K
Khách

Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.

14 tháng 2 2018

=\(\dfrac{\left(\dfrac{7}{2^9}-\dfrac{14}{2^{11}}+\dfrac{21}{768}\right)^2}{\left(\dfrac{5}{2^9}-\dfrac{20}{2^{12}}+\dfrac{25}{1280}\right)^2}\)

= \(\dfrac{\left[7.\left(\dfrac{1}{2^9}-\dfrac{2}{2^{11}}+\dfrac{3}{768}\right)\right]^2}{\left[5.\left(\dfrac{1}{2^9}-\dfrac{4}{2^{12}}+\dfrac{5}{1280}\right)\right]^2}\)

=\(\dfrac{7^2}{5^2}\)

2 tháng 3 2022

49/25

2 tháng 3 2022

adu 

18 tháng 11 2021

\(E=\dfrac{\left|x-3\right|}{\left(x-3\right)\left(x+3\right)}\left(x+3\right)^2=\dfrac{\left|x-3\right|\left(x+3\right)}{x-3}\left(x\ne\pm3\right)\)

Với \(x>3\Leftrightarrow E=x+3\)

Với \(x< 3\Leftrightarrow E=-x-3\)

\(F=\dfrac{x+5\sqrt{x}-10\sqrt{x}-5\sqrt{x}+25}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\left(x\ge0;x\ne25\right)\\ F=\dfrac{\left(\sqrt{x}-5\right)^2}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}=\dfrac{\sqrt{x}-5}{\sqrt{x}+5}\)

3 tháng 7 2018

G=\(\dfrac{\left(-2\right)}{3}+\dfrac{\left(-5\right)}{7}+\dfrac{2}{3}+\dfrac{\left(-2\right)}{7}\)

\(\Rightarrow G=\dfrac{\left(-2\right)}{3}+\dfrac{2}{3}+\dfrac{\left(-5\right)}{7}+\dfrac{\left(-2\right)}{7}\)

\(\Rightarrow G=\dfrac{\left(-2\right)+2}{3}+\dfrac{\left(-5\right)+\left(-2\right)}{7}\)

\(\Rightarrow G=0+\dfrac{-7}{7}\)

\(\Rightarrow G=-1\)

\(H=\dfrac{\left(-5\right)}{7}\cdot\dfrac{2}{11}+\dfrac{\left(-5\right)}{7}\cdot\dfrac{9}{11}\)

\(\Rightarrow H=\dfrac{\left(-5\right)}{7}\cdot\left(\dfrac{2}{11}+\dfrac{9}{11}\right)\)\(\Rightarrow H=\dfrac{\left(-5\right)}{7}\cdot\left(\dfrac{2+9}{11}\right)\)

\(\Rightarrow H=\dfrac{\left(-5\right)}{7}\cdot1\)

\(\Rightarrow H=\dfrac{-5}{7}\)

4 tháng 12 2017

a)\(\left|-0.75\right|+\dfrac{1}{4}-2\dfrac{1}{2}\)

=0.75+0.25-2.5

=1-2.5=-1.5

b)\(15.\dfrac{1}{5}:\left(\dfrac{-5}{7}\right)-2\dfrac{1}{5}.\left(\dfrac{-7}{5}\right)\)

=3.(-1.4)+3.08

=-4.2+3.08=-1.12

c)\(\dfrac{5}{17}+\dfrac{2}{3}-\dfrac{20}{12}+\dfrac{7}{9}+\dfrac{12}{17}\)

=\(\dfrac{49}{51}-\dfrac{5}{3}+\dfrac{7}{9}+\dfrac{12}{17}\)

=\(\dfrac{-12}{17}+\dfrac{7}{9}+\dfrac{12}{17}\)

=\(\dfrac{11}{153}+\dfrac{12}{17}\)

=\(\dfrac{7}{9}\)

d)\(\dfrac{5}{15}+\dfrac{14}{25}-\dfrac{12}{9}+\dfrac{2}{7}+\dfrac{11}{25}\)

=\(\dfrac{67}{75}-\dfrac{4}{3}+\dfrac{2}{7}+\dfrac{11}{25}\)

=-0.44+\(\dfrac{127}{175}\)

=\(\dfrac{2}{7}\)

d) Ta có: \(D=\left(\dfrac{5\sqrt{x}-6}{x-9}-\dfrac{2}{\sqrt{x}+3}\right):\left(1+\dfrac{6}{x-9}\right)\)

\(=\dfrac{5\sqrt{x}-6-2\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}:\dfrac{x-9+6}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

\(=\dfrac{5\sqrt{x}-6-2\sqrt{x}+6}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\cdot\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}{x-3}\)

\(=\dfrac{3\sqrt{x}}{x-3}\)

f) Ta có: \(\left(\dfrac{3}{\sqrt{1+x}}+\sqrt{1-x}\right):\left(\dfrac{3}{\sqrt{1-x^2}}+1\right)\)

\(=\dfrac{3+\sqrt{1-x^2}}{\sqrt{1+x}}:\dfrac{3+\sqrt{1-x^2}}{\sqrt{1-x^2}}\)

\(=\dfrac{\sqrt{1-x^2}}{\sqrt{1+x}}=\sqrt{1-x}\)

20 tháng 12 2022

a: \(=\dfrac{3}{4}-\dfrac{5}{6}+\dfrac{3}{2}=\dfrac{9-10+18}{12}=\dfrac{17}{12}\)

b: \(=\left(\dfrac{1}{9}+\dfrac{6}{9}\right)^2-\dfrac{1}{3}=\dfrac{49}{81}-\dfrac{27}{81}=\dfrac{22}{81}\)

c; \(=\dfrac{5}{11}\left(-\dfrac{3}{7}-\dfrac{5}{7}\right)+\dfrac{-8}{7}\cdot\dfrac{6}{11}=\dfrac{-8}{7}\left(\dfrac{5}{11}+\dfrac{6}{11}\right)=-\dfrac{8}{7}\)

d: \(=\dfrac{2^{26}}{2^{15}\cdot2^{12}}=\dfrac{1}{2}\)