Question

Giải phương trình:

1. \(\sqrt{\dfrac{42}{5-x}}+\sqrt{\dfrac{60}{7-x}}=6\)

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

3. \(x^2+x+12\sqrt{x+1}=36\)

4. \(\sqrt{x+2}-\sqrt{x-6}=2\)

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

6. \(5\sqrt{1+x^3}=2\left(x^2+2\right)\)

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

Answer 1

1.

ĐKXĐ: \(x< 5\)

\(\Leftrightarrow\sqrt{\dfrac{42}{5-x}}-3+\sqrt{\dfrac{60}{7-x}}-3=0\)

\(\Leftrightarrow\dfrac{\dfrac{42}{5-x}-9}{\sqrt{\dfrac{42}{5-x}}+3}+\dfrac{\dfrac{60}{7-x}-9}{\sqrt{\dfrac{60}{7-x}}+3}=0\)

\(\Leftrightarrow\dfrac{9x-3}{\left(5-x\right)\left(\sqrt{\dfrac{42}{5-x}}+3\right)}+\dfrac{9x-3}{\left(7-x\right)\left(\sqrt{\dfrac{60}{7-x}}+3\right)}=0\)

\(\Leftrightarrow\left(9x-3\right)\left(\dfrac{1}{\left(5-x\right)\left(\sqrt{\dfrac{42}{5-x}}+3\right)}+\dfrac{1}{\left(7-x\right)\left(\sqrt{\dfrac{60}{7-x}}+3\right)}\right)=0\)

\(\Leftrightarrow x=\dfrac{1}{3}\)

Answer 2

b.

ĐKXĐ: \(x\ge2\)

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

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

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

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

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

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

\(\Rightarrow x=2\)

Answer 3

3.

ĐKXĐ: \(x\ge-1\)

\(x^2+x-12+12\left(\sqrt{x+1}-2\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(x+4\right)+\dfrac{12\left(x-3\right)}{\sqrt{x+1}+2}=0\)

\(\Leftrightarrow\left(x-3\right)\left(x+4+\dfrac{12}{\sqrt{x+1}+2}\right)=0\)

\(\Leftrightarrow x-3=0\)

\(\Leftrightarrow x=3\)

Answer 4

4.

ĐKXĐ: \(x\ge6\)

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

\(\Leftrightarrow x+2=4+x-6+4\sqrt{x-6}\)

\(\Leftrightarrow\sqrt{x-6}=1\)

\(\Leftrightarrow x=7\)

5.

Đặt \(\left\{{}\begin{matrix}\sqrt[3]{x-1}=a\\\sqrt[3]{x-3}=b\\\sqrt[3]{2}=c\end{matrix}\right.\) ta được hệ:

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

\(\Rightarrow3a^2b-3ab^2=0\)

\(\Leftrightarrow3ab\left(a-b\right)=0\Leftrightarrow\left[{}\begin{matrix}a=0\\b=0\\a=b\end{matrix}\right.\)

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

Answer 5

6.

ĐKXĐ: \(x\ge-1\)

\(\Leftrightarrow5\sqrt{\left(x+1\right)\left(x^2-x+1\right)}=2\left(x^2+2\right)\)

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

\(\Rightarrow5ab=2\left(a^2+b^2\right)\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}2a=b\\a=2b\end{matrix}\right.\)

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

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

\(\Leftrightarrow...\)

Answer 6

6 lần nữa:

ĐKXĐ: \(x\ge-2\)

\(\dfrac{3}{\sqrt[]{x+5}+\sqrt[]{x+2}}\left(1+\sqrt{x^2+7x+10}\right)=3\)

\(\Leftrightarrow1+\sqrt{x^2+7x+10}=\sqrt{x+5}+\sqrt{x+2}\)

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x+5=1\\x+2=1\end{matrix}\right.\)

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

Answer 7

Bài 6 nếu đặt ẩn phụ:

ĐKXĐ...

Đặt \(\left\{{}\begin{matrix}\sqrt{x+5}=a>0\\\sqrt{x+2}=b\ge0\end{matrix}\right.\) 

Pt trở thành:

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

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

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

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

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

\(\Leftrightarrow...\)