Question

\(\sqrt[3]{x+1}=\sqrt{x-3}\)

\(\sqrt[3]{x-2}+\sqrt{X+1}=3\)

Answer 1

Câu 1:

ĐK: \(x\geq 3\)

\(\sqrt[3]{x+1}=\sqrt{x-3}\)

\(\Rightarrow (x+1)^2=(x-3)^3\) (mũ 6 hai vế)

\(\Leftrightarrow x^2+2x+1=x^3-9x^2+27x-27\)

\(\Leftrightarrow x^3-10x^2+25x-28=0\)

\(\Leftrightarrow x^2(x-7)-3x(x-7)+4(x-7)=0\)

\(\Leftrightarrow (x-7)(x^2-3x+4)=0\)

\(\Leftrightarrow (x-7)[(x-\frac{3}{2})^2+\frac{7}{4}]=0\)

Vì \((x-\frac{3}{2})^2+\frac{7}{4}>0, \forall x\) nên $x-7=0$ kéo theo $x=7$

Vậy...........

Answer 2

Câu 2:

Đặt \(\sqrt[3]{x-2}=a; \sqrt{x+1}=b(b\geq 0)\)

Khi đó ta có hệ phương trình:

\(\left\{\begin{matrix} a+b=3\\ b^2-a^3=3\end{matrix}\right.\) \(\Rightarrow \left\{\begin{matrix} b=3-a\\ b^2-a^3=3\end{matrix}\right.\)

\(\Rightarrow (3-a)^2-a^3=3\)

\(\Leftrightarrow a^3+3-(a^2-6a+9)=0\)

\(\Leftrightarrow a^3-a^2+6a-6=0\)

\(\Leftrightarrow a^2(a-1)+6(a-1)=0\Leftrightarrow (a^2+6)(a-1)=0\)

Vì $a^2+6>0$ nên \(a-1=0\Rightarrow a=1\Rightarrow x=a^3+2=3\) (thỏa mãn)

Vậy........