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13 tháng 10 2020

\(\text{a, }\frac{-2}{5}+x=\left(\frac{-1}{3}\right)^2+\frac{2}{3}\)

\(\Leftrightarrow\frac{-2}{5}+x=\frac{1}{9}+\frac{6}{9}\)

\(\Leftrightarrow\text{ }\frac{-2}{5}+x=\frac{7}{9}\)

\(\Leftrightarrow\text{ }x=\frac{7}{9}-\frac{-2}{5}\)

\(\Leftrightarrow\text{ }x=\frac{53}{45}\)

\(\text{Vậy }x=\frac{53}{45}\)

\(\text{Chia hay cộng mình không biết nên mình làm 2 TH, cái nào đúng đề thì bạn nhìn nha:}\)

\(\text{TH 1: Dấu chia}\)

\(\text{b, }\frac{3}{5}-2x=\left(\frac{-3}{5}\right)^2:\frac{9}{25}\)

\(\text{ }\Leftrightarrow\frac{3}{5}-2x=\frac{9}{25}:\frac{9}{25}\)

\(\text{ }\Leftrightarrow\frac{3}{5}-2x=1\)

\(\text{ }\Leftrightarrow2x=\frac{3}{5}-1\)

\(\text{ }\Leftrightarrow2x=\frac{3}{5}-1\)

\(\text{ }\Leftrightarrow2x=\frac{-2}{5}\)

\(\text{ }\Leftrightarrow x=\frac{-2}{5}:2\)

\(\text{ }\Leftrightarrow x=\frac{-1}{5}\)

\(\text{Vậy }\text{​​}x=\frac{-1}{5}\)

\(\text{TH 2:Dấu cộng}\)

\(\text{b, }\frac{3}{5}-2x=\left(\frac{-3}{5}\right)^2+\frac{9}{25}\)

\(\Leftrightarrow\frac{3}{5}-2x=\frac{9}{25}+\frac{9}{25}\)

\(\Leftrightarrow\frac{3}{5}-2x=\frac{18}{25}\)

\(\Leftrightarrow2x=\frac{3}{5}-\frac{18}{25}\)

\(\Leftrightarrow2x=\frac{-3}{25}\)

\(\Leftrightarrow x=\frac{-3}{25}:2\)

\(\Leftrightarrow x=\frac{-3}{50}\)

\(\text{Vậy }x=\frac{-3}{50}\)

\(\text{c, }\left|2x-1\right|=\frac{1}{2}-\frac{-2}{3}\)

\(\Leftrightarrow\left|2x-1\right|=\frac{7}{6}\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-1=\frac{7}{6}\\2x-1=\frac{-7}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=\frac{13}{6}\\2x=\frac{-1}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{13}{12}\\x=\frac{-1}{12}\end{matrix}\right.\)

\(\text{Vậy }x\in\left\{\frac{13}{12};\frac{-1}{12}\right\}\)

\(\text{d, }\left(x-\frac{3}{4}\right).\frac{1}{2}=\left(\frac{-1}{2}\right)^2\)

\(\Leftrightarrow\left(x-\frac{3}{4}\right).\frac{1}{2}=\frac{1}{4}\)

\(\Leftrightarrow x-\frac{3}{4}=\frac{1}{4}:\frac{1}{2}\)

\(\Leftrightarrow x-\frac{3}{4}=\frac{1}{2}\)

\(\Leftrightarrow x=\frac{1}{2}+\frac{3}{4}\)

\(\Leftrightarrow x=\frac{5}{4}\)

\(\text{Vậy }x=\frac{5}{4}\)

Bài 1: Thu gọn a) \(\frac{1}{5}x^4y^3-3x^4y^3\) b) \(5x^2y^5-\frac{1}{4}x^2y^5\) c) \(\frac{1}{7}x^2y^3.\left(-\frac{14}{3}xy^2\right)-\frac{1}{2}xy.\left(x^2y^{\text{4}}\right)\) d) \(\left(3xy\right)^2.\left(-\frac{1}{2}x^3y^2\right)\) e) \(-\frac{1}{4}xy^2+\frac{2}{5}x^2y+\frac{1}{2}xy^2-x^2y\) f) \(\frac{1}{2}x^4y.\left(-\frac{2}{3}x^3y^2\right)-\frac{1}{3}x^7y^3\) g) \(\frac{1}{2}x^2y.\left(-10x^3yz^2\right).\frac{1}{4}x^5y^3z\) h)...
Đọc tiếp

Bài 1: Thu gọn

a) \(\frac{1}{5}x^4y^3-3x^4y^3\)

b) \(5x^2y^5-\frac{1}{4}x^2y^5\)

c) \(\frac{1}{7}x^2y^3.\left(-\frac{14}{3}xy^2\right)-\frac{1}{2}xy.\left(x^2y^{\text{4}}\right)\)

d) \(\left(3xy\right)^2.\left(-\frac{1}{2}x^3y^2\right)\)

e) \(-\frac{1}{4}xy^2+\frac{2}{5}x^2y+\frac{1}{2}xy^2-x^2y\)

f) \(\frac{1}{2}x^4y.\left(-\frac{2}{3}x^3y^2\right)-\frac{1}{3}x^7y^3\)

g) \(\frac{1}{2}x^2y.\left(-10x^3yz^2\right).\frac{1}{4}x^5y^3z\)

h) \(4.\left(-\frac{1}{2}x\right)^2-\frac{3}{2}x.\left(-x\right)+\frac{1}{3}x^2\)

i) \(1\frac{2}{3}x^3y.\left(\frac{-1}{2}xy^2\right)^2-\frac{5}{4}.\frac{8}{15}x^3y.\left(-\frac{1}{2}xy^2\right)^2\)

k) \(-\frac{3}{2}xy^2.\left(\frac{3}{4}x^2y\right)^2-\frac{3}{5}xy.\left(-\frac{1}{3}x^4y^3\right)+\left(-x^2y\right)^2.\left(xy\right)^2\)

n) \(-2\frac{1}{5}xy.\left(-5x\right)^2+\frac{3}{4}y.\frac{2}{3}\left(-x^3\right)-\frac{1}{9}.\left(-x\right)^3.\frac{1}{3}y\)

m) \(\left(-\frac{1}{3}xy^2\right)^2.\left(3x^2y\right)^3.\left(-\frac{5}{2}xy^2z^3\right)^{^2}\)

p) \(-2y.\left|2\right|x^4y^5.\left|-\frac{3}{4}\right|x^3y^2z\)

1
26 tháng 7 2019

Bài 1:

a) \(\frac{1}{5}x^4y^3-3x^4y^3\)

= \(\left(\frac{1}{5}-3\right)x^4y^3\)

= \(-\frac{14}{5}x^4y^3.\)

b) \(5x^2y^5-\frac{1}{4}x^2y^5\)

= \(\left(5-\frac{1}{4}\right)x^2y^5\)

= \(\frac{19}{4}x^2y^5.\)

Mình chỉ làm 2 câu thôi nhé, bạn đăng nhiều quá.

Chúc bạn học tốt!

29 tháng 7 2019

cảm ơn nha

chúc bạn học tốt

HQ
Hà Quang Minh
Giáo viên
19 tháng 9 2023

a)

\(\begin{array}{l}x:{\left( {\frac{{ - 1}}{2}} \right)^3} =  - \frac{1}{2}\\x =  - \frac{1}{2}.{\left( {\frac{{ - 1}}{2}} \right)^3}\\x = {\left( {\frac{{ - 1}}{2}} \right)^4}\\x = \frac{1}{{16}}\end{array}\)              

Vậy \(x = \frac{1}{{16}}\).

 b)

\(\begin{array}{l}x.{\left( {\frac{3}{5}} \right)^7} = {\left( {\frac{3}{5}} \right)^9}\\x = {\left( {\frac{3}{5}} \right)^9}:{\left( {\frac{3}{5}} \right)^7}\\x = {\left( {\frac{3}{5}} \right)^2}\\x = \frac{9}{{25}}\end{array}\)

Vậy \(x = \frac{9}{{25}}\).

c)

\(\begin{array}{l}{\left( {\frac{{ - 2}}{3}} \right)^{11}}:x = {\left( {\frac{{ - 2}}{3}} \right)^9}\\x = {\left( {\frac{{ - 2}}{3}} \right)^{11}}:{\left( {\frac{{ - 2}}{3}} \right)^9}\\x = {\left( {\frac{{ - 2}}{3}} \right)^2}\\x = \frac{4}{9}.\end{array}\)         

Vậy \(x = \frac{4}{9}\).

d)

\(\begin{array}{l}x.{\left( {0,25} \right)^6} = {\left( {\frac{1}{4}} \right)^8}\\x.{\left( {\frac{1}{4}} \right)^6} = {\left( {\frac{1}{4}} \right)^8}\\x = {\left( {\frac{1}{4}} \right)^8}:{\left( {\frac{1}{4}} \right)^6}\\x = {\left( {\frac{1}{4}} \right)^2}\\x = \frac{1}{{16}}\end{array}\)

Vậy \(x = \frac{1}{{16}}\).

20 tháng 10 2016

a) \(\left(\frac{2}{3}:x+\frac{1}{4}\right)^2=\frac{49}{64}\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}\left(\frac{2}{3}:x+\frac{1}{4}\right)^2=\left(\frac{7}{8}\right)^2\\\left(\frac{2}{3}:x+\frac{1}{4}\right)^2=-\left(\frac{7}{8}\right)^2\end{array}\right.\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}\frac{2}{3}:x+\frac{1}{4}=\frac{7}{8}\\\frac{2}{3}:x+\frac{1}{4}=-\frac{7}{8}\end{array}\right.\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}\frac{2}{3}:x=\frac{5}{8}\\\frac{2}{3}:x=-\frac{9}{8}\end{array}\right.\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x=\frac{16}{15}\\x=-\frac{16}{27}\end{array}\right.\)

20 tháng 10 2016

b) \(\frac{2}{5}:\left|x-2\right|-\frac{1}{3}=\frac{1}{2}\)

\(\Leftrightarrow\frac{2}{5}:\left|x-2\right|=\frac{5}{6}\)

\(\Leftrightarrow\left|x-2\right|=\frac{12}{25}\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x-2=\frac{12}{25}\\x-2=-\frac{12}{25}\end{array}\right.\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x=\frac{62}{25}\\x=\frac{38}{25}\end{array}\right.\)