CHUYÊN ĐỀ GIẢI PHƯƠNG TRÌNH
a, \(\sqrt{2x-1}+\sqrt{x^2+3}=4-x\) f, \(2x^2-11x+23=4\sqrt{x+1}\)
b, \(\sqrt{x^2+x+1}=\sqrt{x^2-3x-1}+2x+1\) g, \(\frac{4}{x}+\sqrt{x-\frac{1}{x}}=x+\sqrt{2x-\frac{5}{x}}\)
c, \(\left|x-16\right|^4+\left|x-17\right|^3=1\) h, \(9\left(\sqrt{4x+1}-\sqrt{3x-2}\right)=x+3\)
d, \(\left(x+1\right)\sqrt{x+2}+\left(x+6\right)\sqrt{x+7}=x^2+7x+12\)
e, \(\left(4x^3-x+3\right)^3-x^3=\frac{3}{2}\)
1.
ĐKXĐ: \(x\ge\frac{1}{2}\)
\(\Leftrightarrow\sqrt{2x-1}-1+\sqrt{x^2+3}-2+x-1=0\)
\(\Leftrightarrow\frac{2\left(x-1\right)}{\sqrt{2x-1}+1}+\frac{\left(x-1\right)\left(x+1\right)}{\sqrt{x^2+3}+2}+x-1=0\)
\(\Leftrightarrow\left(x-1\right)\left(\frac{2}{\sqrt{2x-1}+1}+\frac{x+1}{\sqrt{x^2+3}+2}+1\right)=0\)
\(\)\(\Leftrightarrow x=1\)
2.
ĐKXĐ: ...
Đặt \(\left\{{}\begin{matrix}\sqrt{x^2+x+1}=a>0\\\sqrt{x^2-3x-1}=b\ge0\end{matrix}\right.\)
\(\Rightarrow a=b+\frac{1}{2}\left(a^2-b^2\right)\)
\(\Leftrightarrow\left(a-b\right)\left(a+b-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=b\left(1\right)\\a=2-b\left(2\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow x^2+x+1=x^2-3x-1\)
\(\Leftrightarrow x=\frac{1}{2}\)
\(\left(2\right)\Leftrightarrow\sqrt{x^2+x+1}=2-\sqrt{x^2-3x-1}\)
\(\Rightarrow x^2+x+1=x^2-3x+3-4\sqrt{x^2-3x-1}\)
\(\Rightarrow2\sqrt{x^2-3x-1}=1-2x\)
\(\Rightarrow4x^2-12x-4=4x^2-4x+1\)
\(\Rightarrow x=-\frac{5}{8}\)
Do các bước biến đổi ko tương đương nên cần thay nghiệm này vào pt ban đầu để kiểm tra (bạn tự kiểm tra)
3.
- Với \(x=\left\{16;17\right\}\) là 2 nghiệm của pt
- Với \(x< 16\):
\(\left\{{}\begin{matrix}\left|x-16\right|^4>0\\\left|x-17\right|>1\Rightarrow\left|x-17\right|^3>1\end{matrix}\right.\)
\(\Rightarrow\left|x-16\right|^4+\left|x-17\right|^3>1\)
Pt vô nghiệm
- Với \(x>17\Rightarrow\left\{{}\begin{matrix}\left|x-17\right|^3>0\\\left|x-16\right|>1\Rightarrow\left|x-16\right|^4>1\end{matrix}\right.\)
\(\Rightarrow\left|x-16\right|^4+\left|x-17\right|^3>1\)
Pt vô nghiệm
- Với \(16< x< 17\Rightarrow\left\{{}\begin{matrix}0< \left|x-16\right|< 1\\0< \left|17-x\right|< 1\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\left|x-16\right|^4< x-16\\\left|17-x\right|^3< 17-x\end{matrix}\right.\)
\(\Rightarrow\left|x-16\right|^4+\left|x-17\right|^3< x-16+17-x=1\) (vô nghiệm)
Vậy pt có đúng 2 nghiệm \(\left[{}\begin{matrix}x=16\\x=17\end{matrix}\right.\)
4.
ĐKXĐ: ...
\(\Leftrightarrow\left(x+1\right)\left(\sqrt{x+2}-2\right)+\left(x+6\right)\left(\sqrt{x+7}-3\right)=x^2+2x-8\)
\(\Leftrightarrow\frac{\left(x+1\right)\left(x-2\right)}{\sqrt{x+2}+2}+\frac{\left(x+6\right)\left(x-2\right)}{\sqrt{x+7}+3}=\left(x-2\right)\left(x+4\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\\frac{x+1}{\sqrt{x+2}+2}+\frac{x+6}{\sqrt{x+7}+3}=x+4\left(1\right)\end{matrix}\right.\)
Xét (1):
Do \(x\ge-2\Rightarrow VT\le\frac{x+1}{2}+\frac{x+6}{3+\sqrt{5}}< \frac{x+1}{2}+\frac{x+6}{5}< x+4\)
\(\Rightarrow\left(1\right)\) vô nghiệm
5.
Đặt \(4x^3-x+3=t\) ta được:
\(\left\{{}\begin{matrix}4x^3-x+3=t\\t^3-x^3=\frac{3}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}4x^3-x+3=t\\2t^3-2x^3=3\end{matrix}\right.\)
\(\Rightarrow2x^3+2t^3-x-t=0\)
\(\Leftrightarrow\left(x+t\right)\left(2x^2-2xt+2t^2-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+t=0\left(1\right)\\2x^2-2xt+2t^2-1=0\left(2\right)\end{matrix}\right.\)
Xét (1) \(\Leftrightarrow4x^3+3=0\Rightarrow x=-\sqrt[3]{\frac{3}{4}}\)
Xét \(\left(2\right)\) : \(\Delta'=t^2-2\left(2t^2-1\right)\ge0\)
\(\Leftrightarrow t^2\le\frac{2}{3}\) \(\Rightarrow-\sqrt{\frac{8}{27}}\le t^3\le\sqrt{\frac{8}{27}}\)
Tương tự ta có \(-\sqrt{\frac{8}{27}}\le x^3\le\sqrt{\frac{8}{27}}\)
\(\Rightarrow x^3-t^3\le2\sqrt{\frac{8}{27}}< \frac{3}{2}\)
\(\Rightarrow\left(2\right)\) vô nghiệm
6.
ĐKXD: ...
\(\Leftrightarrow2\left(x^2-6x+9\right)+\left(x+5-4\sqrt{x+1}\right)=0\)
\(\Leftrightarrow2\left(x-3\right)^2+\frac{\left(x-3\right)^2}{x+5+4\sqrt{x+1}}=0\)
\(\Leftrightarrow\left(x-3\right)^2\left(2+\frac{1}{x+5+4\sqrt{x+1}}\right)=0\)
\(\Leftrightarrow x=3\)
7.
\(\sqrt{x-\frac{1}{x}}-\sqrt{2x-\frac{5}{x}}+\frac{4}{x}-x=0\)
Đặt \(\left\{{}\begin{matrix}\sqrt{x-\frac{1}{x}}=a\ge0\\\sqrt{2x-\frac{5}{x}}=b\ge0\end{matrix}\right.\) \(\Rightarrow a^2-b^2=\frac{4}{x}-x\)
\(\Rightarrow a-b+a^2-b^2=0\)
\(\Leftrightarrow\left(a-b\right)\left(a+b+1\right)=0\)
\(\Leftrightarrow a=b\Leftrightarrow x-\frac{1}{x}=2x-\frac{5}{x}\)
\(\Leftrightarrow x=\frac{4}{x}\Rightarrow x=\pm2\)
Thế nghiệm lại pt ban đầu để thử (hoặc là bạn tìm ĐKXĐ từ đầu)
8.
ĐKXĐ: \(x\ge\frac{2}{3}\)
\(\Leftrightarrow\frac{9\left(x+3\right)}{\sqrt{4x+1}+\sqrt{3x-2}}=x+3\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\left(l\right)\\\frac{9}{\sqrt{4x+1}+\sqrt{3x-2}}=1\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow\sqrt{4x+1}+\sqrt{3x-2}=9\)
\(\Leftrightarrow\sqrt{4x+1}-5+\sqrt{3x-2}-4=0\)
\(\Leftrightarrow\frac{4\left(x-6\right)}{\sqrt{4x+1}+5}+\frac{3\left(x-6\right)}{\sqrt{3x-2}+4}=0\)
\(\Leftrightarrow\left(x-6\right)\left(\frac{4}{\sqrt{4x+1}+5}+\frac{3}{\sqrt{3x-2}+4}\right)=0\)
\(\Leftrightarrow x=6\)