\(\sqrt{5x^2+14x+9}-\sqrt{x^2-x-20}=5\sqrt{x+1}\)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,ĐK:x\ge-7\\ PT\Leftrightarrow\sqrt{\left(\sqrt{x+7}+1\right)^2}+\sqrt{x+7-\sqrt{x+7}-6}=4\)
Đạt \(\sqrt{x+7}=a\ge0\)
\(PT\Leftrightarrow\sqrt{\left(a+1\right)^2}+\sqrt{a^2-a-6}=4\\ \Leftrightarrow a+1+\sqrt{a^2-a-6}=4\\ \Leftrightarrow\sqrt{a^2-a-6}=3-a\\ \Leftrightarrow a^2-a-6=a^2-6a+9\\ \Leftrightarrow5a=15\Leftrightarrow a=3\\ \Leftrightarrow\sqrt{x+7}=3\\ \Leftrightarrow x+7=9\\ \Leftrightarrow x=2\left(tm\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a. ĐKXĐ: \(x\ge\dfrac{1}{2}\)
Đặt \(\left\{{}\begin{matrix}\sqrt{x^2+2x}=a>0\\\sqrt{2x-1}=b\ge0\end{matrix}\right.\)
\(\Rightarrow a+b=\sqrt{3a^2-b^2}\)
\(\Leftrightarrow\left(a+b\right)^2=3a^2-b^2\)
\(\Leftrightarrow a^2-ab-b^2=0\Leftrightarrow\left(a-\dfrac{1+\sqrt{5}}{2}b\right)\left(a+\dfrac{\sqrt{5}-1}{2}b\right)=0\)
\(\Leftrightarrow a=\dfrac{1+\sqrt{5}}{2}b\Leftrightarrow\sqrt{x^2+2x}=\dfrac{1+\sqrt{5}}{2}\sqrt{2x-1}\)
\(\Leftrightarrow x^2+2x=\dfrac{3+\sqrt{5}}{2}\left(2x-1\right)\)
\(\Leftrightarrow x^2-\left(\sqrt{5}+1\right)x+\dfrac{3+\sqrt{5}}{2}=0\)
\(\Leftrightarrow\left(x-\dfrac{\sqrt{5}+1}{2}\right)^2=0\)
\(\Leftrightarrow x=\dfrac{\sqrt{5}+1}{2}\)
b. ĐKXĐ: \(x\ge5\)
\(\Leftrightarrow\sqrt{5x^2+14x+9}=\sqrt{x^2-x-20}+5\sqrt{x+1}\)
\(\Leftrightarrow5x^2+14x+9=x^2-x-20+25\left(x+1\right)+10\sqrt{\left(x+1\right)\left(x-5\right)\left(x+4\right)}\)
\(\Leftrightarrow2x^2-5x+2=5\sqrt{\left(x^2-4x-5\right)\left(x+4\right)}\)
Đặt \(\left\{{}\begin{matrix}\sqrt{x^2-4x-5}=a\ge0\\\sqrt{x+4}=b>0\end{matrix}\right.\)
\(\Rightarrow2a^2+3b^2=5ab\)
\(\Leftrightarrow\left(a-b\right)\left(2a-3b\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x^2-4x-5}=\sqrt{x+4}\\2\sqrt{x^2-4x-5}=3\sqrt{x+4}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-4x-5=x+4\\4\left(x^2-4x-5\right)=9\left(x+4\right)\end{matrix}\right.\)
\(\Leftrightarrow...\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Lời giải:
ĐKXĐ:.............
PT $\Leftrightarrow \sqrt{5x^2+14x+9}=\sqrt{x^2-x-20}+5\sqrt{x+1}$
$\Rightarrow 5x^2+14x+9=x^2+24x+5+10\sqrt{(x^2-x-20)(x+1)}$
$\Leftrightarrow 4x^2-10x+4=10\sqrt{(x^2-x-20)(x+1)}$
$\Leftrightarrow 2x^2-5x+2=5\sqrt{(x+4)(x-5)(x+1)}$
$\Leftrightarrow 2(x^2-4x-5)+3(x+4)=5\sqrt{(x+4)(x^2-4x-5)}$
Đặt $\sqrt{x^2-4x-5}=a; \sqrt{x+4}=b$ với $a,b\geq 0$
Khi đó: $2a^2+3b^2=5ab$
$\Leftrightarrow (a-b)(2a-3b)=0$
$\Rightarrow a=b$ hoặc $a=1,5b$
Đến đây thì đơn giản rồi.
Đáp số: $x=8$ hoặc $x=\frac{5+\sqrt{61}}{2}$
![](https://rs.olm.vn/images/avt/0.png?1311)
Bạn tham khảo lời giải tại đây:
Câu hỏi của Mai Huy Long - Toán lớp 10 | Học trực tuyến
![](https://rs.olm.vn/images/avt/0.png?1311)
ĐK: $x \ geqslant 5$
\(Pt\Leftrightarrow2x^2-5x+2=5\sqrt{\left(x^2-x-20\right)\left(x+1\right)}\)
Ta có: \(\left(x^2-x-20\right)\left(x+1\right)=\left(x+4\right)\left(x-5\right)\left(x+1\right)=\left(x+4\right)\left(x^2-4x+5\right)\)
\(\Rightarrow2\left(x^2-4x-5\right)+3\left(x+4\right)=5\sqrt{\left(x^2-4x-5\right)\left(x+4\right)}\left(\circledast\right)\)
Đặt \(\left\{{}\begin{matrix}u=x^2-4x-5\\v=x+4\end{matrix}\right.\), \(\left(\circledast\right)\) trở thành: \(2u + 3v = 5\sqrt {uv} \Leftrightarrow \left[ \begin{array}{l} u = v\\ u = \dfrac{9}{4}v \end{array} \right.\)
\(\odot u=v\Rightarrow x^2-4x-5=x+4\Leftrightarrow x^2-5x-9=0\)\(\Leftrightarrow \left[ \begin{array}{l} x = \dfrac{{5 + \sqrt {61} }}{2} \text{(nhận)}\\ x = \dfrac{{5 - \sqrt {61} }}{2} \text{(loại)} \end{array} \right.\)
\(\odot\)\(u=\dfrac{9}{4}v\)\( \Rightarrow {x^2} - 4x - 5 = \dfrac{9}{4}\left( {x + 4} \right) \Leftrightarrow 4{x^2} - 25x - 56 = 0 \Leftrightarrow \left[ \begin{array}{l} x = 8 \text{(nhận)}\\ x=\dfrac{{ - 7}}{4} \text{(loại)} \end{array} \right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
ĐK: \(x\ge5\)
Chuyển vế, bình phương ta đc:
\(\sqrt{5x^2+14x+9}=5\sqrt{\left(x^2-x-20\right)\left(x+1\right)}\)
Nhận xét:
Không tồn tại số \(\alpha,\beta\) để: \(2x^2-5x+2=\alpha\left(x^2-x-20\right)+\beta\left(x+1\right)\)
Ta có: \(\left(x^2-x-20\right)\left(x+1\right)=\left(x+4\right)\left(x-5\right)\left(x+1\right)=\left(x+4\right)\left(x^2-4x-5\right)\)
PT đc vt lại là: \(2\left(x^2-4x-5\right)+3\left(x+4\right)=5\sqrt{\left(x^2-4x-5\right)\left(x+4\right)}\)
Đặt: \(\left\{{}\begin{matrix}u=x^2-4x-5\\v=x+4\end{matrix}\right.\)
Khi đó PT trở thành:
\(2u+3v=5\sqrt{uv}\Leftrightarrow\left[{}\begin{matrix}u=v\\u=\frac{9}{4}v\end{matrix}\right.\)
Xét \(u=v\) ta có PT:
\(x^2-4x-5=x+4\Leftrightarrow x^2-5x+9=0\Leftrightarrow\left[{}\begin{matrix}x=\frac{5+\sqrt{61}}{2}\\x=\frac{5-\sqrt{61}}{2}\left(loại\right)\end{matrix}\right.\)
Xét \(u=\frac{9}{4}v\) ta có PT:
\(x^2-4x-5=\frac{9}{4}\left(x+4\right)\Leftrightarrow4x^2-25x-56=0\Leftrightarrow\left[{}\begin{matrix}x=8\\x=-\frac{7}{4}\left(loại\right)\end{matrix}\right.\)
Vậy PT có 2 nghiệm là \(x=8;x=\frac{5+\sqrt{61}}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) ĐKXĐ: \(x\ge0\)
Ta có: \(\left(x+3\sqrt{x}+2\right)\left(x+9\sqrt{x}+18\right)=168x\)
\(\Leftrightarrow\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)\left(\sqrt{x}+3\right)\left(\sqrt{x}+6\right)=168x\)
\(\Leftrightarrow\left(x+6\right)^2+12\sqrt{x}\left(x+6\right)-133=0\)
\(\Leftrightarrow\left(x+6\right)^2+19\sqrt{x}\left(x+6\right)-7\sqrt{x}\left(x+6\right)-133=0\)
\(\Leftrightarrow\left(x+6\right)\left(x+19\sqrt{x}+6\right)-7\sqrt{x}\left(x+19\sqrt{x}+6\right)=0\)
\(\Leftrightarrow\left(x-7\sqrt{x}+6\right)=0\)
\(\Leftrightarrow\left(\sqrt{x}-1\right)\left(\sqrt{x}-6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=36\end{matrix}\right.\)
Dòng thứ 2 qua dòng thứ 3 anh làm chậm lại được không ạ, tại tắt quá e không hiểu
ĐK: \(x\ge5\)
\(pt\Leftrightarrow\sqrt{5x^2+14x+9}=5\sqrt{x+1}+\sqrt{x^2-x-20}\)
Bình phương 2 vế, ta đc:
\(5x^2+14x+9=25x+5+x^2-x-20+10\sqrt{\left(x+1\right)\left(x^2-x-20\right)}\)
\(\Leftrightarrow5x^2+14x+9-25x-5-x^2+x+20=10\sqrt{\left(x+1\right)\left(x+4\right)\left(x-5\right)}\)
\(\Leftrightarrow4x^2-10x+4=10\sqrt{\left(x+1\right)\left(x-5\right)\left(x+4\right)}\)
\(\Leftrightarrow2x^2-5x+2=5\sqrt{\left(x^2-4x-5\right)\left(x+4\right)}\)
\(\Leftrightarrow2\left(x^2-4x-5\right)+3\left(x+4\right)=5\sqrt{\left(x^2-4x-5\right)\left(x+4\right)}\)
Đặt \(\sqrt{x^2-4x-5}=a\left(a\ge0\right);\sqrt{x+4}=b\left(b\ge3\right)\)
Khi đó,pt trở thành \(2a^2+3b^2=5ab\Leftrightarrow2a^2-2ab+3b^2-3ab=0\)
\(\Leftrightarrow2a\left(a-b\right)+3b\left(b-a\right)=0\Leftrightarrow\left(2a-3b\right)\left(a-b\right)=0\Leftrightarrow\left[{}\begin{matrix}a=b\\2a=3b\end{matrix}\right.\)
Với a=b \(\Rightarrow\sqrt{x^2-4x-5}=\sqrt{x+4}\Leftrightarrow x^2-5x-9=0\Leftrightarrow\left[{}\begin{matrix}x=\frac{5+\sqrt{61}}{2}\left(tmdk\right)\\x=\frac{5-\sqrt{61}}{2}\left(loai\right)\end{matrix}\right.\)
Với 2a=3b \(\Rightarrow2\sqrt{x^2-4x-5}=3\sqrt{x+4}\Leftrightarrow4\left(x^2-4x-5\right)=9\left(x+4\right)\)
\(\Leftrightarrow4x^2-25x-56=0\Leftrightarrow\left[{}\begin{matrix}x=8\left(tmdk\right)\\x=\frac{-7}{4}\left(loai\right)\end{matrix}\right.\)
Vậy ...
đánh nhầm r, dòng 4 vs 5 bạn sửa 25x+5 thành 25x+25 nha, dòng 5 cx -5 thành -25