\(x=\frac{\left(6+3^2.2-6.3^2\right)^2}{\left(3^2+3.3^2-3^4\right)^2}\)
\(x=\frac{15.3^{11}+4.2.7^4}{9^7}\)
Mong các bạn giúp đỡ
Tìm x biết :
a)\(\frac{3}{2}x-\frac{1}{2}=\frac{2}{5}\) b)\(\frac{4}{5}-\left|x-\frac{2}{3}\right|=\frac{2}{3}\)
c)\(x=\left(2-\frac{15}{7}\right)^2+\left|\frac{-6}{7}\right|+\frac{\sqrt{36}}{49}\)
d)\(x=\frac{16^3.3^{10}+120.6^9}{4^6.3^{12}+6^{11}}\)
Các bn lm nhah giúp mik
thanks
tính
a, \(\frac{16^3.3^{10}+120.6^9}{4^6.3^{12}+6^{11}}\)
b , \(\left(\frac{0,4-\frac{8}{9}+\frac{2}{11}}{1,4-\frac{7}{9}+\frac{7}{11}}-\frac{\frac{1}{3}-0,25+\frac{1}{5}}{1\frac{1}{6}-0,875+0,7}\right):\frac{2012}{2013}\)
c, A
= \(1+\frac{1}{2}.\left(1+2\right)+\frac{1}{3}.\left(1+2+3\right)+\frac{1}{4}.\left(1+2+3+...+\frac{1}{20}.\left(1+2+3+....+20\right)\right).155\)
\(a,\frac{16^3.3^{10}+120.6^9}{4^6.3^{12}+6^{11}}\)
\(=\frac{\left(2^4\right)^3.3^{10}+2^3.3.5.\left(2.3\right)^9}{\left(2^2\right)^6.3^{12}+\left(2.3\right)^{11}}\)
\(=\frac{2^{12}.3^{10}+2^3.3.5.2^9.3^9}{2^{12}.3^{12}+2^{11}.3^{11}}\)
\(=\frac{2^{12}.3^{10}+2^{12}.3^{10}.5}{2^{11}.3^{11}\left(2.3+1\right)}\)
\(=\frac{2^{12}.3^{10}\left(1+5\right)}{2^{11}.3^{11}.7}=\frac{2.6}{3.7}=\frac{4}{7}\)
Tính \(A=\frac{2^{12}.3^5-4^6.9^2}{\left(2^2.3\right)^6+8^4.3^5}+\frac{16^3.3^{10}+120.6^9}{4^6.3^{12}+6^{12}}\)
giúp mình vs các bạn ơi
mk ko viết lại đề
\(A=\frac{2^{12}.3^5-2^{12}.3^4}{2^{12}.3^6+2^{12}.3^5}+\frac{2^{12}.3^{10}+2^{12}.3^{10}.5}{2^{12}.3^{12}+2^{12}.3^{12}}\)
\(=\frac{2^{12}.3^4\left(3-1\right)}{2^{12}.3^5\left(3+1\right)}+\frac{2^{12}.3^{10}\left(1+5\right)}{2.\left(2^{12}.3^{12}\right)}\)
\(=\frac{2}{3.4}+\frac{2^{12}.3^{10}.6}{2.2^{12}.3^{12}}=\frac{1}{6}+\frac{1}{3}=\frac{1}{2}\)
Vậy A= \(\frac{1}{2}\)
Các bạn ơi phiền các bạn giải chi tiết giúp mình 4 bài sau nhé, 13h hôm nay mih phải nộp rồi, các bạn giúp mình nhé:
\(\left(\frac{3}{7}\right)^{21}:\left(\frac{9}{49}\right)^6\)
\(4.2^5:\left(2^3.\frac{1}{16}\right)\)
\(3^2.2^5.\left(\frac{2}{3}\right)^2\)
\(\left(x+\frac{1}{2}\right)^4=\frac{1}{16}\)
Phiền các bạn giúp đỡ mình nhé. Cám ơn các bạn nhiều lắm lắm
Hồ Ngọc Minh Châu Võ cho mình hỏi nhưng bài kia mỗi bài 1 dòng hay là cả một bài vậy bạn
a)A=\(\frac{16^3.3^{10}+120.6^9}{4^6.3^{12}+6^{11}}\) b)B=\(\frac{45}{19}-\left(\frac{1}{2}\left(\frac{1}{3}+\left(\frac{1}{4}\right)^{-1}\right)^-\right)^{-1}\) c)C=\(\frac{5.4^{15}.9^9-4.3^{20}.8^9}{5.2^{10}.6^{19}-7.2^{29}.27^6}\)
d)D=\(\frac{2^{21}.3^5-4^6.81}{\left(2^2.3\right)^6+8^4.3^5}\) e) E=\(\left(6^9.2^{10}+12^{10}\right):\left(2^{19}.27^3+15.4^9.9^4\right)\)
f) F=\(\frac{3^6.45^4-15^{13}.5^{-9}}{27^4.24^3+45^6}\) g)G=\(\frac{\left(\frac{2}{5}\right)^7.5^7+\left(\frac{9}{4}\right)^3:\left(\frac{3}{16}\right)^3}{2^7.5^2+512}\) h)H=\(x+\frac{0,2-0,375+\frac{5}{11}}{-0,3+\frac{9}{16}-\frac{15}{22}}\)với x=-1/3
ai nhanh nhất mà trả lời dúng mik tặng 3 k
BT1: Tinh
\(1.A=\left(4-\frac{1}{2}+\frac{2}{3}\right)+\left(5+\frac{4}{3}-\frac{6}{5}\right)-\left(6-\frac{7}{4}+\frac{4}{5}\right)\)
\(2.B=\frac{\left(-1\right)^6.3^5.4^3}{9^2.2^5}\)
\(3.\frac{4}{5}.\frac{11}{3}-\frac{4}{5}.\frac{8}{3}+\frac{1}{5}\)
\(4.\sqrt{289-\sqrt{169+\sqrt{256-\sqrt{196}}}}\)
\(5.\frac{3^{15}.2^{18}.5^4}{6^{14}.10^5}\)
\(2\frac{1}{3}+\left(x-\frac{3}{2}\right)=\left(3-\frac{3}{2}\right)x\)
mong các bạn giúp đỡ
\(2\frac{1}{3}+\left(x-\frac{3}{2}\right)=\left(3-\frac{3}{2}\right)x\)
\(\Rightarrow\frac{7}{3}+x-\frac{3}{2}=\frac{3}{2}x\)
\(\Rightarrow x-\frac{3}{2}x=\frac{3}{2}-\frac{7}{3}\)
\(\Rightarrow\frac{-1}{2}x=-\frac{5}{6}\)
\(\Rightarrow x=\frac{5}{3}\)
Tính: \(D=3\frac{5}{3}.20\frac{3}{41}-7\frac{3}{41}.3\frac{5}{41}+\left(-\frac{1}{2}\right)\)
Tính bằng cách thuận tiện nhất: \(\frac{2^{12}.3^5-4^6.3^6}{2^{12}.9^3+8^4.3^5}\)
Tính: \(F=-\frac{5}{\sqrt{196}}-\frac{5}{\left(2\sqrt{21}\right)^2}-\frac{\sqrt{25}}{204}-\frac{\left(\sqrt{5}\right)^2}{374}\)
3 câu này mình đều tính rồi nhưng kết quả cao hoặc k ra được đáp số, mong các bạn gíup đỡ!
Tính nhanh
a) \(\frac{4}{1.3}.\frac{9}{2.4}.\frac{16}{3.5}.\frac{25}{4.6}.\frac{36}{5.7}\)
b) \(6\frac{4}{11}+\frac{8}{72}-\left(3\frac{8}{22}+\frac{1}{9}\right)\)
Bài 2 :Tìm x
a) \(3\frac{1}{2}x-\frac{x}{2}+x=3,5:1\frac{1}{5}\)
b) \(\left(x-\frac{3}{1.2}\right)+\left(x-\frac{3}{2.3}\right)+...+\left(x-\frac{3}{99.100}\right)=1\)
Các bạn chưa ngủ thì giúp mik nhá mai mik phải nộp cho thầy rồi . Mong các baj hiểu và giúp đỡ
a) \(=\frac{1}{1.3}.\frac{3.3}{2.4}.\frac{4.4}{3.5}.\frac{5.5}{4.6}.\frac{6.6}{5.7}=\frac{6}{2.7}=\frac{3}{7}\)
B) \(=\frac{70}{11}+\frac{1}{9}-\frac{37}{11}-\frac{1}{9}=\left(\frac{70}{11}-\frac{37}{11}\right)+\left(\frac{1}{9}-\frac{1}{9}\right)=\frac{33}{11}+0=3\)
BÀI 2:
A) \(\Leftrightarrow\frac{7}{2}x-\frac{x}{2}+\frac{2x}{2}=\frac{7}{2}.\frac{5}{6}\)
\(\Leftrightarrow\frac{7x-x+2x}{2}=\frac{35}{12}\)
\(\Leftrightarrow\frac{8x}{2}=\frac{35}{12}\)
\(\Leftrightarrow8x.12=35.2\Leftrightarrow96x=70\Leftrightarrow x=\frac{70}{96}=\frac{35}{48}\)
b) \(\left(x-\frac{3}{1.2}\right)+\left(x-\frac{3}{2.3}\right)+...+\left(x-\frac{3}{99.100}\right)=1\)
\(x-\frac{3}{1.2}+x-\frac{3}{2.3}+....x+\frac{3}{99.100}=1\)
\(\Leftrightarrow\left(x+x+x+...+x\right)-3\left(\frac{1}{1.2}+\frac{1}{1.3}+....+\frac{1}{99.100}\right)=1\)
ngoặc 1 có 99 số hạng x
\(\Leftrightarrow99x-3\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{99}-\frac{1}{100}\right)=1\)
\(\Leftrightarrow99x-3\left(1-\frac{1}{100}\right)=1\)
\(\Leftrightarrow99x-3.\frac{99}{100}=1\)
\(\Leftrightarrow99x=1+\frac{3.99}{100}\)
\(\Leftrightarrow99x=\frac{397}{100}\)
\(\Leftrightarrow x=\frac{397}{100.99}=\frac{397}{9900}\)