Rút gọn biểu thức P = a 2 3 a 2
A. a 3 4
B. a 1 3
C. a - 4 3
D. a - 1 2
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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 \(x\in\left\{3;1;5;-1\right\}\)
(a+b+c)3=(a+b)3+3(a+b)2c+3(a+b)c2+c3
=a3+b3+3ab.(a+b)+3(a+b)2c+3(a+b)c2+c3
=a3+b3+c3+3(a+b)(ab+ac+bc+c2)
=a3+b3+c3+3(a+b)[a.(b+c)+c.(b+c)]
=a3+b3+c3+3(a+b)(b+c)(c+a)
=>dpcm
P=12(5^2+1)(5^4+1)(5^8+1)(5^16+1)
=>2P=24(5^2+1)(5^4+1)(5^8+1)(5^16+1)
=(52-1)(52+1)(54+1)(58+1)(516+1)
=(54-1)(54+1)(58+1)(516+1)
=(58-1)(58+1)(516+1)
=(516-1)(516+1)
=532-1
==>P=(532-1)/2
Bài này áp dụng quy tắc dấu ngoặc
Bạn không hiểu từ bước thứ mấy???
Ta có: \(B=\left(a-1\right)+\left(a-2\right)-\left(a+3\right)+\left(a-4\right)\)
\(B=a-1+a-2-a-3+a-4\)
\(B=\left(a+a-a+a\right)-1-2-3-4\)
\(B=2a+\left[\left(-1\right)+\left(-2\right)+\left(-3\right)+\left(-4\right)\right]\)
\(B=2a+\left(-10\right)=2a+2.\left(-5\right)\)
\(B=2\left[a+\left(-10\right)\right]\)
`a)|x-2|=2<=>[(x=4(ko t//m)),(x=0(t//m)):}`
Thay `x=0` vào `A` có: `A=[2\sqrt{0}-3]/[\sqrt{0}-2]=3/2`
`b)` Với `x >= 0,x ne 4` có:
`B=[2(\sqrt{x}-3)+\sqrt{x}(\sqrt{x}+3)-4\sqrt{x}]/[(\sqrt{x}+3)(\sqrt{x}-3)]`
`B=[2\sqrt{x}-6+x+3\sqrt{x}-4\sqrt{x}]/[(\sqrt{x}+3)(\sqrt{x}-3)]`
`B=[x+\sqrt{x}-6]/[(\sqrt{x}+3)(\sqrt{x}-3)]`
`B=[(\sqrt{x}+3)(\sqrt{x}-2)]/[(\sqrt{x}+3)(\sqrt{x}-3)]`
`B=[\sqrt{x}-2]/[\sqrt{x}-3]`
`c)` Với `x >= 0,x ne 4` có:
`C=A.B=[2\sqrt{x}-3]/[\sqrt{x}-2].[\sqrt{x}-2]/[\sqrt{x}-3]=[2\sqrt{x}-3]/[\sqrt{x}-3]`
Có: `C >= 1`
`<=>[2\sqrt{x}-3]/[\sqrt{x}-3] >= 1`
`<=>[2\sqrt{x}-3-\sqrt{x}+3]/[\sqrt{x}-3] >= 0`
`<=>[\sqrt{x}]/[\sqrt{x}-3] >= 0`
Vì `x >= 0=>\sqrt{x} >= 0`
`=>\sqrt{x}-3 > 0`
`<=>x > 9` (t/m đk)
a) \(a^{\dfrac{1}{3}}\cdot a^{\dfrac{1}{2}}\cdot a^{\dfrac{7}{6}}=a^{\dfrac{1}{3}+\dfrac{1}{2}+\dfrac{7}{6}}=a^2\)
b) \(a^{\dfrac{2}{3}}\cdot a^{\dfrac{1}{4}}:a^{\dfrac{1}{6}}=a^{\dfrac{2}{3}+\dfrac{1}{4}-\dfrac{1}{6}}=a^{\dfrac{3}{4}}\)
c) \(\left(\dfrac{3}{2}a^{-\dfrac{3}{2}}\cdot b^{-\dfrac{1}{2}}\right)\left(-\dfrac{1}{3}a^{\dfrac{1}{2}}b^{\dfrac{2}{3}}\right)=\left(\dfrac{3}{2}\cdot-\dfrac{1}{3}\right)\left(a^{-\dfrac{3}{2}}\cdot a^{\dfrac{1}{2}}\right)\left(b^{-\dfrac{1}{2}}\cdot b^{\dfrac{2}{3}}\right)\)
\(=-\dfrac{1}{2}a^{-1}b^{-\dfrac{1}{3}}\)