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\(P=\left(\frac{1}{a-1}+\frac{3\sqrt{a}+5}{a\left(\sqrt{a}-1\right)-\left(\sqrt{a}-1\right)}\right).\frac{\left(\sqrt{a}+1\right)^2}{4a}\)
\(=\left(\frac{\sqrt{a}-1}{\left(a-1\right)\left(\sqrt{a}-1\right)}+\frac{3\sqrt{a}+5}{\left(a-1\right)\left(\sqrt{a}-1\right)}\right).\frac{\left(\sqrt{a}+1\right)^2}{4\sqrt{a}}\)
\(=\left(\frac{4\left(\sqrt{a}+1\right)}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)^2}\right).\frac{\left(\sqrt{a}+1\right)^2}{4\sqrt{a}}\)
\(=\frac{4}{\left(\sqrt{a}-1\right)^2}.\frac{\left(\sqrt{a}+1\right)^2}{4\sqrt{a}}=\frac{\left(\sqrt{a}+1\right)^2}{\sqrt{a}\left(\sqrt{a}-1\right)^2}\)
Không rút gọn được nữa, chắc do bạn ghi sai đề
Ở đằng sau biểu thức là \(\left(\frac{\left(\sqrt{a}+1\right)^2}{4\sqrt{a}}-1\right)\) sẽ hợp lý hơn, khi đó sẽ rút gọn được
a) \(A=\left(\frac{1}{a-\sqrt{a}}+\frac{1}{\sqrt{a}-1}\right):\frac{\sqrt{a}+1}{a-2\sqrt{a}+1}\)
\(=\left[\frac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}+\frac{\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)}\right].\frac{\left(\sqrt{a}-1\right)^2}{\sqrt{a}+1}\)
\(=\frac{\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}.\frac{\left(\sqrt{a}-1\right)^2}{\sqrt{a}+1}=\frac{\sqrt{a}-1}{\sqrt{a}}=1-\frac{1}{\sqrt{a}}\)
b) Ta có : \(\frac{1}{\sqrt{a}}>0\Leftrightarrow-\frac{1}{\sqrt{a}}< 0\Rightarrow\) \(A=1-\frac{1}{\sqrt{a}}< 1\)
a) \(A=\left(\frac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}+\frac{\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)}\right):\frac{\sqrt{a}+1}{\left(\sqrt{a}-1\right)^2}\)\(=\frac{1+\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)}:\frac{\sqrt{a}+1}{\left(\sqrt{a}-1\right)^2}=\frac{\sqrt{a}-1}{\sqrt{a}}\)
b) Do \(\sqrt{a}\ge0\) => \(\sqrt{a}-1< \sqrt{a}\)=> \(\frac{\sqrt{a}-1}{\sqrt{a}}< 1\)
Bài 1:
Ta có:
\(\left(a-b+c\right)^3=a^3-b^3+c^3-3a^2b+3a^2c+3ab^2+3b^2c+3ac^2-3bc^2-6abc\)
\(\Rightarrow\left(\sqrt[3]{\frac{1}{9}}-\sqrt[3]{\frac{2}{9}}+\sqrt[3]{\frac{4}{9}}\right)^3=\frac{1}{9}-\frac{2}{9}+\frac{4}{9}-\frac{1}{3}.\sqrt[3]{2}+\frac{1}{3}.\sqrt[3]{4}+\frac{1}{3}.\sqrt[3]{4}+\frac{2}{3}.\sqrt[3]{2}\)
\(+\frac{2}{3}.\sqrt[3]{2}-\frac{2}{3}.\sqrt[3]{4}-\frac{4}{3}=\sqrt[3]{2}-1\)
\(\Rightarrow\sqrt[3]{\sqrt[3]{2}-1}=\sqrt[3]{\frac{1}{9}}-\sqrt[3]{\frac{2}{9}}+\sqrt[3]{\frac{4}{9}}\)
a: \(P=\dfrac{a+\sqrt{a}+1}{a+1}:\dfrac{a+1-2\sqrt{a}}{\left(\sqrt{a}-1\right)\left(a+1\right)}\)
\(=\dfrac{a+\sqrt{a}+1}{a+1}\cdot\dfrac{\left(a+1\right)}{\sqrt{a}-1}=\dfrac{a+\sqrt{a}+1}{\sqrt{a}-1}\)
b: P<1
=>P-1<0
=>\(\dfrac{a+\sqrt{a}+1-\sqrt{a}+1}{\sqrt{a}-1}< 0\)
=>\(\dfrac{a+2}{\sqrt{a}-1}< 0\)
=>căn a-1<0
=>0<=a<1
c: Khi a=19-8căn 3=(4-căn 3)^2 thì \(P=\dfrac{19-8\sqrt{3}+4-\sqrt{3}+1}{4-\sqrt{3}-1}=\dfrac{24-9\sqrt{3}}{3-\sqrt{3}}=\dfrac{15-\sqrt{3}}{2}\)
a: \(=\dfrac{a+2\sqrt{a}+1-a+2\sqrt{a}-1+4\sqrt{a}\left(a-1\right)}{a-1}\cdot\dfrac{1}{a\sqrt{a}}\)
\(=\dfrac{4\sqrt{a}\left(a-1+1\right)}{a-1}\cdot\dfrac{1}{a\sqrt{a}}=\dfrac{4}{a-1}\)
b: Khi a=2căn 2+1 thì \(A=\dfrac{4}{2\sqrt{2}+1-1}=\sqrt{2}\)
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