\(\frac{2}{\left|x-2\right|+2}=\frac{3}{\left|6-3x\right|+1}\)

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10 tháng 5 2019

What???

11 tháng 5 2019

Nà ní!!!!!!!!!

6 tháng 7 2019

\(a,-\frac{3}{2}-2x+\frac{3}{4}=-2\)

=> \(-\frac{3}{2}+\left(-2x\right)+\frac{3}{4}=-2\)

=> \(\left(-\frac{3}{2}+\frac{3}{4}\right)+\left(-2x\right)=-2\)

=> \(-\frac{3}{4}+\left(-2x\right)=-2\)

=> \(-2x=-2-\left(-\frac{3}{4}\right)=-\frac{5}{4}\)

=> \(x=-\frac{5}{4}:\left(-2\right)=\frac{5}{8}\)

Vậy \(x\in\left\{\frac{5}{8}\right\}\)

\(b,\left(\frac{-2}{3}x-\frac{3}{4}\right)\left(\frac{3}{-2}-\frac{10}{4}\right)=\frac{2}{5}\)

=> \(\left(-\frac{2}{3}x-\frac{3}{4}\right).\left(-4\right)=\frac{2}{5}\)

=> \(-\frac{2}{3}x-\frac{3}{4}=\frac{2}{5}:\left(-4\right)=-\frac{1}{10}\)

=> \(-\frac{2}{3}x=-\frac{1}{10}+\frac{3}{4}=\frac{13}{20}\)

=> \(x=\frac{13}{20}:\left(-\frac{2}{3}\right)=-\frac{39}{40}\)

Vậy \(x\in\left\{-\frac{39}{40}\right\}\)

\(c,\frac{x}{2}-\left(\frac{3x}{5}-\frac{13}{5}\right)=-\left(\frac{7}{5}+\frac{7}{10}x\right)\)

=> \(\frac{x}{2}-\frac{3x}{5}+\frac{13}{5}=-\frac{7}{5}-\frac{7}{10}x\)

=> \(10.\frac{x}{2}-10.\frac{3x}{5}+10.\frac{13}{5}=10.\frac{-7}{5}-10.\frac{7}{10}x\)

( chiệt tiêu )

=> \(5x-6x+26=-14-7x\)

=> \(-x+26=-14-7x\)

=> \(-x+7x=-14-26\)

=> \(6x=-40\)

=> \(x=-40:6=\frac{20}{3}\)

Vậy \(x\in\left\{\frac{20}{3}\right\}\)

\(d,\frac{2x-3}{3}+\frac{-3}{2}=\frac{5-3x}{6}-\frac{1}{3}\)

=> \(6.\frac{2x-3}{3}+6.\frac{-3}{2}=6.\frac{5-3x}{6}-6.\frac{1}{3}\)

( chiệt tiêu )

=> \(2\left(2x-3\right)-9=5-3x-2\)

=> \(4x-6-9=3-3x\)

=> \(4x-15=3-3x\)

=> \(4x+3x=3+15\)

=> \(7x=18\)

=> \(x=18:7=\frac{18}{7}\)

Vậy \(x\in\left\{\frac{18}{7}\right\}\)

\(e,\frac{2}{3x}-\frac{3}{12}=\frac{4}{x}-\left(\frac{7}{x}.2\right)\)

ĐKXĐ : \(x\ne0\)

=> \(\frac{2}{3x}-\frac{1}{4}=\frac{4}{x}-\frac{14}{x}\)

=> \(\frac{2}{3x}-\frac{4}{x}+\frac{14}{x}=\frac{1}{4}\)

=> \(\frac{2}{3x}-\frac{12}{3x}+\frac{42}{3x}=\frac{1}{4}\)

=> \(\frac{32}{3x}=\frac{1}{4}\)

=> \(3x=32.4:1=128\)

=> \(x=128:3=\frac{128}{3}\)

Vậy \(x\in\left\{\frac{128}{3}\right\}\)

\(k,\frac{13}{x-1}+\frac{5}{2x-2}-\frac{6}{3x-3}\)

ĐKXĐ :\(x\ne1;\)

=> \(\frac{13}{x-1}+\frac{5}{2\left(x-1\right)}-\frac{6}{3\left(x-1\right)}\)

=> \(\frac{13}{x-1}+\frac{5}{2\left(x-1\right)}-\frac{1}{x-1}\)

=> \(\frac{2.13}{2\left(x-1\right)}+\frac{5}{2\left(x-1\right)}-\frac{2.1}{2.\left(x-1\right)}\)

=> \(\frac{26+5-2}{2\left(x-1\right)}\)

=> \(\frac{29}{2\left(x-1\right)}\)

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

=> \(\frac{19}{10}:x-\frac{1}{2}=\frac{3}{2}\)

=> \(\frac{19}{10}:x=\frac{3}{2}+\frac{1}{2}=2\)

=> \(x=\frac{19}{10}:2=\frac{19}{20}\)

Vậy \(x\in\left\{\frac{19}{20}\right\}\)

\(n,\left(\frac{3}{2}-\frac{5}{11}-\frac{3}{13}\right)\left(2x-1\right)=\left(\frac{-3}{4}+\frac{5}{22}+\frac{3}{26}\right)\)

=> \(\frac{233}{286}\left(2x-1\right)=-\frac{233}{572}\)

=> \(2x-1=-\frac{233}{572}:\frac{233}{286}=-\frac{1}{2}\)

=> \(2x=-\frac{1}{2}+1=\frac{1}{2}\)

=> \(x=\frac{1}{2}:2=\frac{1}{4}\)

Vậy \(x\in\left\{\frac{1}{4}\right\}\)

18 tháng 5 2016

1) \(x=\frac{99}{196}\)

2) \(x=-2\)

3) \(x\approx-0,59\)

giup mk giải rõ dc ko

12 tháng 10 2020

a) \(\left|\frac{1}{3}x-8\right|+3=15\)

\(\Leftrightarrow\left|\frac{1}{3}x-8\right|=12\)

\(\Leftrightarrow\orbr{\begin{cases}\frac{1}{3}x-8=-12\\\frac{1}{3}x-8=12\end{cases}}\Leftrightarrow\orbr{\begin{cases}\frac{1}{3}x=-4\\\frac{1}{3}x=20\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-12\\x=60\end{cases}}\)

Vậy \(x\in\left\{-12;60\right\}\)

b) \(15-\left|2+3x\right|=8\)

\(\Leftrightarrow\left|2+3x\right|=7\)

\(\Leftrightarrow\orbr{\begin{cases}2+3x=-7\\2+3x=7\end{cases}}\Leftrightarrow\orbr{\begin{cases}3x=-9\\3x=5\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-3\\x=\frac{5}{3}\end{cases}}\)

Vậy \(x\in\left\{-3;\frac{5}{3}\right\}\)

d) \(-1\frac{1}{6}-\left|5-3x\right|=\frac{2}{3}\)

\(\Leftrightarrow\frac{-7}{6}-\left|5-3x\right|=\frac{2}{3}\)

\(\Leftrightarrow\left|5-3x\right|=\frac{-7}{6}-\frac{2}{3}\)

\(\Leftrightarrow\left|5-3x\right|=\frac{-11}{6}\)

Vì \(\left|5-3x\right|\ge0\forall x\)

mà \(\frac{-11}{6}< 0\)\(\Rightarrow\)Vô lý 

Vậy \(x\in\varnothing\)

12 tháng 10 2020

e) \(\left(\frac{3}{7}\right)^{20}:\left(\frac{9}{49}\right)^6=\left(\frac{3}{7}\right)^{20}:\left[\left(\frac{3}{7}\right)^2\right]^6=\left(\frac{3}{7}\right)^{20}:\left(\frac{3}{7}\right)^{2.6}\)

\(=\left(\frac{3}{7}\right)^{20}:\left(\frac{3}{7}\right)^{12}=\left(\frac{3}{7}\right)^8\)

g) \(4.2^5:\left(2^3.1^{16}\right)=2^2.2^5:2^3=2^4=16\)

9 tháng 9 2018

\(\left(4x+3\right)^2=\frac{2}{3}:6\)

\(\left(4x+3\right)^2=\frac{1}{9}\)

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

\(\Rightarrow4x+3=\frac{1}{3}\)

\(4x=-\frac{8}{3}\)

\(x=-\frac{2}{3}\)

9 tháng 9 2018

dòng thứ 4 phải ra 2 trường hợp chứ

3 tháng 8 2019

a,\(\left(x-\frac{2}{3}\right),\left(x+\frac{1}{1}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-\frac{2}{3}\\x+\frac{1}{4}\end{matrix}\right.\)

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

b,\(\left(x-\frac{2}{3}\right)\left(2x-\frac{3}{4}\right)=\left(3x+\frac{1}{2}\right)\left(x+\frac{2}{3}\right)\)

\(\Leftrightarrow2x^2-\frac{3}{4}x-\frac{4}{3}x+\frac{1}{2}=3x^2+2x+\frac{1}{2}x+\frac{1}{3}\)

\(\Leftrightarrow2x^2-\frac{25}{12}x+\frac{1}{2}=3x^2+\frac{5}{2}x+\frac{1}{3}\)

\(\Leftrightarrow24x^2-25x+6=36x^2+30x+4\)

\(\Leftrightarrow24x^2-25x+6-36x^2-30x-4=0\)

\(\Leftrightarrow-12x^2-55x+2=0\)

\(\Leftrightarrow12x^2+55x-2=0\)

\(\Leftrightarrow x=\frac{-55\pm\sqrt{55^2-4.12\left(-2\right)}}{2.12}\)

\(\Leftrightarrow\frac{-55\pm\sqrt{3025+96}}{24}\)

\(\Leftrightarrow\frac{-55\pm\sqrt{3121}}{24}\)

\(\Leftrightarrow\frac{-55+\sqrt{3121}}{24}\)

\(\Leftrightarrow\left[{}\begin{matrix}\frac{-55+\sqrt{3121}}{24}\\\frac{-55-\sqrt{3121}}{24}\end{matrix}\right.\)

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

26 tháng 10 2016

a ) \(\left(\frac{2}{5}-x\right):1\frac{1}{3}+\frac{1}{2}=-4\)

     \(\left(\frac{2}{5}-x\right):\frac{4}{3}+\frac{1}{2}=-4\)

     \(\left(\frac{2}{5}-x\right):\frac{4}{3}=-4-\frac{1}{2}\)

     \(\left(\frac{2}{5}-x\right):\frac{4}{3}=-\frac{9}{2}\)

        \(\frac{2}{5}-x=-\frac{9}{2}.\frac{4}{3}\)

        \(\frac{2}{5}-x=-3\)

                   \(x=\frac{2}{5}-\left(-3\right)\)

                   \(x=\frac{2}{5}+3\)

                   \(x=\frac{3}{5}-\frac{15}{5}\)

                   \(x=-\frac{12}{5}\)

Vay \(x=-\frac{12}{5}\) 

    

  

26 tháng 10 2016

b ) \(\left(-3+\frac{3}{x}-\frac{1}{3}\right):\left(1+\frac{2}{5}+\frac{2}{3}\right)=-\frac{5}{4}\)

     \(\left(-3+\frac{3}{x}-\frac{1}{3}\right):\left(\frac{15}{15}+\frac{6}{15}+\frac{10}{15}\right)=-\frac{5}{4}\)

     \(\left(-3+\frac{3}{x}-\frac{1}{3}\right):\left(\frac{15+6+10}{15}\right)=-\frac{5}{4}\)

     \(\left(-3+\frac{3}{x}-\frac{1}{3}\right):\frac{31}{15}=-\frac{5}{4}\)

     \(\left(-3+\frac{3}{x}-\frac{1}{3}\right)=-\frac{5}{4}.\frac{31}{15}\)

     \(\left(-3+\frac{3}{x}-\frac{1}{3}\right)=-\frac{1}{4}.\frac{31}{3}\)

        \(-3+\frac{3}{x}-\frac{1}{3}=-\frac{31}{12}\)

        \(-3+\frac{3}{x}=-\frac{31}{12}+\frac{1}{2}\)

        \(-3+\frac{3}{x}=-\frac{31}{12}+\frac{6}{12}\)

        \(-3+\frac{3}{x}=\frac{-25}{12}\)

                     \(\frac{3}{x}=\frac{-25}{12}+3\)

                      \(\frac{3}{x}=\frac{-25}{12}+\frac{36}{12}\)

                      \(\frac{3}{x}=\frac{5}{6}\)

                      \(\frac{18}{6x}=\frac{5x}{6x}\)

Đèn dây , bạn tự làm tiếp nhé , de rồi chứ