Rút Gọn:4√21 -4√15-√14+√10 phần 4√6- 2+4√15- √10
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KN
10 tháng 2 2019
\(\frac{7}{8}\div\left(\frac{14}{3}+\frac{14}{4}\right)+\frac{10}{14}\)
\(=\frac{7}{8}\div\left(\frac{56}{12}+\frac{52}{12}\right)+\frac{10}{14}\)
\(=\frac{7}{8}\div\frac{108}{12}+\frac{10}{14}\)
\(=\frac{7}{8}\div9+\frac{5}{7}\)
\(=\frac{7}{8}\times\frac{1}{9}+\frac{5}{7}\)
\(=\frac{7}{72}+\frac{5}{7}\)
\(=\frac{49}{504}+\frac{360}{504}\)
\(=\frac{409}{504}\)
\(\frac{30}{42}\div\left(\frac{5}{7}-\frac{5}{21}\right)+\frac{6}{8}\)
\(=\frac{6}{7}\div\left(\frac{15}{21}-\frac{5}{21}\right)+\frac{3}{4}\)
\(=\frac{6}{7}\div\frac{10}{21}+\frac{3}{4}\)
\(=\frac{6}{7}\times\frac{21}{10}+\frac{3}{4}\)
\(=\frac{126}{70}+\frac{3}{4}\)
\(=\frac{504}{280}+\frac{210}{280}\)
\(=\frac{714}{280}\)
KN
10 tháng 2 2019
\(\left(6-\frac{14}{15}\right).\frac{50}{38}-\frac{8}{18}\)
\(=\left(\frac{90}{15}-\frac{14}{15}\right).\frac{25}{19}-\frac{4}{9}\)
\(=\frac{76}{15}.\frac{25}{19}-\frac{4}{9}\)
\(=\frac{76.25}{15.19}-\frac{4}{9}\)
\(=\frac{19.2.2.5.5}{3.5.19}-\frac{4}{9}\)
\(=\frac{20}{3}-\frac{4}{9}\)
\(=\frac{60}{9}-\frac{4}{9}\)
\(=\frac{56}{9}\)
\(\frac{4}{6}+\frac{5}{21}.\left(\frac{4}{5}-\frac{2}{6}\right)\)
\(=\frac{2}{3}+\frac{5}{21}.\left(\frac{24}{30}-\frac{10}{30}\right)\)
\(=\frac{2}{3}+\frac{5}{21}.\frac{14}{30}\)
\(=\frac{2}{3}+\frac{70}{630}\)
\(=\frac{420}{630}+\frac{70}{630}\)
\(=\frac{490}{630}\)
\(=\frac{7}{9}\)
H9
HT.Phong (9A5)
CTVHS
5 tháng 9 2023
a) \(\left(\sqrt{14}+\sqrt{6}\right)\sqrt{5-\sqrt{21}}\)
\(=\sqrt{14}\cdot\sqrt{5-\sqrt{21}}+\sqrt{6}\cdot\sqrt{5-\sqrt{21}}\)
\(=\sqrt{14\cdot\left(5-\sqrt{21}\right)}+\sqrt{6\cdot\left(5-\sqrt{21}\right)}\)
\(=\sqrt{70-14\sqrt{21}}+\sqrt{30-6\sqrt{21}}\)
\(=\sqrt{7^2-2\cdot7\cdot\sqrt{21}+\left(\sqrt{21}\right)^2}+\sqrt{\left(\sqrt{21}\right)^2-2\cdot3\cdot\sqrt{21}+3^2}\)
\(=\sqrt{\left(7-\sqrt{21}\right)^2}+\sqrt{\left(\sqrt{21}-3\right)^2}\)
\(=\left|7-\sqrt{21}\right|+\left|\sqrt{21}-3\right|\)
\(=7-\sqrt{21}+\sqrt{21}-3\)
\(=4\)
b) \(\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4-\sqrt{15}}\)
\(=\left[4\cdot\left(\sqrt{10}-\sqrt{6}\right)+\sqrt{15}\cdot\left(\sqrt{10}-\sqrt{6}\right)\right]\cdot\sqrt{4-\sqrt{15}}\)
\(=\left(4\sqrt{10}-4\sqrt{6}+\sqrt{150}-\sqrt{90}\right)\sqrt{4-\sqrt{15}}\)
\(=\left(4\sqrt{10}-4\sqrt{6}+5\sqrt{6}-3\sqrt{10}\right)\sqrt{4-\sqrt{15}}\)
\(=\left(\sqrt{10}+\sqrt{6}\right)\left(\sqrt{4-\sqrt{15}}\right)\)
\(=\sqrt{10\cdot\left(4-\sqrt{15}\right)}+\sqrt{6\cdot\left(4-\sqrt{15}\right)}\)
\(=\sqrt{40-10\sqrt{15}}+\sqrt{24-6\sqrt{15}}\)
\(=\sqrt{5^2-2\cdot5\cdot\sqrt{15}+\left(\sqrt{15}\right)^2}+\sqrt{\left(\sqrt{15}\right)^2-2\cdot3\cdot\sqrt{15}+3^2}\)
\(=\sqrt{\left(5-\sqrt{15}\right)^2}+\sqrt{\left(\sqrt{15}-3\right)^2}\)
\(=\left|5-\sqrt{15}\right|+\left|\sqrt{15}-3\right|\)
\(=5-\sqrt{15}+\sqrt{15}-3\)
\(=2\)
31 tháng 12 2015
1. 10 phần 14 = 5/7
2. 5 phần 15 = 1/3
3. 14 phần 22 = 7/11
4. 2 phần 8 = 1/4
5. 4 phần 24= 1/6
6. 2 phần 10 = 1/5
MH
31 tháng 12 2015
1. \(\frac{10}{14}=\frac{10:2}{14:2}=\frac{5}{7}\)
2. \(\frac{5}{15}=\frac{5:5}{15:5}=\frac{1}{3}\)
3. \(\frac{14}{22}=\frac{14:2}{22:2}=\frac{7}{11}\)
4. \(\frac{2}{8}=\frac{2:2}{8:2}=\frac{1}{4}\)
5. \(\frac{4}{24}=\frac{4:4}{24:4}=\frac{1}{6}\)
6. \(\frac{2}{10}=\frac{2:2}{10:2}=\frac{1}{5}\)
3 tháng 1 2018
B=2^10*(2*3)^15+3^14*3*5*(2^2)^13/2^19*(2*3^2)^7*3-3^15*2^25
B=2^10*2^15*3^15+3^15+3^15*5*2^26/2^19*2^7*3^14**3-3^15*2^25
B=2^25*3^15+5*2*3^15*2^25/2*2^25*3^15-3^15*2^25
B=1*2^25*3^15+10*2^25*3^15/2*2^25*3^15-2^25*3^15
B=2^25*3^15*(1+10)/2^25*3^15*(2-1)=11/1=11
- B=2^10*6^15+3^14*15*4^13/2^19*18^7*3-3^15*2^25
XO
21 tháng 7 2019
a) \(\frac{1.3+3.5+5.7+7.9}{3.6+9.10+15.14+21.18}\)
= \(\frac{1.3+3.5+5.7+7.9}{1.3.2.3+3.5.2.3+5.7.2.3+7.9.2.3}\)
= \(\frac{1.3+3.5+5.7+7.9}{1.3.6+3.5.6+5.7.6+7.9.6}\)
= \(\frac{1.3+3.5+5.7+7.9}{6.\left(1.3+3.5+5.7+7.9\right)}=\frac{1}{6}\)
Dấu "." là dấu nhân cấp 2
b) \(\frac{1.2+2.3+3.4+4.5}{3.6+6.9+9.12+12.15}\)
= \(\frac{1.2+2.3+3.4+4.5}{1.2.3.3+2.3.3.3+3.4.3.3+4.5.3.3}\)
= \(\frac{1.2+2.3+3.4+4.5}{1.2.9+2.3.9+3.4.9+4.5.9}\)
= \(\frac{1.2+2.3+3.4+4.5}{9.\left(1.2+2.3+3.4+4.5\right)}=\frac{1}{9}\)
Dấu "." là dấu nhân cấp 2
c) \(\frac{0,3+\frac{3}{7}+\frac{3}{11}}{0,4+\frac{4}{7}+\frac{4}{11}}\)= \(\frac{\frac{3}{10}+\frac{3}{7}+\frac{3}{11}}{\frac{4}{10}+\frac{4}{7}+\frac{4}{11}}\)= \(\frac{3.\left(\frac{1}{10}+\frac{1}{7}+\frac{1}{11}\right)}{4.\left(\frac{1}{10}+\frac{1}{7}+\frac{1}{11}\right)}=\frac{3}{4}\)
HH
1
TD
6 tháng 2 2022
1)\(\dfrac{5}{6}\)
2)\(\dfrac{1}{4}\)
3)\(\dfrac{12}{5}\)
4)\(\dfrac{39}{2}\)
5)\(1\)
S
1
23 tháng 6 2017
What the.....
À hỉu r` nhưng chưa hok tới
=> k giải đc
\//Bn dùng đt nên k cack đc hả
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
13 tháng 10
a; \(\dfrac{9}{27}\) + \(\dfrac{7}{-49}\)
= \(\dfrac{1}{3}\) - \(\dfrac{1}{7}\)
= \(\dfrac{7}{21}\) - \(\dfrac{3}{21}\)
= \(\dfrac{4}{21}\)
b; - \(\dfrac{12}{10}\) + \(\dfrac{-25}{30}\)
= - \(\dfrac{6}{5}\) - \(\dfrac{5}{6}\)
= -\(\dfrac{36}{30}\) - \(\dfrac{25}{30}\)
= \(\dfrac{-61}{30}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
13 tháng 10
c; \(\dfrac{-20}{35}\) + \(\dfrac{-16}{-24}\)
= - \(\dfrac{4}{7}\) + \(\dfrac{2}{3}\)
= - \(\dfrac{12}{21}\) + \(\dfrac{14}{21}\)
= \(\dfrac{2}{21}\)
d; - \(\dfrac{21}{77}\) + \(\dfrac{10}{-35}\)
= - \(\dfrac{3}{11}\) - \(\dfrac{2}{7}\)
= - \(\dfrac{21}{77}\) - \(\dfrac{22}{77}\)
= - \(\dfrac{43}{77}\)