K
Khách

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

AH
Akai Haruma
Giáo viên
28 tháng 12 2023

Lời giải:
a. $\frac{3}{-7}=\frac{-27}{63}$

$\frac{-5}{9}=\frac{-35}{63}$

Do $\frac{27}{63}< \frac{35}{63}$ nên $\frac{-27}{63}> \frac{-35}{63}$

$\Rightarrow \frac{3}{-7}> \frac{-5}{9}$

---------

b.

$-0,625=\frac{-625}{1000}=\frac{-5}{8}=\frac{-125}{200}$

$\frac{-19}{50}=\frac{-76}{200}> \frac{-125}{200}$

$\Rightarrow -0,625> \frac{-19}{50}$

c.

$-2\frac{5}{9}=-(2+\frac{5}{9})=\frac{-23}{9}=-(\frac{-23}{-9})$

a: \(\Leftrightarrow\left(3x-2\right):\dfrac{7}{5}=\dfrac{17}{7}:\dfrac{13}{5}=\dfrac{85}{91}\)

\(\Leftrightarrow3x-2=\dfrac{85}{91}\cdot\dfrac{7}{5}=\dfrac{17}{13}\)

=>3x=43/13

hay x=43/39

b: \(\Leftrightarrow9x+207=121-8x\)

=>19x=-86

hay x=-86/19

c: \(\Leftrightarrow x^2-9=16\)

=>x2=25

=>x=5 hoặc x=-5

d: \(\Leftrightarrow\left|x\right|=\dfrac{1.64\cdot3.11}{8.51}\simeq0,6\)

=>x=0,6 hoặc x=-0,6

20 tháng 1 2018

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,\(\left(\dfrac{-2}{3}\right)^3.9^2+\left(\dfrac{-3}{4}\right)^2.32\)

\(=\dfrac{\left(-2\right)^3}{3^3}.3^4+\dfrac{\left(-3\right)^2}{4^2}.2^5\)

\(=\left(-8\right).3+\dfrac{3^2}{4^2}.2^5\)

\(=\left(-24\right)+2.9\)

\(=\left(-24\right)+18\)

\(=-6\)

20 tháng 1 2018

a,

26 tháng 7 2018

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

8 tháng 4 2017

a)

=

b) =

9 tháng 4 2017

a, \(\left(\dfrac{-2}{3}+\dfrac{3}{7}\right)-\dfrac{5}{21}:\dfrac{4}{5}+\left(\dfrac{-1}{3}+\dfrac{4}{7}\right):\dfrac{4}{5}\\ = -\dfrac{5}{21}:\dfrac{4}{5}+ \left(-\dfrac{5}{21}\right):\dfrac{4}{5}\\ =\left[-\dfrac{5}{21}+\left(-\dfrac{5}{21}\right)\right]:\dfrac{4}{5}\\ -\dfrac{10}{21}:\dfrac{4}{5}\\ =-\dfrac{25}{42}\)

b,

\(\dfrac{5}{9}:\left(\dfrac{1}{11}-\dfrac{5}{22}\right)+\dfrac{5}{9}:\left(\dfrac{1}{15}-\dfrac{2}{3}\right)\\ =\dfrac{5}{9}:\dfrac{-3}{22}+\dfrac{5}{9}:-\dfrac{3}{5}\\ =\dfrac{5}{9}:\left(\dfrac{-3}{22}+-\dfrac{3}{5}\right)\\ =\dfrac{5}{9}:-\dfrac{81}{110}\\ =-\dfrac{550}{729}\)

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 \(x=-\dfrac{26}{21}\)

27 tháng 7 2017

a, \(\left(0,8.7+0,64\right).\left(1,25.7-\dfrac{4}{5}.1,25\right)\)

\(=6,24.1,25\left(7-\dfrac{4}{5}\right)=7,8.6,2=48,36\)

b, \(\dfrac{1}{4}-\dfrac{\dfrac{1}{4}+\dfrac{1}{9}}{\dfrac{1}{9}}=\dfrac{1}{4}-\dfrac{\dfrac{13}{36}}{\dfrac{1}{9}}=\dfrac{1}{4}-\dfrac{13}{4}=-3\)

c, \(2\dfrac{1}{2}-3,4\left(12\right)-\dfrac{4}{3}+\dfrac{1}{3}.\left(\dfrac{1}{2}+0,5-3\dfrac{1}{2}\right)\)

\(=\dfrac{5}{2}-\dfrac{563}{165}-\dfrac{4}{3}+\dfrac{1}{3}.\left(-\dfrac{5}{2}\right)\)

\(=-\dfrac{301}{330}-\dfrac{4}{3}-\dfrac{5}{6}=-\dfrac{508}{165}\)

Chúc bạn học tốt!!!

17 tháng 6 2017

a,\(\dfrac{1}{3}-\dfrac{3}{5}+\dfrac{5}{7}-\dfrac{7}{9}+\dfrac{9}{11}-\dfrac{11}{13}+\dfrac{13}{15}+\dfrac{11}{13}-\dfrac{9}{11}+\dfrac{7}{9}-\dfrac{5}{7}+\dfrac{3}{5}-\dfrac{1}{3}\)

\(=\left(\dfrac{1}{3}-\dfrac{1}{3}\right)+\left(-\dfrac{3}{5}+\dfrac{3}{5}\right)+.....+\left(-\dfrac{11}{13}+\dfrac{11}{13}\right)+\dfrac{13}{15}\)

\(=0+0+...0+0+\dfrac{13}{15}=\dfrac{13}{15}\)

câu b và c xem lại đề nha

Chúc bạn học tốt!!!

17 tháng 6 2017

Đề đúng mà bạn

18 tháng 5 2017

Mình chỉ giải câu a thôi,mấy câu còn lại dễ.

a)Ta có:\(\dfrac{x}{27}=\dfrac{-3}{x}\)

=>\(x^2=-3\cdot27=-81\)(Nhân chéo)

Mà x2>0 với mọi x nên :

Không có giá trị nào thỏa mãn điều kiện của x

18 tháng 5 2017

Tìm x biết :

a) \(\dfrac{x}{27}=-\dfrac{3}{x}\) \(\Rightarrow2x=-3.27\Rightarrow2x=-81\Rightarrow x=-40,5\)

b) \(-\dfrac{9}{x}=-\dfrac{x}{\dfrac{4}{49}}\Rightarrow2x=-9.\left(-\dfrac{4}{9}\right)\Rightarrow2x=4\Rightarrow x=2\)

c) \(\left|7x-\dfrac{5}{3}\right|+\dfrac{7}{19}=-\dfrac{8}{15}\) ( mk nghĩ bn chép sai đề bài câu này )

\(\Rightarrow\left|7x-\dfrac{5}{3}\right|=-\dfrac{8}{15}-\dfrac{7}{19}\)

\(\Rightarrow\left|7x-\dfrac{5}{3}\right|=-\dfrac{257}{285}\)

\(\Rightarrow\left[{}\begin{matrix}7x-\dfrac{5}{3}=-\dfrac{257}{285}\\7x-\dfrac{5}{3}=\dfrac{257}{285}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{218}{1995}\\x=\dfrac{244.}{665}\end{matrix}\right.\)

d) \(\left|\dfrac{1}{23}x\right|+\dfrac{18}{90}=\dfrac{18}{19}-1\dfrac{2}{5}\)

\(\left|\dfrac{1}{23}x\right|+\dfrac{18}{90}=-\dfrac{43}{95}\)

\(\left|\dfrac{1}{23}x\right|=-\dfrac{43}{95}-\dfrac{18}{90}\)

\(\left|\dfrac{1}{23}x\right|=-\dfrac{62}{95}\)

\(\Rightarrow\left[{}\begin{matrix}\dfrac{1}{23}x=\dfrac{62}{95}\\\dfrac{1}{23}x=-\dfrac{62}{95}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=15\dfrac{1}{95}\\x=-15\dfrac{1}{95}\end{matrix}\right.\)

a: \(\Leftrightarrow\left|x-3\right|=12-5x-8=-5x+4\)

\(\Leftrightarrow\left\{{}\begin{matrix}x< =\dfrac{4}{5}\\\left(-5x+4\right)^2=\left(x-3\right)^2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x< =\dfrac{4}{5}\\\left(5x-4-x+3\right)\left(5x-4+x-3\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x< =\dfrac{4}{5}\\\left(4x-1\right)\left(6x-7\right)=0\end{matrix}\right.\Leftrightarrow x=\dfrac{1}{4}\)

b: \(\left(\sqrt{x}+3\right)^{10}=1024\cdot125^2\cdot25^2\)

\(\Leftrightarrow\left(\sqrt{x}+3\right)^{10}=2^{10}\cdot5^6\cdot5^4=10^{10}\)

\(\Leftrightarrow\sqrt{x}+3=10\)

hay x=49

c: \(\dfrac{3-0.2x}{5}=\dfrac{7}{15}+1.4x\)

\(\Leftrightarrow\dfrac{9-0.6x}{15}=\dfrac{7}{15}+\dfrac{21x}{15}\)

=>21x+7=9-0,6x

=>21,6x=-2

hay x=-5/54

d: \(\Leftrightarrow\left(\dfrac{4}{3}\right)^{3x}=\dfrac{5^9\cdot7^9\left(4\cdot7-5^2\right)}{5^9\cdot7^9\cdot4}\)

\(\Leftrightarrow\left(\dfrac{4}{3}\right)^{3x}=\dfrac{28-25}{4}=\dfrac{3}{4}\)

=>3x=-1

hay x=-1/3