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Ta có: \(P=\left(\dfrac{\sqrt{x}-1}{3\sqrt{x}-1}-\dfrac{1}{3\sqrt{x}+1}+\dfrac{8\sqrt{x}}{9x-1}\right):\left(1-\dfrac{3\sqrt{x}-2}{3\sqrt{x}+1}\right)\)
\(=\dfrac{3x+\sqrt{x}-3\sqrt{x}-1-3\sqrt{x}+1+8\sqrt{x}}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}:\dfrac{3\sqrt{x}+1-3\sqrt{x}+2}{3\sqrt{x}+1}\)
\(=\dfrac{3x+3\sqrt{x}}{3\sqrt{x}-1}\cdot\dfrac{1}{3}\)
\(=\dfrac{x+\sqrt{x}}{3\sqrt{x}-1}\)
\(\left(\dfrac{\sqrt{x}-1}{3\sqrt{x}-1}-\dfrac{1}{3\sqrt{x}+1}+\dfrac{8\sqrt{x}}{9x-1}\right):\left(1-\dfrac{3\sqrt{x}-2}{3\sqrt{x}+1}\right)=\dfrac{\left(\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)-\left(3\sqrt{x}-1\right)+8\sqrt{x}}{9x-1}:\dfrac{3\sqrt{x}+1-3\sqrt{x}+2}{3\sqrt{x}+1}=\dfrac{3x+3\sqrt{x}-1}{9x-1}.\dfrac{3\sqrt{x}+1}{3}=\dfrac{3x+3\sqrt{x}-1}{9\sqrt{x}-3}\)
a) ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne\dfrac{1}{9}\end{matrix}\right.\)
Ta có: \(P=\left(\dfrac{\sqrt{x}-1}{3\sqrt{x}-1}-\dfrac{1}{3\sqrt{x}+1}+\dfrac{5\sqrt{x}}{9x-1}\right):\left(1-\dfrac{3\sqrt{x}-2}{3\sqrt{x}+1}\right)\)
\(=\dfrac{\left(\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)-3\sqrt{x}+1+5\sqrt{x}}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}:\left(\dfrac{3\sqrt{x}+1-3\sqrt{x}+2}{3\sqrt{x}+1}\right)\)
\(=\dfrac{3x+\sqrt{x}-3\sqrt{x}-1-3\sqrt{x}+1+5\sqrt{x}}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}\cdot\dfrac{3\sqrt{x}+1}{3}\)
\(=\dfrac{3x}{3\sqrt{x}-1}\cdot\dfrac{1}{3}\)
\(=\dfrac{x}{3\sqrt{x}-1}\)
b) Ta có: \(9x^2-10x+1=0\)
\(\Leftrightarrow\left(9x-1\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{9}\left(loại\right)\\x=1\left(nhận\right)\end{matrix}\right.\)
Thay x=1 vào P, ta được:
\(P=\dfrac{1}{3-1}=\dfrac{1}{2}\)
c) Thay \(x=8-2\sqrt{7}\) vào P, ta được:
\(P=\dfrac{8-2\sqrt{7}}{3\left(\sqrt{7}-1\right)-1}=\dfrac{8-2\sqrt{7}}{3\sqrt{7}-4}\)
\(=\dfrac{-10+16\sqrt{7}}{47}\)
a)
\(P=\dfrac{\left(\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)-\left(3\sqrt{x}-4\right)+5\sqrt{x}}{\left(3\sqrt{x}+1\right)\left(3\sqrt{x}-1\right)}.\dfrac{3\sqrt{x}+1}{3}\)
\(P=\dfrac{3x-2\sqrt{x}-1-3\sqrt{x}+4+5\sqrt{x}}{\left(3\sqrt{x}+1\right)\left(3\sqrt{x}-1\right)}.\dfrac{3\sqrt{x}+1}{3}\)
\(P=\dfrac{3\left(x+1\right)}{\left(3\sqrt{x}+1\right)\left(3\sqrt{x}-1\right)}.\dfrac{3\sqrt{x}+1}{3}\)
\(P=\dfrac{x+1}{3\sqrt{x}-1}\)
\(a,\)
\(=\left(\dfrac{\sqrt{x}-1}{3\sqrt{x}-1}-\dfrac{1}{3\sqrt{x}+1}+\dfrac{8\sqrt{x}}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}\right):\left(\dfrac{3\sqrt{x}+1-3\sqrt{x}+2}{3\sqrt{x}+1}\right)\)
\(=\left(\dfrac{\left(\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)-3\sqrt{x}+1+8\sqrt{x}}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}\right):\left(\dfrac{3}{3\sqrt{x}+1}\right)\)
\(=\dfrac{3x+\sqrt{x}-3\sqrt{x}-1-3\sqrt{x}+1+8\sqrt{x}}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}.\dfrac{3\sqrt{x}+1}{3}\)
\(=\dfrac{3\sqrt{x}+3x}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}.\dfrac{3\sqrt{x}+1}{3}\)
\(=\dfrac{3\sqrt{x}\left(\sqrt{x}+1\right)}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}.\dfrac{3\sqrt{x}+1}{3}\)
\(=\dfrac{3\sqrt{x}+1}{3\sqrt{x}-1}\)
Vậy \(P=\dfrac{3\sqrt{x}+1}{3\sqrt{x}-1}\)
\(b,\)Thay \(P=\dfrac{6}{5}\) vào pt, ta có :
\(\dfrac{3\sqrt{x}+1}{3\sqrt{x}-1}=\dfrac{6}{5}\)
\(\Leftrightarrow5\left(3\sqrt{x}+1\right)=6\left(3\sqrt{x}-1\right)\)
\(\Leftrightarrow15\sqrt{x}+5-18\sqrt{x}+6=0\)
\(\Leftrightarrow-3\sqrt{x}+11=0\)
\(\Leftrightarrow-3\sqrt{x}=-11\)
\(\Leftrightarrow\sqrt{x}=\dfrac{11}{3}\)
\(\Leftrightarrow x=\left(\dfrac{11}{3}\right)^2\)
\(\Leftrightarrow x=\dfrac{121}{9}\)
Vậy \(x=\dfrac{121}{9}\) thì \(P=\dfrac{6}{5}\)
Bài 1:
a) \(\dfrac{a+\sqrt{a}}{\sqrt{a}}=\sqrt{a}+1\)
b) \(\dfrac{\sqrt{\left(x-3\right)^2}}{3-x}=\dfrac{\left|x-3\right|}{3-x}=\pm1\)
Bài 2:
a) \(\dfrac{\sqrt{9x^2-6x+1}}{9x^2-1}=\dfrac{\left|3x-1\right|}{\left(3x-1\right)\left(3x+1\right)}=\pm\dfrac{1}{3x+1}\)
b) \(4-x-\sqrt{x^2-4x+4}=4-x-\left|x-2\right|=\left[{}\begin{matrix}6-2x\left(x\ge2\right)\\2\left(x< 2\right)\end{matrix}\right.\)
1) \(\dfrac{1}{\sqrt{3}+1}+\dfrac{1}{\sqrt{3}-1}-2\sqrt{3}=\dfrac{\sqrt{3}-1+\sqrt{3}+1}{3-1}-2\sqrt{3}=\sqrt{3}-2\sqrt{3}=-\sqrt{3}\)
+) \(ĐKXĐ:\left\{{}\begin{matrix}x>0\\x\ne1\\x\ne4\end{matrix}\right.\)
\(P=\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{1}{\sqrt{x}}\right):\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-2}-\dfrac{\sqrt{x}+2}{\sqrt{x}-1}\right)\)
\(P=\dfrac{\sqrt{x}-\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}:\dfrac{x-1-x+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\)
\(P=\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}{3}\)
\(P=\dfrac{\sqrt{x}-2}{3\sqrt{x}}\)
2) \(\sqrt{3-2\sqrt{2}}+\dfrac{1}{\sqrt{2}-1}=\sqrt{2}-1+\sqrt{2}+1=2\sqrt{2}\)
+) \(ĐKXĐ:\left\{{}\begin{matrix}a>0\\a\ne4\end{matrix}\right.\)
\(M=\left(\dfrac{\sqrt{a}}{\sqrt{a}-2}+\dfrac{\sqrt{a}}{\sqrt{a}+2}\right)\cdot\dfrac{a-4}{\sqrt{4a}}\)
\(M=\dfrac{a+2\sqrt{a}+a-2\sqrt{a}}{a-4}\cdot\dfrac{a-4}{2\sqrt{a}}\)
\(M=\dfrac{2a}{2\sqrt{a}}=\sqrt{a}\)
+) \(ĐKXĐ:\left\{{}\begin{matrix}x\ge0\\x\ne4\end{matrix}\right.\)
\(N=\left(1-\dfrac{\sqrt{x}}{\sqrt{x}+1}\right):\left(\dfrac{\sqrt{x}+2}{\sqrt{x}+3}+\dfrac{\sqrt{x}-3}{2-\sqrt{x}}+\dfrac{\sqrt{x}-2}{x+\sqrt{x}-6}\right)\)
\(N=\dfrac{\sqrt{x}+1-\sqrt{x}}{\sqrt{x}+1}:\left(\dfrac{\sqrt{x}+2}{\sqrt{x}+3}-\dfrac{\sqrt{x}-3}{\sqrt{x}-2}+\dfrac{\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\right)\)
\(N=\dfrac{1}{\sqrt{x}+1}:\dfrac{x-4-x+9+\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\)
\(N=\dfrac{1}{\sqrt{x}+1}\cdot\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}{\sqrt{x}+3}\)
\(N=\dfrac{\sqrt{x}-2}{\sqrt{x}+1}\)
+) \(ĐKXĐ:\left\{{}\begin{matrix}x\ge0\\x\ne9\\x\ne4\end{matrix}\right.\)
\(Q=\left(1-\dfrac{\sqrt{x}}{\sqrt{x}+1}\right):\left(\dfrac{\sqrt{x}+2}{\sqrt{x}-3}+\dfrac{\sqrt{x}-8}{x-5\sqrt{x}+6}+\dfrac{\sqrt{x}+3}{2-\sqrt{x}}\right)\)
\(Q=\dfrac{\sqrt{x}+1-\sqrt{x}}{\sqrt{x}+1}:\left(\dfrac{\sqrt{x}+2}{\sqrt{x}-3}+\dfrac{\sqrt{x}-8}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}-\dfrac{\sqrt{x}+3}{\sqrt{x}-2}\right)\)
\(Q=\dfrac{1}{\sqrt{x}+1}:\dfrac{x-4+\sqrt{x}-8-x+9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\)
\(Q=\dfrac{1}{\sqrt{x}+1}\cdot\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}{\sqrt{x}-3}\)
\(Q=\dfrac{\sqrt{x}-2}{\sqrt{x}+1}\)
p/s: sorry tại n' câu wa nên mình ko làm chi tiết đc =(( lần sau nhớ chia các câu ra cho dễ nhìn hơn nha, đánh hơi mỏi tay :'( có j ko hỉu cmt dưới nha
Câu 1:
Sửa đề: \(B=\left(\dfrac{x}{x+3\sqrt{x}}+\dfrac{1}{\sqrt{x}+3}\right):\left(1-\dfrac{2}{\sqrt{x}}+\dfrac{6}{x+3\sqrt{x}}\right)\)
Ta có: \(B=\left(\dfrac{x}{x+3\sqrt{x}}+\dfrac{1}{\sqrt{x}+3}\right):\left(1-\dfrac{2}{\sqrt{x}}+\dfrac{6}{x+3\sqrt{x}}\right)\)
\(=\left(\dfrac{x}{\sqrt{x}\left(\sqrt{x}+3\right)}+\dfrac{1}{\sqrt{x}+3}\right):\left(\dfrac{x+3\sqrt{x}-2\left(\sqrt{x}+3\right)+6}{\sqrt{x}\left(\sqrt{x}+3\right)}\right)\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}+3}:\dfrac{x+3\sqrt{x}-2\sqrt{x}-6+6}{\sqrt{x}\left(\sqrt{x}+3\right)}\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}+3}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)}{x+\sqrt{x}}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}=1\)
Câu 3:
Ta có: \(Q=\left(\dfrac{a}{a-2\sqrt{a}}+\dfrac{a}{\sqrt{a}-2}\right):\dfrac{\sqrt{a}+1}{a-4\sqrt{a}+4}\)
\(=\left(\dfrac{a}{\sqrt{a}\left(\sqrt{a}-2\right)}+\dfrac{a}{\sqrt{a}-2}\right):\dfrac{\sqrt{a}+1}{\left(\sqrt{a}-2\right)^2}\)
\(=\dfrac{a+\sqrt{a}}{\sqrt{a}-2}\cdot\dfrac{\sqrt{a}-2}{\sqrt{a}+1}\cdot\dfrac{\sqrt{a}-2}{1}\)
\(=\sqrt{a}\left(\sqrt{a}-2\right)\)
\(=a-2\sqrt{a}\)
a) Đk: \(x>0;x\ne9;x\ne25\)
Đặt \(A=\left(\dfrac{\sqrt{x}}{3+\sqrt{x}}+\dfrac{2x}{9-x}\right):\left(\dfrac{\sqrt{x}-1}{x-3\sqrt{x}}-\dfrac{2}{\sqrt{x}}\right)\)
\(=\left[\dfrac{\sqrt{x}\left(3-\sqrt{x}\right)}{\left(3+\sqrt{x}\right)\left(3-\sqrt{x}\right)}+\dfrac{2x}{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}\right]\)\(:\left[\dfrac{\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-3\right)}-\dfrac{2\left(\sqrt{x}-3\right)}{\sqrt{x}\left(\sqrt{x}-3\right)}\right]\)
\(=\dfrac{\sqrt{x}\left(3-\sqrt{x}\right)+2x}{\left(3+\sqrt{x}\right)\left(3-\sqrt{x}\right)}:\dfrac{\sqrt{x}-1-2\left(\sqrt{x}-3\right)}{\sqrt{x}\left(\sqrt{x}-3\right)}\)
\(=\dfrac{3\sqrt{x}+x}{\left(3+\sqrt{x}\right)\left(3-\sqrt{x}\right)}:\dfrac{-\sqrt{x}+5}{\sqrt{x}\left(\sqrt{x}-3\right)}\)
\(=\dfrac{\sqrt{x}\left(3+\sqrt{x}\right)}{\left(3+\sqrt{x}\right)\left(3-\sqrt{x}\right)}.\dfrac{\sqrt{x}\left(3-\sqrt{x}\right)}{\sqrt{x}-5}\)
\(=\dfrac{x}{\sqrt{x}-5}\)
b) Đk: \(x\ge0;x\ne1;x\ne25\)
Biểu thức
\(=\left[\dfrac{\sqrt{x}-2}{\sqrt{x}+5}+\dfrac{\sqrt{x}}{\sqrt{x}-5}-\dfrac{x+9}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\right]:\dfrac{1-\sqrt{x}}{5+\sqrt{x}}\)
\(=\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-5\right)+\sqrt{x}\left(\sqrt{x}+5\right)-x-9}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}.\dfrac{\sqrt{x}+5}{1-\sqrt{x}}\)
\(=\dfrac{x-7\sqrt{x}+10+x+5\sqrt{x}-x-9}{\left(\sqrt{x}-5\right)\left(1-\sqrt{x}\right)}\)
\(=\dfrac{x-2\sqrt{x}+1}{\left(\sqrt{x}-5\right)\left(1-\sqrt{x}\right)}\)\(=\dfrac{\left(1-\sqrt{x}\right)^2}{\left(\sqrt{x}-5\right)\left(1-\sqrt{x}\right)}=\dfrac{1-\sqrt{x}}{\sqrt{x}-5}\)
a) Ta có: \(A=3\sqrt{2x}-5\sqrt{8x}+7\sqrt{18x}+30\)
\(=3\sqrt{2x}-10\sqrt{2x}+21\sqrt{2x}+30\)
\(=14\sqrt{2x}+30\)
b) Ta có: \(B=4\sqrt{\dfrac{25x}{4}}-\dfrac{8}{3}\sqrt{\dfrac{9x}{4}}-\dfrac{4}{3x}\cdot\sqrt{\dfrac{9x^3}{64}}\)
\(=4\cdot\dfrac{5\sqrt{x}}{2}-\dfrac{8}{3}\cdot\dfrac{3\sqrt{x}}{2}-\dfrac{4}{3x}\cdot\dfrac{3x\sqrt{x}}{8}\)
\(=10\sqrt{x}-4\sqrt{x}-\dfrac{1}{2}\sqrt{x}\)
\(=\dfrac{11}{2}\sqrt{x}\)
c) Ta có: \(\dfrac{y}{2}+\dfrac{3}{4}\sqrt{9y^2-6y+1}-\dfrac{3}{2}\)
\(=\dfrac{1}{2}y+\dfrac{3}{4}\left(1-3y\right)-\dfrac{3}{2}\)
\(=\dfrac{1}{2}y+\dfrac{3}{4}-\dfrac{9}{4}y-\dfrac{3}{2}\)
\(=-\dfrac{7}{4}y-\dfrac{3}{4}\)
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