Giup minh bai nay` vs

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01263812493

phuong287 said:
Giải phương trình sau:
[TEX]\sqrt[]{x+2}+\sqrt[]{3-x}=x^3+x^2-4x-1[/TEX]
;):D:)
Bấm để xem đầy đủ nội dung ...
[TEX]\blue: \ Dk: \ -2 \leq x \leq 3[/TEX]

[TEX]\blue pt \leftrightarrow \sqrt{x+2}-2+\sqrt{3-x}-1=x^3+x^2-4x-4[/TEX]

[TEX]\blue \leftrightarrow \frac{x-2}{\sqrt{x+2}+2}+ \frac{2-x}{\sqrt{3-1}+1}=(x-2)(x+1)(x+2)[/TEX]

[TEX]\blue \leftrightarrow \left[x=2\\ \frac{1}{\sqrt{x+2}+2}-\frac{1}{\sqrt{3-x}+1}=(x+1)(x+2)('')[/TEX]

[TEX]('')\blue \leftrightarrow \frac{1}{\sqrt{x+2}+2}-\frac{1}{3}-\frac{1}{\sqrt{3-x}+1}+\frac{1}{3}=(x+1)(x+2)[/TEX]

[TEX]\blue \leftrightarrow \frac{-(1+x)}{3(\sqrt{x+2}+2)(1+\sqrt{x+2})}- \frac{1+x}{3(\sqrt{3-x}+1)(2+\sqrt{3-x})}=(x+1)(x+2)[/TEX]

[TEX]\blue \Rightarrow \left[x=-1 \\ \frac{-1}{3(\sqrt{x+2}+2)(1+\sqrt{x+2})}- \frac{1}{3(\sqrt{3-x}+1)(2+\sqrt{3-x})}=x+2[/TEX]

Dễ thấy do Đk nên VT <0 < VP
Pt có 2 nghiệm x=2; x=-1
Mỏi tay :)|
Bấm để xem đầy đủ nội dung ...
 
0

0915549009

01263812493 said:
[TEX]\blue: \ Dk: \ -2 \leq x \leq 3[/TEX]

[TEX]\blue pt \leftrightarrow \sqrt{x+2}-2+\sqrt{3-x}-1=x^3+x^2-4x-4[/TEX]

[TEX]\blue \leftrightarrow \frac{x-2}{\sqrt{x+2}+2}+ \frac{2-x}{\sqrt{3-1}+1}=(x-2)(x+1)(x+2)[/TEX]

[TEX]\blue \leftrightarrow \left[x=2\\ \frac{1}{\sqrt{x+2}+2}-\frac{1}{\sqrt{3-x}+1}=(x+1)(x+2)('')[/TEX]

[TEX]('')\blue \leftrightarrow \frac{1}{\sqrt{x+2}+2}-\frac{1}{3}-\frac{1}{\sqrt{3-x}+1}+\frac{1}{3}=(x+1)(x+2)[/TEX]

[TEX]\blue \leftrightarrow \frac{-(1+x)}{3(\sqrt{x+2}+2)(1+\sqrt{x+2})}- \frac{1+x}{3(\sqrt{3-x}+1)(2+\sqrt{3-x})}=(x+1)(x+2)[/TEX]

[TEX]\blue \Rightarrow \left[x=-1 \\ \frac{-1}{3(\sqrt{x+2}+2)(1+\sqrt{x+2})}- \frac{1}{3(\sqrt{3-x}+1)(2+\sqrt{3-x})}=x+2[/TEX]

Dễ thấy do Đk nên VT <0 < VP
Pt có 2 nghiệm x=2; x=-1
Bấm để xem đầy đủ nội dung ...
Làm đến bước xét [TEX]x=2[/TEX] thì dừng lại nên ...... :">
[TEX]\sqrt[3]{2x+1} +1 = x^3+3x^2+2x[/TEX]
 
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