Question

Giải phương trình sau

1) x³+1=2\(\sqrt[3]{2x-1}\)

2) 2(x²+2)=5\(\sqrt{x^3+1}\)

3) \(\sqrt[3]{\left(3x+1\right)^2}\)+\(\sqrt[3]{\left(3x-1\right)^2}\)+\(\sqrt[3]{9x^2-1}\)=1

4)2( x²-3x+2) =3\(\sqrt{x^3+8}\)

5)\(\sqrt{\frac{1}{2}-x}\)+\(\sqrt{\frac{1}{2}+x}\)=1

6) \(\sqrt{x-1}\)+\(\sqrt{x^3+x^2+x+1}\)=1+\(\sqrt{x^4-1}\)

7) (2x+7)\(\sqrt{2x+7}\)=x²+9x+7

8) \(\sqrt[3]{x+10}\)+\(\sqrt[3]{17-x}\)=3

Answer 1

1.

Đặt \(\sqrt[3]{2x-1}=a\Rightarrow a^3+1=2x\)

Ta được hệ: \(\left\{{}\begin{matrix}x^3+1=2a\\a^3+1=2x\end{matrix}\right.\)

\(\Rightarrow x^3-a^3=2\left(a-x\right)\)

\(\Leftrightarrow\left(x-a\right)\left(x^2+a^2+ax\right)+2\left(x-a\right)=0\)

\(\Leftrightarrow\left(x-a\right)\left(x^2+a^2+ax+2\right)=0\)

\(\Leftrightarrow x-a=0\Leftrightarrow x=a\)

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

\(\Leftrightarrow x^3-2x+1=0\)

\(\Leftrightarrow\left(x-1\right)\left(x^2+x-1\right)=0\)

\(\Leftrightarrow...\)

Answer 2

2.

ĐKXĐ: \(x\ge-1\)

\(\Leftrightarrow2\left(x^2+2\right)=5\sqrt{\left(x+1\right)\left(x^2-x+1\right)}\)

Đặt \(\left\{{}\begin{matrix}\sqrt{x+1}=a\ge0\\\sqrt{x^2-x+1}=b>0\end{matrix}\right.\)

\(\Leftrightarrow2\left(a^2+b^2\right)=5ab\)

\(\Leftrightarrow2a^2-5ab+2b^2=0\)

\(\Leftrightarrow\left(a-2b\right)\left(2a-b\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}4x+4=x^2-x+1\\x+1=4x^2-4x+4\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2-5x-3=0\\4x^2-5x+3=0\end{matrix}\right.\) \(\Leftrightarrow...\)

Answer 3

5.

ĐKXĐ: \(-\frac{1}{2}\le x\le\frac{1}{2}\)

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

\(\Leftrightarrow\sqrt{\left(\frac{1}{2}-x\right)\left(\frac{1}{2}+x\right)}=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{2}\\x=-\frac{1}{2}\end{matrix}\right.\)

6.

ĐKXĐ: \(x\ge1\)

\(\Leftrightarrow\sqrt{x-1}+\sqrt{x^3+x^2+x+1}=1+\sqrt{\left(x^2-1\right)\left(x^2+1\right)}\)

\(\Leftrightarrow\sqrt{x-1}+\sqrt{x^3+x^2+x+1}=1+\sqrt{\left(x-1\right)\left(x+1\right)\left(x^2+1\right)}\)

\(\Leftrightarrow\sqrt{\left(x-1\right)\left(x^3+x^2+x+1\right)}-\sqrt{x-1}-\left(\sqrt{x^3+x^2+x+1}-1\right)=0\)

\(\Leftrightarrow\sqrt{x-1}\left(\sqrt{x^3+x^2+x+1}-1\right)-\left(\sqrt{x^3+x^2+x+1}-1\right)=0\)

\(\Leftrightarrow\left(\sqrt{x-1}-1\right)\left(\sqrt{x^3+x^2+x+1}-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-1}=1\\\sqrt{x^3+x^2+x+1}=1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x^3+x^2+x=0\left(vn\right)\end{matrix}\right.\)

Answer 4

3.

Đặt \(\left\{{}\begin{matrix}\sqrt[3]{3x+1}=a\\\sqrt[3]{3x-1}=b\end{matrix}\right.\) ta được hệ:

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

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

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

\(\Rightarrow\left(b+2\right)^2+b^2+b\left(b+2\right)=1\)

\(\Leftrightarrow3b^2+6b+3=0\)

\(\Leftrightarrow b=-1\Rightarrow a=1\)

\(\Rightarrow\sqrt[3]{3x+1}=1\Rightarrow x=0\)

Answer 5

4.

ĐKXĐ: \(x\ge-2\)

\(\Leftrightarrow2\left(x^2-3x+2\right)=3\sqrt{\left(x+2\right)\left(x^2-2x+4\right)}\)

Đặt \(\left\{{}\begin{matrix}\sqrt{x^2-2x+4}=a>0\\\sqrt{x+2}=b\ge0\end{matrix}\right.\)

\(\Leftrightarrow2\left(a^2-b^2\right)=3ab\)

\(\Leftrightarrow2a^2-3ab-2b^2=0\)

\(\Leftrightarrow\left(a-2b\right)\left(2a+b\right)=0\)

\(\Leftrightarrow a-2b=0\Leftrightarrow a=2b\)

\(\Leftrightarrow\sqrt{x^2-2x+4}=2\sqrt{x+2}\)

\(\Leftrightarrow x^2-2x+4=4x+8\)

\(\Leftrightarrow x^2-6x-4=0\)

\(\Leftrightarrow...\)

Answer 6

7.

ĐKXĐ: \(x\ge-\frac{7}{2}\)

\(\Leftrightarrow x^2+9x+7-\left(2x+7\right)\sqrt{2x+7}=0\)

\(\Leftrightarrow-x^2+2x+7+\left(2x+7\right)\left(x-\sqrt{2x+7}\right)=0\)

\(\Leftrightarrow-x^2+2x+7+\frac{\left(2x+7\right)\left(x^2-2x-7\right)}{x+\sqrt{2x+7}}=0\)

\(\Leftrightarrow\left(-x^2+2x+7\right)\left(1-\frac{2x+7}{x+\sqrt{2x+7}}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}-x^2+2x+7=0\Leftrightarrow x=...\\2x+7=x+\sqrt{2x+7}\left(1\right)\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow x+7=\sqrt{2x+7}\)

\(\Leftrightarrow x^2+14x+49=2x+7\)

\(\Leftrightarrow x^2+12x+42=0\) (vô nghiệm)

Answer 7

8.

Đặt \(\left\{{}\begin{matrix}\sqrt[3]{x+10}=a\\\sqrt[3]{17-x}=b\end{matrix}\right.\) ta được hệ:

\(\left\{{}\begin{matrix}a+b=3\\a^3+b^3=27\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a+b=3\\\left(a+b\right)\left(a^2+b^2-ab\right)=27\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a+b=3\\a^2+b^2-ab=9\end{matrix}\right.\)

\(\Rightarrow a^2+\left(3-a\right)^2-a\left(3-a\right)-9=0\)

\(\Leftrightarrow3a^2-9a=0\)

\(\Leftrightarrow\left[{}\begin{matrix}a=0\\a=3\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\sqrt[3]{x+10}=0\\\sqrt[3]{x+10}=3\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-10\\x=17\end{matrix}\right.\)