Question

[Mai Phạm Quỳnh]

Giải phương trình sau:

\(1,\sqrt{x-2}-\sqrt{x+1}=\sqrt{2\text{x}-1}-\sqrt{x+3}\)

\(2,x^2-6\text{x}+26=6\sqrt{2\text{x}+1}\)

\(3,\left(\sqrt{x+5}-\sqrt{x-2}\right)\left(1+\sqrt{x^2+7\text{x}+10}\right)=3\)

4,\(\sqrt[3]{x-4}-\sqrt{9-x}=-1\)

5,\(\left(x+1\right)\sqrt{16\text{x}+17}=8\text{x}^2-15\text{x}-23\)

Giúp mình với ạ mình đang cần gấp <3

Answer 1

[Nguyễn Việt Lâm]

Câu 3: đề là \(\sqrt{x+5}-\sqrt{x-2}\) hay \(\sqrt{x+5}-\sqrt{x+2}\)?

Câu 4:

ĐKXĐ: \(x\le9\)

Đặt \(\left\{{}\begin{matrix}\sqrt[3]{x-4}=a\\\sqrt{9-x}=b\end{matrix}\right.\) ta có hệ:

\(\left\{{}\begin{matrix}a-b=-1\\a^3+b^2=5\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}b=a+1\\a^3+b^2=5\end{matrix}\right.\)

\(\Rightarrow a^3+\left(a+1\right)^2=5\)

\(\Leftrightarrow a^3+a^2+2a-4=0\) \(\Rightarrow a=1\)

\(\Rightarrow\sqrt[3]{x-4}=1\Rightarrow x-4=1\Rightarrow x=5\)

5.

ĐKXĐ: \(x\ge-\frac{17}{16}\)

\(\Leftrightarrow8x^2-15x-23-\left(x+1\right)\sqrt{16x+17}=0\)

\(\Leftrightarrow\left(x+1\right)\left(8x-23\right)-\left(x+1\right)\sqrt{16x+17}=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\8x-23=\sqrt{16x+17}\left(1\right)\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow16x+17-2\sqrt{16x+17}-63=0\)

Đặt \(\sqrt{16x+17}=t\ge0\)

\(\Rightarrow t^2-2t-63=0\Rightarrow\left[{}\begin{matrix}t=9\\t=-7\left(l\right)\end{matrix}\right.\)

\(\Rightarrow\sqrt{16x+17}=9\Leftrightarrow x=\frac{32}{3}\)

Answer 2

[Mai Phạm Quỳnh]

mình cần phần 3 4 5 nữa thui ạ