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19 tháng 1 2022
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 \(x\in\left\{3;1;5;-1\right\}\)
NV
Nguyễn Việt Lâm
Giáo viên
26 tháng 12 2022
1,
\(A=\dfrac{4x^2}{\left(x-2\right)\left(x+2\right)}+\dfrac{x-2}{\left(x-2\right)\left(x+2\right)}-\dfrac{x+2}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{4x^2+x-2-\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{4x^2-4}{\left(x-2\right)\left(x+2\right)}\)
\(x=4\Rightarrow A=\dfrac{4.x^2-4}{\left(4-2\right)\left(4+2\right)}=...\)
2.
\(A=\dfrac{x\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}+\dfrac{3\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}+\dfrac{3-5x}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{x\left(x+1\right)+3\left(x-1\right)+3-5x}{\left(x-1\right)\left(x+1\right)}=\dfrac{x^2-2x+1}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{\left(x-1\right)^2}{\left(x-1\right)\left(x+1\right)}=\dfrac{x-1}{x+1}\)
3.
Đề lỗi, thiếu dấu trước \(\dfrac{6+5x}{4-x^2}\)
NV
Nguyễn Việt Lâm
Giáo viên
26 tháng 12 2022
4.
\(A=\dfrac{2x}{\left(x-5\right)\left(x+5\right)}-\dfrac{5\left(x+5\right)}{\left(x-5\right)\left(x+5\right)}-\dfrac{x-5}{\left(x-5\right)\left(x+5\right)}\)
\(=\dfrac{2x-5\left(x+5\right)-\left(x-5\right)}{\left(x-5\right)\left(x+5\right)}=\dfrac{-4x-20}{\left(x-5\right)\left(x+5\right)}\)
\(=\dfrac{-4\left(x+5\right)}{\left(x-5\right)\left(x+5\right)}=\dfrac{-4}{x-5}\)
\(x=\dfrac{4}{5}\Rightarrow A=\dfrac{-4}{\dfrac{4}{5}-5}=\dfrac{20}{21}\)
5.
\(M=\dfrac{x^2}{x\left(x+2\right)}+\dfrac{2x}{x\left(x+2\right)}+\dfrac{2\left(x+2\right)}{x\left(x+2\right)}\)
\(=\dfrac{x^2+2x+2\left(x+2\right)}{x\left(x+2\right)}=\dfrac{x^2+4x+4}{x\left(x+2\right)}\)
\(=\dfrac{\left(x+2\right)^2}{x\left(x+2\right)}=\dfrac{x+2}{x}\)
\(x=-\dfrac{3}{2}\Rightarrow M=\dfrac{-\dfrac{3}{2}+2}{-\dfrac{3}{2}}=-\dfrac{1}{3}\)
SA
1
TN
1
LT
1
\(A=\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{63}+1\right).\)
\(=\frac{\left(3+1\right)\left(3-1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{63}+1\right)}{2}\)
\(=\frac{\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{63}+1\right)}{2}\)
\(=\frac{\left(3^{64}-1\right)\left(3^{63}+1\right)}{2}\left(\text{bn xem lại chỗ }3^{63}\text{ nhé!! ko thì ko lm đc tiếp đâu}\right)\)
mk viết nhầm 3 ^ 64