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a.\(ĐKXĐ:\hept{\begin{cases}x^2-2x\ne0\\x-2\ne0\\x\left(x+1\right)\ne0\end{cases}\Leftrightarrow\hept{\begin{cases}x\left(x-2\right)\ne0\\x-2\ne0\\x\left(x+1\right)\ne0\end{cases}\Leftrightarrow}\hept{\begin{cases}x\ne0\\x\ne2\\x\ne-1\end{cases}}}\)
b.\(M=\left(\frac{1}{x^2-2x}+\frac{2}{x-2}\right)\div\frac{2x+1}{x\left(x+1\right)}\)
\(=\left(\frac{1}{x\left(x-2\right)}+\frac{2}{x-2}\right)\div\frac{2x+1}{x\left(x+1\right)}\)
\(=\left(\frac{1}{x\left(x-2\right)}+\frac{2x}{x\left(x-2\right)}\right)\div\frac{2x+1}{x\left(x+1\right)}\)
\(=\frac{2x+1}{x\left(x-2\right)}\div\frac{2x+1}{x\left(x+1\right)}\)
\(=\frac{2x+1}{x\left(x-2\right)}.\frac{x\left(x+1\right)}{2x+1}=\frac{x\left(2x+1\right)\left(x+1\right)}{x\left(x-2\right)\left(2x+1\right)}=\frac{x+1}{x-2}\)
c.Để \(M>1\)thì
\(\frac{x+1}{x-2}>1\)
c, Ta có : \(M>1\Rightarrow\frac{x+1}{x-2}>1\Leftrightarrow\frac{x+1}{x-2}-1>0\)
\(\Leftrightarrow\frac{x+1-x+2}{x-2}>0\Leftrightarrow\frac{3}{x-2}>0\)
\(\Rightarrow x-2>0\Leftrightarrow x>2\)vì 3 > 0
d, Để M nguyên khi \(x+1⋮x-2\Leftrightarrow x-2+3⋮x-2\)ĐK : \(x\ne2\)
\(\Leftrightarrow3⋮x-2\Rightarrow x-2\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)
| x - 2 | 1 | -1 | 3 | -3 |
| x | 3 | 1 | 5 | -1 |
\(a,M=1:\left(\frac{x^2+2}{x^3-1}+\frac{x+1}{x^2+x+1}-\frac{1}{x-1}\right)\)
\(=1:\left[\frac{x^2+2}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{x+1}{x^2+x+1}+\frac{-1}{x-1}\right]\)
\(=1:\left[\frac{\left(x^2+2\right)+\left(x+1\right)\left(x-1\right)+\left(-1\right)\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\right]\)
\(=1:\left[\frac{x^2+2+x^2-1-x^2-x-1}{\left(x-1\right)\left(x^2+x+1\right)}\right]\)
\(=1:\left[\frac{x^2-x}{\left(x-1\right)\left(x^2+x+1\right)}\right]=1:\left[\frac{x\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\right]\)
\(=1:\frac{x}{x^2+x+1}=\frac{x^2+x+1}{x}\)
\(a.\left(x^3-16x\right)=0\)
\(\Leftrightarrow x\left(x^2-16\right)=0\)
\(\Leftrightarrow x\left(x-4\right)\left(x+4\right)=0\)
\(\Leftrightarrow\hept{\begin{cases}x=0\\x-4=0\\x+4=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=0\\x=4\\x=-4\end{cases}}}\)
Uầy lười lm waa
. Hãy nhiệt tình lên :>> Chúng ta là công dân cùng một nước,phải giúp đỡ nhau a~~~
1/
a, (x-3)2+(4+x)(4-x)=10
<=>x2-6x+9+(16-x2)=10
<=>-6x+25=10
<=>-6x=-15
<=>x=5/2
còn lại tương tự a
2/
a, \(a^2\left(a+1\right)+2a\left(a+1\right)=\left(a^2+2a\right)\left(a+1\right)=a\left(a+1\right)\left(a+2\right)\)
Vì a(a+1)(a+2) là tích 3 nguyên liên tiếp nên a(a+1)(a+2) chia hết cho 2,3
Mà (2,3)=1
=>a(a+1)(a+2) chia hết cho 6 (đpcm)
b, \(x^2+2x+2=\left(x^2+2x+1\right)+1=\left(x+1\right)^2+1\)
Vì \(\left(x+1\right)^2\ge0\Rightarrow\left(x+1\right)^2+1\ge1>0\left(đpcm\right)\)
c, \(x^2-x+1=\left(x^2-x+\frac{1}{4}\right)+\frac{3}{4}=\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\)
Vì \(\left(x-\frac{1}{2}\right)^2\ge0\Rightarrow\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}>0\)(đpcm)
d, \(-x^2+4x-5=-\left(x^2-4x+4\right)-1=-\left(x-2\right)^2-1\)
Vì \(-\left(x-2\right)^2\le0\Rightarrow-\left(x-2\right)^2-1\le-1< 0\) (đpcm)
g,\(-4\left(x-1\right)^2+\left(2x+1\right)\left(2x-1\right)=-3\)
\(\Leftrightarrow-4\left(x^2-2x+1\right)+4x^2-1=-3\)
\(\Leftrightarrow-4x^2+8x-4+4x^2-1=-3\)
\(\Leftrightarrow8x=2\)
\(\Leftrightarrow x=\frac{1}{4}\)
bn xem lại đi nha