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\(A=\dfrac{6x^2+8x+7+x^2-x-6x^2-6x-6}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\dfrac{x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{1}{x-1}\)
Để 4A=x-1 thì \(\dfrac{4}{x-1}=x-1\)
=>x-1=2 hoặc x-1=-2
=>x=3(loại) hoặc x=-1(nhận)
\(A=\frac{6x^2+8x+7}{x^3-1}+\frac{x}{x^2+x+1}+\frac{6}{1-x}\)
<=>\(A=\frac{6x^2+8x+7}{x^3-1}+\frac{\left(x-1\right)x}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{\left(-6\right)\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)
<=>\(A=\frac{6x^2+8x+7}{x^3-1}+\frac{x^2-x}{x^3-1}+\frac{-6x^2-6x-6}{x^3-1}\)
<=>\(A=\frac{x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}\)<=>\(A=\frac{1}{x-1}\)<=>\(4A=\frac{4}{x-1}\)
Theo đề bài 4A=x-1 => \(4A=\frac{4}{x-1}=x-1\Rightarrow\left(x-1\right)^2=4\Rightarrow\orbr{\begin{cases}x-1=-2\\x-1=2\end{cases}}\Rightarrow\orbr{\begin{cases}x=-1\\x=3\end{cases}}\)
Vì x<0 nên x=-1
\(P=\left(\frac{8}{\left(x+4\right)\left(x-4\right)}+\frac{1}{x+4}\right):\frac{1}{x^2-2x-8}\)
\(P=\left(\frac{8}{\left(x+4\right)\left(x-4\right)}+\frac{x-4}{\left(x-4\right)\left(x+4\right)}\right)\cdot\frac{x^2-2x-8}{1}\)
\(P=\left(\frac{x+4}{\left(x+4\right)\left(x-4\right)}\right)\cdot x^2-2x-8\)
\(P=\frac{1}{x-4}\cdot x^2-2x-8\)
P\(P=\frac{x^2+2x-4x+8}{x-4}\)
\(P=\frac{x\left(x+2\right)-4\left(x+2\right)}{x-4}\)
\(P=\frac{\left(x-4\right)\left(x+2\right)}{x-4}\)
\(P=x+2\)
2 ,\(x^2-9x+20=0\)
\(\Rightarrow x^2-4x-5x+20=0\)
\(\Rightarrow x\left(x-4\right)-5\left(x-4\right)=0\)
\(\Rightarrow\left(x-5\right)\left(x-4\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-5=0\\x-4=0\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x=5\\x=4\end{cases}}\)
\(\orbr{\begin{cases}x=5\Rightarrow\\x=4\Rightarrow\end{cases}}\orbr{\begin{cases}P=7\\P=6\end{cases}}\)
a)
DK:tồn tại P \(\hept{\begin{cases}x\ne0\\x\ne-+6\\x\ne3\end{cases}}\)
\(P=\left(\frac{x}{\left(x-6\right)\left(x+6\right)}-\frac{x-6}{x\left(x+6\right)}\right).\frac{x\left(x+6\right)}{2\left(x-3\right)}\\ \)
\(P=\left(\frac{x^2-\left(x-6\right)\left(x-6\right)}{x\left(x-6\right)\left(x+6\right)}\right).\frac{x\left(x+6\right)}{2\left(x-3\right)}\)
\(P=\left(\frac{x^2-\left(x^2-12x+36\right)}{x\left(x-6\right)\left(x+6\right)}\right).\frac{x\left(x+6\right)}{2\left(x-3\right)}\)
\(P=\left(\frac{12\left(x-3\right)}{x\left(x-6\right)\left(x+6\right)}\right).\frac{x\left(x+6\right)}{2\left(x-3\right)}=\frac{6}{x-6}\)
b)6/(x-6)=1=> x-6=6=> x=12
c)x-6<0=> x<6
2) \(ĐKXĐ:x\notin\left\{-2;-3;-4\right\}\)
PT <=> \(x+\frac{x}{x+2}+\frac{x+3}{x^2+3x+2x+6}+\frac{x+4}{x^2+4x+2x+8}-1=0\)
<=>\(x+\frac{x}{x+2}+\frac{x+3}{x\left(x+3\right)+2\left(x+3\right)}+\frac{x+4}{x\left(x+4\right)+2\left(x+4\right)}-1=0\)
<=>\(x+\frac{x}{x+2}+\frac{x+3}{\left(x+2\right)\left(x+3\right)}+\frac{x+4}{\left(x+2\right)\left(x+4\right)}-1=0\)
<=>\(x+\frac{x}{x+2}+\frac{1}{x+2}+\frac{1}{x+2}-1=0\)
<=>\(x+\frac{x+1+1}{x+2}-1=0\)
<=>\(x+\frac{x+2}{x+2}-1=0\Leftrightarrow x+1-1=0\Leftrightarrow x=0\)
Vậy x=0 thì thỏa mãn PT
\(A=\frac{6x^2+8x+7+x\left(x-1\right)-6\left(x^2+x+2\right)}{\left(x-1\right)\left(x^2+x+1\right)} \)
\(A=\frac{x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}=\frac{1}{x-1}\Leftrightarrow\frac{1}{4A}\)
Ta có: \(A=\frac{1}{4A}\)
\(4A^2=1\)
\(A^2=\frac{1}{4}\)
\(\Rightarrow A=\sqrt{\frac{1}{4}}=\frac{1}{2}\\ \)
hoặc \(=-\frac{1}{2}\)
mình nhầm phần đầu
phải là: \(A=\frac{6x^2+8x+7+x\left(x-1\right)-6\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)mới đúng
cho mình sorry
Sửa đề: \(x+\dfrac{1}{x}=a\)
\(A=x^3+\dfrac{1}{x^3}=\left(x+\dfrac{1}{x}\right)^3-3\left(x+\dfrac{1}{x}\right)=a^3-3a\\ B=x^6+\dfrac{1}{x^6}=\left(x^3+\dfrac{1}{x^3}\right)^2-2=\left(a^3-3a\right)^2-2=a^6-6a^4+9a^2-2\\ C=x^7+\dfrac{1}{x^7}=\left(x^3+\dfrac{1}{x^3}\right)\left(x^4+\dfrac{1}{x^4}\right)-\left(x+\dfrac{1}{x}\right)\)
Mà \(x^4+\dfrac{1}{x^4}=\left(x^2+\dfrac{1}{x^2}\right)^2-2=\left[\left(x+\dfrac{1}{x}\right)^2-2\right]^2-2=\left(a^2-2\right)^2-2=a^4-4a^2+2\)
\(\Leftrightarrow C=\left(a^3-3a\right)\left(a^4-4a^2+2\right)-a=...\)
