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a/ \(\frac{2}{a}.\frac{4\left|a\right|}{3}=\frac{-8a}{3a}=-\frac{8}{3}\)
b/ \(\frac{3}{a-1}\sqrt{\frac{4\left(a-1\right)^2}{25}}=\frac{3}{\left(a-1\right)}.\frac{2\left|a-1\right|}{5}=\frac{6\left(a-1\right)}{5\left(a-1\right)}=\frac{6}{5}\)
c/ \(\frac{3\sqrt{9a^2b^4}}{\sqrt{a^2b^2}}=\frac{9.\left|a\right|.b^2}{\left|a\right|\left|b\right|}=9\left|b\right|\)
d/ \(\left(1+\frac{\sqrt{a}\left(\sqrt{a}+1\right)}{\sqrt{a}+1}\right)\left(1-\frac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}\right)=\left(1+\sqrt{a}\right)\left(1-\sqrt{a}\right)=1-a\)
a/ \(=\frac{2}{a}.\frac{4\left|a\right|}{3}=\frac{2}{a}.\frac{-4a}{3}=\frac{-8}{3}\)
b/ \(=\frac{3}{a-1}.\frac{\left|2a-2\right|}{5}=\frac{3}{a-1}.\frac{2\left(a-1\right)}{5}=\frac{6}{5}\)
c/ \(=\sqrt{\frac{162a^2b^4}{2a^2b^2}}=\sqrt{81b^2}=9\left|b\right|\)
d/ \(=\left(1+\frac{\sqrt{a}\left(\sqrt{a}+1\right)}{\sqrt{a}+1}\right)\left(1-\frac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}\right)\)
\(=\left(1+\sqrt{a}\right)\left(1-\sqrt{a}\right)=1-a\)
Bài 1 :
a) \(P=\left(\frac{1}{x-\sqrt{x}}+\frac{1}{\sqrt{x}-1}\right):\frac{\sqrt{x}}{x-2\sqrt{x}+1}\)
\(P=\left(\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}+\frac{1}{\sqrt{x}-1}\right).\frac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}}\)
\(P=\frac{1+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}.\frac{\sqrt{x}-1}{\sqrt{x}}\)
\(P=\frac{\sqrt{x}+1}{x}\)
b) \(P>\frac{1}{2}\)
\(\Leftrightarrow\frac{\sqrt{x}+1}{x}>\frac{1}{2}\)
\(\Leftrightarrow\frac{\sqrt{x}+1}{x}-\frac{1}{2}>0\)
\(\Leftrightarrow\frac{\sqrt{x}+1-2x}{x}>0\)
\(\Leftrightarrow\sqrt{x}-2x+1>0\left(x>0\right)\)
\(\Leftrightarrow\sqrt{x}+x^2-2x+1-x^2>0\)
\(\Leftrightarrow\sqrt{x}+x^2+\left(x-1\right)^2>0\left(\forall x>0\right)\)
Vậy P > 1/2 với mọi x> 0 ; x khác 1
Bài 2 :
a) \(K=\left(\frac{\sqrt{a}}{\sqrt{a}-1}-\frac{1}{a-\sqrt{a}}\right):\left(\frac{1}{\sqrt{a}+a}+\frac{2}{a-1}\right)\)
\(K=\left(\frac{\sqrt{a}}{\sqrt{a}-1}-\frac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\right):\left(\frac{1}{\sqrt{a}\left(\sqrt{a}+1\right)}+\frac{2}{a-1}\right)\)
\(K=\frac{a-1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\frac{a-1+2\sqrt{a}\left(\sqrt{a}+1\right)}{\sqrt{a}\left(a-1\right)\left(\sqrt{a}+1\right)}\)
\(K=\frac{a-1}{\sqrt{a}\left(\sqrt{a}-1\right)}.\frac{\sqrt{a}\left(a-1\right)\left(\sqrt{a}-1\right)}{a-1+2a+2\sqrt{a}}\)
\(K=\frac{\left(a-1\right)^2}{3a+2\sqrt{a}-1}\)
b) \(a=3+2\sqrt{2}=2+2\sqrt{2}+1=\left(\sqrt{2}+1\right)^2\)( thỏa mãn ĐKXĐ )
Thay a vào biểu thức K , ta có :
\(K=\frac{\left(3+2\sqrt{2}-1\right)^2}{3\left(3+2\sqrt{2}\right)+2\sqrt{\left(\sqrt{2}+1\right)^2}-1}\)
\(K=\frac{\left(2+2\sqrt{2}\right)^2}{9+6\sqrt{2}+2\left|\sqrt{2}+1\right|-1}\)
\(K=\frac{\left(2+2\sqrt{2}\right)^2}{8+6\sqrt{2}+2\sqrt{2}+2}\)
\(K=\frac{\left(2+2\sqrt{2}\right)^2}{10+8\sqrt{2}}\)
\(A=\left(1-\frac{\sqrt{a}\left(\sqrt{a}-3\right)}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-3\right)}\right):\left(\frac{\left(\sqrt{a}-2\right)^2-\left(\sqrt{a}-3\right)\left(\sqrt{a}+3\right)-\left(9-a\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+3\right)}\right)\)
\(=\left(\frac{\sqrt{a}+3-\sqrt{a}}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-3\right)}\right):\left(\frac{\left(\sqrt{a}-2\right)^2-a+9-9+a}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+3\right)}\right)\)
\(=\frac{3}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-3\right)}:\left(\frac{\left(\sqrt{a}-2\right)^2}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+3\right)}\right)\)
\(=\frac{3}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-3\right)}.\frac{\left(\sqrt{a}+3\right)}{\left(\sqrt{a}-2\right)}=\frac{3}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-3\right)}\)
Để \(A+\left|A\right|\ne0\Rightarrow\left|A\right|\ne-A\Rightarrow A>0\)
\(\Rightarrow\left[{}\begin{matrix}\sqrt{a}< 2\\\sqrt{a}>3\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}a< 4\\a>9\end{matrix}\right.\)
Kết hợp điều kiện \(\Rightarrow\left[{}\begin{matrix}0\le a< 4\\a>9\end{matrix}\right.\)
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,
\(=\left(a-b\right)\left(\sqrt{\frac{a+b}{a-b}}-1\right)\left(a-b\right)\left(\sqrt{\frac{a+b}{a-b}}+1\right)\)
\(=\left(a-b\right)^2\left(\frac{a+b}{a-b}-1\right)\)
\(=\left(a-b\right)^2\cdot\frac{a+b-a+b}{a-b}\)
\(=\left(a-b\right)2b=2ab-2b^2\)
\(B=\frac{9-x}{\sqrt{x}+3}-\frac{x-6\sqrt{x}+9}{\sqrt{x}-3}-6\)(đk: x ≥ 0 và x ≠ 9)
\(B=\frac{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}{\sqrt{x}+3}-\frac{\left(\sqrt{x}-3\right)^2}{\sqrt{x}-3}-6\)
\(B=\left(3-\sqrt{x}\right)-\left(\sqrt{x}-3\right)-6\)
\(B=3-\sqrt{x}-\sqrt{x}+3-6\)
\(B=-2\sqrt{x}\)
\(A=\frac{\sqrt{x}}{\sqrt{x}-6}-\frac{3}{\sqrt{x}+6}+\frac{x}{36-x}\)(đk: x ≥ 0 và x ≠ 36)
\(=\frac{\sqrt{x}}{\sqrt{x}-6}-\frac{3}{\sqrt{x}+6}-\frac{x}{x-36}\)
\(=\frac{\sqrt{x}}{\sqrt{x}-6}-\frac{3}{\sqrt{x}+6}-\frac{x}{x-36}\)
\(=\frac{\sqrt{x}\left(\sqrt{x}+6\right)-3\left(\sqrt{x-6}\right)-x}{(\sqrt{x}-6)\left(\sqrt{x}+6\right)}\)
\(=\frac{x+6\sqrt{x}-3\sqrt{x}+18-x}{(\sqrt{x}-6)\left(\sqrt{x}+6\right)}\)
\(=\frac{3\sqrt{x}+18}{(\sqrt{x}-6)\left(\sqrt{x}+6\right)}\)
\(=\frac{3(\sqrt{x}+6)}{(\sqrt{x}-6)\left(\sqrt{x}+6\right)}\)
\(=\frac{3}{\sqrt{x}-6}\)
Ta có : \(Q=\left(\frac{1}{\sqrt{a}-1}-\frac{1}{\sqrt{a}}\right):\left(\frac{\sqrt{a}+1}{\sqrt{a}-2}-\frac{\sqrt{a}+2}{\sqrt{a}-1}\right)\)
=> \(Q=\left(\frac{\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)}-\frac{\sqrt{a}-1}{\sqrt{a}\left(\sqrt{a}-1\right)}\right):\left(\frac{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}-\frac{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}{\left(\sqrt{a}-1\right)\left(\sqrt{a}-2\right)}\right)\)
=> \(Q=\left(\frac{\sqrt{a}-\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}\right):\left(\frac{a-1}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}-\frac{a-4}{\left(\sqrt{a}-1\right)\left(\sqrt{a}-2\right)}\right)\)
=> \(Q=\left(\frac{\sqrt{a}-\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}\right):\left(\frac{a-1-a+4}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\right)\)
=> \(Q=\left(\frac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\right):\left(\frac{3}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\right)\)
=> \(Q=\frac{\left(\sqrt{a}-1\right)\left(\sqrt{a}-2\right)}{3\sqrt{a}\left(\sqrt{a}-1\right)}\)
=> \(Q=\frac{\sqrt{a}-2}{3\sqrt{a}}\)