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a. ĐK: \(x\ge0,x\ne49\)
\(M=\frac{3\left(\sqrt{x}+7\right)-\left(\sqrt{x}-7\right)}{\left(\sqrt{x}-7\right)\left(\sqrt{x}+7\right)}:\frac{2\sqrt{x}+6}{x-49}\)
\(=\frac{2\sqrt{x}+28}{x-49}.\frac{x-49}{2\sqrt{x}+6}=\frac{2\sqrt{x}+28}{2\sqrt{x}+6}\)
b. M nguyên \(\Leftrightarrow\frac{2\sqrt{x}+28}{2\sqrt{x}+6}\in Z\Rightarrow\frac{2\sqrt{x}+6+22}{2\sqrt{x}+6}\in Z\Rightarrow1+\frac{22}{2\sqrt{x}+6}\in Z\Rightarrow\frac{22}{2\sqrt{x}+6}\in Z\Rightarrow\left(2\sqrt{x}+6\right)\inƯ\left(22\right)\)
Đến đây đã rất dễ dàng rồi nhé ^^
đề không cho tìm x NGUYÊN để m nguyên mà chỉ tìm các điểm x để m nguyên thôi
a: \(A=\dfrac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\sqrt{a}-\sqrt{b}}-\dfrac{\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{ab}}\)
\(=\sqrt{a}-\sqrt{b}-\sqrt{a}-\sqrt{b}=-2\sqrt{b}\)
b: \(B=\dfrac{2\sqrt{x}-x-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\dfrac{x+\sqrt{x}+1}{x-1}\)
\(=\dfrac{-2x+\sqrt{x}-1}{\sqrt{x}-1}\cdot\dfrac{1}{x-1}\)
c: \(C=\dfrac{x-9-x+3\sqrt{x}}{x-9}:\left(\dfrac{3-\sqrt{x}}{\sqrt{x}-2}+\dfrac{\sqrt{x}-2}{\sqrt{x}+3}+\dfrac{x-9}{x+\sqrt{x}-6}\right)\)
\(=\dfrac{3\left(\sqrt{x}-3\right)}{x-9}:\dfrac{9-x+x-4\sqrt{x}+4+x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{3}{\sqrt{x}+3}\cdot\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}{x-4\sqrt{x}+4}\)
\(=\dfrac{3}{\sqrt{x}-2}\)
1) Khi x = 49 thì:
\(A=\frac{4\sqrt{49}}{\sqrt{49}-1}=\frac{4\cdot7}{7-1}=\frac{28}{6}=\frac{14}{3}\)
2) Ta có:
\(B=\frac{1}{\sqrt{x}+1}+\frac{\sqrt{x}}{\sqrt{x}-1}+\frac{2}{x-1}\)
\(B=\frac{\sqrt{x}-1+x+\sqrt{x}+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(B=\frac{x+2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(B=\frac{\sqrt{x}+1}{\sqrt{x}-1}\)
c) \(P=A\div B=\frac{4\sqrt{x}}{\sqrt{x}-1}\div\frac{\sqrt{x}+1}{\sqrt{x}-1}=\frac{4\sqrt{x}}{\sqrt{x}+1}\)
Ta có: \(P\left(\sqrt{x}+1\right)=x+4+\sqrt{x-4}\)
\(\Leftrightarrow\frac{4\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}+1}=x+4+\sqrt{x-4}\)
\(\Leftrightarrow4\sqrt{x}=x+4+\sqrt{x-4}\)
\(\Leftrightarrow\left(\sqrt{x}-2\right)^2+\sqrt{x-4}=0\)
Mà \(VT\ge0\left(\forall x\ge0,x\ne1\right)\)
\(\Rightarrow\hept{\begin{cases}\left(\sqrt{x}-2\right)^2=0\\\sqrt{x-4}=0\end{cases}}\Leftrightarrow\hept{\begin{cases}\sqrt{x}=2\\x-4=0\end{cases}}\Rightarrow x=4\)
Vậy x = 4
1.
= -(13 + 3 căn7 ) / 2 + -(7 + 3 căn7 ) / 2
= -7 + 3 căn7
bài 2 : ĐKXĐ : \(x\ge0\) và \(x\ne1\)
Rút gọn :\(B=\frac{\sqrt{x}+1}{\sqrt{x}-1}-\frac{\sqrt{x}-1}{\sqrt{x}+1}-\frac{5\sqrt{x}-1}{x-1}\)
\(B=\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{5\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(B=\frac{x+2\sqrt{x}+1-x+2\sqrt{x}-1-5\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(B=\frac{-\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(B=\frac{-1}{\sqrt{x}+1}\)
Bài 1:
a, \(4\sqrt{3+2\sqrt{2}}-\sqrt{57+40\sqrt{2}}\)
\(=4\sqrt{\left(\sqrt{2}+1\right)^2}-\sqrt{\left(4\sqrt{2}+5\right)^2}\)
\(=4\left(\sqrt{2}+1\right)-4\sqrt{2}-5\)
\(=4\sqrt{2}+4-4\sqrt{2}-5=-1\)
b, \(B=\sqrt{1100}-7\sqrt{44}+2\sqrt{176}-\sqrt{1331}\)
\(=10\sqrt{11}-14\sqrt{11}+8\sqrt{11}-11\sqrt{11}=-7\sqrt{11}\)
c, \(C=\sqrt{\left(1-\sqrt{2002}\right)^2}.\sqrt{2003+2\sqrt{2002}}\)
\(=\left(1-\sqrt{2002}\right).\sqrt{\left(\sqrt{2002}+1\right)^2}\)
\(=\left(1-\sqrt{2002}\right).\left(\sqrt{2002}+1\right)=-2001\)
Câu d bạn kiểm tra lại đề bài nhé.
Bài 2:
\(A=\frac{1}{2\sqrt{x}-2}-\frac{1}{2\sqrt{2}+2}+\frac{\sqrt{x}}{1-x}\)
a, ĐK: \(x\ge0,x\ne1\)
b, ĐK: \(x\ge0,x\ne1\)
\(A=\frac{1}{2\sqrt{x}-2}-\frac{1}{2\sqrt{2}+2}+\frac{\sqrt{x}}{1-x}\)
\(=\frac{1}{2\sqrt{x}-2}-\frac{1}{2\sqrt{2}+2}-\frac{\sqrt{x}}{x-1}\)
\(=\frac{1}{2\left(\sqrt{x}-1\right)}-\frac{1}{2\left(\sqrt{x}+1\right)}-\frac{\sqrt{x}}{x-1}\)
\(=\frac{2\sqrt{x}+2-2\sqrt{x}+2}{4\left(x-1\right)}-\frac{\sqrt{x}}{x-1}\)
\(=\frac{4-4\sqrt{x}}{4\left(x-1\right)}=\frac{4\left(1-\sqrt{x}\right)}{4\left(1-x\right)}=\frac{1-\sqrt{x}}{1-x}\)
Thay \(x=3\left(TM\right)\)vào A ta có: \(A=\frac{1-\sqrt{3}}{3-1}=\frac{1-\sqrt{3}}{2}\)
Vậy với \(x=3\)thì \(A=\frac{1-\sqrt{3}}{2}\)
c, \(\left|A\right|=\frac{1}{2}\Leftrightarrow\orbr{\begin{cases}A=\frac{1}{2}\\A=-\frac{1}{2}\end{cases}}\)
TH1: \(A=\frac{1}{2}\)\(\Leftrightarrow\frac{1-\sqrt{x}}{x-1}=\frac{1}{2}\Leftrightarrow2-2\sqrt{x}=x-1\)\(\Leftrightarrow x-1-2+2\sqrt{x}=0\)\(\Leftrightarrow x+2\sqrt{x}-3=0\Leftrightarrow\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)=0\)\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x}-1=0\\\sqrt{x}+3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1\left(TM\right)\\\sqrt{x}=-3\left(L\right)\end{cases}}}\)
TH2: \(A=-\frac{1}{2}\Leftrightarrow\frac{1-\sqrt{x}}{x-1}=-\frac{1}{2}\)\(\Leftrightarrow2-2\sqrt{x}=1-x\Leftrightarrow-x+1-2+2\sqrt{x}=0\)\(\Leftrightarrow-x-1+2\sqrt{x}=0\Leftrightarrow x-2\sqrt{x}+1=0\)\(\Leftrightarrow\left(\sqrt{x}+1\right)^2=0\Leftrightarrow\sqrt{x}=-1\left(L\right)\)
Vậy với \(x=1\)thì \(\left|A\right|=\frac{1}{2}\)
\(đkxđ\Leftrightarrow\hept{\begin{cases}x\ge0\\x\ne49\end{cases}}\)
\(B=\left(\frac{\sqrt{x}}{x-49}-\frac{\sqrt{x}-7}{x+7\sqrt{x}}\right):\)\(\frac{2\sqrt{x}-7}{x+7\sqrt{x}}+\frac{\sqrt{x}}{7-\sqrt{x}}\)
\(=\left(\frac{\sqrt{x}}{\left(\sqrt{x}-7\right)\left(\sqrt{x}+7\right)}-\frac{\left(\sqrt{x}-7\right)^2}{\sqrt{x}\left(\sqrt{x}+7\right)\left(\sqrt{x}-7\right)}\right)\)\(:\frac{2\sqrt{x}-7}{\sqrt{x}\left(\sqrt{x}+7\right)}-\frac{\sqrt{x}}{\sqrt{x}-7}\)
\(\frac{x-x+14\sqrt{x}-49}{\sqrt{x}\left(\sqrt{x}-7\right)\left(\sqrt{x}+7\right)}:\frac{2\sqrt{x}-7}{\sqrt{x}\left(\sqrt{x}+7\right)}\)\(-\frac{\sqrt{x}}{\sqrt{x}-7}\)
\(=\frac{7\left(2\sqrt{x}-7\right)\sqrt{x}\left(\sqrt{x}+7\right)}{\sqrt{x}\left(\sqrt{x}+7\right)\left(\sqrt{x}-7\right)\left(2\sqrt{x}-7\right)}\)\(-\frac{\sqrt{x}}{\sqrt{x}-7}\)
\(=\frac{7}{\sqrt{x}-7}-\frac{\sqrt{x}}{\sqrt{x}-7}=\frac{7-\sqrt{x}}{\sqrt{x}-7}=-1\)