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a)\(P=\left(\frac{1}{a-\sqrt{a}}+\frac{1}{\sqrt{a}-1}\right):\frac{\sqrt{a}-1}{a-2\sqrt{a}+1}\)
\(P=\left(\frac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}+\frac{1}{\sqrt{a}-1}\right):\frac{\sqrt{a}-1}{a-2\sqrt{a}+1}\)
\(P=\left(\frac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}+\frac{\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)}\right):\frac{\sqrt{a}-1}{\left(\sqrt{a}-1\right)^2}\)
\(P=\left(\frac{1+\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)}\right):\frac{1}{\left(\sqrt{a}-1\right)}\)
\(P=\frac{\sqrt{a}+1}{\sqrt{a}}\)
b) Để \(P=\frac{1}{4}\Leftrightarrow\frac{\sqrt{a}+1}{\sqrt{a}}=\frac{1}{4}\)
\(\Rightarrow4\left(\sqrt{a}+1\right)=\sqrt{a}\)
\(\Leftrightarrow3\sqrt{a}+1=0\)
<=> a ko có giá trị
P/s tha m khảo nha
c,\(\left(\frac{\sqrt{1+a}}{\sqrt{1+a}-\sqrt{1-a}}+\frac{1-a}{\sqrt{1-a^2}-1+a}\right)\left(\sqrt{\frac{1}{a^2}-1}-\frac{1}{a}\right)\)
\(=\left(\frac{\sqrt{1+a}}{\sqrt{1+a}-\sqrt{1-a}}+\frac{\sqrt{1-a}.\sqrt{1-a}}{\sqrt{1-a}\left(\sqrt{1+a}-\sqrt{1-a}\right)}\right)\left(\frac{\sqrt{1-a^2}-1}{a}\right)\)
\(=\frac{\left(\sqrt{1+a}+\sqrt{1-a}\right)^2}{\left(1+a\right)-\left(1-a\right)}.\frac{\left(\sqrt{1-a^2}-1\right)}{a}=-1\)
M chỉ làm tiếp thôi nha, ko chép lại đề với đk đâu
a,
\(=\frac{a+2\sqrt{ab}+b-4\sqrt{ab}}{\sqrt{a}-\sqrt{b}}-\)\(\frac{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{ab}}\)
\(=\frac{a-2\sqrt{ab}+b}{\sqrt{a}-\sqrt{b}}-\left(\sqrt{a}-\sqrt{b}\right)\)
\(=\frac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\sqrt{a}-\sqrt{b}}-\sqrt{a}+\sqrt{b}\)
\(=\sqrt{a}-\sqrt{b}-\sqrt{a}+\sqrt{b}\)
\(=0\)
b,
\(=\left(a-b\right)\left(\sqrt{\frac{a+b}{a-b}}-1\right)\left(a-b\right)\left(\sqrt{\frac{a+b}{a-b}}+1\right)\)
\(=\left(a-b\right)^2\left(\frac{a+b}{a-b}-1\right)\)
\(=\left(a-b\right)^2\cdot\frac{a+b-a+b}{a-b}\)
\(=\left(a-b\right)2b=2ab-2b^2\)
Ta có: a√a = √(a².a) = (√a)³
=> 1 - a√a = 1 - (√a)³ = (1 - √a)(a + √a + 1) (1)
Tương tự: 1 + a√a = 1 + (√a)³ = (1 + √a)(a - √a + 1) (2)
Từ (1) và (2) => [ (1-a√a/1-√a+√a).(1+a√a/1+√a-√a) + 1 ].
= [(1 - √a)(a + √a + 1)/(1 - √a) + √a].[(1 + √a)(a - √a + 1)/(1 + √a) - √a ] +1
=(a + √a + 1 + √a)(a - √a + 1- √a) + 1
= (a + 2√a + 1)(a - 2√a + 1) + 1
= (√a + 1)²(√a - 1)² +1
= [(√a + 1)(√a - 1)]² + 1
= (a - 1)² + 1
= a² - 2a + 1 + 1
= a² - 2a + 2
=> [ (1-a√a/1-√a+√a).(1+a√a/1+√a-√a) + 1 ] = a² - 2a + 2 (3)
Áp dụng (3) vào A ta được A = [(1 - a)²]/(a² - 2a + 2)
<=> A = (a² - 2a + 1)/(a² - 2a + 2)
\(A=\left[\frac{1}{\sqrt{x}\left(\sqrt{x}+1\right)}-\frac{1}{\sqrt{x}+1}\right]\div\frac{\sqrt{x}-1}{\left(\sqrt{x}+1\right)^2}\)
\(A=\frac{1-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\times\frac{\left(\sqrt{x}+1\right)^2}{\sqrt{x}-1}\)
\(A=\frac{-\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}{\sqrt{x}\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(A=\frac{-\left(\sqrt{x}+1\right)}{\sqrt{x}}=\frac{-\sqrt{x}-x}{x}\)
\(=\left(\frac{\sqrt{a}+a+\sqrt{a}-a}{1-a}\right)\div\frac{\sqrt{a}}{a-1}\)
\(=\frac{2\sqrt{a}}{1-a}\div\frac{\sqrt{a}}{a-1}\)
\(=\frac{2\sqrt{a}}{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)}.\frac{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}{\sqrt{a}}\)
\(=\frac{2\left(\sqrt{a}-1\right)}{\left(1-\sqrt{a}\right)}\)\(=-2\)
\(Với:a>0;a\ne0\)
Rút gọn \(C=-2\)
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