rút gọn \(P=\left(\frac{x-4}{x^3-1}+\frac{1}{x-1}\right):\left(1-\frac{x-8}{x^2+x+1}\right)\left(x\ne1\right)\)
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(2xy-x-y=2\)
\(\Rightarrow4xy-2x-2y=4\)
\(\Rightarrow2x\left(2y-1\right)-\left(2y-1\right)=5\)
\(\Rightarrow\left(2y-1\right)\left(2x-1\right)=5=5.1=1.5=\left(-1\right).\left(-5\right)=\left(-5\right).\left(-1\right)\)
\(\text{Bạn tự lập bảng nhé}\)
2xy-x-y=2
4xy-2x-2y=4
2x.(2y-1)-2y=4
2x.(2y-1)-(2y-1)=3
(2x-1).(2y-1)=3
còn lại lập bảng tự làm
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Trl:
\(\left(3x-2^4\right).7^{10}=2.7^{10}\)
\(\Rightarrow\left(3x-16\right)=2.7^{10}:7^{10}\)
\(\Rightarrow\left(3x-16\right)=2.1\)
\(\Rightarrow\left(3x-16\right)=2\)
\(\Rightarrow3x=16+2\)
\(\Rightarrow3x=18\)
\(\Rightarrow x=18:3\)
\(\Rightarrow x=6\)
Hc tốt
\(\left(3x-2^4\right)\times7^{10}=2\times7^{10}\)
\(3x-2^4=2\times7^{10}:7^{10}\)
\(3x-2^4=2\times1\)
\(3x-16=2\)
\(3x=18\)
\(x=6\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\text{b) Ta có: MD vuông góc với BE}\)
\(\text{ BE vuông góc với EN}\)
Xét tam giác MDI và tam giác IEN ta có:
MD=EN(vì tam giác MBD = tam giác CEN)
góc MDI = góc IEN(=90 độ)
góc DMI = góc INE(cmt)
=>tam giác MDI = tam giác IEN(CGV-GN)
=>IM=IN(ctư)
=>đường thẳng BC cắt MN tại trung điểm I của MN
\(P=\left(\frac{x-4}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}\right):\left(\frac{x^2+x+1}{x^2+x+1}-\frac{x-8}{x^2+x+1}\right)\)
\(=\left(\frac{x-4+x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}\right):\left(\frac{x^2+x+1-x+8}{x^2+x+1}\right)\)
\(=\left(\frac{x^2+2x-3}{\left(x-1\right)\left(x^2+x+1\right)}\right):\left(\frac{x^2+9}{x^2+x+1}\right)\)
\(=\left(\frac{\left(x+3\right)\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\right).\left(\frac{x^2+x+1}{x^2+9}\right)\)
\(=\frac{\left(x+3\right)\left(x-1\right)}{\left(x-1\right)\left(x^2+9\right)}\)
\(=\frac{x+3}{x^2+9}\)với \(x\ne1\)
Ta có: P = \(\left(\frac{x-4}{x^3-1}+\frac{1}{x-1}\right):\left(1-\frac{x-8}{x^2+x+1}\right)\)
P = \(\left(\frac{x-4}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}\right):\left(\frac{x^2+x+1-x+8}{x^2+x+1}\right)\)
P = \(\left(\frac{x-4+x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}\right):\left(\frac{x^2+9}{x^2+x+1}\right)\)
P = \(\frac{x^2+2x-3}{\left(x-1\right)\left(x^2+x+1\right)}\cdot\frac{x^2+x+1}{x^2+9}\)
P = \(\frac{x^2+3x-x-3}{\left(x-1\right)\left(x^2+x+1\right)}\cdot\frac{x^2+x+1}{x^2+9}\)
P = \(\frac{\left(x+3\right)\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\cdot\frac{x^2+x+1}{x^2+9}\)
P = \(\frac{x+3}{x^2+9}\)