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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: \(A=\left(\dfrac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right)\left(\dfrac{1-\sqrt{a}}{1-a}\right)^2\)
\(=\left(\dfrac{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}+a\right)}{1-\sqrt{a}}+\sqrt{a}\right)\left(\dfrac{1}{\sqrt{a}+1}\right)^2\)
\(=\left(\sqrt{a}+1\right)^2\cdot\dfrac{1}{\left(\sqrt{a}+1\right)^2}\)
\(=1\)
Bài 1:
a) Ta có: \(Q=\left(\dfrac{1}{\sqrt{a}-1}-\dfrac{1}{\sqrt{a}}\right):\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-1}\right)\)
\(=\left(\dfrac{\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)}-\dfrac{\sqrt{a}-1}{\sqrt{a}\left(\sqrt{a}-1\right)}\right):\left(\dfrac{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}-\dfrac{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\right)\)
\(=\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\dfrac{a-1-a+4}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\)
\(=\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\dfrac{3}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\)
\(=\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}{3}\)
\(=\dfrac{\sqrt{a}-2}{3\sqrt{a}}\)
b) Để Q dương thì \(\dfrac{\sqrt{a}-2}{3\sqrt{a}}>0\)
mà \(3\sqrt{a}>0\forall a\) thỏa mãn ĐKXĐ
nên \(\sqrt{a}-2>0\)
\(\Leftrightarrow\sqrt{a}>2\)
hay a>4
Kết hợp ĐKXĐ,ta được: a>4
Vậy: Để Q dương thì a>4
a) \(A=\dfrac{3\left(\sqrt{a}-1\right)}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}-\dfrac{\sqrt{a}+1}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}-\dfrac{\sqrt{a}-3}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\)
\(=\dfrac{3\sqrt{a}-3-\sqrt{a}-1-\sqrt{a}+3}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}=\dfrac{\sqrt{a}-1}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}=\dfrac{1}{\sqrt{a}+1}\)
b) Với \(a=3-2\sqrt{2}\)(tmđk)
\(A=\dfrac{1}{\sqrt{a}+1}=\dfrac{\sqrt{a}-1}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}=\dfrac{\sqrt{a}-1}{a-1}\)
\(=\dfrac{\sqrt{3-2\sqrt{2}}-1}{3-2\sqrt{2}-1}=\dfrac{\sqrt{\left(\sqrt{2}-1\right)^2}-1}{2-2\sqrt{2}}=\dfrac{\left|\sqrt{2}-1\right|-1}{2-2\sqrt{2}}=\dfrac{\sqrt{2}-1-1}{2-2\sqrt{2}}=\dfrac{\sqrt{2}-2}{2-2\sqrt{2}}=\dfrac{\sqrt{2}\left(1-\sqrt{2}\right)}{2\left(1-\sqrt{2}\right)}=\dfrac{\sqrt{2}}{2}\)
a: Ta có: \(P=\left(\dfrac{4a}{\sqrt{a}-1}-\dfrac{\sqrt{a}}{a-\sqrt{a}}\right)\cdot\dfrac{\sqrt{a}-1}{a^2}\)
\(=\dfrac{4a-1}{\sqrt{a}-1}\cdot\dfrac{\sqrt{a}-1}{a^2}\)
\(=\dfrac{4a-1}{a^2}\)
b: Để P=3 thì \(4a-1=3a^2\)
\(\Leftrightarrow3a^2-4a+1=0\)
\(\Leftrightarrow\left(3a-1\right)\left(a-1\right)=0\)
hay \(a=\dfrac{1}{9}\)
a) ĐK: a>0; a≠1
Ta có: \(P=\left(\dfrac{4a}{\sqrt{a}-1}-\dfrac{\sqrt{a}}{a-\sqrt{a}}\right).\dfrac{\sqrt{a}-1}{a^2}\)
\(=\left(\dfrac{4a}{\sqrt{a}-1}-\dfrac{1}{\sqrt{a}-1}\right).\dfrac{\sqrt{a}-1}{a^2}\)
\(=\dfrac{4a-1}{\sqrt{a}-1}.\dfrac{\sqrt{a}-1}{a^2}=\dfrac{4a-1}{a^2}\)
b) Ta có: \(P=3\Leftrightarrow\dfrac{4a-1}{a^2}=3\Leftrightarrow3a^2=4a-1\Leftrightarrow3a^2-4a+1=0\)
\(\Leftrightarrow\left(a-1\right)\left(3a-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=1\left(loại\right)\\a=\dfrac{1}{3}\left(tm\right)\end{matrix}\right.\)
a, Để A nhận giá trị dương thì \(A>0\)hay \(x-1>0\Leftrightarrow x>1\)
b, \(B=2\sqrt{2^2.5}-3\sqrt{3^2.5}+4\sqrt{4^2.5}\)
\(=4\sqrt{5}-9\sqrt{5}+16\sqrt{5}=\left(4-9+16\right)\sqrt{5}=11\sqrt{5}\)
( theo công thức \(A\sqrt{B}=\sqrt{A^2B}\))
c, Với \(a\ge0;a\ne1\)
\(C=\left(\frac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right)\left(\frac{1-\sqrt{a}}{1-a}\right)^2\)
\(=\left(\frac{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}+a\right)}{1-\sqrt{a}}+\sqrt{a}\right)\left(\frac{1-\sqrt{a}}{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)}\right)^2\)
\(=\left(\sqrt{a}+1\right)^2.\frac{1}{\left(\sqrt{a}+1\right)^2}=1\)
\(a,A=4\sqrt{3}-5\sqrt{3}+2-\sqrt{3}=2-2\sqrt{3}\\ B=\dfrac{x+2\sqrt{x}+8+2\sqrt{x}-8}{\left(\sqrt{x}-4\right)\left(\sqrt{x}+4\right)}=\dfrac{\sqrt{x}\left(\sqrt{x}+4\right)}{\left(\sqrt{x}-4\right)\left(\sqrt{x}+4\right)}=\dfrac{\sqrt{x}}{\sqrt{x}-4}\\ b,B-\dfrac{1}{2}A=\dfrac{\sqrt{x}}{\sqrt{x}-4}-\dfrac{1}{2}\left(2-2\sqrt{3}\right)=0\\ \Leftrightarrow\dfrac{\sqrt{x}}{\sqrt{x}-4}=1+\sqrt{3}\\ \Leftrightarrow\sqrt{x}=\left(1+\sqrt{3}\right)\left(\sqrt{x}-4\right)\Leftrightarrow\sqrt{x}=\sqrt{x}-4\sqrt{3}+\sqrt{3x}-4\\ \Leftrightarrow\sqrt{3x}=4\sqrt{3}+4\\ \Leftrightarrow\sqrt{x}=\dfrac{4\sqrt{3}+4}{\sqrt{3}}\\ \Leftrightarrow\sqrt{x}=\dfrac{12+4\sqrt{3}}{3}\\ \Leftrightarrow x=\dfrac{192+96\sqrt{3}}{9}=\dfrac{64+32\sqrt{3}}{3}\)
\(a,P=\dfrac{\sqrt{x}+2+\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\cdot\dfrac{2-\sqrt{x}}{\sqrt{x}}=\dfrac{-2\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)}=\dfrac{-2}{\sqrt{x}+2}\\ P=-\dfrac{3}{5}\Leftrightarrow\dfrac{2}{\sqrt{x}+2}=\dfrac{3}{5}\\ \Leftrightarrow3\sqrt{x}+6=10\Leftrightarrow\sqrt{x}=\dfrac{4}{3}\Leftrightarrow x=\dfrac{16}{9}\left(tm\right)\)
\(P=-\dfrac{3}{5}\) sao suy ra đc \(\dfrac{2}{\sqrt{x}+2}=\dfrac{3}{5}\) thế
a: Ta có: \(P=\left(\dfrac{1}{1-\sqrt{a}}-\dfrac{1}{1+\sqrt{a}}\right)\left(\dfrac{1}{\sqrt{a}}+1\right)\)
\(=\dfrac{1+\sqrt{a}-1+\sqrt{a}}{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)}\cdot\dfrac{1+\sqrt{a}}{\sqrt{a}}\)
\(=\dfrac{2}{1-\sqrt{a}}\)
a) \(H=\left(\dfrac{a-3\sqrt{a}}{a-2\sqrt{a}-3}-\dfrac{2a}{a-1}\right):\dfrac{1-\sqrt{a}}{a-2\sqrt{a}+1}\)
\(H=\left[\dfrac{\sqrt{a}\left(\sqrt{a}-3\right)}{\left(\sqrt{a}-3\right)\left(\sqrt{a}+1\right)}-\dfrac{2a}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\right]:\dfrac{1-\sqrt{a}}{\left(\sqrt{a}-1\right)^2}\)
\(H=\left[\dfrac{\sqrt{a}\left(\sqrt{a}-1\right)}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}-\dfrac{2a}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\right]:\dfrac{-\left(\sqrt{a}-1\right)}{\left(\sqrt{a}-1\right)^2}\)
\(H=\dfrac{a-\sqrt{a}-2a}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}:\dfrac{-1}{\sqrt{a}-1}\)
\(H=\dfrac{-a-\sqrt{a}}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\cdot-\left(\sqrt{a}-1\right)\)
\(H=\dfrac{-\sqrt{a}\left(\sqrt{a}+1\right)}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\cdot-\left(\sqrt{a}-1\right)\)
\(H=\dfrac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}\)
\(H=\sqrt{a}\)
b) Thay x = 2023 vào ta có:
\(H=\sqrt{2023}\)