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PT
1
CM
11 tháng 4 2018
a) Với a > 0 và a ≠ 1 ta có:
P = a − 1 ( a − 1 ) ( a − 1 ) + 3 a + 5 ( a − 1 ) ( a − 1 ) . ( a + 2 a + 1 ) − 4 a 4 a = 4 a + 4 ( a − 1 ) 2 ( a + 1 ) . a − 2 a + 1 4 a = 4 ( a − 1 ) 2 . ( a − 1 ) 2 4 a = 1 a
b, Có Q = a − a + 1 a
Xét Q − 1 = a − 2 a + 1 a = ( a − 1 ) 2 a
Vì ( a − 1 ) 2 > 0 , a > 0 , ∀ a > 0 , a ≠ 1 ⇒ Q − 1 > 0 ⇒ Q > 1
CM
26 tháng 10 2018
Với 0 < a < 1 ta có:
P = 1 + a 1 + a − 1 − a + 1 − a 2 1 − a 1 + a − 1 − a 2 1 − a 2 a 2 − 1 a = 1 + a 1 + a − 1 − a + 1 − a 2 1 − a 1 + a − 1 − a ( 1 − a ) ( 1 + a ) a 2 − 1 a = 1 + a 1 + a − 1 − a + 1 − a 1 + a − 1 − a 1 − a . 1 + a a 2 − 1 a = 1 + a + 1 − a 1 + a − 1 − a . 2 1 − a . 1 + a − ( 1 − a ) − ( 1 + a ) 2 a = 1 + a + 1 − a 1 + a − 1 − a . − 1 + a − 1 − a 2 2 a = − 1 + a + 1 − a 1 + a − 1 − a 2 a = − 1 + a − 1 + a 2 a = − 2 a 2 a = − 1
CM
21 tháng 12 2017
Biến đổi vế trái:
= (-√7 - √5)(√7 - √5)
= -(√7 + √5)(√7 - √5)
= -(7 - 5) = -2 = VP (đpcm)
= (1 + √a)(1 - √a)
= 1 - (√a)2 = 1 - a = VP (đpcm)
5 tháng 7 2021
Bài 1 :
a, ĐKXĐ : \(\dfrac{1}{2-x}\ge0\)
Mà 1 > 0
\(\Rightarrow2-x>0\)
\(\Rightarrow x< 2\)
Vậy ...
b, Ta có : \(\sqrt[3]{125}.\sqrt[3]{216}-\sqrt[3]{512}.\sqrt[3]{\dfrac{1}{8}}\)
\(=5.6-\dfrac{8.1}{2}=26\)
AT
5 tháng 7 2021
1a) Để căn thức bậc 2 có nghĩa thì \(\dfrac{1}{2-x}\ge0\Rightarrow2-x>0\Rightarrow x< 2\)
b) \(\sqrt[3]{125}.\sqrt[3]{-216}-\sqrt[3]{512}.\sqrt[3]{\dfrac{1}{8}}=\sqrt[3]{5^3}.\sqrt[3]{\left(-6\right)^3}-\sqrt[3]{8^3}.\sqrt[3]{\left(\dfrac{1}{2}\right)^3}\)
\(=5.\left(-6\right)-8.\dfrac{1}{2}=-34\)
\(\dfrac{\sqrt{ab}-b}{b}-\sqrt{\dfrac{a}{b}}=\dfrac{\sqrt{b}\left(\sqrt{a}-\sqrt{b}\right)}{\left(\sqrt{b}\right)^2}-\dfrac{\sqrt{a}}{\sqrt{b}}=\dfrac{\sqrt{a}-\sqrt{b}}{\sqrt{b}}-\dfrac{\sqrt{a}}{\sqrt{b}}\)
\(=-\dfrac{\sqrt{b}}{\sqrt{b}}=-1< 0\)
1 tháng 9 2021
Bài 2:
Ta có: \(P=\dfrac{15\sqrt{x}-11}{x+2\sqrt{x}-3}-\dfrac{3\sqrt{x}-2}{\sqrt{x}-1}-\dfrac{2\sqrt{x}+3}{\sqrt{x}+3}\)
\(=\dfrac{15\sqrt{x}-11-3x-9\sqrt{x}+2\sqrt{x}+6-2x+2\sqrt{x}-3\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{-5x+7\sqrt{x}-2}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{-5\sqrt{x}+1}{\sqrt{x}+3}\)
HP
22 tháng 10 2021
a.
A = \(\left(\dfrac{x-2}{x+2\sqrt{x}}+\dfrac{1}{\sqrt{x}+2}\right)+\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
A = \(\left(\dfrac{x-2}{\sqrt{x}\left(\sqrt{x}+2\right)}+\dfrac{1}{\sqrt{x}+2}\right)+\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
A = \(\left(\dfrac{x-2}{\sqrt{x}\left(\sqrt{x}+2\right)}+\dfrac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)}\right)+\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
A = \(\dfrac{x-2+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)}+\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
A = \(\dfrac{\left(x-2+\sqrt{x}\right).\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}+\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)}{\sqrt{x}\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)
A = \(\dfrac{\left(x-2+\sqrt{x}\right)\left(\sqrt{x}-1\right)+\sqrt{x}\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)}{\sqrt{x}\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)
A = \(\dfrac{x\sqrt{x}-x-2\sqrt{x}+2+x-\left(x+\sqrt{x}\right)\left(\sqrt{x}+2\right)}{\sqrt{x}\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)
A = \(\dfrac{x\sqrt{x}-x-2\sqrt{x}+2+x-\left(x\sqrt{x}+2x+x+2\sqrt{x}\right)}{\sqrt{x}\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)
A = \(\dfrac{x\sqrt{x}-x-2\sqrt{x}+2+x-x\sqrt{x}-2x-x-2\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)
A = \(\dfrac{-3x-4\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)
A = \(\dfrac{-\left(3x+4\sqrt{x}\right)}{\sqrt{x}\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)
A = \(\dfrac{-\sqrt{x}\left(3\sqrt{x}+4\right)}{\sqrt{x}\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)
A = \(\dfrac{3\sqrt{x}+4}{x-\sqrt{x}+2\sqrt{x}-2}\)
A = \(\dfrac{3\sqrt{x}+4}{x-\sqrt{x}-2}\)