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ĐK: a > 0, a khác 1
\(M=\dfrac{a-1}{\sqrt{a}-1}=\dfrac{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}{\sqrt{a}-1}=\sqrt{a}+1\)
\(N=\dfrac{a-1}{\sqrt{a}+1}=\dfrac{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}{\sqrt{a}+1}=\sqrt{a}-1\)
\(P=\dfrac{a\sqrt{a}-1}{\sqrt{a}-1}=\dfrac{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}{\sqrt{a}-1}=a+\sqrt{a}+1\)
\(Q=\dfrac{a\sqrt{a}+1}{\sqrt{a}+1}=\dfrac{\left(\sqrt{a}+1\right)\left(a-\sqrt{a}+1\right)}{\sqrt{a}+1}=a-\sqrt{a}+1\)
\(A=\dfrac{\sqrt{x}}{\sqrt{x}+1}+\dfrac{3}{\sqrt{x}}-\dfrac{5\sqrt{x}+3}{x+\sqrt{x}}\)
\(=\dfrac{\sqrt{x}.\sqrt{x}+3\left(\sqrt{x}+1\right)-\left(5\sqrt{x}+3\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(=\dfrac{x+3\sqrt{x}+3-5\sqrt{x}-3}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(=\dfrac{x-2\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}=\dfrac{\sqrt{x}-2}{\sqrt{x}+1}\)
\(A=\dfrac{\sqrt{x}}{\sqrt{x}+1}+\dfrac{3}{\sqrt{x}}-\dfrac{5\sqrt{x}+3}{x+\sqrt{x}}\\ ĐKXĐ:x>0;x\ne1\\ \Rightarrow A=\dfrac{x}{\sqrt{x}\left(\sqrt{x}+1\right)}+\dfrac{3\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}-\dfrac{5\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}+1\right)}\\ =\dfrac{x+3\sqrt{x}+3-5\sqrt{x}-3}{\sqrt{x}\left(\sqrt{x}+1\right)}\\ =\dfrac{x-2\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\\ =\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}\\ =\dfrac{\sqrt{x}-2}{\sqrt{x}+1}\)
Vậy \(A=\dfrac{\sqrt{x}-2}{\sqrt{x}+1}\) với \(=\dfrac{\sqrt{x}-2}{\sqrt{x}+1}\)
a: \(A=\dfrac{a\left(\sqrt{a}+1\right)}{a-1}+\dfrac{\sqrt{a}-1}{\sqrt{a}\left(a-1\right)}-\dfrac{a+1}{\sqrt{a}}\)
\(=\dfrac{a^2+a\sqrt{a}+\sqrt{a}-1-a^2+1}{\sqrt{a}\left(a-1\right)}\)
\(=\dfrac{a\left(\sqrt{a}+1\right)}{\sqrt{a}\left(a-1\right)}=\dfrac{\sqrt{a}}{\sqrt{a}-1}\)
b: Để M>2 thì M-2>0
\(\Leftrightarrow\dfrac{\sqrt{a}-2\sqrt{a}+2}{\sqrt{a}-1}>0\)
\(\Leftrightarrow\dfrac{\sqrt{a}-2}{\sqrt{a}-1}< 0\)
=>1<a<4
c: Để M=-1 thì \(\sqrt{a}=-\sqrt{a}+1\)
=>a=1/4
a)
Đặt
\(\sqrt{1+x}=a; \sqrt{1-x}=b\Rightarrow \left\{\begin{matrix} ab=\sqrt{(1+x)(1-x)}=\sqrt{1-x^2}\\ a\geq b\\ a^2+b^2=2\end{matrix}\right.\)
Khi đó:
\(A=\frac{\sqrt{1-\sqrt{1-x^2}}(\sqrt{(1+x)^3}+\sqrt{(1-x)^3})}{2-\sqrt{1-x^2}}\)
\(=\frac{\sqrt{\frac{a^2+b^2}{2}-ab}(a^3+b^3)}{a^2+b^2-ab}=\frac{\sqrt{\frac{a^2+b^2-2ab}{2}}(a+b)(a^2-ab+b^2)}{a^2+b^2-ab}\)
\(=\sqrt{\frac{a^2-2ab+b^2}{2}}(a+b)=\sqrt{\frac{(a-b)^2}{2}}(a+b)=\frac{1}{\sqrt{2}}|a-b|(a+b)\)
\(=\frac{1}{\sqrt{2}}(a-b)(a+b)=\frac{1}{\sqrt{2}}(a^2-b^2)=\frac{1}{\sqrt{2}}[(1+x)-(1-x)]=\sqrt{2}x\)
Sửa đề: \(\frac{25}{(x+z)^2}=\frac{16}{(z-y)(2x+y+z)}\)
Ta có:
Áp dụng tính chất dãy tỉ số bằng nhau thì:
\(k=\frac{a}{x+y}=\frac{5}{x+z}=\frac{a+5}{2x+y+z}=\frac{5-a}{z-y}\) ($k$ là một số biểu thị giá trị chung)
Khi đó:
\(\frac{16}{(z-y)(2x+y+z)}=\frac{25}{(x+z)^2}=(\frac{5}{x+z})^2=k^2\)
Mà: \(k^2=\frac{a+5}{2x+y+z}.\frac{5-a}{z-y}=\frac{25-a^2}{(2x+y+z)(z-y)}\)
Do đó: \(\frac{16}{(z-y)(2x+y+z)}=\frac{25-a^2}{(2x+y+z)(z-y)}\Rightarrow 16=25-a^2\)
\(\Rightarrow a^2=9\Rightarrow a=\pm 3\)
Suy ra:
\(Q=\frac{a^6-2a^5+a-2}{a^5+1}=\frac{a^5(a-2)+(a-2)}{a^5+1}=\frac{(a-2)(a^5+1)}{a^5+1}=a-2=\left[\begin{matrix}
1\\
-5\end{matrix}\right.\)
a: \(P=\dfrac{a+5\sqrt{a}+6-a+3\sqrt{a}-2-4\sqrt{a}+4}{a-4}\)
\(=\dfrac{4\sqrt{a}+8}{a-4}=\dfrac{4}{\sqrt{a}-2}\)
b: Khi a=1/9 thì \(P=\dfrac{4}{\dfrac{1}{3}-2}=4:\dfrac{-5}{3}=-\dfrac{12}{5}\)
c: Để P=2 thì \(2\sqrt{a}-4=4\)
=>2căn a=8
=>căn a=4
hay a=16
\(A=\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-1}-\dfrac{\sqrt{a}-1}{\sqrt{a}+1}+4\sqrt{a}\right)\left(\sqrt{a}-\dfrac{1}{\sqrt{a}}\right)\)
\(=\left(\dfrac{\left(\sqrt{a}+1\right)^2-\left(\sqrt{a}-1\right)^2}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}+4\sqrt{a}\right)\cdot\dfrac{a-1}{\sqrt{a}}\)
\(=\left(\dfrac{2\cdot2\sqrt{a}}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}+4\sqrt{a}\right)\cdot\dfrac{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}{\sqrt{a}}\)
\(=\left(\dfrac{4\sqrt{a}}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}+4\sqrt{a}\right)\cdot\dfrac{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}{\sqrt{a}}\)
\(=\dfrac{4\sqrt{a}+4\sqrt{a}\cdot\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\cdot\dfrac{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}{\sqrt{a}}\)
\(=4\sqrt{a}\cdot\left(1+\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)\right)\cdot\dfrac{1}{\sqrt{a}}\)
\(=4\left(1+a-1\right)\)
\(=4a\)
Để \(\sqrt{a}>A\) thì \(\sqrt{a}>4a\)
\(\Leftrightarrow a>\sqrt{4a}\left(đk:a\ge0\right)\)
\(\Leftrightarrow a>2\sqrt{a}\)
\(\Leftrightarrow2\sqrt{a}< a\)
\(\Leftrightarrow\left\{{}\begin{matrix}2\sqrt{a}< a\left(đk:a\ge0\right)\\2\sqrt{a}< a\left(đk:a< 0\right)\end{matrix}\right.\)
\(A=\left[1:\left(1-\frac{\sqrt{a}}{1+\sqrt{a}}\right)\right]\left[\frac{1}{\sqrt{a}-1}-\frac{2\sqrt{a}}{a\sqrt{a}-a+\sqrt{a}-1}\right]\)
\(=\left[1:\left(\frac{1+\sqrt{a}-\sqrt{a}}{1+\sqrt{a}}\right)\right]\left[\frac{1}{\sqrt{a}-1}-\frac{2\sqrt{a}}{\left(a+1\right)\left(\sqrt{a}-1\right)}\right]\)
\(=\left(1:\frac{1}{1+\sqrt{a}}\right).\frac{a+1-2\sqrt{a}}{\left(\sqrt{a}-1\right)\left(a+1\right)}\)
\(=\left(\sqrt{a}+1\right).\frac{\left(\sqrt{a}-1\right)^2}{\left(\sqrt{a}-1\right)\left(a+1\right)}\)
\(=\frac{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}{a+1}=\frac{a-1}{a+1}\)
a: \(A=\left(\dfrac{\sqrt{3}\left(x-\sqrt{3}\right)+3}{\left(x-\sqrt{3}\right)\left(x^2+x\sqrt{3}+3\right)}\right)\cdot\dfrac{x^2+3+x\sqrt{3}}{x\sqrt{3}}\)
\(=\dfrac{x\sqrt{3}}{\left(x-\sqrt{3}\right)\left(x^2+x\sqrt{3}+3\right)}\cdot\dfrac{x^2+x\sqrt{3}+3}{x\sqrt{3}}\)
\(=\dfrac{1}{x-\sqrt{3}}\)
b: \(B=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{x+\sqrt{x}+1}-\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{x-\sqrt{x}+1}+x+1\)
\(=x-\sqrt{x}-x-\sqrt{x}+x+1\)
\(=x-2\sqrt{x}+1\)
c: \(C=\left(\dfrac{\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2}-\dfrac{\sqrt{x}-2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right)\cdot\dfrac{x\left(\sqrt{x}+1\right)-\left(\sqrt{x}+1\right)}{\sqrt{x}}\)
\(=\dfrac{x+\sqrt{x}-2-\left(x-\sqrt{x}-2\right)}{\left(\sqrt{x}+1\right)^2\cdot\left(\sqrt{x}-1\right)}\cdot\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}{\sqrt{x}}\)
\(=\dfrac{2\sqrt{x}}{\sqrt{x}}=2\)
đk a ≠ 1 ; a ≥ 0
\(=\left(1+\dfrac{\sqrt{a}\left(\sqrt{a}+1\right)}{\sqrt{a}+1}\right)\left(1+\dfrac{\sqrt{a}\left(\sqrt{a}-1\right)}{1-\sqrt{a}}\right)\\ =\left(1+\sqrt{a}\right)\left(1+\dfrac{-\sqrt{a}\left(1-\sqrt{a}\right)}{1-\sqrt{a}}\right)\\ =\left(1+\sqrt{a}\right)\left(1-\sqrt{a}\right)=1-a\)