - 8^2 = (8 + 4)^2 - 8^2 = 8^2 + 2.4.8 + 4^2 - 8^2 = 2.4.8 + 4^2 > 4^2

c) 7^10 = 7.7^9 = (6+1).7^9 = 6.7^9 + 7^9 = 6.7^9 + 7.7^8 > 7^8 + 7^9

Okok cảm ơn nhiều :)

Đề yêu cầu gì vậy em?

đây là so sánh hay quy đồng v 

\(\dfrac{3}{5}v\text{à}\dfrac{7}{3}\\ MSC:15\\ \dfrac{3}{5}=\dfrac{3\times3}{5\times3}=\dfrac{9}{15};\dfrac{7}{3}=\dfrac{7\times5}{5\times3}=\dfrac{35}{15}\\ 9< 35\\ \Rightarrow\dfrac{3}{5}< \dfrac{7}{5}\\ \dfrac{9}{10}v\text{à}\dfrac{5}{4}\\MSC:20\\ \dfrac{9\times2}{10\times2}=\dfrac{18}{20};\dfrac{5}{4}=\dfrac{5\times5}{4\times5}=\dfrac{25}{20}\\ 18< 25\\ \Rightarrow\dfrac{9}{10}< \dfrac{5}{4}\\ \dfrac{2}{7}v\text{à}\dfrac{5}{4}\\ MSC:28\\ \dfrac{2}{7}=\dfrac{2\times4}{7\times4}=\dfrac{8}{28};\dfrac{5}{4}=\dfrac{5\times7}{4\times7}=\dfrac{35}{28}\\ 8< 35\\ \dfrac{2}{7}< \dfrac{5}{4}\\ \dfrac{8}{9}v\text{à}\dfrac{11}{6}\\ MSC:18\\ \dfrac{8\times2}{9\times2}=\dfrac{16}{18};\dfrac{11}{6}=\dfrac{11\times3}{6\times3}=\dfrac{33}{18}\\ 16< 33\\ \Rightarrow\dfrac{8}{9}< \dfrac{11}{6}\)

Quy đồng hả ?

\(\dfrac{3}{5}và\dfrac{7}{3}=\dfrac{3}{5}=\dfrac{3\times3}{5\times3}=\dfrac{9}{15};\dfrac{7}{3}=\dfrac{7\times5}{3\times5}=\dfrac{35}{15}\)

\(\dfrac{9}{10}và\dfrac{5}{4}=\dfrac{9}{10}=\dfrac{9\times2}{10\times2}=\dfrac{18}{20};\dfrac{5}{4}=\dfrac{5\times5}{4\times5}=\dfrac{25}{20}\)

\(\dfrac{2}{7}và\dfrac{5}{4}=\dfrac{2}{7}=\dfrac{2\times4}{7\times4}=\dfrac{8}{28};\dfrac{5}{4}=\dfrac{5\times7}{4\times7}=\dfrac{35}{28}\)

\(\dfrac{8}{9}và\dfrac{11}{6}=\dfrac{8}{9}=\dfrac{8\times2}{9\times2}=\dfrac{16}{18};\dfrac{11}{6}=\dfrac{11\times3}{6\times3}=\dfrac{33}{18}\)

\(\dfrac{6}{5}và\dfrac{13}{8}=\dfrac{6\times8}{5\times8}=\dfrac{48}{40};\dfrac{13}{8}=\dfrac{13\times5}{8\times5}=\dfrac{65}{40}\)

\(\dfrac{4}{5}và\dfrac{3}{2}=\dfrac{4}{5}=\dfrac{4\times2}{5\times2}=\dfrac{8}{10};\dfrac{3}{2}=\dfrac{3\times5}{2\times5}=\dfrac{15}{10}\)

\(\dfrac{5}{12}và\dfrac{7}{8}=\dfrac{5}{12}=\dfrac{5\times2}{12\times2}=\dfrac{10}{24};\dfrac{7}{8}=\dfrac{7\times3}{8\times3}=\dfrac{21}{24}\)

\(\dfrac{11}{4}và\dfrac{13}{6}=\dfrac{11}{4}=\dfrac{11\times3}{4\times3}=\dfrac{33}{12};\dfrac{13}{6}=\dfrac{13\times2}{6\times2}=\dfrac{26}{12}\)

em trả lời ccaua này hi vọng thầy còn nhớ em

a) -9/4<`1/3

c) 9/-5<7/-10

a) \(\dfrac{-9}{4}< 0\)

\(0< \dfrac{1}{3}\)

Do đó: \(\dfrac{-9}{4}< \dfrac{1}{3}\)

a >

b <

c >

d <

>

<

>

<

\(a:ta.c\text{ó}:BCNN:12\\ \dfrac{2}{3}=\dfrac{2\cdot4}{3\cdot4}=\dfrac{8}{12};\dfrac{1}{4}=\dfrac{1\cdot3}{4\cdot3}=\dfrac{3}{12}\\ v\text{ì }\dfrac{8}{12}< \dfrac{3}{12}n\text{ê}n\dfrac{2}{3}< \dfrac{1}{4}\\ b:ta.c\text{ó}:\\ 10=2\cdot5\\ 8=2^3\\ \Rightarrow BCNN=2^3\cdot5=8\cdot5=40\\ \dfrac{7}{10}=\dfrac{7\cdot4}{10\cdot4}=\dfrac{28}{40};\dfrac{7}{8}=\dfrac{7\cdot5}{8\cdot5}=\dfrac{35}{40}\\ v\text{ì }\dfrac{28}{40}< \dfrac{35}{40}n\text{ê}n\dfrac{7}{10}< \dfrac{7}{8}\\ c:ta.c\text{ó}:\\ 7=7;5=5\\ \Rightarrow BCNN=7\cdot5=35\\ \dfrac{6}{7}=\dfrac{6\cdot5}{7\cdot5}=\dfrac{30}{35};\dfrac{3}{5}=\dfrac{3\cdot7}{5\cdot7}=\dfrac{21}{35}\\ v\text{ì }\dfrac{30}{35}>\dfrac{21}{35}n\text{ê}n\dfrac{6}{7}>\dfrac{3}{5}\\ d:ta.c\text{ó}:\\ 21=3\cdot7\\ 72=2^3\cdot3^2\\ \Rightarrow BCNN=2^3\cdot3^2\cdot7=504\\ \dfrac{14}{21}=\dfrac{14\cdot24}{21\cdot24}=\dfrac{336}{504};\dfrac{60}{72}=\dfrac{60\cdot7}{72\cdot7}=\dfrac{420}{504}\\ v\text{ì }\dfrac{336}{504}< \dfrac{420}{504}n\text{ê}n\dfrac{14}{21}< \dfrac{60}{72}\)

Bài 1

a: 11/12=1-1/12

23/24=1-1/24

mà -1/12>-1/24

nên 11/12>23/24

b: -3/20=-9/60

-7/12=-35/60

mà -9>-35

nên -3/20>-7/12

c.

(\sqrt{5}-\sqrt{3})-(\sqrt{10}-\sqrt{7})=(\sqrt{5}+\sqrt{7})-(\sqrt{3}+\sqrt{10})

Mà:

\((\sqrt{5}+\sqrt{7})^2=12+\sqrt{35}< 12+\sqrt{36}=18\)

\((\sqrt{3}+\sqrt{10})^2=13+\sqrt{30}>13+\sqrt{25}=18\)

\(\Rightarrow \sqrt{3}+\sqrt{10}> \sqrt{5}+\sqrt{7}\Rightarrow \sqrt{5}-\sqrt{3}< \sqrt{10}-\sqrt{7}\)

Lời giải:

a.

$5+\sqrt{2}>5+\sqrt{1}=6$

$4+\sqrt{3}< 4+\sqrt{4}=6$

$\Rightarrow 5+\sqrt{2}>4+\sqrt{3}$

b.

$\sqrt{8}-\sqrt{2}=2\sqrt{2}-\sqrt{2}=\sqrt{2}$

$\sqrt{5}-\sqrt{3}=\frac{5-3}{\sqrt{5}+\sqrt{3}}=\frac{2}{\sqrt{5}+\sqrt{3}}< \frac{2}{\sqrt{2}}=\sqrt{2}$

Vậy $\sqrt{8}-\sqrt{2}>\sqrt{5}-\sqrt{2}$

a) Dấu hiệu là điểm bài thi học kì của 100 học sinh lớp 7 của một trường Trung học Cơ Sở Hòa Bình. Số các dấu hiệu là 100
b) Bảng tần số
 

Nhận xét: Giá trị lớn nhất là 19, giá trị nhỏ nhất là 1; tần số lớn nhất là 13, tần số nhỏ nhất là 1.

a) Ta có: \(\dfrac{-5}{7}\left(\dfrac{14}{5}-\dfrac{7}{10}\right):\left|-\dfrac{2}{3}\right|-\dfrac{3}{4}\left(\dfrac{8}{9}+\dfrac{16}{3}\right)+\dfrac{10}{3}\left(\dfrac{1}{3}+\dfrac{1}{5}\right)\)

\(=\dfrac{-5}{7}\cdot\dfrac{3}{2}\cdot\dfrac{21}{10}-\dfrac{3}{4}\cdot\dfrac{56}{3}+\dfrac{10}{3}\cdot\dfrac{8}{15}\)

\(=\dfrac{-9}{4}-14+\dfrac{16}{9}\)

\(=\dfrac{-1621}{126}\)

b) Ta có: \(\dfrac{17}{-26}\cdot\left(\dfrac{1}{6}-\dfrac{5}{3}\right):\dfrac{17}{13}-\dfrac{20}{3}\left(\dfrac{2}{5}-\dfrac{1}{4}\right)+\dfrac{2}{3}\left(\dfrac{6}{5}-\dfrac{9}{2}\right)\)

\(=\dfrac{-17}{26}\cdot\dfrac{13}{17}\cdot\dfrac{-3}{2}-\dfrac{20}{3}\cdot\dfrac{3}{20}+\dfrac{2}{3}\cdot\dfrac{-33}{10}\)

\(=\dfrac{3}{4}-1-\dfrac{11}{5}\)

\(=-\dfrac{49}{20}\)