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\(a,PT\Leftrightarrow\left|x+3\right|=3x-6\\ \Leftrightarrow\left[{}\begin{matrix}x+3=3x-6\left(x\ge-3\right)\\x+3=6-3x\left(x< -3\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{9}{2}\left(tm\right)\\x=\dfrac{3}{4}\left(ktm\right)\end{matrix}\right.\\ \Leftrightarrow x=\dfrac{9}{2}\\ b,PT\Leftrightarrow\left|x-1\right|=\left|2x-1\right|\\ \Leftrightarrow\left[{}\begin{matrix}x-1=2x-1\\1-x=2x-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{2}{3}\end{matrix}\right.\)
\(c,ĐK:x\le\dfrac{2}{5}\\ PT\Leftrightarrow4-5x=25x^2-20x+4\\ \Leftrightarrow25x^2-15x=0\\ \Leftrightarrow5x\left(5x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\left(tm\right)\\x=\dfrac{3}{5}\left(ktm\right)\end{matrix}\right.\Leftrightarrow x=0\\ d,ĐK:x\le\dfrac{2}{5}\\ PT\Leftrightarrow4-5x=2-5x\\ \Leftrightarrow x\in\varnothing\)
Điều kiện xác định: \(\left\{{}\begin{matrix}5x^2+4x\ge0\\x^2-3x-18\ge0\\x\ge0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\left(5x+4\right)\ge0\\\left(x-6\right)\left(x+3\right)\ge0\\x\ge0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x\ge0\\x\le\dfrac{-4}{5}\end{matrix}\right.\\\left[{}\begin{matrix}x\ge6\\x\le-3\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow x\ge6\) (*)
Khi đó phương trình \(\Leftrightarrow\) \(\sqrt{5x^2+4x}=\sqrt{x^2-3x-18}+5\sqrt{x}\)
\(\Leftrightarrow5x^2+4x=x^2+22x-18+10\sqrt{x\left(x^2-3x-18\right)}\\ \Leftrightarrow4x^2-18x+18=10\sqrt{x\left(x^2-3x-18\right)}\\ \Leftrightarrow5\sqrt{x\left(x-6\right)\left(x+3\right)}=2x^2-9x+9\\ \Leftrightarrow5\sqrt{\left(x^2-6x\right)\left(x+3\right)}=2\left(x^2-6x\right)+3\left(x+3\right)\left(1\right)\)
Đặt \(\left\{{}\begin{matrix}a=\sqrt{x^2-6x}\ge0\\b=\sqrt{x+3}\ge0\end{matrix}\right.\)
Khi đó pt \(\left(1\right)\) trở thành: \(2a^2+3b^2-5ab=0\\ \Leftrightarrow\left(a-b\right)\left(2a-3b\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}a=b\\2a=3b\end{matrix}\right.\)
- TH1: \(a=b\Rightarrow x^2-6x=x+3\Leftrightarrow x^2-7x-3=0\\ \Leftrightarrow\left[{}\begin{matrix}\dfrac{7+\sqrt{61}}{2}\left(tm\right)\\\dfrac{7-\sqrt{61}}{2}\left(ktm\right)\end{matrix}\right.\)
-TH2: \(2a=3b\Leftrightarrow4a^2=9b^2\\ \Leftrightarrow4\left(x^2-6x\right)=9\left(x+3\right)\\ \Leftrightarrow4x^2-33x-27=0\\ \Leftrightarrow\left[{}\begin{matrix}x=9\left(tm\right)\\x=\dfrac{-3}{4}\left(ktm\right)\end{matrix}\right.\)
Vậy pt có 2 nghiệm \(x=\dfrac{7+\sqrt{61}}{2};x=9\)
a: Ta có: \(\sqrt{4x+20}-3\sqrt{x+5}+\dfrac{4}{3}\sqrt{9x+45}=6\)
\(\Leftrightarrow2\sqrt{x+5}-3\sqrt{x+5}+4\sqrt{x+5}=6\)
\(\Leftrightarrow3\sqrt{x+5}=6\)
\(\Leftrightarrow x+5=4\)
hay x=-1
b: Ta có: \(\dfrac{1}{2}\sqrt{x-1}-\dfrac{3}{2}\sqrt{9x-9}+24\sqrt{\dfrac{x-1}{64}}=-17\)
\(\Leftrightarrow\dfrac{1}{2}\sqrt{x-1}-\dfrac{9}{2}\sqrt{x-1}+3\sqrt{x-1}=-17\)
\(\Leftrightarrow\sqrt{x-1}=17\)
\(\Leftrightarrow x-1=289\)
hay x=290
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 \(\left\{{}\begin{matrix}\sqrt{x^2-4x-5}=a\ge0\\\sqrt{x+4}=b>0\end{matrix}\right.\)
\(\Rightarrow2a^2+3b^2=5ab\)
\(\Leftrightarrow\left(a-b\right)\left(2a-3b\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x^2-4x-5}=\sqrt{x+4}\\2\sqrt{x^2-4x-5}=3\sqrt{x+4}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-4x-5=x+4\\4\left(x^2-4x-5\right)=9\left(x+4\right)\end{matrix}\right.\)
\(\Leftrightarrow...\)
1. ĐKXĐ: $x\geq \frac{-3}{5}$
PT $\Leftrightarrow 5x+3=3-\sqrt{2}$
$\Leftrightarrow x=\frac{-\sqrt{2}}{5}$
2. ĐKXĐ: $x\geq \sqrt{7}$
PT $\Leftrightarrow (\sqrt{x}-7)(\sqrt{x}+7)=4$
$\Leftrightarrow x-49=4$
$\Leftrightarrow x=53$ (thỏa mãn)
a: Ta có: \(\sqrt{4x^2+4x+3}=8\)
\(\Leftrightarrow4x^2+4x+1+2-64=0\)
\(\Leftrightarrow4x^2+4x-61=0\)
\(\Delta=4^2-4\cdot4\cdot\left(-61\right)=992\)
Vì Δ>0 nên phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}x_1=\dfrac{-4-4\sqrt{62}}{8}=\dfrac{-1-\sqrt{62}}{2}\\x_2=\dfrac{-4+4\sqrt{62}}{8}=\dfrac{-1+\sqrt{62}}{2}\end{matrix}\right.\)
Câu 4:
Giả sử điều cần chứng minh là đúng
\(\Rightarrow x=y\), thay vào điều kiện ở đề bài, ta được:
\(\sqrt{x+2014}+\sqrt{2015-x}-\sqrt{2014-x}=\sqrt{x+2014}+\sqrt{2015-x}-\sqrt{2014-x}\) (luôn đúng)
Vậy điều cần chứng minh là đúng
2) \(\sqrt{x^2-5x+4}+2\sqrt{x+5}=2\sqrt{x-4}+\sqrt{x^2+4x-5}\)
⇔ \(\sqrt{\left(x-4\right)\left(x-1\right)}-2\sqrt{x-4}+2\sqrt{x+5}-\sqrt{\left(x+5\right)\left(x-1\right)}=0\)
⇔ \(\sqrt{x-4}.\left(\sqrt{x-1}-2\right)-\sqrt{x+5}\left(\sqrt{x-1}-2\right)=0\)
⇔ \(\left(\sqrt{x-4}-\sqrt{x+5}\right)\left(\sqrt{x-1}-2\right)=0\)
⇔ \(\left[{}\begin{matrix}\sqrt{x-4}-\sqrt{x+5}=0\\\sqrt{x-1}-2=0\end{matrix}\right.\)
⇔ \(\left[{}\begin{matrix}\sqrt{x-4}=\sqrt{x+5}\\\sqrt{x-1}=2\end{matrix}\right.\)
⇔ \(\left[{}\begin{matrix}x\in\varnothing\\x=5\end{matrix}\right.\)
⇔ x = 5
Vậy S = {5}