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vào đây : https://coccoc.com/search/math#query=(+2x-5)%5E9%3D(2x-5)%5E6

22 tháng 9 2018

* Trả lời:

\(\left(1\right)\) \(-3\left(1-2x\right)-4\left(1+3x\right)=-5x+5\)

\(\Leftrightarrow-3+6x-4-12x=-5x+5\)

\(\Leftrightarrow6x-12x+5x=3+4+5\)

\(\Leftrightarrow x=12\)

\(\left(2\right)\) \(3\left(2x-5\right)-6\left(1-4x\right)=-3x+7\)

\(\Leftrightarrow6x-15-6+24x=-3x+7\)

\(\Leftrightarrow6x+24x+3x=15+6+7\)

\(\Leftrightarrow33x=28\)

\(\Leftrightarrow x=\dfrac{28}{33}\)

\(\left(3\right)\) \(\left(1-3x\right)-2\left(3x-6\right)=-4x-5\)

\(\Leftrightarrow1-3x-6x+12=-4x-5\)

\(\Leftrightarrow-3x-6x+4x=-1-12-5\)

\(\Leftrightarrow-5x=-18\)

\(\Leftrightarrow x=\dfrac{18}{5}\)

\(\left(4\right)\) \(x\left(4x-3\right)-2x\left(2x-1\right)=5x-7\)

\(\Leftrightarrow4x^2-3x-4x^2+2x=5x-7\)

\(\Leftrightarrow-x-5x=-7\)

\(\Leftrightarrow-6x=-7\)

\(\Leftrightarrow x=\dfrac{7}{6}\)

\(\left(5\right)\) \(3x\left(2x-1\right)-6x\left(x+2\right)=-3x+4\)

\(\Leftrightarrow6x^2-3x-6x^2-12x=-3x+4\)

\(\Leftrightarrow-15x+3x=4\)

\(\Leftrightarrow-12x=4\)

\(\Leftrightarrow x=-\dfrac{1}{3}\)

13 tháng 1 2016

4x-2x+x-27:9=33 

2x+x-27:9=33

3x-27:9=33

3x-27=33×9=297

3x=297+27=324

x=324÷3=108

23 tháng 11 2017

a,\(\left|9+x\right|=2x\)

\(\Leftrightarrow\left[{}\begin{matrix}9+x=2x\\9x+x=-2x\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}9=x\\9=-3x\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=9\\x=-3\end{matrix}\right.\)

Vậy...

Trường hợp 2 chưa chắc chắn lắm!!!

23 tháng 11 2017

a) \(\left|9+x\right|=2x\)

Xét trường hợp 1:

\(9+x=2x\)

\(\Leftrightarrow9+x-2x=0\)

\(\Leftrightarrow9-x=0\)

\(\Leftrightarrow x=9\)

Xét trường hợp 2:

\(9+x=-2x\)

\(\Leftrightarrow9+x-\left(-2x\right)=0\)

\(\Leftrightarrow9+x+2x=0\)

\(\Leftrightarrow9+3x=0\)

\(\Leftrightarrow3x=-9\)

\(\Leftrightarrow x=-9:3\)

\(\Leftrightarrow x=-3\)

Vậy x=9 hoặc x=-3

b) \(\left|x+6\right|-9=2x\)

\(\Leftrightarrow\left|x+6\right|=2x+9\)

Xét trường hợp 1:

\(x+6=2x+9\)

\(\Leftrightarrow x+6-\left(2x+9\right)=0\)

\(\Leftrightarrow x+6-2x-9=0\)

\(\Leftrightarrow-3-x=0\)

\(\Leftrightarrow x=-3\)

Xét trường hợp 2:

\(x+6=-\left(2x+9\right)\)

\(\Leftrightarrow x+6-\left[-\left(2x+9\right)\right]=0\)

\(\Leftrightarrow x+6+\left(2x+9\right)=0\)

\(\Leftrightarrow x+6+2x+9=0\)

\(\Leftrightarrow3x+15=0\)

\(\Leftrightarrow3x=-15\)

\(\Leftrightarrow x=-15:3\)

\(\Leftrightarrow x=-5\)

Vậy x=-3 hoặc x=-5

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

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

\(=\dfrac{x^2-25+2x^2-12x-3x^2+2x+9}{2x\left(x-5\right)}\)

\(=\dfrac{-10x-16}{2x\left(x-5\right)}=\dfrac{-5x-8}{x\left(x-5\right)}\)

2 tháng 7 2018

a) \(\left(2x+3\right)^3=\left(2x+3\right)^8\)

TH1 \(2x+3=1\)

\(2x=1-3=-2\)

\(x=-1\)

TH2 \(2x+3=0\)

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

b) ? sai đề

c) \(\left|5-3\right|=\left|11+2x\right|\Rightarrow\left|2\right|=\left|11+2x\right|\)

\(\hept{\begin{cases}11+2x=-2\\11+2x=2\end{cases}\Rightarrow}\hept{\begin{cases}2x=13\\2x=9\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}x=\frac{13}{2}\\x=\frac{9}{2}\end{cases}}\)

d) \(\left(x-5\right)^4=\left(x-5\right)^6\Rightarrow\hept{\begin{cases}x-5=0\\x-5=1\end{cases}}\Rightarrow\hept{\begin{cases}x=5\\x=6\end{cases}}\)

2 tháng 7 2018

Cảm ơn bạn. nhưng đúng ko z bn

25 tháng 8 2020

a) \(2x+\frac{3}{15}=\frac{7}{5}\) 

=> \(2x=\frac{7}{5}-\frac{3}{15}=\frac{21}{15}-\frac{3}{15}=\frac{18}{15}\)

=> \(x=\frac{18}{15}:2=\frac{18}{15}\cdot\frac{1}{2}=\frac{9}{15}\cdot\frac{1}{1}=\frac{9}{15}\)

b) \(x-\frac{2}{9}=\frac{8}{3}\)

=> \(x=\frac{8}{3}+\frac{2}{9}\)

=> \(x=\frac{24}{9}+\frac{2}{9}=\frac{26}{9}\)

c) \(\frac{-8}{x}=\frac{-x}{18}\)

=> x(-x) = (-8).18

=> -x2 = -144

=> x2 = 144(bỏ dấu âm)

=> x = \(\pm\)12

d) \(\frac{2x+3}{6}=\frac{x-2}{5}\)

=> 5(2x + 3) = 6(x - 2)

=> 10x + 15 = 6x - 12

=> 10x + 15 - 6x + 12 = 0

=> 4x + 27 = 0

=> 4x = -27

=> x = -27/4

e) \(\frac{x+1}{22}=\frac{6}{x}\)

=> x(x + 1) = 132

=> x(x + 1) = 11.12

=> x = 11

f) \(\frac{2x-1}{2}=\frac{5}{x}\)

=> x(2x - 1) = 10

=> 2x2 - x = 10

=> 2x2 - x - 10 = 0

tới đây tự làm đi nhé

g) \(\frac{2x-1}{21}=\frac{3}{2x+1}\)

=> (2x - 1)(2x + 1) = 63

=> 4x2 - 1 = 63

=> 4x2 = 64

=> x2 = 16

=> x = \(\pm\)4

h) Tương tự

25 tháng 8 2020

a) \(\frac{2x+3}{15}=\frac{7}{5}\Leftrightarrow10x+15=105\Leftrightarrow10x=90\Rightarrow x=9\)

b) \(\frac{x-2}{9}=\frac{8}{3}\Leftrightarrow3x-6=72\Leftrightarrow3x=78\Rightarrow x=26\)

c) \(\frac{-8}{x}=\frac{-x}{18}\Leftrightarrow x^2=144\Leftrightarrow\orbr{\begin{cases}x=12\\x=-12\end{cases}}\)

d) \(\frac{2x+3}{6}=\frac{x-2}{5}\Leftrightarrow10x+15=12x-12\Leftrightarrow2x=27\Rightarrow x=\frac{27}{2}\)

e) \(\frac{x+1}{22}=\frac{6}{x}\Leftrightarrow x^2+x-132=0\Leftrightarrow\left(x-11\right)\left(x+12\right)=0\Leftrightarrow\orbr{\begin{cases}x=11\\x=-12\end{cases}}\)

f) \(\frac{2x-1}{2}=\frac{5}{x}\Leftrightarrow2x^2-x-10=0\Leftrightarrow\left(x-2\right)\left(2x+5\right)=0\Leftrightarrow\orbr{\begin{cases}x=2\\x=-\frac{5}{2}\end{cases}}\)

g) \(\frac{2x-1}{21}=\frac{3}{2x+1}\Leftrightarrow4x^2=64\Leftrightarrow x^2=16\Rightarrow\orbr{\begin{cases}x=4\\x=-4\end{cases}}\)

h) \(\frac{10x+5}{6}=\frac{5}{x+1}\Leftrightarrow10x^2+15x-25=0\Leftrightarrow5\left(x-1\right)\left(2x+5\right)=0\Leftrightarrow\orbr{\begin{cases}x=1\\x=-\frac{5}{2}\end{cases}}\)

3 tháng 11 2018

\(a,x=3x^2\Rightarrow x-3x^2=0\Rightarrow x\left(1-3x\right)=0\Rightarrow\orbr{\begin{cases}x=0\\1-3x=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=\frac{1}{3}\end{cases}}\)

\(b,\left(2x-6\right)\left(x+4\right)+2\left(2x-6\right)=0\)

\(\Rightarrow\left(2x-6\right)\left(x+4+2\right)=0\)

\(\Rightarrow\left(2x-6\right)\left(x+6\right)=0\)

\(\Rightarrow\orbr{\begin{cases}2x-6=0\\x+6=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=3\\x=-6\end{cases}}\)

\(c,\left(2x-5\right)\left(x+9\right)+6x-15=0\)

\(\Rightarrow\left(2x-5\right)\left(x+9\right)+3\left(2x-5\right)=0\)

\(\Rightarrow\left(2x-5\right)\left(x+9+3\right)=0\)

\(\Rightarrow\left(2x-5\right)\left(x+12\right)=0\)

\(\Rightarrow\orbr{\begin{cases}2x-5=0\\x+12=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{5}{2}\\x=-12\end{cases}}\)

a: =x^4-3x^5+4x^8

b: =2x^3+2x^2+4x

c: =4x^2+8x-5

d: =2x+3x^2+7x^4