Câu hỏi của Lê Quỳnh Chi Phạm

Question

1) Cho biểu thức A= ($\dfrac{3\sqrt{x}}{\sqrt{x}-1}$-$\dfrac{1}{\sqrt{x}+1}$- 3) . $\dfrac{\sqrt{x}+1}{\sqrt{x}+2}$ với x≥0 và x≠1a) rút gọn Ab) tìm x để A<0

Toru · Accepted Answer

$a,A=\left(\dfrac{3\sqrt{x}}{\sqrt{x}-1}-\dfrac{1}{\sqrt{x}+1}-3\right)\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\left(đk:x\ge0;x\ne1\right)$$=\left[\dfrac{3\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\dfrac{\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\dfrac{3\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right]\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}+2}$$=\dfrac{3x+3\sqrt{x}-\sqrt{x}+1-3\left(x-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}+2}$$=\dfrac{3x+2\sqrt{x}+1-3x+3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}$$=\dfrac{2\sqrt{x}+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}$$=\dfrac{2\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}$$=\dfrac{2}{\sqrt{x}-1}$$---$$b,A< 0\Leftrightarrow\dfrac{2}{\sqrt{x}-1}< 0$$\Leftrightarrow\sqrt{x}-1< 0$$\Leftrightarrow\sqrt{x}< 1$$\Leftrightarrow x< 1$Kết hợp với điều kiện của $x$, ta được:$0\le x< 1$Vậy: ...$Toru$Câu trả lời: $a,A=\left(\dfrac{3\sqrt{x}}{\sqrt{x}-1}-\dfrac{1}{\sqrt{x}+1}-3\right)\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\left(đk:x\ge0;x\ne1\right)$$=\left[\dfrac{3\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\dfrac{\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\dfrac{3\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right]\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}+2}$$=\dfrac{3x+3\sqrt{x}-\sqrt{x}+1-3\left(x-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}+2}$$=\dfrac{3x+2\sqrt{x}+1-3x+3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}$$=\dfrac{2\sqrt{x}+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}$$=\dfrac{2\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}$$=\dfrac{2}{\sqrt{x}-1}$$---$$b,A< 0\Leftrightarrow\dfrac{2}{\sqrt{x}-1}< 0$$\Leftrightarrow\sqrt{x}-1< 0$$\Leftrightarrow\sqrt{x}< 1$$\Leftrightarrow x< 1$Kết hợp với điều kiện của $x$, ta được:$0\le x< 1$Vậy: ...$Toru$

HT.Phong (9A5) · Answer

a) $A=\left(\dfrac{3\sqrt{x}}{\sqrt{x}-1}-\dfrac{1}{\sqrt{x}+1}-3\right)\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}+2}$$A=\left[\dfrac{3\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\dfrac{\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\dfrac{3\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}+1\right)}\right]\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}+2}$$A=\dfrac{3x+3\sqrt{x}-\sqrt{x}+1-3x+3}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}+2}$$A=\dfrac{2\sqrt{x}+4}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}+2}$$A=\dfrac{2\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}+2}$$A=\dfrac{2}{\sqrt{x}-1}$b) $A< 0$ khi$\dfrac{2}{\sqrt{x}-1}< 0\Leftrightarrow\sqrt{x}-1< 0$$\Leftrightarrow\sqrt{x}< 1$$\Leftrightarrow x< 1$Kết hợp với đk:$0\le x< 1$Câu trả lời: a) $A=\left(\dfrac{3\sqrt{x}}{\sqrt{x}-1}-\dfrac{1}{\sqrt{x}+1}-3\right)\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}+2}$$A=\left[\dfrac{3\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\dfrac{\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\dfrac{3\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}+1\right)}\right]\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}+2}$$A=\dfrac{3x+3\sqrt{x}-\sqrt{x}+1-3x+3}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}+2}$$A=\dfrac{2\sqrt{x}+4}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}+2}$$A=\dfrac{2\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}+2}$$A=\dfrac{2}{\sqrt{x}-1}$b) $A< 0$ khi$\dfrac{2}{\sqrt{x}-1}< 0\Leftrightarrow\sqrt{x}-1< 0$$\Leftrightarrow\sqrt{x}< 1$$\Leftrightarrow x< 1$Kết hợp với đk:$0\le x< 1$

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