Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(A=17\dfrac{2}{31}-\left(\dfrac{15}{17}+6\dfrac{2}{31}\right)=17\dfrac{2}{31}-\dfrac{15}{17}-6\dfrac{2}{31}\)
\(=11-\dfrac{15}{17}=\dfrac{172}{17}\)
\(B=\left(31\dfrac{6}{13}+5\dfrac{9}{41}\right)-36\dfrac{6}{12}=36\dfrac{363}{533}-36\dfrac{6}{12}=\dfrac{193}{1066}\)
\(C=27\dfrac{51}{59}-\left(7\dfrac{51}{59}-\dfrac{1}{3}\right)=27\dfrac{51}{59}-7\dfrac{51}{59}+\dfrac{1}{3}=20+\dfrac{1}{3}=\dfrac{61}{3}\)
\(A=17\dfrac{2}{31}-\left(\dfrac{15}{17}+6\dfrac{2}{31}\right)=17\dfrac{2}{31}-\dfrac{15}{17}-6\dfrac{2}{31}\)
\(=\left(17\dfrac{2}{31}-6\dfrac{2}{31}\right)-\dfrac{15}{17}=11-\dfrac{15}{17}=\dfrac{172}{17}\)
\(B=\left(31\dfrac{6}{13}+5\dfrac{9}{41}\right)-36\dfrac{6}{12}=36\dfrac{363}{533}-36\dfrac{1}{2}=\dfrac{193}{1066}\) (Casio :>)
\(C=27\dfrac{51}{59}-\left(7\dfrac{51}{59}-\dfrac{1}{3}\right)=27\dfrac{51}{59}-7\dfrac{51}{59}+\dfrac{1}{3}\)
\(=20+\dfrac{1}{3}=\dfrac{61}{3}\)
a. = \(\dfrac{-1}{24}-\left\{\dfrac{1}{4}-\dfrac{-3}{8}\right\}\)
= \(\dfrac{-1}{24}-\left\{\dfrac{1}{4}+\dfrac{3}{8}\right\}\)
= \(\dfrac{-1}{24}-\dfrac{5}{8}\)
= \(\dfrac{-2}{3}\)
b. = \(12\dfrac{7}{88}-3\dfrac{5}{11}\)
= \(8\dfrac{5}{8}\)
c. = \(\dfrac{-28}{9}+\dfrac{-413}{9}\)
= \(-49\)
d. = \(\dfrac{8}{35}:\dfrac{2}{11}+\dfrac{-8}{35}:\dfrac{2}{11}\)
= \(\dfrac{2}{11}:\left(\dfrac{8}{35}+\dfrac{-8}{35}\right)\)
= 0
Bài1:
Ta có:
a)\(\sqrt{\dfrac{3^2}{5^2}}=\sqrt{\dfrac{9}{25}}=\dfrac{3}{5}\)
b)\(\dfrac{\sqrt{3^2}+\sqrt{42^2}}{\sqrt{5^2}+\sqrt{70^2}}=\dfrac{\sqrt{9}+\sqrt{1764}}{\sqrt{25}+\sqrt{4900}}=\dfrac{3+42}{5+70}=\dfrac{45}{75}=\dfrac{3}{5}\)
c)\(\dfrac{\sqrt{3^2}-\sqrt{8^2}}{\sqrt{5^2}-\sqrt{8^2}}=\dfrac{\sqrt{9}-\sqrt{64}}{\sqrt{25}-\sqrt{64}}=\dfrac{3-8}{5-8}=\dfrac{-5}{-3}=\dfrac{5}{3}\)
Từ đó, suy ra: \(\dfrac{3}{5}=\sqrt{\dfrac{3^2}{5^2}}=\dfrac{\sqrt{3^2}+\sqrt{42^2}}{\sqrt{5^2}+\sqrt{70^2}}\)
Bài 2:
Không có đề bài à bạn?
Bài 3:
a)\(\sqrt{x}-1=4\)
\(\Rightarrow\sqrt{x}=5\)
\(\Rightarrow x=\sqrt{25}\)
\(\Rightarrow x=5\)
b)Vd:\(\sqrt{x^4}=\sqrt{x.x.x.x}=x^2\Rightarrow\sqrt{x^4}=x^2\)
Từ Vd suy ra:\(\sqrt{\left(x-1\right)^4}=16\)
\(\Rightarrow\left(x-1\right)^2=16\)
\(\Rightarrow\left(x-1\right)^2=4^2\)
\(\Rightarrow x-1=4\)
\(\Rightarrow x=5\)
a)
\(\left[\dfrac{3}{8}+\left(-\dfrac{3}{4}+\dfrac{7}{12}\right)\right].\dfrac{5}{6}+\dfrac{1}{2}\)
\(=\dfrac{5}{24}.\dfrac{5}{6}+\dfrac{1}{2}\)
\(=\dfrac{25}{144}+\dfrac{1}{2}\)
\(=\dfrac{97}{144}\)
b)
\(\dfrac{1}{2}+\dfrac{3}{4}-\left(\dfrac{3}{4}+\dfrac{1}{2}\right)\)
\(=\dfrac{1}{2}+\dfrac{3}{4}-\dfrac{3}{4}-\dfrac{1}{2}\)
\(=0\)
c)
\(\dfrac{6}{\dfrac{5}{12}}:\dfrac{2}{\dfrac{3}{4}}+\dfrac{11}{\dfrac{1}{4}}\left(\dfrac{1}{3}-\dfrac{1}{5}\right)\)
\(=\dfrac{27}{5}+\dfrac{11}{\dfrac{1}{4}}.\dfrac{2}{15}\)
\(=\dfrac{27}{5}+\dfrac{88}{15}\)
\(=\dfrac{169}{15}\)
a,\(\dfrac{3}{5}+\dfrac{2}{7}=\dfrac{31}{35}\)
b,\(\dfrac{-4}{3}:\dfrac{2}{15}=\dfrac{-60}{6}=-10\)
c,\(\dfrac{3}{7}.\dfrac{2}{9}+\dfrac{7}{9}.\dfrac{3}{7}=\dfrac{3}{7}.\left(\dfrac{2}{9}+\dfrac{7}{9}\right)=\dfrac{3}{7}\)
d,\(\left(\dfrac{-2}{3}\right)^3.9^2+\left(\dfrac{-3}{4}\right)^2.32\)
\(=\dfrac{\left(-2\right)^3}{3^3}.3^4+\dfrac{\left(-3\right)^2}{4^2}.2^5\)
\(=\left(-8\right).3+\dfrac{3^2}{4^2}.2^5\)
\(=\left(-24\right)+2.9\)
\(=\left(-24\right)+18\)
\(=-6\)
a, \(\dfrac{5}{6}-\left|2-x\right|=\dfrac{1}{3}\Rightarrow\dfrac{5}{6}-\dfrac{1}{3}=\left|2-x\right|\)
<=> \(\dfrac{1}{2}=\left|2-x\right|\) \(\Leftrightarrow\left[{}\begin{matrix}2-x=\dfrac{1}{2}\\2-x=\dfrac{-1}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{5}{2}\end{matrix}\right.\)
==================
Mấy câu sau tương tự thôi
a)\(\dfrac{3}{2}hay\dfrac{-3}{2}\)
b)\(\dfrac{13}{20}hay\dfrac{-13}{20}\)
c)\(\dfrac{11}{6}hay\dfrac{-11}{6}\)
d)\(\dfrac{4}{3}hay\dfrac{-4}{3}\)
e)\(\dfrac{1}{5}hay\dfrac{-1}{5}\)
Đây là câu trả lời của mình
Hay có nghĩa là hoặc
a) \(\dfrac{-2}{3}+\dfrac{3}{4}-\dfrac{-1}{6}+\dfrac{-2}{5}=\dfrac{1}{12}-\dfrac{-1}{6}+\dfrac{-2}{5}=\dfrac{1}{4}+\dfrac{-2}{5}=\dfrac{-3}{20}\)
b) \(\dfrac{-2}{3}+\dfrac{-1}{5}+\dfrac{3}{4}-\dfrac{5}{6}-\dfrac{-7}{10}=\left(\dfrac{-2}{3}-\dfrac{5}{6}\right)+\left(\dfrac{-1}{5}-\dfrac{-7}{10}\right)+\dfrac{3}{4}\)
\(=\dfrac{-3}{2}+\dfrac{1}{2}-\dfrac{3}{4}\)
= \(=-1-\dfrac{3}{4}\)
\(=\dfrac{-1}{4}\)
c)\(\dfrac{1}{2}-\dfrac{-2}{5}+\dfrac{1}{3}+\dfrac{5}{7}-\dfrac{-1}{6}+\dfrac{-4}{35}+\dfrac{1}{41}\)
= \(\left(\dfrac{1}{2}-\dfrac{-1}{6}+\dfrac{1}{3}\right)+\left(\dfrac{-4}{35}+\dfrac{5}{7}-\dfrac{-2}{5}\right)+\dfrac{1}{41}\)
= \(1+1+\dfrac{1}{41}\)
= \(\dfrac{83}{41}\)
d)\(\dfrac{1}{100.99}-\dfrac{1}{99.98}-\dfrac{1}{98.97}-...-\dfrac{1}{3.2}-\dfrac{1}{2.1}\)
= \(\dfrac{1}{100}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{98}+...+\dfrac{1}{3}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{1}\)
= \(\dfrac{1}{100}-\dfrac{1}{1}\)
= \(\dfrac{-99}{100}\)
d đảo 1/1.2.1/2.3 ... 1/99.1000
=1/1 -1/2 +1/2-1/3 ... -1/99 - 1/1000
=1/1 -1/1000
=999/1000
Thực hiện phép tính
a) \(1-\dfrac{1}{2}:\dfrac{-3}{2}+\dfrac{2}{3}=1-\dfrac{1}{2}.\dfrac{2}{-3}+\dfrac{2}{3}=\)
\(=1+\dfrac{1}{3}+\dfrac{2}{3}=\dfrac{3+1+2}{3}=\dfrac{6}{3}=2\)
b) \(\dfrac{-3}{4}-\dfrac{-7}{2}+\dfrac{2}{-3}=\dfrac{-3}{4}+\dfrac{7}{2}+\dfrac{-2}{3}=\)
\(=\dfrac{-9+42-8}{12}=\dfrac{25}{12}\)
c) \(0,25-3\dfrac{1}{2}:\dfrac{1}{2}+\dfrac{-3}{4}.\dfrac{2}{-3}=0,25-\dfrac{7}{2}.2+\dfrac{1}{2}=\)
\(=0,25-7+0,5=-\dfrac{25}{4}\)
b: \(\left(\dfrac{2}{5}-\dfrac{7}{10}x\right):\dfrac{5}{3}=-\dfrac{3}{4}\)
\(\Leftrightarrow\dfrac{2}{5}-\dfrac{7}{10}x=\dfrac{-3}{4}\cdot\dfrac{5}{3}=\dfrac{-5}{4}\)
\(\Leftrightarrow x\cdot\dfrac{7}{10}=\dfrac{2}{5}+\dfrac{5}{4}=\dfrac{8+25}{20}=\dfrac{33}{20}\)
\(\Leftrightarrow x=\dfrac{33}{20}:\dfrac{7}{10}=\dfrac{33}{20}\cdot\dfrac{10}{7}=\dfrac{33}{14}\)
c: \(\dfrac{7}{16}:\left(\dfrac{1}{4}x+\dfrac{9}{2}\right)-\dfrac{11}{6}=0\)
\(\Leftrightarrow\dfrac{7}{16}:\left(\dfrac{1}{4}x+\dfrac{9}{2}\right)=\dfrac{11}{6}\)
\(\Leftrightarrow x\cdot\dfrac{1}{4}+\dfrac{9}{2}=\dfrac{11}{6}:\dfrac{7}{16}=\dfrac{88}{21}\)
\(\Leftrightarrow x\cdot\dfrac{1}{4}=\dfrac{88}{21}-\dfrac{9}{2}=-\dfrac{13}{42}\)
hay \(x=-\dfrac{26}{21}\)
a. \(\dfrac{5}{3}+\left(\dfrac{-2}{7}\right)-\left(-1.2\right)\)
= \(\dfrac{5}{3}-\dfrac{2}{7}+\dfrac{6}{5}\)
= \(\dfrac{175}{105}-\dfrac{30}{105}+\dfrac{126}{105}\)
= \(\dfrac{271}{105}\)
b.\(\dfrac{-4}{7}+\left(\dfrac{-5}{6}\right)-\dfrac{17}{4}\)
= \(\dfrac{-4}{9}-\dfrac{5}{6}-\dfrac{17}{4}\)
= \(\dfrac{-16}{36}-\dfrac{30}{36}-\dfrac{153}{36}\)
= \(\dfrac{-199}{36}\)
a)
\(-12:\left(\dfrac{3}{4}-\dfrac{5}{6}\right)^2\)
\(=-12:\left(\dfrac{18}{24}-\dfrac{20}{24}\right)^2\)
\(=-12:\left(\dfrac{-1}{12}\right)^2\)
\(=-12:\dfrac{1}{144}\)
\(=-12\times\dfrac{144}{1}\)
\(=-1728\)
b)
\(\left(2^2:\dfrac{4}{3}-\dfrac{1}{2}\right)\times\dfrac{6}{5}-17\)
\(=\left(4\times\dfrac{3}{4}-\dfrac{1}{2}\right)\times\dfrac{6}{5}-17\)
\(=\left(3-\dfrac{1}{2}\right)\times\dfrac{6}{5}-17\)
\(=\dfrac{5}{2}\times\dfrac{6}{5}-17\)
\(=3-17\)
\(=-14\)
a)\(=-12:\left(-\dfrac{1}{12}\right)^2\)
\(=-12:\dfrac{1}{144}\)\(=-12.144=-1728\)
b)\(=\left(8:\dfrac{4}{3}-\dfrac{1}{2}\right).\dfrac{6}{5}-17\)
\(=\left(6-\dfrac{1}{2}\right).\dfrac{6}{5}-17\)
\(=\dfrac{11}{2}.\dfrac{6}{5}-17=\dfrac{33}{5}-17=\dfrac{33}{5}-\dfrac{85}{5}=-\dfrac{2}{5}\)