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Đặt biểu thức đã cho là A.
Ta có: 2A = (3 - 1) * (3 + 1) * (3^2 + 1) * .... * (3^64 + 1)
= (3^2 - 1) * (3^2 + 1) * ... * (3^64 + 1) (hằng đẳng thức a^2 - b^ 2 = (a+b)(a-b))
Rút gọn triệt tiêu ta được 2A=3^64 - 1
=> A = (3^64 - 1)/2
![](https://rs.olm.vn/images/avt/0.png?1311)
đk : x >= 0 ; x khác 4
\(B=\left(\dfrac{\sqrt{x}-2+\sqrt{x}+2}{x-4}\right).\dfrac{\sqrt{x}-2}{2}=\dfrac{2\sqrt{x}\left(\sqrt{x}-2\right)}{2\left(x-4\right)}=\dfrac{\sqrt{x}}{\sqrt{x}+2}\)
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\(A=\left(\frac{1}{1-x}+\frac{2}{x+1}-\frac{5-x}{1-x^2}\right):\frac{1-2x}{x^2-1}\left(x\ne\pm1;x\ne\frac{1}{2}\right)\)
\(\Leftrightarrow A=\left(\frac{-1}{x-1}+\frac{2}{x+1}+\frac{5-x}{x^2-1}\right)\cdot\frac{\left(x-1\right)\left(x+1\right)}{1-2x}\)
\(\Leftrightarrow A=\left[\frac{-x-1}{\left(x-1\right)\left(x+1\right)}+\frac{2x-2}{\left(x-1\right)\left(x+1\right)}+\frac{5-x}{\left(x-1\right)\left(x+1\right)}\right]\cdot\frac{\left(x-1\right)\left(x+1\right)}{1-2x}\)
\(\Leftrightarrow A=\frac{-x-1+2x-2+5-x}{\left(x-1\right)\left(x+1\right)}\cdot\frac{\left(x-1\right)\left(x+1\right)}{2}\)
\(\Leftrightarrow A=\frac{2\left(x-1\right)\left(x+1\right)}{2\left(x-1\right)\left(x+1\right)}=1\)
vậy \(A=1\left(x\ne\pm1;x\ne\frac{1}{2}\right)\)
\(A=\left(\frac{1}{1-x}+\frac{2}{x+1}-\frac{5-x}{1-x^2}\right):\frac{1-2x}{x^2-1}.\)
\(A=\left(\frac{x+1}{\left(1-x\right)\left(x+1\right)}+\frac{2\left(1-x\right)}{\left(x+1\right)\left(1-x\right)}-\frac{5-x}{1-x^2}\right):\frac{1-2x}{x^2-1}.\)
\(A=\left(\frac{x+1}{\left(1-x\right)\left(x+1\right)}+\frac{2\left(1-x\right)}{\left(x+1\right)\left(1-x\right)}-\frac{5-x}{1-x^2}\right):\frac{1-2x}{x^2-1}.\)
\(A=\left(\frac{x+1+2-2x-5+x}{1-x^2}\right):\frac{1-2x}{x^2-1}.\)
\(A=\left(\frac{-2}{1-x^2}\right):\frac{1-2x}{x^2-1}.\)
\(A=\frac{2}{x^2-1}:\frac{1-2x}{x^2-1}.\)
\(A=\frac{2}{x^2-1}\cdot\frac{^2-1}{1-2x}=\frac{2}{1-2x}\)ĐK: x khác 1/2
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(\dfrac{\dfrac{x}{x+1}}{\dfrac{x^2}{x^2+x+1}}-\dfrac{2x+1}{x^2+x}\right)\dfrac{x^2-1}{x-1}\)ĐK : \(x\ne\pm1\)
\(=\left(\dfrac{x}{x+1}.\dfrac{x^2+x+1}{x^2}-\dfrac{2x+1}{x\left(x+1\right)}\right)\left(x+1\right)=\left(\dfrac{x^2+x-1}{x^2+x}-\dfrac{2x+1}{x\left(x+1\right)}\right)\left(x+1\right)\)
\(=\left(\dfrac{x^2+x-1-2x-1}{x\left(x+1\right)}\right)\left(x+1\right)=\dfrac{x^2-3x-2}{x}\)
à xin lỗi mình nhầm dòng cuối
\(=\dfrac{x^2-x-2}{x}=\dfrac{\left(x+1\right)\left(x-2\right)}{x}\)
Để biểu thức trên nhận giá trị dương khi
\(\dfrac{\left(x+1\right)\left(x-2\right)}{x}>0\)bạn tự xét TH cả tử và mẫu nhé, mình đánh trên này bị lỗi
![](https://rs.olm.vn/images/avt/0.png?1311)
\(=x^6-6x^4+12x^2-8-x^3+x+6x^2-18x\\ =x^6-6x^4-x^3+18x^2-17x-8\)
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Bài 1:
a: \(A=\dfrac{x^2-3+x+3}{\left(x-3\right)\left(x+3\right)}\cdot\dfrac{x+3}{x}=\dfrac{x\left(x+1\right)}{x\left(x-3\right)}=\dfrac{x+1}{x-3}\)
b: Để A=3 thì 3x-9=x+1
=>2x=10
hay x=5
Bài 2:
a: \(A=\dfrac{x+x-2-2x-4}{\left(x-2\right)\left(x+2\right)}:\dfrac{x+2-x}{x+2}\)
\(=\dfrac{-6}{x-2}\cdot\dfrac{1}{2}=\dfrac{-3}{x-2}\)
b: Để A nguyên thì \(x-2\in\left\{1;-1;3;-3\right\}\)
hay \(x\in\left\{3;1;5;-1\right\}\)
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\(\left(x-1\right)\left(x+2\right)+\left(x+1\right)x=x^2+2x-x-2+x^2+x=\left(x^2+x^2\right)+\left(2x-x+x\right)-2=2x^2+2x-2=2\left(x^2+x-1\right)\)
\(\left(x-1\right)\left(x+2\right)+\left(x+1\right)\)
\(=x^2+2x-x-2+x+1\)
\(=x^2+2x-x+x-2+1\)
\(=x^2+2x-1\)
tíc mình nha