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a: \(A=\left(\dfrac{2\left(2x+1\right)}{2\left(2x+4\right)}-\dfrac{x}{3x-6}-\dfrac{2x^3}{3x^3-12x}\right):\dfrac{6x+13x^2}{24x-12x^2}\)
\(=\left(\dfrac{2x+1}{2\left(x+2\right)}-\dfrac{x}{3\left(x-2\right)}-\dfrac{2x^3}{3x\left(x^2-4\right)}\right):\dfrac{x\left(13x+6\right)}{x\left(24-12x\right)}\)
\(=\left(\dfrac{2x+1}{2\left(x+2\right)}-\dfrac{x}{3\left(x-2\right)}-\dfrac{2x^2}{3\left(x-2\right)\left(x+2\right)}\right):\dfrac{13x+6}{-12\left(x-2\right)}\)
\(=\dfrac{3\left(2x+1\right)\left(x-2\right)-2x\left(x+2\right)-4x^2}{6\left(x+2\right)\left(x-2\right)}\cdot\dfrac{-12\left(x-2\right)}{13x+6}\)
\(=\dfrac{3\left(2x^2-3x-2\right)-2x^2-4x-4x^2}{x-2}\cdot\dfrac{-2}{13x+6}\)
\(=\dfrac{6x^2-9x-6-6x^2-4x}{x-2}\cdot\dfrac{-2}{13x+6}\)
\(=\dfrac{-\left(13x+6\right)\cdot\left(-2\right)}{\left(13x+6\right)\left(x-2\right)}=\dfrac{2}{x-2}\)
b: Để A>0 thì x-2>0
hay x>2
Để A>-1 thì A+1>0
\(\Leftrightarrow\dfrac{2+x-2}{x-2}>0\)
=>x/x-2>0
=>x>2 hoặc x<0
\(x^2-6x+9=\left(x-3\right)^2\)
\(4x^2+4x+1=\left(2x+1\right)^2\)
\(x^3+6x^2+12x+8=x^3+3.x^2.2+3.x.2^2+2^3=\left(x+2\right)^3\)
\(27x^3-27x^2+9x-1=\left(3x-1\right)^3\)( sửa đề chút )
\(\left(3x+2\right)\left(9x^2-6x+4\right)=\left(3x\right)^3+2^3\)( sửa đề chút )
Tham khảo nhé~
\(b,\frac{x-3}{x-2}=\frac{5}{\left(x-2\right)\left(x+3\right)}\)ĐKXĐ : \(x\ne2;\ne-3\)
\(\Leftrightarrow\frac{x^2-9}{\left(x-2\right)\left(x+3\right)}=\frac{5}{\left(x-2\right)\left(x+3\right)}\)
\(\Leftrightarrow x^2-9=5\)
\(\Leftrightarrow x^2=14\)
\(x=\sqrt{14}\)
.....
a) \(\left(x+3\right)^2-\left(x-3\right)^2=6x\Leftrightarrow\left(x^2+6x+9\right)-\left(x^2-6x+9\right)=6x\)
\(\Leftrightarrow x^2+6x+9-x^2+6x-9=6x\Leftrightarrow12x=6x\)\(\Leftrightarrow12x-6x=0\Leftrightarrow6x=0\Leftrightarrow x=0\)
Vậy phương trình có tập nghiệm S = { 0 }
b)\(-ĐKXĐ:\hept{\begin{cases}x-2\ne0\\\left(x-2\right)\left(x+3\right)\ne0\end{cases}}\Leftrightarrow\hept{\begin{cases}x-2\ne0\\x+3\ne0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ne2\\x\ne-3\end{cases}}\)
- Ta có : \(\frac{x-3}{x-2}=\frac{5}{\left(x-2\right)\left(x+3\right)}\Leftrightarrow\frac{x-3}{x-2}-\frac{5}{\left(x-2\right)\left(x+3\right)}=0\)
\(\Leftrightarrow\frac{\left(x-3\right)\left(x+3\right)-5}{\left(x-2\right)\left(x+3\right)}=0\Leftrightarrow\left(x-3\right)\left(x+3\right)=0\)\(\Leftrightarrow\orbr{\begin{cases}x-3=0\\x+3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\left(thoaman\right)\\x=-3\left(kothoaman\right)\end{cases}}\)
Vậy phương trình có tập nghiệm S = { 3 }
b: \(=\dfrac{4x\left(x-1\right)\left(x+1\right)}{6x\left(x-1\right)}=\dfrac{2\left(x+1\right)}{3}\)
c: \(=\dfrac{\left(5-x-1\right)\left(5+x+1\right)}{\left(x+6\right)^2}=\dfrac{\left(4-x\right)\left(x+6\right)}{\left(x+6\right)^2}=\dfrac{4-x}{x+6}\)
d: \(=\dfrac{\left(x+2\right)\left(x+3\right)}{\left(x+2\right)^2}=\dfrac{x+3}{x+2}\)
\(a,=3x\left(y-z\right)-y\left(y-z\right)=\left(3x-y\right)\left(y-z\right)\\ b,=x^3\left(x-1\right)+x\left(x-1\right)=x\left(x^2+1\right)\left(x-1\right)\\ c,=x\left(y+z\right)+y\left(y+z\right)=\left(x+y\right)\left(y+z\right)\\ d,=\left(x-3\right)^2\\ e,=\left(x+2\right)^3\\ f,=\left(2x-x+y\right)\left(2x+x-y\right)=\left(x+y\right)\left(3x-y\right)\\ g,=\left(y+1\right)\left(5x-2\right)\\ h,=\left(x+2\right)^2\\ i,=x^2\left(x^2-2\right)\\ k,=3x\left(x-4y\right)\)
Mình thử nha :33
ĐKXĐ : \(x\ne-3,x\ne-26,x\ne-6,x\ne1\)
Ta có :
\(A=\left[\frac{3}{2}-\left(\frac{x^4\left(x^2+1\right)-x^4-1}{x^2+1}\right)\cdot\frac{x^3-4x^2+\left(x-4\right)}{x^6\left(x+6\right)-\left(x+6\right)}\right]:\frac{\left(x+3\right)\left(x+26\right)}{3\left(x-2\right)\left(x+6\right)}\)
\(=\left[\frac{3}{2}-\left(\frac{x^6-1}{x^2+1}\right)\cdot\frac{\left(x-4\right)\left(x^2+1\right)}{\left(x+6\right)\left(x^6-1\right)}\right]\cdot\frac{3\left(x-2\right)\left(x+6\right)}{\left(x+3\right)\left(x+26\right)}\)
\(=\left[\frac{3}{2}-\frac{x-4}{x+6}\right]\cdot\frac{3\left(x-2\right)\left(x+6\right)}{\left(x+3\right)\left(x+26\right)}\)
\(=\frac{x+26}{2\left(x+6\right)}\cdot\frac{3\left(x-2\right)\left(x+6\right)}{\left(x+3\right)\left(x+26\right)}\)
\(=\frac{3\left(x-2\right)}{2\left(x+3\right)}\)
Vậy : \(A=\frac{3\left(x-2\right)}{2\left(x+3\right)}\left(x\ne-3,x\ne-26,x\ne-6,x\ne1\right)\)
