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mọi ng ơi mk viết thiếu dấu ngoặc nha.thiếu ngoặc lownns nha. đóng ngoắc ở trước dấu chia
ĐKXĐ: \(x\ge0;\)\(x\ne1\)
\(P=\left(\frac{\sqrt{x}}{\sqrt{x}-1}-\frac{1}{x-\sqrt{x}}\right):\left(\frac{1}{\sqrt{x}+1}+\frac{2}{x-1}\right)\)
\(=\left(\frac{x}{\sqrt{x} \left(\sqrt{x}-1\right)}-\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}\right):\left(\frac{\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}+\frac{2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right)\)
\(=\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}.\left(\sqrt{x}-1\right)}:\frac{\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(=\frac{\sqrt{x}+1}{\sqrt{x}}:\frac{1}{\sqrt{x}-1}\)
\(=\frac{x-1}{\sqrt{x}}\)
1) \(ĐKXĐ:\hept{\begin{cases}x\ge0\\x\ne4\end{cases}}\)
\(P=\frac{2+\sqrt{x}}{2-\sqrt{x}}-\frac{2-\sqrt{x}}{2+\sqrt{x}}-\frac{4x}{x-4}\)
\(\Leftrightarrow P=\frac{\left(2+\sqrt{x}\right)^2-\left(2-\sqrt{x}\right)^2+4x}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}\)
\(\Leftrightarrow P=\frac{4+4\sqrt{x}+x-4+4\sqrt{x}-x+4x}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}\)
\(\Leftrightarrow P=\frac{4x+8\sqrt{x}}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}\)
\(\Leftrightarrow P=\frac{4\sqrt{x}}{2-\sqrt{x}}\)
2) Để \(P=2\)
\(\Leftrightarrow\frac{4\sqrt{x}}{2-\sqrt{x}}=2\)
\(\Leftrightarrow4\sqrt{x}=4-2\sqrt{x}\)
\(\Leftrightarrow6\sqrt{x}=4\)
\(\Leftrightarrow\sqrt{x}=\frac{2}{3}\)
\(\Leftrightarrow x=\frac{4}{9}\)
Vậy để \(P=2\Leftrightarrow x=\frac{4}{9}\)
3) Khi \(\left(\sqrt{x}-2\right)\left(2\sqrt{x}-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x}-2=0\\2\sqrt{x}-1==0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x}=2\\\sqrt{x}=\frac{1}{2}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=4\left(ktm\right)\\x=\frac{1}{4}\left(tm\right)\end{cases}}\)
Thay \(x=\frac{1}{4}\)vào P, ta được :
\(\Leftrightarrow P=\frac{4\sqrt{\frac{1}{4}}}{2-\sqrt{\frac{1}{4}}}=\frac{4\cdot\frac{1}{2}}{2-\frac{1}{2}}=\frac{2}{\frac{3}{2}}=\frac{4}{3}\)
4) Để \(P=\frac{\sqrt{x}+3}{2\sqrt{x}-1}\)
\(\Leftrightarrow\frac{4\sqrt{x}}{2-\sqrt{x}}=\frac{\sqrt{x}+3}{2\sqrt{x}-1}\)
\(\Leftrightarrow8x-4\sqrt{x}=-x-\sqrt{x}+6\)
\(\Leftrightarrow9x-3\sqrt{x}-6=0\)
\(\Leftrightarrow3x-\sqrt{x}-2=0\)
\(\Leftrightarrow\sqrt{x}=3x-2\)
\(\Leftrightarrow x=9x^2-12x+4\)
\(\Leftrightarrow9x^2-13x+4=0\)
\(\Leftrightarrow\left(9x-4\right)\left(x-1\right)=0\)
\(\Leftrightarrow\hept{\begin{cases}9x-4=0\\x-1=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{4}{9}\\x=1\end{cases}}\)
Thử lại ta được kết quá : \(x=\frac{4}{9}\left(ktm\right)\); \(x=1\left(tm\right)\)
Vậy để \(P=\frac{\sqrt{x}+3}{2\sqrt{x}-1}\Leftrightarrow x=1\)
5) Để biểu thức nhận giá trị nguyên
\(\Leftrightarrow\frac{4\sqrt{x}}{2-\sqrt{x}}\inℤ\)
\(\Leftrightarrow4\sqrt{x}⋮2-\sqrt{x}\)
\(\Leftrightarrow-4\left(2-\sqrt{x}\right)+8⋮2-\sqrt{x}\)
\(\Leftrightarrow8⋮2-\sqrt{x}\)
\(\Leftrightarrow2-\sqrt{x}\inƯ\left(8\right)=\left\{\pm1;\pm2;\pm4;\pm8\right\}\)
\(\Leftrightarrow\sqrt{x}\in\left\{1;3;0;4;-2;6;-6;10\right\}\)
Ta loại các giá trị < 0
\(\Leftrightarrow\sqrt{x}\in\left\{1;3;0;4;6;10\right\}\)
\(\Leftrightarrow x\in\left\{1;9;0;16;36;100\right\}\)
Vậy để \(P\inℤ\Leftrightarrow x\in\left\{1;9;0;16;36;100\right\}\)
\(\)
\(1,\frac{\sqrt{x}+1}{\sqrt{x}-3}=\frac{\sqrt{x}-3+4}{\sqrt{x}-3}=1+\frac{4}{\sqrt{x}-3}\)
Để \(\frac{\sqrt{x}+1}{\sqrt{x}-3}\in Z\Rightarrow\frac{4}{\sqrt{x}-3}\in Z\)
\(\Rightarrow\sqrt{x}-3\in\left(1;4;-1;-4\right)\)
\(\Rightarrow\sqrt{x}\in\left(4;7;2;-1\right)\)
\(\Rightarrow\sqrt{x}=4\Leftrightarrow x=2\)
\(4,A=x+\sqrt{x}+1\)
\(A=\left(\sqrt{x}\right)^2+2.\frac{1}{2}.\sqrt{x}+\left(\frac{1}{2}\right)^2+\frac{3}{4}\)
\(A=\left(\sqrt{x}+\frac{1}{2}\right)^2+\frac{3}{4}\)
\(\Rightarrow A\ge\frac{3}{4}.\left(\sqrt{x}+\frac{1}{2}\right)^2\ge0\)
Dấu "=" xảy ra khi :
\(\sqrt{x}+\frac{1}{2}=0\Leftrightarrow\sqrt{x}=-\frac{1}{2}\)
Vậy Min A = 3/4 khi căn x = -1/2
Bài 1
1)
Đkxđ \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)
Ta có \(4-2\sqrt{3}=\left(\sqrt{3}-1\right)^2\)
Khi đó A=\(\frac{\sqrt{3}-1-1}{\sqrt{3}-1+1}=\frac{\sqrt{3}-2}{\sqrt{3}}\)
2) Đề là \(5-2\sqrt{6}\)sẽ hợp lý hơn nha bn
Đkxđ\(\left\{{}\begin{matrix}x\ge0\\\sqrt{x}-\sqrt{2}\ne0\end{matrix}\right.\)
Ta có \(5-2\sqrt{6}=\left(1-\sqrt{6}\right)^2\)
Khi đó
B= \(\frac{1-\sqrt{6}}{1-\sqrt{6}-\sqrt{2}}\)
1)
đk: \(\left\{{}\begin{matrix}x\ge0\\x\ne4\end{matrix}\right.\)
Rgọn
A=\(\frac{x+12}{x-4}+\frac{1}{\sqrt{x}+2}-\frac{4}{\sqrt{x}-2}\)
= \(\frac{x+12+\sqrt{x}-2-\left(4\sqrt{x}+8\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
=\(\frac{x-3\sqrt{x}+2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
=\(\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
=\(\frac{\sqrt{x}-1}{\sqrt{x}+2}\)
2)
B=\(\frac{3\sqrt{x}-1}{\sqrt{x}+2}-\frac{\sqrt{x}+1}{\sqrt{x}-2}+\frac{10\sqrt{x}}{x-4}\) đk \(\left\{{}\begin{matrix}x\ge0\\x\ne4\end{matrix}\right.\)
= \(\frac{\left(3\sqrt{x}-1\right)\left(\sqrt{x}-2\right)-\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)+10\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
= \(\frac{3x-5\sqrt{x}-2-\left(x+3\sqrt{x}+2\right)+10\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
=\(\frac{3x-5\sqrt{x}-2-x-3\sqrt{x}-2+10\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
=\(\frac{2x+2\sqrt{x}-4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
=\(\frac{\left(2x+2\sqrt{x}\right)-\left(4\sqrt{x}+4\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
=\(\frac{2\sqrt{x}\left(\sqrt{x}+2\right)-4\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
=\(\frac{\left(\sqrt{x}+2\right)2\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=2\)
Chúc bn học tốt
Nhớ tích cho mk nhé