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Lời giải:
Yêu cầu 1:
\(\frac{5+3\sqrt{5}}{\sqrt{5}}+\frac{3+\sqrt{3}}{\sqrt{3}+1}-(\sqrt{5}+3)=\frac{\sqrt{5}(\sqrt{5}+3)}{\sqrt{5}}+\frac{\sqrt{3}(\sqrt{3}+1)}{\sqrt{3}+1}-(\sqrt{5}+3)\)
\(=\sqrt{5}+3+\sqrt{3}-(\sqrt{5}+3)=\sqrt{3}\) (đpcm)
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Yêu cầu 2:
\(P=a-\frac{\sqrt{a}+\sqrt{a-1}-\sqrt{a}+\sqrt{a-1}}{(\sqrt{a}-\sqrt{a-1})(\sqrt{a}+\sqrt{a-1})}=a-\frac{2\sqrt{a-1}}{a-(a-1)}=a-2\sqrt{a-1}\)
\(=(a-1)-2\sqrt{a-1}+1=(\sqrt{a-1}-1)^2\geq 0\) với mọi $a\geq 1$
Ta có đpcm.
\(T=a-\left(\frac{\sqrt{a}+\sqrt{a-1}-\sqrt{a}+\sqrt{a-1}}{\left(\sqrt{a}-\sqrt{a-1}\right)\left(\sqrt{a}+\sqrt{a-1}\right)}\right)\)
\(=a-\left(\frac{2\sqrt{a-1}}{a-a+1}\right)=a-2\sqrt{a-1}\)
\(=a-1-2\sqrt{a-1}+1\)
\(=\left(\sqrt{a-1}\right)^2-2\sqrt{a-1}+1\)
\(=\left(\sqrt{a-1}-1\right)^2\)
\(T=\left(\sqrt{a-1}-1\right)^2\ge0,\forall a\in R\)
a) Ta có: \(\left(\frac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right)\left(\frac{1-\sqrt{a}}{1-a}\right)\)
\(=\left[\frac{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}+a\right)}{1-\sqrt{a}}+\sqrt{a}\right]\cdot\frac{1-\sqrt{a}}{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)}\)
\(=\left(1+\sqrt{a}+a+\sqrt{a}\right)\cdot\frac{1}{1+\sqrt{a}}\)
\(=\left(1+\sqrt{a}\right)^2\cdot\frac{1}{1+\sqrt{a}}\)
\(=1+\sqrt{a}\) Bằng 1 kiểu gì đây._.?
a) Xin lỗi sửa lại phần a:
Ta có: \(\left(\frac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right)\left(\frac{1-\sqrt{a}}{1-a}\right)^2\)
\(=\left(1+\sqrt{a}\right)^2\cdot\frac{1}{\left(1+\sqrt{a}\right)^2}\)
\(=1\)
b) Ta có: \(\left(\sqrt{3}-\sqrt{2}\right)\sqrt{5+2\sqrt{6}}\)
\(=\left(\sqrt{3}-\sqrt{2}\right)\sqrt{3+2\sqrt{6}+2}\)
\(=\left(\sqrt{3}-\sqrt{2}\right)\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}\)
\(=\left(\sqrt{3}-\sqrt{2}\right)\left(\sqrt{3}+\sqrt{2}\right)=3-2=1\)
\(VT=\sqrt{a}\left(\sqrt{a}-1\right)-\sqrt{a}\left(\sqrt{a}+1\right)+a+1\)
\(=a-\sqrt{a}-a-\sqrt{a}+a+1\)
\(=a-2\sqrt{a}+1=\left(\sqrt{a}-1\right)^2=VP\)
Đề của bạn bị sai, mình sửa lại đề ở dưới nhé!
\(\frac{a^2-\sqrt{a}}{a+\sqrt{a}+1}-\frac{a^2+\sqrt{a}}{a-\sqrt{a}+1}+a+1\)
\(=\frac{\left(a^2-\sqrt{a}\right)\left(a+\sqrt{a}+1\right)-\left(a^2+\sqrt{a}\right)\left(a+\sqrt{a}\right)\left(a+\sqrt{a}+1\right)}{\left(a+\sqrt{a}+1\right)\left(a-\sqrt{a}+1\right)}+a+1\)
\(=\frac{\left(a^2-\sqrt{a}\right)\left(a-\sqrt{a}+1\right)-\left(a^2+\sqrt{a}\right)\left(a+\sqrt{a}+1\right)+\left(a+1\right)\left(a+\sqrt{a}+1\right)\left(a-\sqrt{a}+1\right)}{\left(a+1\right)^2-\left(\sqrt{a}\right)^2}\)
\(=\frac{\left(a^2-\sqrt{a}\right)\left(a-\sqrt{a}+1\right)-\left(a^2+\sqrt{a}\right)\left(a+\sqrt{a}+1\right)\left(a+1\right)\left[\left(a+1\right)^2-a\right]}{\left(a-1\right)^2-a}\)
\(=\frac{a^3-a^2\sqrt{x}+a^2-a\sqrt{a}+a-\sqrt{a}-a^3-a^2\sqrt{a}-a^2-a\sqrt{a}-a-\sqrt{a}+\left(a+1\right)\left[\left(a+1\right)^2-a\right]}{a^2+2a+1-a}\)
\(=\frac{-2a^2\sqrt{a}-2a\sqrt{a}-2\sqrt{a}+\left(a+1\right)\left(a^2+2a+1-a\right)}{a^2+a+1}\)
\(=\frac{-2\sqrt{a}\left(a^2+a+1\right)+\left(a+1\right)\left(a^2+a+1\right)}{a^2+a+1}\)
\(=\frac{\left(a^2+a+1\right)\left[-2\sqrt{x}+\left(x+1\right)\right]}{a^2+a+1}\)
\(=x-1-2\sqrt{x}\)
\(=\left(\sqrt{x}-1\right)^2\)
(Chúc you học giỏi nhoa!)
Ui ui mình ghi lộn, xin lỗi nhoa, vì x dễ làm hơn nên mình ghi lộn a thành x, mong bạn thông cảm hihi!
a/ĐK: \(a\ge0;a\ne1\)
Ta có: P\(=\frac{a+3\sqrt{a}+2}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-1\right)}:\left(\frac{1}{\sqrt{a}+1}+\frac{1}{\sqrt{a}-1}\right)=\frac{\left(\sqrt{a}+2\right)\left(\sqrt{a}+1\right)}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-1\right)}:\frac{\sqrt{a}-1+\sqrt{a}+1}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}=\frac{\sqrt{a}+1}{\sqrt{a}-1}\times\frac{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}{2\sqrt{a}}=\frac{\left(\sqrt{a}+1\right)^2}{2\sqrt{a}}\)
