Question

Cho A=\(\frac{2\sqrt{x}-9}{x-5\sqrt{x}+6}+\frac{2\sqrt{x}+1}{\sqrt{x}-3}+\frac{\sqrt{x}+3}{2-\sqrt{x}}\)

Tìm x để A có giá trị âm

 

Answer 1

\(A=\frac{2\sqrt{x}-9}{x-5\sqrt{x}+6}+\frac{2\sqrt{x}+1}{\sqrt{x}-3}+\frac{\sqrt{x}+3}{2-\sqrt{x}}\)(ĐKXĐ: \(x\ge0;x\ne4;x\ne9\))

\(A=\frac{2\sqrt{x}-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}+\frac{2\sqrt{x}+1}{\sqrt{x}-3}+\frac{\sqrt{x}+3}{2-\sqrt{x}}\)

\(A=\frac{2\sqrt{x}-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}+\frac{2x-3\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\frac{x-9}{\left(\sqrt{x}-2\right)\cdot\left(\sqrt{x}-3\right)}\)

\(A=\frac{x-\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}=\frac{x+\sqrt{x}-2\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)

\(A=\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}=\frac{\sqrt{x}+1}{\sqrt{x}-3}\)

\(A< 0\Leftrightarrow\frac{\sqrt{x}+1}{\sqrt{x}-3}< 0\Leftrightarrow\orbr{\begin{cases}\sqrt{x}+1< 0\\\sqrt{x}-3< 0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x< 1\\x< 9\end{cases}}\)

Vậy với \(x< 1\)thì  \(A\)nhận giá trị âm.

Answer 2

Nhưng \(x< 1\) lại không thỏa mãn ĐKXĐ của A

Vậy thì các giá trị của x để A nhận giá trị âm phải là \(0\le x< 9\)và x khác 4

Bạn sửa đi nhé !

Answer 3

A=x−5x​+62x​−9​+x​−32x​+1​+2−x​x​+3​(ĐKXĐ: x\ge0;x\ne4;x\ne9x≥0;x̸​=4;x̸​=9)

A=\frac{2\sqrt{x}-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}+\frac{2\sqrt{x}+1}{\sqrt{x}-3}+\frac{\sqrt{x}+3}{2-\sqrt{x}}A=(x​−2)(x​−3)2x​−9​+x​−32x​+1​+2−x​x​+3​

A=\frac{2\sqrt{x}-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}+\frac{2x-3\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\frac{x-9}{\left(\sqrt{x}-2\right)\cdot\left(\sqrt{x}-3\right)}A=(x​−2)(x​−3)2x​−9​+(x​−2)(x​−3)2x−3x​−2​−(x​−2)⋅(x​−3)x−9​

A=\frac{x-\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}=\frac{x+\sqrt{x}-2\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}A=(x​−2)(x​−3)x−x​−2​=(x​−2)(x​−3)x+x​−2x​−2​

A=\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}=\frac{\sqrt{x}+1}{\sqrt{x}-3}A=(x​−2)(x​−3)(x​+1)(x​−2)​=x​−3x​+1​

\(A< 0\Leftrightarrow\frac{\sqrt{x}+1}{\sqrt{x}-3}< 0\Leftrightarrow\orbr{\begin{cases}\sqrt{x}+1< 0\\\sqrt{x}-3< 0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x< 1\\x< 9\end{cases}}\)

Vậy với x&lt; 1x<1thì  AAnhận giá trị âm.