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a: \(=\dfrac{x\sqrt{x}+26\sqrt{x}-19-2x-6\sqrt{x}+x-4\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{x\sqrt{x}-x+16\sqrt{x}-16}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}=\dfrac{x+16}{\sqrt{x}+3}\)
b: Để A=4 thì \(x+16-4\sqrt{x}-12=0\)
=>x=4
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a.\(A=\dfrac{x^2-4x+4}{x^3-2x^2-\left(4x-8\right)}=\dfrac{\left(x-2\right)^2}{x^2\left(x-2\right)-4\left(x-2\right)}=\dfrac{\left(x-2\right)^2}{\left(x^2-4\right)\left(x-2\right)}=\dfrac{x-2}{\left(x-2\right)\left(x+2\right)}=\dfrac{1}{x+2}\)
\(A=\dfrac{\left(x-2\right)^2}{x^2\left(x-2\right)-4\left(x-2\right)}\left(x\ne\pm2\right)\\ A=\dfrac{\left(x-2\right)^2}{\left(x-2\right)^2\left(x+2\right)}=\dfrac{1}{x+2}\\ B=\dfrac{x+2-x+\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\cdot\dfrac{4\sqrt{x}}{3}\left(x>0\right)\\ B=\dfrac{4\sqrt{x}\left(\sqrt{x}+1\right)}{3\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}=\dfrac{4\sqrt{x}}{3\left(x-\sqrt{x}+1\right)}\)
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Sửa đề: \(C=\dfrac{x+4}{\sqrt{x}}+\dfrac{x\sqrt{x}-8}{\sqrt{x}\left(\sqrt{x}-2\right)}-\dfrac{x\sqrt{x}+8}{\sqrt{x}\left(\sqrt{x}+2\right)}\)
\(=\dfrac{x+4+x+2\sqrt{x}+4-x+2\sqrt{x}-4}{\sqrt{x}}\)
\(=\dfrac{x+4\sqrt{x}+4}{\sqrt{x}}=\dfrac{\left(\sqrt{x}+2\right)^2}{\sqrt{x}}\)
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Tử số của phân số đầu phải là \(\sqrt{x}+2\) chứ không phải \(\sqrt{x+2}\), vì cái \(\sqrt{x}+2\) nó mới logic để rút gọn: )
\(Q=\left(\dfrac{\left(\sqrt{x}+2\right)^2}{\sqrt{x}^3+8}-\dfrac{x-\sqrt{x}}{\sqrt{x}^3+8}\right)\left(\dfrac{5x-10\sqrt{x}+20}{5\sqrt{x}+4}\right)\\ =\left(\dfrac{x+4\sqrt{x}+4-x+\sqrt{x}}{\sqrt{x}^3+8}\right)\left(\dfrac{5x-10\sqrt{x}+20}{5\sqrt{x}+4}\right)\\ =\dfrac{\left(5\sqrt{x}+4\right).5.\left(x-2\sqrt{x}+4\right)}{\left(\sqrt{x}+2\right)\left(x-2\sqrt{x}+4\right)\left(5\sqrt{x}+4\right)}\\ =\dfrac{5}{\sqrt{x}+2}\)
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Với x >= 0 ; x khác 16
\(B=\dfrac{2x+8+x-4\sqrt{x}-8\sqrt{x}-8}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-4\right)}=\dfrac{3x-12\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-4\right)}=\dfrac{3\sqrt{x}}{\sqrt{x}+1}\)
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ĐK: \(x\ne\pm3\)
Khi đó:
\(C=\dfrac{2\left(x-3\right)}{x^2-9}+\dfrac{1\left(x+3\right)}{x^2-9}-\dfrac{8}{x^2-9}\\ =\dfrac{2x-6}{x^2-9}+\dfrac{x+3}{x^2-9}-\dfrac{8}{x^2-9}\\ =\dfrac{2x-6+x+3-8}{x^2-9}\\ =\dfrac{3x-11}{x^2-9}\)
Thế x = 4 vào C được:
\(C=\dfrac{3.4-11}{4^2-9}=\dfrac{12-11}{16-9}=\dfrac{1}{7}\)
\(\dfrac{8}{\sqrt{x}-2}-\dfrac{\sqrt{x}+26}{x-4}\\ đk:x\ne4;x>0\\ =\dfrac{8\left(\sqrt{x}+2\right)-\sqrt{x}-26}{x-4}\\ =\dfrac{8\sqrt{x}+16-\sqrt{x}-26}{x-4}\\ =\dfrac{7\sqrt{x}-10}{x-4}\)