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TH1: |20x-1/2|^17=1 và |20x+1/2|^17=0

=>(20x-1/2=1 hoặc 20x-1/2=-1) và (20x+1/2=0)

=>x=-1/40

TH2: |20x-1/2|^17=0 và |20x+1/2|^17=1

=>20x-1/2=0 hoặc (20x+1/2=1 hoặc 20x+1/2=-1)

=>x=1/40

27 tháng 9 2021

\(E=\dfrac{98:\left(\dfrac{4}{5}\cdot\dfrac{5}{4}\right)}{\dfrac{16}{25}-\dfrac{1}{25}}+\dfrac{\left(\dfrac{27}{25}-\dfrac{2}{25}\right)\cdot\dfrac{7}{4}}{\left(\dfrac{59}{9}-\dfrac{13}{4}\right)\cdot\dfrac{36}{17}}\\ E=\dfrac{98}{\dfrac{3}{5}}+\dfrac{\dfrac{7}{4}}{\dfrac{119}{36}\cdot\dfrac{36}{17}}\\ E=\dfrac{490}{3}+\dfrac{\dfrac{7}{4}}{7}=\dfrac{490}{3}+\dfrac{1}{4}=\dfrac{1963}{12}\)

27 tháng 9 2021

bạn ơi chỗ kia mik nhìn hơi loạn tí bạn giải thích giúp mik với

 

Bài 1:1/\(\left(-\dfrac{25}{13}\right)+\left(-\dfrac{19}{17}\right)+\dfrac{12}{13}+\left(-\dfrac{25}{17}\right)\)                       6/ \(2\dfrac{2}{15}.\dfrac{9}{17}.\dfrac{3}{32}:\left(-\dfrac{3}{17}\right)\)2/\(\dfrac{1}{2}-\left(-\dfrac{1}{3}\right)+\dfrac{1}{23}+\dfrac{1}{6}\)                                           ...
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Bài 1:

1/\(\left(-\dfrac{25}{13}\right)+\left(-\dfrac{19}{17}\right)+\dfrac{12}{13}+\left(-\dfrac{25}{17}\right)\)                       6/ \(2\dfrac{2}{15}.\dfrac{9}{17}.\dfrac{3}{32}:\left(-\dfrac{3}{17}\right)\)

2/\(\dfrac{1}{2}-\left(-\dfrac{1}{3}\right)+\dfrac{1}{23}+\dfrac{1}{6}\)                                             7/\(\left(\dfrac{-3}{4}+\dfrac{2}{5}\right):\dfrac{3}{7}+\left(\dfrac{3}{5}+\dfrac{-1}{4}\right):\dfrac{3}{7}\)

3/\(\left(-\dfrac{3}{7}\right).\dfrac{5}{11}+\left(-\dfrac{5}{14}\right).\dfrac{5}{11}\)                                     8/\(\left(-\dfrac{1}{3}\right).\left(-\dfrac{15}{19}\right).\dfrac{38}{45}\)

4/\(\left(-\dfrac{5}{11}\right).\dfrac{7}{15}.\dfrac{11}{-5}.\left(-30\right)\)                                         9/\(\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+......+\dfrac{1}{19.20}\)

5/\(\left(-\dfrac{5}{9}\right).\dfrac{3}{11}+\left(-\dfrac{13}{18}\right).\dfrac{3}{11}\)                                 10/\(\dfrac{1}{9.10}-\dfrac{1}{8.9}-\dfrac{1}{7.8}-......-\dfrac{1}{2.3}-\dfrac{1}{1.2}\)

0
8 tháng 9 2017

\(A=\dfrac{\left(1+17\right).\left(1+\dfrac{17}{2}\right)..........\left(1+\dfrac{17}{19}\right)}{\left(1+19\right).\left(1+\dfrac{19}{2}\right)..........\left(1+\dfrac{19}{17}\right)}\)

\(=\dfrac{18.\dfrac{19}{2}.............\dfrac{36}{19}}{20.\dfrac{21}{2}..........\dfrac{36}{17}}\)

\(=\dfrac{18.19.20.......36}{1.2.3...19}:\dfrac{20.21.....36}{1.2.3...17}\)

\(=\dfrac{1.2.3......36}{1.2.....36}\)

\(=1\)

18 tháng 10 2021

\(\dfrac{\dfrac{4}{5}:\left(\dfrac{4}{5}\cdot\dfrac{5}{4}\right)}{\dfrac{16}{25}-\dfrac{1}{25}}+\dfrac{\left(\dfrac{27}{25}-\dfrac{2}{25}\right):\dfrac{4}{7}}{\left(\dfrac{59}{9}-\dfrac{13}{4}\right)\cdot\dfrac{36}{17}}+\left(\dfrac{6}{5}\cdot\dfrac{1}{2}\right):\dfrac{4}{5}\)

\(=\dfrac{4}{5}:\dfrac{3}{5}+\dfrac{7}{4}:7+\dfrac{3}{5}:\dfrac{4}{5}\)

\(=\dfrac{4}{3}+\dfrac{1}{4}+\dfrac{3}{4}\)

\(=\dfrac{7}{3}\)

19 tháng 10 2021

thanks bn nhìuvuiyeu

a: \(\left(2x-1\right)^{2008}+\left(y-\dfrac{2}{5}\right)^{2008}+\left|x+y-z\right|=0\)

nên \(\left\{{}\begin{matrix}2x-1=0\\y-\dfrac{2}{5}=0\\x+y-z=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=\dfrac{2}{5}\\z=x+y=\dfrac{9}{10}\end{matrix}\right.\)

b: Bạn xem lại đề, nghiệm rất xấu

 

 

4 tháng 10 2021

ừ bài nâng cao mà bạn ơi :)))

4 tháng 10 2021

\(P=\dfrac{1}{3}-\left(\dfrac{1}{3}\right)^2+\left(\dfrac{1}{3}\right)^3-\left(\dfrac{1}{3}\right)^4+...+\left(\dfrac{1}{3}\right)^{19}-\left(\dfrac{1}{3}\right)^{20}\)

\(=\left(\dfrac{1}{3}-\left(\dfrac{1}{3}\right)^2\right)+\left(\left(\dfrac{1}{3}\right)^3-\left(\dfrac{1}{4}\right)^4\right)+...+\left(\left(\dfrac{1}{3}\right)^{19}-\left(\dfrac{1}{3}\right)^{20}\right)\)

\(=\dfrac{1}{3}.\dfrac{2}{3}+\left(\dfrac{1}{3}\right)^3.\dfrac{2}{3}+...+\left(\dfrac{1}{3}\right)^{19}.\dfrac{2}{3}\)

\(=\dfrac{2}{3}.\left[\dfrac{1}{3}+\left(\dfrac{1}{3}\right)^3+...+\left(\dfrac{1}{3}\right)^{19}\right]\)

Bài 1: 

a: \(\left(2x-15\right)^5=\left(2x-15\right)^3\)

\(\Leftrightarrow\left(2x-15\right)^3\cdot\left[\left(2x-15\right)^2-1\right]=0\)

\(\Leftrightarrow\left(2x-15\right)^3\cdot\left(2x-16\right)\left(2x-14\right)=0\)

hay \(x\in\left\{8;7;\dfrac{15}{2}\right\}\)

b: \(\left(x-1\right)^3=\left(x-1\right)\)

\(\Leftrightarrow\left(x-1\right)\left[\left(x-1\right)^2-1\right]=0\)

=>x(x-1)(x-2)=0

hay \(x\in\left\{0;1;2\right\}\)

c: \(\left(x-1\right)^{x+2}=\left(x-1\right)^2\)

\(\Leftrightarrow\left(x-1\right)^x\cdot\left(x-1\right)^2-\left(x-1\right)^2=0\)

\(\Leftrightarrow\left(x-1\right)^2\cdot\left[\left(x-1\right)^x-1\right]=0\)

hay x=1

\(u_1=\dfrac{1}{\sqrt{2}};q=\dfrac{1}{\sqrt{2}}\)

\(S_{99}=\dfrac{\dfrac{1}{\sqrt{2}}\cdot\left(\dfrac{1}{\sqrt{2}}^{99}-1\right)}{\dfrac{1}{\sqrt{2}}-1}=\dfrac{1}{\sqrt{2}}\cdot\left(\dfrac{1-2^{49}\cdot\sqrt{2}}{2^{49}\cdot\sqrt{2}}\right):\dfrac{1-\sqrt{2}}{\sqrt{2}}\)

\(=\dfrac{1}{1-\sqrt{2}}\cdot\dfrac{1-2^{49}\cdot\sqrt{2}}{2^{49}\cdot\sqrt{2}}\)

24 tháng 11 2023

\(\dfrac{17}{-26}\cdot\left(\dfrac{1}{6}-\dfrac{5}{3}\right):\dfrac{17}{13}-\dfrac{20}{3}\cdot\left(\dfrac{2}{5}-\dfrac{1}{4}\right)+\dfrac{2}{3}\left(\dfrac{6}{5}-\dfrac{9}{2}\right)\)

\(=\dfrac{-17}{26}\cdot\dfrac{13}{17}\cdot\dfrac{1-10}{6}-\dfrac{20}{3}\cdot\dfrac{8-5}{20}+\dfrac{2}{3}\cdot\dfrac{12-45}{10}\)

\(=\dfrac{-1}{2}\cdot\dfrac{-9}{6}-\dfrac{20}{3}\cdot\dfrac{3}{20}+\dfrac{2}{3}\cdot\dfrac{-33}{10}\)

\(=\dfrac{3}{4}-1-\dfrac{66}{30}=\dfrac{-1}{4}-\dfrac{11}{5}=\dfrac{-5-44}{20}=-\dfrac{49}{20}\)