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Ta có:
\(A=x-\left(\frac{1}{\sqrt{x}-\sqrt{x-1}}-\frac{1}{\sqrt{x}+\sqrt{x-1}}\right)\)
\(A=x-\frac{\sqrt{x}+\sqrt{x-1}-\sqrt{x}+\sqrt{x-1}}{\left(\sqrt{x}-\sqrt{x-1}\right)\left(\sqrt{x}+\sqrt{x-1}\right)}\)
\(A=x-\frac{2\sqrt{x-1}}{x-x+1}\)
\(A=x-2\sqrt{x-1}\)
\(A=\left(x-1\right)-2\sqrt{x-1}+1\)
\(A=\left(\sqrt{x-1}-1\right)^2\ge0\left(\forall x\ge1\right)\)
=> đpcm
a/ĐK: \(a\ge0;a\ne1\)
Ta có: P\(=\frac{a+3\sqrt{a}+2}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-1\right)}:\left(\frac{1}{\sqrt{a}+1}+\frac{1}{\sqrt{a}-1}\right)=\frac{\left(\sqrt{a}+2\right)\left(\sqrt{a}+1\right)}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-1\right)}:\frac{\sqrt{a}-1+\sqrt{a}+1}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}=\frac{\sqrt{a}+1}{\sqrt{a}-1}\times\frac{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}{2\sqrt{a}}=\frac{\left(\sqrt{a}+1\right)^2}{2\sqrt{a}}\)
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,
\(=\left(a-b\right)\left(\sqrt{\frac{a+b}{a-b}}-1\right)\left(a-b\right)\left(\sqrt{\frac{a+b}{a-b}}+1\right)\)
\(=\left(a-b\right)^2\left(\frac{a+b}{a-b}-1\right)\)
\(=\left(a-b\right)^2\cdot\frac{a+b-a+b}{a-b}\)
\(=\left(a-b\right)2b=2ab-2b^2\)
\(T=a-\left(\frac{\sqrt{a}+\sqrt{a-1}-\sqrt{a}+\sqrt{a-1}}{\left(\sqrt{a}-\sqrt{a-1}\right)\left(\sqrt{a}+\sqrt{a-1}\right)}\right)\)
\(=a-\left(\frac{2\sqrt{a-1}}{a-a+1}\right)=a-2\sqrt{a-1}\)
\(=a-1-2\sqrt{a-1}+1\)
\(=\left(\sqrt{a-1}\right)^2-2\sqrt{a-1}+1\)
\(=\left(\sqrt{a-1}-1\right)^2\)
\(T=\left(\sqrt{a-1}-1\right)^2\ge0,\forall a\in R\)
Câu 2:
Ta có: \(M=\left(\dfrac{a+\sqrt{a}}{\sqrt{a}+1}+1\right)\left(1+\dfrac{a-\sqrt{a}}{1-\sqrt{a}}\right)\)
\(=\left(\dfrac{\sqrt{a}\left(\sqrt{a}+1\right)}{\sqrt{a}+1}+1\right)\left(1-\dfrac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}\right)\)
\(=\left(1+\sqrt{a}\right)\left(1-\sqrt{a}\right)\)
\(=1-a\)
Câu 1:
Ta có: \(A=\left(\dfrac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right)\left(\dfrac{1-\sqrt{a}}{1-a}\right)^2\)
\(=\left(\dfrac{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}+a\right)}{1-\sqrt{a}}+\sqrt{a}\right)\left(\dfrac{1}{\sqrt{a}+1}\right)^2\)
\(=\left(\sqrt{a}+1\right)^2\cdot\dfrac{1}{\left(\sqrt{a}+1\right)^2}\)
\(=1\)
\(P=a-\left(\frac{1}{\sqrt{a}-\sqrt{a-1}}-\frac{1}{\sqrt{a}+\sqrt{a-1}}\right)\)
\(P=a-\frac{\sqrt{a}+\sqrt{a-1}-\sqrt{a}+\sqrt{a-1}}{\left(\sqrt{a}-\sqrt{a-1}\right)\left(\sqrt{a}+\sqrt{a-1}\right)}=a-\frac{2\sqrt{a-1}}{a-a+1}=a-2\sqrt{a-1}\)
Ta có:
\(a-2\sqrt{a-1}=a-1-2\sqrt{a-1}+1=\left(\sqrt{a-1}-1\right)^2\ge0\)(đúng)
ủa sao bằng \(\left(\sqrt{a-1}-1\right)^2\) được v bạn ? phải là : \(\left(\sqrt{a-1}-1\right)\left(\sqrt{a-1}+1\right)\) chứ??