Thực hiện các phép tính sau:
b) 1 b 4 a 2 b 4 với a > 0, b khác 0
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CM
30 tháng 6 2017
c) 16 a 4 b 6 128 a 6 b 6 với a < 0, b khác 0
= 1 8 a 2 = 1 2 2 a = - 1 2 2 a
H9
HT.Phong (9A5)
CTVHS
30 tháng 8 2023
Bài 1:
a) \(\dfrac{5^{16}\cdot27^7}{125^5\cdot9^{11}}\)
\(=\dfrac{5^{16}\cdot\left(3^3\right)^7}{\left(5^3\right)^5\cdot\left(3^2\right)^{11}}\)
\(=\dfrac{5^{16}\cdot3^{21}}{5^{15}\cdot3^{22}}\)
\(=\dfrac{5}{3}\)
b) \(\left(0,2\right)^2\cdot5-\dfrac{2^3\cdot27}{4^6\cdot9^5}\)
\(=0,2\cdot5\cdot0,2-\dfrac{2^3\cdot3^3}{\left(2^2\right)^6\cdot\left(3^2\right)^5}\)
\(=\dfrac{1}{5}-\dfrac{2^3\cdot3^3}{2^{12}\cdot3^{10}}\)
\(=\dfrac{1}{5}-\dfrac{1}{2^9\cdot3^7}\)
\(=\dfrac{2^9\cdot3^7}{2^9\cdot3^7\cdot5}-\dfrac{5}{2^9\cdot3^7\cdot5}\)
\(=\dfrac{2^9\cdot3^7-5}{2^9\cdot3^7\cdot5}\)
c) \(\dfrac{5^6+2^2\cdot25^3+2^3\cdot125^2}{26\cdot5^6}\)
\(=\dfrac{5^6\cdot\left(1+2^2+2^3\right)}{26\cdot5^6}\)
\(=\dfrac{1+2^2+2^3}{26}\)
\(=\dfrac{1+4+8}{26}\)
\(=\dfrac{13}{26}\)
\(=\dfrac{1}{2}\)
Bài 2:
Theo đề ta có:
\(\left(a\cdot\dfrac{1}{2}+\dfrac{3}{4}\right):-\dfrac{1}{4}=-\dfrac{15}{4}\)
\(\Rightarrow\left(a\cdot\dfrac{1}{2}+\dfrac{3}{4}\right)=-\dfrac{15}{4}\cdot-\dfrac{1}{4}\)
\(\Rightarrow a\cdot\dfrac{1}{2}+\dfrac{3}{4}=\dfrac{15}{16}\)
\(\Rightarrow a\cdot\dfrac{1}{2}=\dfrac{15}{16}-\dfrac{3}{4}\)
\(\Rightarrow a\cdot\dfrac{1}{2}=\dfrac{3}{16}\)
\(\Rightarrow a=\dfrac{3}{16}:\dfrac{1}{2}\)
\(\Rightarrow a=\dfrac{3}{8}\)
30 tháng 8 2023
1:
a: \(=\dfrac{5^{16}\cdot3^{21}}{3^{22}\cdot5^{15}}=\dfrac{1}{3}\cdot5=\dfrac{5}{3}\)
b: \(=0.04\cdot5-\dfrac{2^3\cdot3^3}{3^6\cdot2^{12}}\)
\(=0.2-\dfrac{1}{3^3\cdot2^9}=\dfrac{1}{5}-\dfrac{1}{3^3\cdot2^9}=\dfrac{3^3\cdot2^9-5}{5\cdot3^3\cdot2^9}\)
c: \(=\dfrac{5^6+4\cdot5^6+2^3\cdot5^6}{26\cdot5^6}=\dfrac{1+4+8}{26}=\dfrac{13}{26}=\dfrac{1}{2}\)
2:
Theo đề, ta có:
\(\left(a\cdot\dfrac{1}{2}+\dfrac{3}{4}\right):\dfrac{-1}{4}=\dfrac{-15}{4}\)
=>\(\dfrac{1}{2}a+\dfrac{3}{4}=\dfrac{15}{16}\)
=>1/2a=15/16-12/16=3/16
=>a=3/8
CM
9 tháng 3 2018
a) Quy đồng mẫu thức và sử dụng hằng đẳng thức rồi rút gọn thu được x + 1 2 ( x − 1 )
b) Tương tự a) thu được 2 2 − y
13 tháng 11 2018
a) 0,(1) + 0,(13) - 0,(123)
=0,(24)-0,(123)
=0,(119301)
b) 4,(14) + 2,(133)
\(\approx6,2745\)
H9
HT.Phong (9A5)
CTVHS
8 tháng 8 2023
b) \(B=\left(\dfrac{a\sqrt{a}+b\sqrt{b}}{\sqrt{a}+\sqrt{b}}\right):\left(a-b\right)+\dfrac{2\sqrt{b}}{\sqrt{a}+\sqrt{b}}\)
\(B=\left[\dfrac{\left(\sqrt{a}\right)^3+\left(\sqrt{b}\right)^3}{\sqrt{a}+\sqrt{b}}\right]:\left(a-b\right)+\dfrac{2\sqrt{b}}{\sqrt{a}+\sqrt{b}}\)
\(B=\left[\dfrac{\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+\sqrt{b}\right)}{\sqrt{a}+\sqrt{b}}\right]:\left(a-b\right)+\dfrac{2\sqrt{b}}{\sqrt{a}+\sqrt{b}}\)
\(B=\left(a-\sqrt{ab}+\sqrt{b}\right):\left(a-b\right)+\dfrac{2\sqrt{b}}{\sqrt{a}+\sqrt{b}}\)
\(B=\dfrac{a-\sqrt{ab}+b}{a-b}+\dfrac{2\sqrt{b}\left(\sqrt{a}-\sqrt{b}\right)}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}\)
\(B=\dfrac{a-\sqrt{ab}+b}{a-b}+\dfrac{2\sqrt{ab}-2b}{a-b}\)
\(B=\dfrac{a-\sqrt{ab}+b+2\sqrt{ab}-2b}{a-b}\)
\(B=\dfrac{a+\sqrt{ab}-b}{a-b}\)
G
8 tháng 8 2023
a) \(\sqrt{2}A=\sqrt{2x-2\sqrt{x-2}.\sqrt{x+2}}+\sqrt{2x+2\sqrt{x-2}.\sqrt{x+2}}\) (\(x\ge2\) )
\(=\sqrt{\left(x+2\right)-2\sqrt{x+2}.\sqrt{x-2}+\left(x-2\right)}+\sqrt{\left(x+2\right)+2\sqrt{x+2}.\sqrt{x-2}+\left(x-2\right)}\)
\(=\sqrt{\left(\sqrt{x+2}-\sqrt{x-2}\right)^2}+\sqrt{\left(\sqrt{x+2}+\sqrt{x-2}\right)^2}\)
\(=\left|\sqrt{x+2}-\sqrt{x-2}\right|+\sqrt{x+2}+\sqrt{x-2}\)
\(=\sqrt{x+2}-\sqrt{x-2}+\sqrt{x+2}+\sqrt{x-2}\) ( do \(x+2>x-2\ge0\Leftrightarrow\sqrt{x+2}>\sqrt{x-2}\) )
\(=2\sqrt{x+2}\)
\(\Leftrightarrow A=\sqrt{2}.\sqrt{x+2}\)
Vậy...
b) \(B=\left(\dfrac{a\sqrt{a}+b\sqrt{b}}{\sqrt{a}+\sqrt{b}}\right):\left(a-b\right)+\dfrac{2\sqrt{b}}{\sqrt{a}+\sqrt{b}}\)
\(=\dfrac{\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)}{\sqrt{a}+\sqrt{b}}.\dfrac{1}{a-b}+\dfrac{2\sqrt{b}}{\sqrt{a}+\sqrt{b}}\)
\(=\dfrac{a-\sqrt{ab}+b}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}+\dfrac{2\sqrt{b}\left(\sqrt{a}-\sqrt{b}\right)}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}\)
\(=\dfrac{a-\sqrt{ab}+b+2\sqrt{ab}-2b}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}\)
\(=\dfrac{a+\sqrt{ab}-b}{a-b}\)
Vậy...
CM
3 tháng 4 2019
a) Ta được: ( x + 2 ) 2 6 ( x + 2 ) = x + 2 6 ;
b) Ta được: 5 a + 9 5 a 2 b .
1 b 4 a 2 b 4 với a > 0, b khác 0
= 1 b . 2 a b 2 = 2 a b