Tìm x:
a) \(\left(5x-6\right)^2-\frac{1}{\sqrt{5x-7}}=x^2-\frac{1}{\sqrt{x-1}}\)
b) \(4x^3+x-\left(x+1\right)\sqrt{2x+1}=0\)
c) \(\frac{\sqrt{x+1}-2}{\sqrt[3]{2x+1}-3}=\frac{1}{x+2}\)
d) \(-2x^3+10x^2-17x+8=2x^2\sqrt[3]{5x-x^2}\)
e) \(9x^2-28x+21=\sqrt{x-1}\)
f) \(3x\left(2+\sqrt{9x^2+3}\right)+\left(4x+2\right)\sqrt{1+x+x^2}+1=0\)
Mng giúp em với ạ, em cảm ơn
1. ĐKXĐ: \(x>\frac{7}{5}\)
Đặt \(\left\{{}\begin{matrix}\sqrt{5x-7}=a>0\\\sqrt{x-1}=b>0\end{matrix}\right.\)
\(\Rightarrow\left(a^2+1\right)^2-\frac{1}{a}=\left(b^2+1\right)^2-\frac{1}{b}\)
\(\Leftrightarrow\left(a^2+1\right)^2-\left(b^2+1\right)^2+\frac{1}{b}-\frac{1}{a}=0\)
\(\Leftrightarrow\left(a^2+b^2+2\right)\left(a-b\right)\left(a+b\right)+\frac{a-b}{ab}=0\)
\(\Leftrightarrow\left(a-b\right)\left[\left(a^2+b^2+2\right)\left(a+b\right)+\frac{1}{ab}\right]=0\)
\(\Leftrightarrow a=b\)
\(\Leftrightarrow5x-7=x-1\)
\(\Leftrightarrow x=?\)
2.
ĐKXĐ: \(x\ge-\frac{1}{2}\)
\(\Leftrightarrow8x^3+2x-\left(2x+2\right)\sqrt{2x+1}=0\)
Đặt \(\left\{{}\begin{matrix}2x=a\\\sqrt{2x+1}=b\end{matrix}\right.\)
\(\Rightarrow a^3+a-\left(b^2+1\right)b=0\)
\(\Leftrightarrow a^3-b^3+a-b=0\)
\(\Leftrightarrow\left(a-b\right)\left(a^2+ab+b^2+1\right)=0\)
\(\Leftrightarrow a=b\)
\(\Leftrightarrow2x=\sqrt{2x+1}\) (\(x\ge0\))
\(\Leftrightarrow4x^2=2x+1\)
\(\Leftrightarrow x=?\)
3.
ĐKXĐ: \(x\ge-1;x\ne13\)
\(\left(x+2\right)\left(\sqrt{x+1}-2\right)=\sqrt[3]{2x+1}-3\)
\(\Leftrightarrow\left(x+2\right)\sqrt{x+1}-2x-4=\sqrt[3]{2x+1}-3\)
\(\Leftrightarrow\left(x+1\right)\sqrt{x+1}+x+1-\left(2x+1\right)-\sqrt[3]{2x+1}=0\)
Đặt \(\left\{{}\begin{matrix}\sqrt{x+1}=a\\\sqrt[3]{2x+1}=b\end{matrix}\right.\)
\(\Rightarrow a^3+a-b^3-b=0\)
\(\Leftrightarrow\left(a-b\right)\left(a^2+ab+b^2+1\right)=0\)
\(\Leftrightarrow a=b\)
\(\Leftrightarrow\sqrt{x+1}=\sqrt[3]{2x+1}\) (\(x\ge-\frac{1}{2}\))
\(\Leftrightarrow\left(x+1\right)^3=\left(2x+1\right)^2\)
\(\Leftrightarrow x=?\)
5.
ĐKXĐ: \(x\ge1\)
\(\Leftrightarrow9x^2-27x+20=x-1+\sqrt{x-1}\)
\(\Leftrightarrow\left(3x-5\right)^2+3x-5=x-1+\sqrt{x-1}\)
Đặt \(\left\{{}\begin{matrix}3x-5=a\\\sqrt{x-1}=b\end{matrix}\right.\)
\(\Rightarrow a^2+a=b^2+b\)
\(\Leftrightarrow\left(a-b\right)\left(a+b+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=b\\b=-1-a\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-1}=3x-5\left(x\ge\frac{5}{3}\right)\\\sqrt{x-1}=4-3x\left(x\le\frac{4}{3}\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=9x^2-30x+25\left(x\ge\frac{5}{3}\right)\\x-1=9x^2-24x+16\left(x\le\frac{4}{3}\right)\end{matrix}\right.\)
\(\Leftrightarrow...\)
6.
Hình như bài này bạn ghi sai đề, giải nửa tiếng ko ra.
Nhưng thêm 1 dấu ngoặc thành:
\(3x\left(2+\sqrt{9x^2+3}\right)+\left(4x+2\right)\left(\sqrt{1+x+x^2}+1\right)=0\)
Thì lại tìm được cách giải đơn giản:
\(\Leftrightarrow3x\left(2+\sqrt{\left(3x\right)^2+3}\right)+\left(2x+1\right)\left(\sqrt{4x^2+4x+4}+2\right)=0\)
\(\Leftrightarrow3x\left(\sqrt{\left(3x\right)^2+3}+2\right)+\left(2x+1\right)\left(\sqrt{\left(2x+1\right)^2+3}+2\right)=0\)
- Nếu \(x\ge0\Rightarrow VT>0\) vô nghiệm
- Nếu \(x\le-\frac{1}{2}\Rightarrow VT< 0\) vô nghiệm
- Với \(-\frac{1}{2}< x< 0\)
Đặt \(\left\{{}\begin{matrix}3x=a\\-\left(2x+1\right)=b\end{matrix}\right.\)
\(\Leftrightarrow a\left(\sqrt{a^2+3}+2\right)-b\left(\sqrt{b^2+3}+2\right)=0\)
\(\Leftrightarrow a\sqrt{a^2+3}-b\sqrt{b^2+3}+2\left(a-b\right)=0\)
\(\Leftrightarrow\frac{a^4+3a^2-b^4-3b^2}{a\sqrt{a^2+3}+b\sqrt{b^2+3}}+2\left(a-b\right)=0\)
\(\Leftrightarrow\left(a-b\right)\left(\frac{\left(a+b\right)\left(a^2+b^2+3\right)}{a\sqrt{a^2+3}+b\sqrt{b^2+3}}+2\right)=0\)
\(\Rightarrow a=b\Rightarrow3x=-2x-1\)
