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29 tháng 8 2016

\(\frac{2x-1}{3}=\frac{-27}{1-2x}\)

\(\Rightarrow\left(2x-1\right)\left(1-2x\right)=\left(-27\right).3\)

\(\Rightarrow2x-4x^2-1-2x=-81\)

\(\Rightarrow4x^2-1=-81\)

\(\Rightarrow4x^2=-81+1\)

\(\Rightarrow4x^2=-80\)

Vì \(x^2\ge0\)mà \(4x^2=-80\)

\(\Rightarrow x\in\theta\)

tíc mình nha

30 tháng 11 2016

\(MTC:\left(x-3\right)^2\left(x^2+3x+9\right)\)

\(\frac{x}{x^3-27}=\frac{x}{\left(x-3\right)\left(x^2+3x+9\right)}=\frac{x\left(x-3\right)}{\left(x-3\right)^2\left(x^2+3x+9\right)}\)

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

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

\(MTC:2\left(x-1\right)\left(x+1\right)\)

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

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

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

\(MTC:2\left(x+1\right)\left(x^2-x+1\right)\)

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

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

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

=\(\frac{-1}{9}\)chuẩn cmnr

7 tháng 1 2016

bạn giúp mình giải 3 câu này nhé

 

27 tháng 7 2018

\(\frac{2x-1}{-27}=\frac{3}{1-2x}\)

\(\Rightarrow\left(2x-1\right).\left(1-2x\right)=3.\left(-27\right)\)

\(-4x^2=-81\) ( chỗ này bn tự phân tích ra nha!)

\(x^2=\frac{-81}{-4}=\frac{81}{4}=\left(\frac{9}{2}\right)^2=\left(-\frac{9}{2}\right)^2\)

=>  x = 9/2 hoặc x = -9/2

22 tháng 9 2018

lớp 5 thì có 

24 tháng 3 2020

\(1+\frac{2x}{x+4}+\frac{27}{2x^2+7x-4}=\frac{6}{2x-1}\left(x\ne-4;x\ne\frac{1}{2}\right)\)

\(\Leftrightarrow1+\frac{2x}{x+4}+\frac{27}{\left(x+4\right)\left(2x-1\right)}-\frac{6}{2x-1}=0\)

\(\Leftrightarrow\frac{2x^2+7x-4}{\left(x+4\right)\left(2x-1\right)}+\frac{2x\left(2x-1\right)}{\left(x+4\right)\left(2x-1\right)}+\frac{27}{\left(x+4\right)\left(2x-1\right)}-\frac{6\left(x+4\right)}{\left(x+4\right)\left(2x-1\right)}=0\)

\(\Leftrightarrow\frac{2x^2+7x-4}{\left(x+4\right)\left(2x-1\right)}+\frac{4x^2-2x}{\left(x+4\right)\left(2x-1\right)}+\frac{27}{\left(x+4\right)\left(2x-1\right)}-\frac{6x+24}{\left(x+4\right)\left(2x-1\right)}=0\)

\(\Leftrightarrow\frac{2x^2+7x-4+4x^2-2x+27-6x-24}{\left(x+4\right)\left(2x-1\right)}=0\)

\(\Leftrightarrow\frac{6x^2-x-1}{\left(x+4\right)\left(2x-1\right)}=0\)

\(\Leftrightarrow6x^2-x-1=0\)

\(\Leftrightarrow6x^2+2x-3x-1=0\)

<=> 2x(3x+1)-(3x+1)=0

<=> (3x+1)(2x-1)=0

\(\Leftrightarrow\orbr{\begin{cases}3x+1=0\\2x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{-1}{3}\left(tm\right)\\x=\frac{1}{2}\left(ktm\right)\end{cases}}}\)

Vậy pt có nghiệm \(x=\frac{-1}{3}\)

24 tháng 3 2020

\(ĐKXĐ:x\ne-4;x\ne\frac{1}{2}\)

\(1+\frac{2x}{x+4}+\frac{27}{2x^2+7x-4}=\frac{6}{2x-1}\)

\(\Leftrightarrow\frac{\left(x+4\right)\left(2x-1\right)}{\left(x+4\right)\left(2x-1\right)}+\frac{2x\left(2x-1\right)}{\left(x+4\right)\left(2x-1\right)}+\frac{27}{\left(x+4\right)\left(2x-1\right)}-\frac{6\left(x+4\right)}{\left(x+4\right)\left(2x-1\right)}=0\)

\(\Leftrightarrow\frac{2x^2+7x-4+4x^2-2x+27-6x-24}{\left(x+4\right)\left(2x-1\right)}=0\)

\(\Leftrightarrow6x^2-x-1=0\)

\(\Leftrightarrow\left(2x-1\right)\left(3x+1\right)=0\)

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