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\(2\sqrt{5}+\sqrt{\left(1-\sqrt{5}\right)^2}\\ =2\sqrt{5}+\left|1-\sqrt{5}\right|\\ =2\sqrt{5}+\sqrt{5}-1\\ =3\sqrt{5}-1\)
\(\dfrac{1}{\sqrt{3}+1}+\dfrac{1}{\sqrt{3}-1}2\sqrt{3}\\ =\dfrac{1}{\sqrt{3}+1}+\dfrac{2\sqrt{3}}{\sqrt{3}-1}\\ =\dfrac{\sqrt{3}-1+2\sqrt{3}\left(\sqrt{3}+1\right)}{\sqrt{3}^2-1^2}\\ =\dfrac{\sqrt{3}-1+6+2\sqrt{3}}{2}\\ =\dfrac{3\sqrt{3}+5}{2}\)
Bài 2:
a: ĐKXĐ: 1/x+1>=0
=>x+1>0
=>x>-1
B: ĐKXĐ: (x+1)(x-1)>=0
=>x>=1 hoặc x<=-1
\(1.\sqrt{\dfrac{4}{\left(2-\sqrt{5}\right)^2}}-\sqrt{\dfrac{4}{\left(2+\sqrt{5}\right)^2}}=\dfrac{2}{\sqrt{5}-2}-\dfrac{2}{\sqrt{5}+2}=2\left(\sqrt{5}+2\right)-2\left(\sqrt{5}-2\right)=8\) \(2.a,b.A=\left(\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}+\dfrac{x^2-4x-1}{x^2-1}\right).\dfrac{x+2003}{x}\) ( x # 0 ; x # -1 ; x # 1 )
\(A=\dfrac{x^2+2x+1-x^2+2x-1+x^2-4x-1}{x^2-1}.\dfrac{x+2003}{x}\)
\(A=\dfrac{x^2-1}{x^2-1}.\dfrac{x+2003}{x}=\dfrac{x+2003}{x}\)
c. \(A=1+\dfrac{2003}{x}\)
Để A ∈ Z ⇒ x ∈ { 1 ; -1 ; 2003 ; - 2003 )
KL...............
a/ ĐKXĐ: \(x\ge0;x\ne1\)
= \(\dfrac{x+1+\sqrt{x}}{x+1}:\left[\dfrac{1}{\sqrt{x}-1}-\dfrac{2\sqrt{x}}{\left(x+1\right)\left(\sqrt{x}-1\right)}\right]-1\)
= \(\dfrac{x+1+\sqrt{x}}{x+1}:\dfrac{x+1-2\sqrt{x}}{\left(x+1\right)\left(\sqrt{x}-1\right)}-1\)
= \(\dfrac{x+1+\sqrt{x}}{x+1}:\dfrac{\left(\sqrt{x}-1\right)^2}{\left(x+1\right)\left(\sqrt{x}-1\right)}-1\)
\(=\dfrac{\left(x+1+\sqrt{x}\right)\left(x+1\right)\left(\sqrt{x}-1\right)}{\left(x+1\right)\left(\sqrt{x}-1\right)^2}-1\)
= \(\dfrac{x+1+\sqrt{x}}{\sqrt{x}-1}-1=\dfrac{x+2}{\sqrt{x}-1}\)
b/ Ta có:
\(Q=P-\sqrt{x}\)
= \(\dfrac{x+2}{\sqrt{x}-1}-\sqrt{x}\)
= \(\dfrac{\sqrt{x}+2}{\sqrt{x}-1}=\dfrac{\left(\sqrt{x}-1\right)+3}{\sqrt{x}-1}=1+\dfrac{3}{\sqrt{x}-1}\)
Để Q nhận giá trị nguyên thì \(1+\dfrac{3}{\sqrt{x}-1}\in Z\)
\(\Leftrightarrow\dfrac{3}{\sqrt{x}-1}\in Z\) ( vì 1\(\in Z\) )
\(\Leftrightarrow\sqrt{x}-1\inƯ_{\left(3\right)}\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}-1=3\\\sqrt{x}-1=-3\\\sqrt{x}-1=1\\\sqrt{x}-1=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=4\\\sqrt{x}=-2\\\sqrt{x}=2\\\sqrt{x}=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=16\left(tm\right)\\\\x=4\left(tm\right)\\x=0\left(tm\right)\end{matrix}\right.\)
Vậy để biểu thức \(Q=P-\sqrt{x}\) nhận giá trị nguyên thì x=\(\left\{16;4;0\right\}\)
\(a.\dfrac{\sqrt{x-2\sqrt{x-1}}+\sqrt{x+2\sqrt{x-1}}}{\sqrt{x^2-4\left(x-1\right)}}\left(1-\dfrac{1}{x-1}\right)=\dfrac{\sqrt{x-1-2\sqrt{x-1}+1}+\sqrt{x-1+2\sqrt{x-1}+1}}{\sqrt{x^2-4x+4}}.\dfrac{x-2}{x-1}=\dfrac{\left|\sqrt{x-1}-1\right|+\left|\sqrt{x-1}+1\right|}{\left|x-2\right|}.\dfrac{x-2}{x-1}\left(x>1\right)\)
Tới đây dễ r , bạn tự chia TH ra làm nhé :D
\(b.\dfrac{1}{\sqrt{x}+\sqrt{x-1}}-\dfrac{1}{\sqrt{x}-\sqrt{x-1}}-\dfrac{\sqrt{x^3}-x}{1-\sqrt{x}}=\dfrac{\sqrt{x}-\sqrt{x-1}-\sqrt{x}-\sqrt{x-1}}{\left(\sqrt{x}+\sqrt{x-1}\right)\left(\sqrt{x}-\sqrt{x-1}\right)}+\dfrac{x\sqrt{x}-x}{\sqrt{x}-1}=-2\sqrt{x-1}+x\left(x\ge1\right)\)
Bạn ơi câu a có vẻ có vấn đề ý. Nếu bạn áp dụng HĐT thì phải là√(x-2)2 chứ nhỉ. Mong bạn giải đáp
a: ĐKXĐ: x=0; x<>1
\(M=\left(2+\sqrt{x}\right)\left(1-2\sqrt{x}-x+1+\sqrt{x}+x\right)\)
\(=\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)=4-x\)
b: Sửa đề: P=1/M
P=1/4-x=-1/x-4
Để P nguyên thì x-4 thuộc {1;-1}
=>x thuộc {5;3}
a/ đkxđ: x > 0; x≠1
b/ \(A=\left(\dfrac{\sqrt{x}}{2}-\dfrac{1}{2\sqrt{x}}\right):\left(\dfrac{x-\sqrt{x}}{\sqrt{x}+1}-\dfrac{x+\sqrt{x}}{\sqrt{x}-1}\right)\)
\(=\dfrac{x-1}{2\sqrt{x}}\cdot\left(\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)^2-\sqrt{x}\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right)\)
\(=\dfrac{x-1}{2\sqrt{x}}\cdot\dfrac{x\sqrt{x}-2x+\sqrt{x}-x\sqrt{x}-2x-\sqrt{x}}{x-1}\)
\(=\dfrac{-4x}{2\sqrt{x}}=-2\sqrt{x}\)
a) \(x>0,x\ne1\)
b) \(P=\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{1}{x-\sqrt{x}}\right):\left(\dfrac{1}{\sqrt{x}+1}+\dfrac{2}{x-1}\right)\)
\(=\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}\right):\left(\dfrac{1}{\sqrt{x}+1}+\dfrac{2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\)
\(=\dfrac{x-1}{\sqrt{x}\left(\sqrt{x}-1\right)}:\dfrac{\sqrt{x}-1+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}:\dfrac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}}:\dfrac{1}{\sqrt{x}-1}=\dfrac{\sqrt{x}+1}{\sqrt{x}}.\left(\sqrt{x}-1\right)=\dfrac{x-1}{\sqrt{x}}\)
c) \(P< 0\Rightarrow\dfrac{x-1}{\sqrt{x}}< 0\) mà \(\sqrt{x}>0\Rightarrow x-1< 0\Rightarrow x< 1\Rightarrow0< x< 1\)
\(\left(\dfrac{x-3\sqrt{x}}{x-9}-1\right):\left(\dfrac{9-x}{x+\sqrt{x}-6}+\dfrac{\sqrt{x}-3}{\sqrt{x}-2}-\dfrac{\sqrt{x}-2}{\sqrt{x}+3}\right)\left(x\ge0;x\ne3;x\ne-3;x\ne9;x\ne4\right)\)
\(=\left(\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}-1\right):\left(\dfrac{9-x}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}+\dfrac{\sqrt{x}-3}{\sqrt{x}-2}-\dfrac{\sqrt{x}-2}{\sqrt{x}+3}\right)\\ =\dfrac{\sqrt{x}-\sqrt{x}-3}{\sqrt{x}+3}:\dfrac{9-x+\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)-\left(\sqrt{x}-2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\\ =\dfrac{-3}{\sqrt{x}+3}:\dfrac{9-x+x-9-x+4\sqrt{x}-4}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\\ =\dfrac{-3}{\sqrt{x}+3}\cdot\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}{-\left(\sqrt{x}-2\right)^2}\\ =\dfrac{3}{\sqrt{x}-2}\)
Tick hộ nha 😘
a) điều kiện để M có nghĩa là \(x\ge0;x\ne4\)
b) \(M=\left(\dfrac{\sqrt{x}}{\sqrt{x}+2}-\dfrac{\sqrt{x}}{\sqrt{x}-2}+\dfrac{4\sqrt{x}-1}{x-4}\right):\dfrac{1}{x-4}\)
\(M=\left(\dfrac{\sqrt{x}}{\sqrt{x}+2}-\dfrac{\sqrt{x}}{\sqrt{x}-2}+\dfrac{4\sqrt{x}-1}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right):\dfrac{1}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(M=\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)-\sqrt{x}\left(\sqrt{x}+2\right)+4\sqrt{x}-1}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}:\dfrac{1}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(M=\dfrac{x-2\sqrt{x}-x-2\sqrt{x}+4\sqrt{x}-1}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}.\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{1}\)
\(M=\dfrac{-1}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}.\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{1}=\dfrac{-1}{1}=-1\)