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a.
\(M=\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\frac{3\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{6\sqrt{x}-4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\\ =\frac{x+\sqrt{x}+3\sqrt{x}-3-6\sqrt{x}+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\\ =\frac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\\ =\frac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\\ =\frac{\sqrt{x}-1}{\sqrt{x}+1}\)
b.
\(M< \frac{1}{2}\Leftrightarrow\frac{\sqrt{x}-1}{\sqrt{x}+1}< \frac{1}{2}\\ \Leftrightarrow\frac{\sqrt{x}-1}{\sqrt{x}+1}-\frac{1}{2}< 0\\ \Leftrightarrow\frac{2\left(\sqrt{x}-1\right)-\left(\sqrt{x}+1\right)}{2\left(\sqrt{x}+1\right)}< 0\\ \Leftrightarrow\frac{\sqrt{x}-3}{2\left(\sqrt{x}+1\right)}< 0\\ \Leftrightarrow\sqrt{x}-3< 0\Leftrightarrow x< 9\)
Vậy với \(0\le x< 9;x\ne1\) thì ..........
\(đkxđ\Leftrightarrow\hept{\begin{cases}x\ge0\\\sqrt{x}-1\ne0\end{cases}\Rightarrow\hept{\begin{cases}x\ge0\\\sqrt{x}\ne1\end{cases}\Rightarrow}\hept{\begin{cases}x\ge0\\x\ne1\end{cases}}}\)
\(M=\frac{\sqrt{x}}{\sqrt{x}-1}+\frac{3}{\sqrt{x}+1}-\frac{6\sqrt{x}-4}{x-1}.\)
\(=\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{x-1}+\frac{3\left(\sqrt{x}-1\right)}{x-1}-\frac{6\sqrt{x}-4}{x-1}\)
\(=\frac{x+\sqrt{x}+3\sqrt{x}-3-6\sqrt{x}+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{\sqrt{x}-1}{\sqrt{x}+1}\)
\(b,M< \frac{1}{2}\Leftrightarrow\frac{\sqrt{x}-1}{\sqrt{x}+1}< \frac{1}{2}\)
\(\Rightarrow\frac{\sqrt{x}-1}{\sqrt{x}+1}-\frac{1}{2}< 0\)\(\Rightarrow\frac{2\left(\sqrt{x}-1\right)}{2\left(\sqrt{x}+1\right)}-\frac{\sqrt{x}+1}{2\left(\sqrt{x}+1\right)}< 0\)
\(\Rightarrow\frac{2\sqrt{x}-1-\sqrt{x}-1}{2\left(\sqrt{x}+1\right)}< 0\)\(\Rightarrow\frac{\sqrt{x}-2}{2\left(\sqrt{x}+1\right)}< 0\)
Vì \(2\left(\sqrt{x}+1\right)>0\Rightarrow\sqrt{x}-2>0\Rightarrow\sqrt{x}>2\)
\(\Rightarrow\sqrt{x}>\sqrt{4}\Leftrightarrow x>4\)
\(M=\frac{\sqrt{x}}{\sqrt{x}-1}+\frac{3}{\sqrt{x}+1}-\frac{6\sqrt{x}-4}{x-1}\left(x\ge0;x\ne1\right)\)
\(M=\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\frac{3\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{6\sqrt{x}-4}{x-1}\)
\(M=\frac{x+\sqrt{x}+3\sqrt{x}-3}{\left(\sqrt{x}\right)^2-1^2}-\frac{6\sqrt{x}-4}{x-1}\)
\(M=\frac{x-2\sqrt{x}+1}{x-1}\)
\(M=\frac{\left(\sqrt{x}-1\right)^2}{x-1}\)
Lời giải:
a)
\(A=\frac{\sqrt{3}-1+\sqrt{3}+1}{(\sqrt{3}+1)(\sqrt{3}-1)}+2-\sqrt{3}=\frac{2\sqrt{3}}{3-1}+2-\sqrt{3}=\sqrt{3}+2-\sqrt{3}=2\)
b)
\(B=\left(\frac{1}{\sqrt{x}(\sqrt{x}-1)}+\frac{\sqrt{x}}{\sqrt{x}(\sqrt{x}-1)}\right):\frac{\sqrt{x}}{(\sqrt{x}-1)^2}\)
\(=\frac{\sqrt{x}+1}{\sqrt{x}.(\sqrt{x}-1)}.\frac{(\sqrt{x}-1)^2}{\sqrt{x}}=\frac{(\sqrt{x}-1)(\sqrt{x}+1)}{x}=\frac{x-1}{x}\)
Ta có: \(B=\frac{\left(\sqrt{x}+3\right)\left(\sqrt{x}+1\right)+5\left(\sqrt{x}+1\right)+4}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(=\frac{x+2\sqrt{x}-3+5\sqrt{x}+5+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+6\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{\sqrt{x}+6}{\sqrt{x}-1}\)
do đó \(P=\frac{\sqrt{x}-1}{\sqrt{x}+1}.\frac{\sqrt{x}-6}{\sqrt{x}-1}=\frac{\sqrt{x}-6}{\sqrt{x}+1}=1-\frac{7}{\sqrt{x}+1}\)
Vì \(x\ge0\Rightarrow0< \frac{7}{\sqrt{x}+1}\le7\)
Để P nguyên thì \(\frac{7}{\sqrt{x}+1}\in Z\)
do đó \(\frac{7}{\sqrt{x}+1}\in\left\{1,2,3,4,5,6,7\right\}\)
Đến đây xét từng TH là ra
rút gọn B ta có B=\(\frac{\sqrt{x}+6}{\sqrt{x}-1}\)\(\Rightarrow\)\(AB=\frac{\sqrt{x}+6}{\sqrt{x}+1}\in Z\)
=\(1+\frac{5}{\sqrt{x}+1}\)
Vì 1\(\in Z\) nên để P thuộc Z thì \(\frac{5}{\sqrt{x}+1}\in Z\)
\(\Rightarrow\left(\sqrt{x}+1\right)\inƯ\left(5\right)=\pm1;\pm5\)
Đến đây thì ez rồi
Ai trả lời nhanh và chính xác mình k
Lê Thị Tuyết
a) \(\sqrt{12}-3\sqrt{75}+0,5\sqrt{\left(-6\right)^2\cdot3}\)
\(=2\sqrt{3}-15\sqrt{3}+0,5\sqrt{108}\)
\(=-13\sqrt{3}+3\sqrt{3}\)
\(=-10\sqrt{3}\)
b) \(3\sqrt{\left(\sqrt{2}-\sqrt{3}\right)^2}-\sqrt{4+2\sqrt{3}}\)
\(=3\left|\sqrt{2}-\sqrt{3}\right|-\sqrt{\left(\sqrt{3}+1\right)^2}\)
\(=3\left(\sqrt{3}-\sqrt{2}\right)-\left|\sqrt{3}+1\right|\)
\(=3\sqrt{3}-3\sqrt{2}-\sqrt{3}-1\)
\(=2\sqrt{3}-3\sqrt{2}-1\)
c) \(\left(\frac{2x+1}{x\sqrt{x}-1}-\frac{\sqrt{x}}{x+\sqrt{x}+1}\right)\div\frac{1}{x-2\sqrt{x}+1}\)
\(=\frac{2x+1-\left(\sqrt{x}-1\right)\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\div\frac{1}{\left(\sqrt{x}-1\right)^2}\)
\(=\frac{2x+1-x+\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\left(\sqrt{x}-1\right)^2\)
\(=\frac{x+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\left(\sqrt{x}-1\right)^2\)
\(=\sqrt{x}-1\)
1: Ta có: \(Q=\left(\frac{2\sqrt{x}+x}{x\sqrt{x}-1}-\frac{1}{\sqrt{x}-1}\right):\left(1-\frac{\sqrt{x}+2}{x+\sqrt{x}+1}\right)\)
\(=\left(\frac{\left(2\sqrt{x}+x\right)\left(\sqrt{x}-1\right)}{\left(x\sqrt{x}-1\right)\left(\sqrt{x}-1\right)}-\frac{x\sqrt{x}-1}{\left(x\sqrt{x}-1\right)\left(\sqrt{x}-1\right)}\right):\left(\frac{x+\sqrt{x}+1}{x+\sqrt{x}+1}-\frac{\sqrt{x}+2}{x+\sqrt{x}+1}\right)\)
\(=\frac{x-2\sqrt{x}+x\sqrt{x}-x\sqrt{x}+1}{\left(x\sqrt{x}-1\right)\left(\sqrt{x}-1\right)}:\frac{x+\sqrt{x}+1-\sqrt{x}-2}{x+\sqrt{x}+1}\)
\(=\frac{x-2\sqrt{x}+1}{\left(x\sqrt{x}-1\right)\left(\sqrt{x}-1\right)}:\frac{x-1}{x+\sqrt{x}+1}\)
\(=\frac{\left(\sqrt{x}-1\right)^2}{\left(x\sqrt{x}-1\right)\left(\sqrt{x}-1\right)}\cdot\frac{x+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{x+\sqrt{x}+1}{\left(x\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
2: Ta có: \(\frac{1}{Q}=4\sqrt{x}-4\)
\(\Leftrightarrow Q=\frac{1}{4\sqrt{x}-4}\)
\(\Leftrightarrow\frac{x+\sqrt{x}+1}{\left(x\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{1}{4\sqrt{x}-4}\)
\(\Leftrightarrow\left(x\sqrt{x}-1\right)\left(\sqrt{x}+1\right)=\left(x+\sqrt{x}+1\right)\left(4\sqrt{x}-4\right)\)
\(\Leftrightarrow x+x\sqrt{x}-\sqrt{x}-1=4x\sqrt{x}-4\)
\(\Leftrightarrow x+x\sqrt{x}-\sqrt{x}-1-4x\sqrt{x}+4=0\)
\(\Leftrightarrow x-3x\sqrt{x}-\sqrt{x}+3=0\)
\(\Leftrightarrow\sqrt{x}\left(\sqrt{x}-1\right)-\left(3x\sqrt{x}-3\right)=0\)
\(\Leftrightarrow\sqrt{x}\left(\sqrt{x}-1\right)-3\left(x\sqrt{x}-1\right)=0\)
\(\Leftrightarrow\sqrt{x}\left(\sqrt{x}-1\right)-3\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)=0\)
\(\Leftrightarrow\left(\sqrt{x}-1\right)\left[\sqrt{x}-3\left(x+\sqrt{x}+1\right)\right]=0\)
\(\Leftrightarrow\left(\sqrt{x}-1\right)\left(\sqrt{x}-3x-3\sqrt{x}-3\right)=0\)
\(\Leftrightarrow\left(\sqrt{x}-1\right)\left(-3x-2\sqrt{x}-3\right)=0\)
\(\Leftrightarrow\left(\sqrt{x}-1\right)=0\)(vì \(-3x-2\sqrt{x}-3\ne0\forall x\) thỏa mãn ĐKXĐ)
\(\Leftrightarrow\sqrt{x}=1\)
hay x=1(không thỏa mãn ĐKXĐ)
Vậy: Không có giá trị nào của x thỏa mãn \(\frac{1}{Q}=4\sqrt{x}-4\)
rút gọn giúp mình nha mình quên ghi
\(A=\frac{\sqrt{x}}{\sqrt{x}-1}+\frac{3}{\sqrt{x}+1}-\frac{6\sqrt{x}-4}{x-1}\)
\(A=\frac{\sqrt{x}}{\sqrt{x}-1}+\frac{3}{\sqrt{x}+1}-\frac{6\sqrt{x}-4}{\left(\sqrt{x}-1\right).\left(\sqrt{x}+1\right)}\)
\(A=\frac{\sqrt{x}.\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right).\left(\sqrt{x}+1\right)}+\frac{3\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right).\left(\sqrt{x}+1\right)}-\frac{6\sqrt{x}-4}{\left(\sqrt{x}-1\right).\left(\sqrt{x}+1\right)}\)
\(A=\frac{\sqrt{x}.\left(\sqrt{x}+1\right)+3.\left(\sqrt{x}-1\right)-6\sqrt{x}+4}{\left(\sqrt{x}-1\right).\left(\sqrt{x}+1\right)}=\frac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right).\left(\sqrt{x}+1\right)}\)
\(A=\frac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right).\left(\sqrt{x}+1\right)}=\frac{\sqrt{x}-1}{\sqrt{x}+1}\)