Question

giải các phương trình sau

1) 15.\(\sqrt{x^3-1}\)=4x²+8

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

3) \(\sqrt{x-\frac{1}{x}}\)+\(\frac{4}{x}\)=x+\(\sqrt{2x-\frac{5}{x}}\)

4)\(\sqrt{5x-1}\)-\(\sqrt{x+2}\)=\(\frac{4x-3}{5}\)

5) \(\sqrt{2x-\frac{3}{x}}\)-1=\(\frac{3}{2x}\)-\(\sqrt{\frac{6}{x}-2x}\)

Answer 1

1/

ĐKXĐ: \(x\ge1\)

\(\Leftrightarrow15\sqrt{\left(x-1\right)\left(x^2+x+1\right)}=4\left(x^2+2\right)\)

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

\(\Rightarrow15ab=4\left(b^2-a^2\right)\)

\(\Leftrightarrow4a^2+15ab-4b^2=0\)

\(\Leftrightarrow\left(a+4b\right)\left(4a-b\right)=0\)

\(\Leftrightarrow4a=b\Leftrightarrow4\sqrt{x-1}=\sqrt{x^2+x+1}\)

\(\Leftrightarrow x^2-15x+17=0\) (bấm máy)

Answer 2

b/ ĐKXĐ: \(x\ge\frac{3}{2}\)

\(\Leftrightarrow\sqrt{2x-3}-\sqrt{x}=2x-6\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\Rightarrow x=3\\\frac{1}{\sqrt{2x-3}+\sqrt{x}}=2\left(1\right)\end{matrix}\right.\)

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

Do \(x\ge\frac{3}{2}\Rightarrow\left\{{}\begin{matrix}\sqrt{2x-3}\ge0\\\sqrt{x}>1\end{matrix}\right.\)

\(\Rightarrow VT>1>\frac{1}{2}\Rightarrow\left(1\right)\) vô nghiệm

Vậy pt có nghiệm duy nhất \(x=3\)

Answer 3

3/

\(\Leftrightarrow\sqrt{2x-\frac{5}{x}}-\sqrt{x-\frac{1}{x}}+x-\frac{4}{x}=0\)

Đặt \(\left\{{}\begin{matrix}\sqrt{2x-\frac{5}{x}}=a\ge0\\\sqrt{x-\frac{1}{x}}=b\ge0\end{matrix}\right.\)

\(\Rightarrow x-\frac{4}{x}=a^2-b^2\)

Phương trình trở thành:

\(a-b+a^2-b^2=0\)

\(\Leftrightarrow\left(a-b\right)\left(a+b+1\right)=0\)

\(\Leftrightarrow a=b\Leftrightarrow\sqrt{2x-\frac{5}{x}}=\sqrt{x-\frac{1}{x}}\)

\(\Rightarrow2x-\frac{5}{x}=x-\frac{1}{x}\)

\(\Rightarrow x^2=4\Rightarrow x=\pm2\)

Do ko tìm ĐKXĐ nên bạn cần thay nghiệm vào pt ban đầu để thử lại

Answer 4

4/

ĐKXĐ: \(x\ge\frac{1}{5}\)

\(\Leftrightarrow\frac{4x-3}{\sqrt{5x-1}+\sqrt{x+2}}=\frac{4x-3}{5}\)

\(\Leftrightarrow\left[{}\begin{matrix}4x-3=0\Rightarrow x=\frac{3}{4}\\\sqrt{5x-1}+\sqrt{x+2}=5\left(1\right)\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow\sqrt{5x-1}-3+\sqrt{x+2}-2=0\)

\(\Leftrightarrow\frac{5\left(x-2\right)}{\sqrt{5x-1}+3}+\frac{x-2}{\sqrt{x+2}+2}=0\)

\(\Leftrightarrow\left(x-2\right)\left(\frac{5}{\sqrt{5x-1}+3}+\frac{1}{\sqrt{x+2}+2}\right)=0\)

\(\Leftrightarrow x=2\)

Answer 5

5/

Đặt \(\left\{{}\begin{matrix}\sqrt{2x-\frac{3}{x}}=a\ge0\\\sqrt{\frac{6}{x}-2x}=b\ge0\end{matrix}\right.\) \(\Rightarrow a^2+b^2=\frac{3}{x}\)

Pt trở thành:

\(a-1=\frac{a^2+b^2}{2}-b\)

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

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

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

\(\Leftrightarrow\left\{{}\begin{matrix}a=1\\b=1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{2x-\frac{3}{x}}=1\\\sqrt{\frac{6}{x}-2x}=1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x^2-x-3=0\\2x^2+x-6=0\end{matrix}\right.\) \(\Rightarrow x=\frac{3}{2}\)