\(\frac{121.75.130.169}{39.60.11.198}=\frac{11.11.25.3.10.13.13.13}{3.13.10.6.11.11.18}=\frac{5.5.13.13}{6.18}=\frac{4225}{108}\)

b) 

\(\frac{3^{10}.\left(-5\right)^{21}}{\left(-5\right)^{20}.3^{12}}=\frac{\left(-5\right)}{3^2}=\frac{\left(-5\right)}{9}\)

mk ko gửi ảnh bài làm được

2.

\(\frac{2^5.7+2^5}{2^5.5^2-2^5.3}=\frac{2^5.\left(7+1\right)}{2^5.\left(25-3\right)}=\frac{8}{22}=\frac{4}{11}\)

2.Rút gọn

\(2^5.7+\frac{2^5}{2^5.5^2}-2^5.3=2^5.7+\frac{1}{25}-2^5.3=2^5.\left(7-5\right)+\frac{1}{25}=32.2+\frac{1}{25}=64+\frac{1}{25}=\frac{1600}{25}+\frac{1}{25}=\frac{1601}{25}\)

1600/25

\(\frac{2106}{7320}=\frac{351}{1220}\)

\(\frac{4212}{14604}=\frac{351}{1217}\)

\(\frac{6318}{21960}=\frac{351}{1220}\)

=> \(\frac{351}{1217}>\frac{351}{1220}\);\(\frac{351}{1217}>\frac{351}{1220}\)

Vậy : \(\frac{2106}{7320}< \frac{6318}{21960}\);\(\frac{4212}{14604}>\frac{6318}{21960}\)

Ta có:

\(\frac{2106}{7320}=\frac{2106:3}{7320:3}=\frac{720}{2440}\)   (1)

\(\frac{4212}{14604}=\frac{4212:6}{14604:6}=\frac{702}{2434}\)

\(\frac{6218}{21960}=\frac{6218:9}{21960:9}=\frac{702}{2440}\)    (2)

Từ (1) và (2)=>\(\frac{2106}{7320}=\frac{6318}{21960}\)

a) Ta có :

\(\frac{1717}{2929}=\frac{17.101}{29.101}=\frac{17}{29}\)

\(\frac{171717}{292929}=\frac{17.10101}{29.10101}=\frac{17}{29}\)

Vì \(\frac{17}{29}=\frac{17}{29}\) nên \(\frac{1717}{2929}=\frac{171717}{292929}\)

b) Ta có:

\(\frac{6420-68}{8340-82}=\frac{2.3210-2.34}{2.4170-2.41}=\frac{2.\left(3210-34\right)}{2.\left(4170-41\right)}=\frac{3210-34}{4170-41}\)

Vì \(\frac{3210-34}{4170-41}=\frac{3210-34}{4170-41}\) nên \(\frac{3210-34}{4170-41}=\frac{6420-68}{8340-82}\)

help me m cần gấp

\(\frac{4212}{14640}=\frac{4212:2}{14640:2}=\frac{2106}{7320}\)

\(\frac{6318}{21960}=\frac{6318:3}{21960:3}=\frac{2106}{7320}\)

Vậy\(\frac{2106}{7320}=\frac{4212}{14640}=\frac{6318}{21960}\)

tớ chỉ biết làm câu a thôi

a) 1717/2929 và 171717/292929

1717/2929 = 17x101/29x101    = 17/29                 và 17x10101/29x10101 = 17/29

Vậy hai phân số bằng nhau

a. ĐK: a, b, c khác 0.

 \(\frac{a^2+b^2-c^2}{2ab}+\frac{b^2+c^2-a^2}{2bc}+\frac{c^2+a^2-b^2}{2ac}=1\)

\(\Leftrightarrow\left[\frac{a^2+b^2-c^2}{2ab}-1\right]+\left[\frac{b^2+c^2-a^2}{2bc}+\frac{c^2+a^2-b^2}{2ca}\right]=0\)

\(\Leftrightarrow\frac{\left(a-b\right)^2-c^2}{2ab}+\frac{1}{2c}\left[\frac{c^2-\left(a^2-b^2\right)}{b}+\frac{c^2+\left(a^2-b^2\right)}{a}\right]=0\)

\(\Leftrightarrow\frac{\left(a-b\right)^2-c^2}{2ab}+\frac{1}{2c}\left[\frac{c^2\left(a+b\right)-\left(a^2-b^2\right)\left(a-b\right)}{ab}\right]=0\)

\(\Leftrightarrow\frac{\left(a-b\right)^2-c^2}{2ab}+\frac{\left(a+b\right)\left(c^2-\left(a-b\right)^2\right)}{2abc}=0\)

\(\Leftrightarrow\left[\left(a-b\right)^2-c^2\right]\left(1-\frac{a+b}{c}\right)=0\)

\(\Leftrightarrow\left(a-b-c\right)\left(a-b+c\right)\left(c-a-b\right)=0\)

\(\Leftrightarrow a=b+c\)hoặc \(b=a+c\)hoặc \(c=a+b\).

b) Không mất tính tổng quả. G/s: a = b + c

Khi đó ta có:

\(\frac{a^2+b^2-c^2}{2ab}=\frac{\left(b+c\right)^2+b^2-c^2}{2\left(b+c\right)b}=1\)

\(\frac{b^2+c^2-a^2}{2bc}=\frac{b^2+c^2-\left(b+c\right)^2}{2bc}=-1\)

\(\frac{c^2+a^2-b^2}{2ca}=\frac{c^2+\left(b+c\right)^2-b^2}{2\left(b+c\right)c}=1\)

=> Điều phải chứng minh.

Đáp án A