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2xy2.(-1/3x2y3)
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\(A=\left(\sqrt{2}-8\sqrt{32}+2\sqrt{450}\right):\left(-3\sqrt{8}\right)\)
\(=\left(\sqrt{2}-32\sqrt{2}+30\sqrt{2}\right):\left(-6\sqrt{2}\right)\)
\(=\sqrt{2}\left[\left(1-32+30\right):\left(-6\right)\right]\)
\(=\sqrt{2}\left[\left(-1\right):\left(-6\right)\right]\)
\(=\sqrt{2}.\dfrac{1}{6}\)
\(=\dfrac{\sqrt{2}}{6}\)
Bài 1 :
Ta có :
2\(n+1=n-3+4\)chia hết cho \(n-3\)\(\Rightarrow\)\(4\)chia hết cho \(n-3\)\(\Rightarrow\)\(\left(n-3\right)\inƯ\left(4\right)\)
Mà \(Ư\left(4\right)=\left\{1;-1;2;-2;4;-4\right\}\)
Do đó :
\(n\in\left\{4;2;5;1;7;-1\right\}\)
Bài 2 :
- Dãy số : \(1;3;5;...;19\)
Số số hạng là : \(\left(19-1\right):2+1=10\)
Tổng là : \(\left(19+1\right).10:2=100\)
- Dãy số : \(21;23;25;...;39\)
\(\left(39-21\right):2+1=10\)
Tổng là : \(\left(39+21\right).10:2=3000\)
\(\Rightarrow\)\(\frac{1+3+5+...+19}{21+23+25+...+39}=\frac{100}{3000}=\frac{1}{30}\)
(2x + 1)(x - 2) - (2x - 1)2
= (2x + 1)(x - 2) - (4x2 - 4x + 1)
= 2x2 - 3x - 2x - 4x2 + 4x - 1
= -2x2 + x - 3
Với `x >= 0,x \ne 1` có:
`C=A/B=A:B=[\sqrt{x}+1]/[x+\sqrt{x}+1]:(\sqrt{x}/[x\sqrt{x}-1]+1/[\sqrt{x}-1])`
`C=[\sqrt{x}+1]/[x+\sqrt{x}+1]:[\sqrt{x}+x+\sqrt{x}+1]/[(\sqrt{x}-1)(x+\sqrt{x}+1)]`
`C=[\sqrt{x}+1]/[x+\sqrt{x}+1].[(\sqrt{x}-1)(x+\sqrt{x}+1)]/[x+2\sqrt{x}+1]`
`C=[\sqrt{x}+1]/[x+\sqrt{x}+1].[(\sqrt{x}-1)(x+\sqrt{x}+1)]/[(\sqrt{x}+1)^2]`
`C=[\sqrt{x}-1]/[\sqrt{x}+1]`
1.Thế \(x=4\) vào A, ta được:
\(A=\dfrac{\sqrt{4}+1}{4+\sqrt{4}+1}=\dfrac{2+1}{4+2+1}=\dfrac{3}{7}\)
2.
\(B=\dfrac{\sqrt{x}}{x\sqrt{x}-1}+\dfrac{1}{\sqrt{x}-1}\)
\(B=\dfrac{\sqrt{x}}{\sqrt{x}^3-1}+\dfrac{1}{\sqrt{x}-1}\)
\(B=\dfrac{\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\dfrac{1}{\sqrt{x}-1}\)
\(B=\dfrac{\sqrt{x}+\left(x+\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(B=\dfrac{x+2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(B=\dfrac{\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(C=\dfrac{A}{B}\)
\(C=\dfrac{\sqrt{x}+1}{x+\sqrt{x}+1}:\dfrac{\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(C=\dfrac{\sqrt{x}+1}{x+\sqrt{x}+1}.\dfrac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)^2}\)
\(C=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\)
\(B=\dfrac{2\sqrt{x}-6\sqrt{x}+x+3\sqrt{x}-3x-9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{-3\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}=\dfrac{-3}{\sqrt{x}-3}\)
\(B=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\dfrac{3\sqrt{x}-9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\\ B=\dfrac{3\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\dfrac{3}{\sqrt{x}+3}\)
Bài 8:
a: Ta có: \(E=\left(\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}\right):\left(\dfrac{1}{x+1}+\dfrac{x}{x-1}+\dfrac{2}{x^2-1}\right)\)
\(=\dfrac{x^2+2x+1-x^2+2x-1}{\left(x-1\right)\left(x+1\right)}:\dfrac{x-1+x^2+x+2}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{4x}{x^2+2x+1}\)
b: Thay x=3 vào E, ta được:
\(E=\dfrac{4\cdot3}{\left(3+1\right)^2}=\dfrac{12}{4^2}=\dfrac{3}{4}\)
Thay x=-3 vào E, ta được:
\(E=\dfrac{4\cdot\left(-3\right)}{\left(-3+1\right)^2}=\dfrac{-12}{4}=-3\)
2xy2.(-1/3x2y3)
= -2/3x3y5