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\(A=4\sqrt{x}-\frac{x+6\sqrt{x}+9}{x-9}\)
\(=4\sqrt{x}-\frac{\left(\sqrt{x}+3\right)^2}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=4\sqrt{x}-\frac{\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)}\)
\(=\frac{4\sqrt{x}\left(\sqrt{x}-3\right)}{\sqrt{x}-3}-\frac{\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)}\)
\(=\frac{4x-12\sqrt{x}-\sqrt{x}-3}{\sqrt{x}-3}\)
\(=\frac{4x-13\sqrt{x}-3}{\sqrt{x}-3}\)
C.Tham khảo ở dây:Câu hỏi của Đặng Phương Thảo - Toán lớp 9 - Học toán với OnlineMath
\(B=\frac{5\sqrt{x}-\left(x-10\sqrt{x}+25\right)\left(\sqrt{x}+5\right)}{x-25}\)
\(=\frac{5\sqrt{x}-\left(\sqrt{x}-5\right)^2\left(\sqrt{x}+5\right)}{x-25}\)
\(=\frac{5\sqrt{x}-\left(\sqrt{x}-5\right)\left(x-25\right)}{x-25}\)
\(=\frac{5\sqrt{x}-\left(x\sqrt{x}-25\sqrt{x}-5x+125\right)}{x-25}\)
\(=\frac{5\sqrt{x}-x\sqrt{x}+25\sqrt{x}+5x-125}{x-25}\)
\(=\frac{-x\sqrt{x}+30\sqrt{x}+5x-125}{x-25}\)
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\(A=3\left(x+2\sqrt{x}\right)-\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)\)
\(=3x+6\sqrt{x}-\left(x-1\right)\)
\(=3x+6\sqrt{x}-x+1\)
\(=2x+6\sqrt{x}+1\)
\(B=\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)-2\left(\sqrt{x}-1\right)^2\)
\(=x+3\sqrt{x}+\sqrt{x}+3-2\left(x-2\sqrt{x}+1\right)\)
\(=x+4\sqrt{x}+3-2x+4\sqrt{x}-2\)
\(=-x+8\sqrt{x}+1\)
\(C=3x-3\sqrt{x}-2+\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)\)
\(=3x-3\sqrt{x}-2+\left(\sqrt{x^2}-1\right)\)
\(=3x-3\sqrt{x}-2+x-1\)
\(=4x-3\sqrt{x}-3\)
\(D=\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)-\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)\)
\(=x-9-\left(2x-3\sqrt{x}-2\right)\)
\(=x-9-2x+3\sqrt{x}+2\)
\(=-x+3\sqrt{x}-7\)
\(E=\left(\sqrt{x}+4\right)\left(\sqrt{x}-4\right)-2\left(2\sqrt{x}-1\right)\left(\sqrt{x}+2\right)\)
\(=\sqrt{x^2}-2^2-2\left(2x+4\sqrt{x}-\sqrt{x}-2\right)\)
\(=x-4-2\left(2x+3\sqrt{x}-2\right)\)
\(=x-4-4x-6\sqrt{x}+4\)
\(=-3-6\sqrt{x}\)
1. ĐKXĐ: $x>0; x\neq 9$
\(A=\frac{\sqrt{x}+3+\sqrt{x}-3}{(\sqrt{x}-3)(\sqrt{x}+3)}.\frac{\sqrt{x}-3}{\sqrt{x}}=\frac{2\sqrt{x}}{(\sqrt{x}-3)(\sqrt{x}+3)}.\frac{\sqrt{x}-3}{\sqrt{x}}=\frac{2}{\sqrt{x}+3}\)
2. ĐKXĐ: $x\geq 0; x\neq 4$
\(B=\left[\frac{\sqrt{x}(\sqrt{x}+2)+\sqrt{x}-2}{(\sqrt{x}-2)(\sqrt{x}+2)}+\frac{6-7\sqrt{x}}{(\sqrt{x}-2)(\sqrt{x}+2)}\right](\sqrt{x}+2)\)
\(=\frac{x+3\sqrt{x}-2+6-7\sqrt{x}}{(\sqrt{x}-2)(\sqrt{x}+2)}.(\sqrt{x}+2)=\frac{x-4\sqrt{x}+4}{\sqrt{x}-2}=\frac{(\sqrt{x}-2)^2}{\sqrt{x}-2}=\sqrt{x}-2\)
a) \(\sqrt{36\left(x-5\right)^2}\left(x\ge5\right)=6\left|x-5\right|=6\left(x-5\right)=6x-30\)
b) \(\sqrt{\dfrac{1}{4}\left(1-x\right)^2}\left(x>1\right)=\dfrac{1}{2}\left|1-x\right|=\dfrac{1}{2}\left(x-1\right)=\dfrac{1}{2}x-\dfrac{1}{2}\)
c) \(\sqrt{x^2\left(2x-4\right)^2}\left(x\ge2\right)=\left|x\left(2x-4\right)\right|=x\left(2x-4\right)=2x^2-4x\)
d) \(\dfrac{1}{x}\sqrt{x^2\left(1+x\right)^2}\left(x< -1\right)=\dfrac{1}{x}\left|x\left(1+x\right)\right|=\dfrac{1}{x}x\left(1+x\right)=1+x\)
a.\(A=\dfrac{x^2-4x+4}{x^3-2x^2-\left(4x-8\right)}=\dfrac{\left(x-2\right)^2}{x^2\left(x-2\right)-4\left(x-2\right)}=\dfrac{\left(x-2\right)^2}{\left(x^2-4\right)\left(x-2\right)}=\dfrac{x-2}{\left(x-2\right)\left(x+2\right)}=\dfrac{1}{x+2}\)
\(A=\dfrac{\left(x-2\right)^2}{x^2\left(x-2\right)-4\left(x-2\right)}\left(x\ne\pm2\right)\\ A=\dfrac{\left(x-2\right)^2}{\left(x-2\right)^2\left(x+2\right)}=\dfrac{1}{x+2}\\ B=\dfrac{x+2-x+\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\cdot\dfrac{4\sqrt{x}}{3}\left(x>0\right)\\ B=\dfrac{4\sqrt{x}\left(\sqrt{x}+1\right)}{3\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}=\dfrac{4\sqrt{x}}{3\left(x-\sqrt{x}+1\right)}\)
a) Ta có: \(A=\left(\dfrac{1}{\sqrt{a}+2}+\dfrac{1}{\sqrt{a}-2}\right):\dfrac{\sqrt{a}}{a-4}\)
\(=\dfrac{\sqrt{a}-2+\sqrt{a}+2}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}\cdot\dfrac{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}{\sqrt{a}}\)
=2
b) Ta có: \(B=\left(\dfrac{4x}{\sqrt{x}-1}-\dfrac{\sqrt{x}-2}{x-3\sqrt{x}+2}\right)\cdot\dfrac{\sqrt{x}-1}{x^2}\)
\(=\dfrac{4x-1}{\sqrt{x}-1}\cdot\dfrac{\sqrt{x}-1}{x^2}\)
\(=\dfrac{4x-1}{x^2}\)
a: Ta có: \(A=\left(\dfrac{1}{\sqrt{x}+2}+\dfrac{1}{\sqrt{x}-2}\right)\cdot\dfrac{x-4}{3\sqrt{x}}\)
\(=\dfrac{\sqrt{x}-2+\sqrt{x}+2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\cdot\dfrac{x-4}{3\sqrt{x}}\)
\(=\dfrac{2}{3}\)
\(a,C=\dfrac{2x^2-x-x-1+2-x^2}{x-1}\left(x\ne1\right)\\ C=\dfrac{x^2-2x+1}{x-1}=\dfrac{\left(x-1\right)^2}{x-1}=x-1\\ b,D=\dfrac{1+\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{\left(\sqrt{a}-1\right)^2}{\sqrt{a}+1}\left(a>0;a\ne1\right)\\ D=\dfrac{\sqrt{a}-1}{\sqrt{a}}\)
Có
ĐKXĐ: \(x>0;x\ne1\)
\(A=\left(\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)^2\left(\sqrt{x}+1\right)}-\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)^2\left(\sqrt{x}+1\right)}\right).\dfrac{\left(x-1\right)^2}{4x}\)
\(=\left(\dfrac{2\sqrt{x}}{\left(\sqrt{x}-1\right)^2\left(\sqrt{x}+1\right)}\right).\dfrac{\left(\sqrt{x}-1\right)^2\left(\sqrt{x}+1\right)^2}{4x}\)
\(=\dfrac{\sqrt{x}+1}{2\sqrt{x}}\)
b.
\(\left|x-5\right|=4\Rightarrow\left[{}\begin{matrix}x-5=4\\x-5=-4\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=9\\x=1\left(loại\right)\end{matrix}\right.\)
\(\Rightarrow A=\dfrac{\sqrt{9}+1}{2\sqrt{9}}=\dfrac{2}{3}\)
a) \(A=4\sqrt{x^2+1}-2\sqrt{16\left(x^2+1\right)}+5\sqrt{25\left(x^2+1\right).}\)
\(=4\sqrt{x^2+1}-2.4\sqrt{x^2+1}+5.5\sqrt{x^2+1}\)
\(=4\sqrt{x^2+1}-8\sqrt{x^2+1}+25\sqrt{x^2+1}\)
\(=\left(4-8+25\right)\sqrt{x^2+1}\)
\(=21\sqrt{x^2+1}\)
b) \(B=\frac{2}{x+y}\sqrt{\frac{3\left(x+y\right)^2}{4}}\)
\(B=\frac{2}{x+y}.\frac{\sqrt{3}\left(x+y\right)}{2}\)
\(B=\frac{\sqrt{3}\left(x+y\right)}{x+y}\)
\(B=\sqrt{3}\)