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\(\frac{\sqrt{x}-1}{\sqrt{x}+1}< \frac{1}{2}\Leftrightarrow\frac{\sqrt{x}-1}{\sqrt{x}+1}-\frac{1}{2}< 0\)
\(\Leftrightarrow\frac{2\sqrt{x}-2-\sqrt{x}-1}{2\left(\sqrt{x}+1\right)}< 0\Leftrightarrow\frac{\sqrt{x}-3}{2\left(\sqrt{x}+1\right)}< 0\)
\(\Leftrightarrow\sqrt{x}-3< 0\Leftrightarrow x< 9\Rightarrow0\le x< 9\)
anh đi anh nhớ quê nha
nhớ canh rau muống nhớ cà dầm tương
nhớ thằng đẩy bố xuống mương
bố mà bắt được bố tương vỡ mồm
\(A=\sqrt{6+3\sqrt{3}}-\sqrt{14-3\sqrt{3}}-2\sqrt{2}\)
\(A\sqrt{2}=\sqrt{12+6\sqrt{3}}-\sqrt{28-6\sqrt{3}}-4\)
\(=\sqrt{9+2.3.\sqrt{3}+3}-\sqrt{27-2.3\sqrt{3}.1+1}-4\)
\(=\sqrt{\left(3+\sqrt{3}\right)^2}-\sqrt{\left(3\sqrt{3}-1\right)^2}-4\)
\(=3+\sqrt{3}-3\sqrt{3}+1-4\)
\(=-2\sqrt{3}\)
Suy ra \(A=-\sqrt{6}\).
ta có
\(A=\frac{\sqrt{x}-\sqrt{x-1}-\left(\sqrt{x}+\sqrt{x-1}\right)}{\left(\sqrt{x}+\sqrt{x-1}\right)\left(\sqrt{x}-\sqrt{x-1}\right)}-\frac{0}{1-\sqrt{x}}\)
\(=-\frac{2\sqrt{x-1}}{x-\left(x-1\right)}=-2\sqrt{x-1}\) dễ thấy \(A\le0\) với mọi x
a) \(A=\frac{a+2\sqrt{ab}+b}{\sqrt{a}+\sqrt{b}}-\frac{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{ab}}\)
\(=\sqrt{a}+\sqrt{b}-\sqrt{a}+\sqrt{b}=2\sqrt{b}\)
b) \(B=\left(\frac{\sqrt{x}-\sqrt{y}\left(x+\sqrt{xy}+y\right)}{\sqrt{x}-\sqrt{y}}+\sqrt{xy}\right):\left(x-y\right)-\frac{2\sqrt{y}}{\sqrt{x}-\sqrt{y}}\)
\(=\frac{x+y+2\sqrt{xy}}{x-y}-\frac{2\sqrt{y}}{\sqrt{x}-\sqrt{y}}\)\(=\frac{\sqrt{x}+\sqrt{y}}{\sqrt{x}-\sqrt{y}}-\frac{2\sqrt{y}}{\sqrt{x}-\sqrt{y}}=\frac{\sqrt{x}-\sqrt{y}}{\sqrt{x}-\sqrt{y}}=1\)
\(ĐKXĐ:x\ge0;x\ne1\)
\(\frac{1}{\sqrt{x}-1}-\frac{2\sqrt{x}}{x\sqrt{x}-x+\sqrt{x}-1}\)
\(\frac{1}{\sqrt{x}-1}-\frac{2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+1\right)}\)
\(\frac{x+1-2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+1\right)}=\frac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(x+1\right)}=\frac{\sqrt{x}-1}{x+1}\)
\(=\frac{1}{\sqrt{x}-1}-\frac{2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+1\right)}\)
\(=\frac{x+1-2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+1\right)}=\frac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(x+1\right)}=\frac{\sqrt{x}-1}{x+1}\)
\(=\frac{x+\sqrt{x}}{x\left(\sqrt{x}+1\right)+\left(\sqrt{x}+1\right)}+\frac{1}{x+1}\)
\(=\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(x+1\right)}+\frac{1}{x+1}\)
\(=\frac{\sqrt{x}}{x+1}+\frac{1}{x+1}=\frac{\sqrt{x}+1}{x+1}\)
Xét d cắt với Ox khi đó \(y=0\Rightarrow-4x+3=0\Leftrightarrow x=\frac{3}{4}\) Vậy giao với Ox tại điểm \(\left(\frac{3}{4};0\right)\)
d cắt với Oy khi đó : \(x=0\Rightarrow y=-4.0+3=3\) vậy giao với Oy tại điểm \(\left(0,3\right)\)
\(M=\left(\frac{1}{\sqrt{x}-1}+\frac{\sqrt{x}}{x-1}\right)\div\left(\frac{\sqrt{x}}{\sqrt{x}-1}-1\right)\)
\(=\frac{\sqrt{x}+1+\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\div\frac{\sqrt{x}-\sqrt{x}+1}{\sqrt{x}-1}\)
\(=\frac{2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}.\frac{\sqrt{x-1}}{1}\)
\(=\frac{2\sqrt{x}+1}{\sqrt{x}+1}\)
\(M< \frac{3}{2}\Leftrightarrow\frac{2\sqrt{x}+1}{\sqrt{x}+1}< \frac{3}{2}\Leftrightarrow2\left(2\sqrt{x}+1\right)< 3\left(\sqrt{x}+1\right)\)
\(\Leftrightarrow4\sqrt{x}+2< 3\sqrt{x}+3\Leftrightarrow\sqrt{x}< 1\Leftrightarrow0\le x< 1\).