Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a: \(=\left(1+\dfrac{4}{23}-\dfrac{4}{23}\right)+\left(\dfrac{5}{21}+\dfrac{16}{21}\right)+\dfrac{1}{2}\)
\(=1+1+\dfrac{1}{2}=2+\dfrac{1}{2}=\dfrac{5}{2}\)
b: \(=\left(\dfrac{1}{25}+\dfrac{5}{25}+\dfrac{25}{25}\right):\left(\dfrac{1}{25}-\dfrac{5}{25}-\dfrac{25}{25}\right)\)
\(=\dfrac{31}{25}:\dfrac{-29}{25}=\dfrac{-31}{29}\)
c: \(=\dfrac{\dfrac{1}{9}-\dfrac{1}{7}-\dfrac{1}{11}}{\dfrac{4}{9}-\dfrac{4}{7}-\dfrac{4}{11}}+\dfrac{\dfrac{3}{5}-\dfrac{3}{25}-\dfrac{3}{125}-\dfrac{3}{625}}{\dfrac{4}{5}-\dfrac{4}{25}-\dfrac{4}{125}-\dfrac{4}{625}}\)
=1/4+3/4
=1
Ta có: \(\left|x-\dfrac{1}{3}\right|\)+0,8 = \(\left|-3,2+0,4\right|\)
---> \(\left|x-\dfrac{1}{3}\right|\)+0,8 =2,8 ---> \(\left|x-\dfrac{1}{3}\right|\)= 2,8-0,8=2 --> TH1: x-\(\dfrac{1}{3}\) = 2 --> x= 2+ \(\dfrac{1}{3}\)= \(\dfrac{7}{3}\) TH2: x-\(\dfrac{1}{3}\) = -2---> x= -2+\(\dfrac{1}{3}\)=\(\dfrac{-5}{3}\) Vạy x =\(\dfrac{-5}{3}\) hoặc x= \(\dfrac{7}{3}\) Tick cho mik nha!!
1.Tính
a.\(\dfrac{7}{23}\left[(-\dfrac{8}{6})-\dfrac{45}{18}\right]=\dfrac{7}{23}.-\dfrac{12}{6}=-\dfrac{7}{6}\)
b.\(\dfrac{1}{5}\div\dfrac{1}{10}-\dfrac{1}{3}(\dfrac{6}{5}-\dfrac{9}{4})=2-(-\dfrac{7}{20})=\dfrac{47}{20}\)
c.\(\dfrac{3}{5}.(-\dfrac{8}{3})-\dfrac{3}{5}\div(-6)=-\dfrac{3}{2}\)
d.\(\dfrac{1}{2}.(\dfrac{4}{3}+\dfrac{2}{5})-\dfrac{3}{4}.(\dfrac{8}{9}+\dfrac{16}{3})=-\dfrac{19}{5}\)
e.\(\dfrac{6}{7}\div(\dfrac{3}{26}-\dfrac{3}{13})+\dfrac{6}{7}.(\dfrac{1}{10}-\dfrac{8}{5})=-\dfrac{61}{7}\)
Bài 2
a.\(1^2_5x+\dfrac{3}{7}=\dfrac{4}{5}\)
\(x=\dfrac{13}{49}\)
b.\(\left|x-1,5\right|=2\)
Xảy ra 2 trường hợp
TH1
\(x-1,5=2\)
\(x=3,5\)
TH2
\(x-1,5=-2\)
\(x=-0,5\)
Vậy \(x=3,5\) hoặc \(x=-0,5\) .
Ngại làm quá trời ơi,lần sau bn tách ra nhá làm vậy mỏi tay quá.
\(B=\frac{1-\frac{1}{\sqrt{49}}+\frac{1}{49}-\frac{1}{\left(7\sqrt{7}\right)^2}}{\frac{\sqrt{64}}{2}-\frac{4}{7}+\frac{2^2}{7^2}-\frac{4}{343}}\)
\(B=\frac{1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}}{\frac{8}{2}-\frac{4}{7}+\frac{4}{49}-\frac{4}{343}}\)
\(B=\frac{\frac{343}{343}-\frac{49}{343}+\frac{7}{343}-\frac{1}{343}}{4-\frac{4}{7}+\frac{28}{343}-\frac{4}{343}}\)
\(B=\frac{\frac{300}{343}}{\frac{28}{7}-\frac{4}{7}+\frac{24}{343}}\)
\(B=\frac{\frac{300}{343}}{\frac{24}{7}+\frac{24}{343}}\)
\(B=\frac{\frac{300}{343}}{\frac{1323}{343}+\frac{24}{343}}\)
\(B=\frac{300}{343}:\frac{1347}{343}\)
\(B=\frac{100}{449}\)
\(A=\frac{2^{12}.3^5-4^6.9^2}{\left(2^2.3\right)^6+8^4.3^5}-\frac{5^{10}.7^3-25^5.49^2}{\left(125.7\right)^3+5^9.14^3}\)
\(A=\frac{2^{12}.3^5-2^{12}.3^6}{2^{12}.3^6+2^{12}.3^5}-\frac{5^{10}.7^3-5^{10}.7^6}{5^9.7^3+5^9.2^3.7^3}\)
\(A=\frac{2^{12}.3^5\left(1-3\right)}{2^{12}.3^5.\left(3+1\right)}-\frac{5^{10}.7^3.\left(1-7^3\right)}{5^9.7^3.\left(1+8\right)}\)
\(A=\frac{-2}{4}-\frac{5.\left(-342\right)}{9}\)
\(A=\frac{-1}{2}+\frac{1710}{9}\)
\(A=\frac{-1}{2}+190\)
\(A=\frac{-1}{2}+\frac{380}{2}\)
\(A=\frac{379}{2}\)
\(A=\dfrac{2^{12}.3^5-4^6.9^2}{\left(2^2.3\right)^6+8^4.3^5}-\dfrac{5^{10}.7^3-25^5.49^2}{\left(125.7\right)^3+5^9.14^3}\)
\(=\dfrac{2^{12}.3^5-\left(2^2\right)^6.\left(3^2\right)^2}{2^{12}.3^6+\left(2^3\right)^4.3^5}-\dfrac{5^{10}.7^3-\left(5^2\right)^5.\left(7^2\right)^2}{\left(5^3\right)^3.7^3+5^9.\left(2.7\right)^3}\)
\(=\dfrac{2^{12}.3^5-2^{12}.3^4}{2^{12}.3^6+2^{12}.3^5}-\dfrac{5^{10}.7^3-5^{10}.7^4}{5^9.7^3+5^9.2^3.7^3}\)
= \(\dfrac{2^{12}.3^4\left(3-1\right)}{2^{12}.3^5\left(3+1\right)}-\dfrac{5^{10}.7^3\left(1-7\right)}{5^9.7^3\left(1+2^3\right)}\)
= \(\dfrac{2}{3.4}-\dfrac{5\left(-6\right)}{9}\)
= \(\dfrac{7}{2}\)
Bạn tính hai vế à.!? Hay tính vế thứ nhất rồi với vế thứ 2.!???
\(A=\dfrac{2^{12}.3^5-\left(2^2\right)^6.\left(3^2\right)^2}{\left(2^2\right)^6.3+\left(2^3\right)^4.3^5}-\dfrac{5^{10}+7^3-\left(5^2\right)^5.\left(7^2\right)^2}{(5^3).7^3+5^9.\left(7^2\right)^3}\)
\(A=\dfrac{2^4.1-2.3^3}{1.1+1.1}-\dfrac{5+1-5.1}{1.1+1.7}\)
\(A=\dfrac{2^4-54}{2}-\dfrac{1.1}{2\cdot7}\)
\(A=(2^3-54)-\dfrac{1}{14}\)
\(A=\left(-46\right)-\dfrac{1}{14}\)
iúp mình vs help me >3
mk làm rùi nên mn k cần giúp nx đâu.Hihi