Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a.
ĐKXĐ: \(x>0\)
\(\sqrt{x\left(x+3\right)}+2\sqrt{x+2}=2x+\sqrt{\dfrac{\left(x+2\right)\left(x+3\right)}{x}}\)
\(\Leftrightarrow\sqrt{x}\left(2\sqrt{x}-\sqrt{x+3}\right)+\sqrt{\dfrac{x+2}{x}}\left(\sqrt{x+3}-2\sqrt{x}\right)=0\)
\(\Leftrightarrow\sqrt{x}\left(\dfrac{4x-x-3}{2\sqrt{x}+\sqrt{x+3}}\right)-\sqrt{\dfrac{x+2}{x}}\left(\dfrac{4x-x-3}{\sqrt{x+3}+2\sqrt{x}}\right)=0\)
\(\Leftrightarrow\dfrac{3\left(x-1\right)}{2\sqrt{x}+\sqrt{x+3}}\left(\sqrt{x}-\sqrt{\dfrac{x+2}{x}}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{x+2}{x}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\\x=-1\left(loại\right)\end{matrix}\right.\)
b.
ĐKXĐ: \(x\ge-\dfrac{1}{2};x\ne1-\sqrt{2}\)
\(x+2+x\sqrt{2x+1}=x\sqrt{x+2}+\sqrt{\left(x+2\right)\left(2x+1\right)}\)
\(\Leftrightarrow\sqrt{x+2}\left(\sqrt{2x+1}-\sqrt{x+2}\right)-x\left(\sqrt{2x+1}-\sqrt{x+2}\right)=0\)
\(\Leftrightarrow\left(\sqrt{2x+1}-\sqrt{x+2}\right)\left(\sqrt{x+2}-x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{2x+1}=\sqrt{x+2}\\\sqrt{x+2}=x\left(x\ge0\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+1=x+2\\x^2-x-2=0\left(x\ge0\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\\x=-1\left(loại\right)\end{matrix}\right.\)
a. ĐKXĐ: \(\left[{}\begin{matrix}x\ge1\\-1\le x< 0\end{matrix}\right.\)
Do \(x\ne0\) nên pt tương đương:
\(x+2\sqrt{x-\dfrac{1}{x}}=3+\dfrac{1}{x}\)
\(\Leftrightarrow x-\dfrac{1}{x}+2\sqrt{x-\dfrac{1}{x}}-3=0\)
Đặt \(\sqrt{x-\dfrac{1}{x}}=t\ge0\)
\(\Rightarrow t^2+2t-3=0\Rightarrow\left[{}\begin{matrix}t=1\\t=-3\left(loại\right)\end{matrix}\right.\)
\(\Rightarrow x-\dfrac{1}{x}=1\)
\(\Rightarrow x^2-x-1=0\Rightarrow x=\dfrac{1\pm\sqrt{5}}{2}\)
b.
ĐKXĐ: \(x\ge0\)
\(x+\sqrt{x}-\sqrt{x+3}=0\)
\(\Leftrightarrow x-1+\sqrt{x}-1-\left(\sqrt{x+3}-2\right)=0\)
\(\Leftrightarrow x-1+\dfrac{x-1}{\sqrt{x}+1}-\dfrac{x-1}{\sqrt{x+3}+2}=0\)
\(\Leftrightarrow\left(x-1\right)\left(1+\dfrac{1}{\sqrt{x}+1}-\dfrac{1}{\sqrt{x+3}+2}\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(\dfrac{1}{\sqrt{x}+1}+\dfrac{\sqrt{x+3}+1}{\sqrt{x+3}+2}\right)=0\)
\(\Leftrightarrow x-1=0\)
\(\Leftrightarrow x\left(\sqrt{1+x}+\sqrt{1-x}\right)+\dfrac{1}{2}\left(\sqrt{1-x}+\sqrt{1+x}\right)=x\)
\(\Leftrightarrow x\left(\sqrt{1+x}+\sqrt{1-x}\right)+\dfrac{1}{2}.\dfrac{1-x-1-x}{\sqrt{1-x}+\sqrt{1+x}}=x\)
\(\Leftrightarrow x\left(\sqrt{1+x}+\sqrt{1-x}\right)-\dfrac{x}{\sqrt{1-x}+\sqrt{1+x}}=x\)
\(x=0\) la nghiem cua pt
\(x\ne0\Rightarrow pt:\sqrt{1+x}+\sqrt{1-x}-\dfrac{1}{\sqrt{1-x}+\sqrt{1+x}}=1\)
\(u=\sqrt{1+x}+\sqrt{1-x}\Rightarrow pt:u-\dfrac{1}{u}=1\)
\(\Leftrightarrow u^2-u-1=0\Leftrightarrow\left[{}\begin{matrix}u=\dfrac{1+\sqrt{5}}{2}\\u=\dfrac{1-\sqrt{5}}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{1+x}+\sqrt{1-x}=\dfrac{1+\sqrt{5}}{2}\\\sqrt{1+x}+\sqrt{1-x}=\dfrac{1-\sqrt{5}}{2}\end{matrix}\right.\)
\(\sqrt{1+x}+\sqrt{1-x}=\dfrac{1+\sqrt{5}}{2}\Leftrightarrow2+2\sqrt{1-x^2}=\dfrac{3+\sqrt{5}}{2}\)
\(\Leftrightarrow1-x^2=\left(\dfrac{\sqrt{5}-1}{4}\right)^2\Leftrightarrow x=\pm\sqrt{\dfrac{5+\sqrt{5}}{8}}\left(tm\right)\)
Nghiệm còn lại tự xét nhé :v
P/s: Ý tưởng thuộc về Ck iu , em tag anh rồi nhé ck :v
Ơ mà này, dạo này chả thấy anh Lâm onl nhờ bà nhỉ? Bà biết ảnh bay đâu r ko? Muốn hỏi bài mà mãi chả thấy hiện hồn :v
xài nhầm nick thông cảm :v
c.
ĐKXĐ: \(\left[{}\begin{matrix}x>1\\x< -2\end{matrix}\right.\)
\(\Leftrightarrow x+4-2\sqrt[]{\left(\dfrac{x+2}{x-1}\right)^2\left(\dfrac{x-1}{x+2}\right)}=0\)
\(\Leftrightarrow x+4-2\sqrt[]{\dfrac{x+2}{x-1}}=0\)
\(\Leftrightarrow x+4=2\sqrt[]{\dfrac{x+2}{x-1}}\) (\(x\ge-4\))
\(\Leftrightarrow x^2+8x+16=\dfrac{4\left(x+2\right)}{x-1}\)
\(\Rightarrow x^3+7x^2+4x-24=0\)
\(\Leftrightarrow\left(x+3\right)\left(x^2+4x-8\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-2+2\sqrt{3}\\x=-2-2\sqrt{3}\left(loại\right)\end{matrix}\right.\)
a.
\(\Leftrightarrow2x^2-11x+21=3\sqrt[3]{4\left(x-1\right)}\)
Do \(2x^2-11x+21=2\left(x-\dfrac{11}{4}\right)^2+\dfrac{47}{8}>0\Rightarrow3\sqrt[3]{4\left(x-1\right)}>0\Rightarrow x-1>0\)
Ta có:
\(VT=2x^2-11x+21-3\sqrt[3]{4x-4}=2\left(x^2-6x+9\right)+x+3-3\sqrt[3]{4\left(x-1\right)}\)
\(=2\left(x-3\right)^2+x+3-3\sqrt[3]{4\left(x-1\right)}\)
\(\Rightarrow VT\ge x+3-3\sqrt[3]{4\left(x-1\right)}=\left(x-1\right)+2+2-3\sqrt[3]{4\left(x-1\right)}\)
\(\Rightarrow VT\ge3\sqrt[3]{\left(x-1\right).2.2}-3\sqrt[3]{4\left(x-1\right)}=0\)
Đẳng thức xảy ra khi và chỉ khi:
\(\left\{{}\begin{matrix}\left(x-3\right)^2=0\\x-1=2\\\end{matrix}\right.\) \(\Leftrightarrow x=3\)
Vậy pt có nghiệm duy nhất \(x=3\)
ĐKXĐ: \(\left[{}\begin{matrix}-1\le x< 0\\x\ge\sqrt{\dfrac{5}{2}}\end{matrix}\right.\)
\(x-\dfrac{4}{x}+\sqrt{2x-\dfrac{5}{x}}-\sqrt{x-\dfrac{1}{x}}=0\)
\(\Leftrightarrow x-\dfrac{4}{x}+\dfrac{x-\dfrac{4}{x}}{\sqrt{2x-\dfrac{5}{x}}+\sqrt{x-\dfrac{1}{x}}}=0\)
\(\Leftrightarrow\left(x-\dfrac{4}{x}\right)\left(1+\dfrac{1}{\sqrt{2x-\dfrac{5}{x}}+\sqrt{x-\dfrac{1}{x}}}\right)=0\)
\(\Leftrightarrow x-\dfrac{4}{x}=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\left(loại\right)\end{matrix}\right.\)
a.
ĐKXĐ: \(x\ge0\)
\(\sqrt{2x^2+13x+5}-5\sqrt{x}+\sqrt{2x^2-3x+5}-3\sqrt{x}=0\)
\(\Leftrightarrow\dfrac{2x^2-12x+5}{\sqrt{2x^2+13x+5}+5\sqrt{x}}+\dfrac{2x^2-12x+5}{\sqrt{2x^2-3x+5}+3\sqrt{x}}=0\)
\(\Leftrightarrow\left(2x^2-12x+5\right)\left(\dfrac{1}{\sqrt{2x^2+13x+5}+5\sqrt{x}}+\dfrac{1}{\sqrt{2x^2-3x+5}+3\sqrt{x}}\right)=0\)
\(\Leftrightarrow2x^2-12x+5=0\)
\(\Leftrightarrow...\)
b.
ĐKXĐ: \(x^2\ge\dfrac{4}{3}\)
\(\sqrt{x^2-\dfrac{4}{3}}+\sqrt{4x^2-4}-x=0\)
\(\Leftrightarrow\sqrt{\dfrac{3x^2-4}{3}}+\dfrac{3x^2-4}{\sqrt{4x^2-4}+x}=0\)
\(\Leftrightarrow\sqrt{3x^2-4}\left(\dfrac{1}{\sqrt{3}}+\dfrac{\sqrt{3x^2-4}}{\sqrt{4x^2-4}+x}\right)=0\)
\(\Leftrightarrow3x^2-4=0\)
\(\Leftrightarrow...\)
a.
ĐKXĐ: \(x\ge0\)
\(\Leftrightarrow x^2+4-3\sqrt{x\left(x^2+4\right)}+2x=0\)
Đặt \(\left\{{}\begin{matrix}\sqrt{x^2+4}=a\\\sqrt{x}=b\end{matrix}\right.\)
\(\Rightarrow a^2-3ab+2b^2=0\)
\(\Leftrightarrow\left(a-b\right)\left(a-2b\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=b\\a=2b\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x^2+4}=\sqrt{x}\\\sqrt{x^2+4}=2\sqrt{x}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+4=x^2\left(vn\right)\\x^2+4=4x\end{matrix}\right.\)
\(\Leftrightarrow x=2\)
b,
ĐKXĐ: \(x\ge0\)
\(\Leftrightarrow x^2+1-\sqrt{\dfrac{x\left(x^2+1\right)}{2}}-x=0\)
Đặt \(\left\{{}\begin{matrix}\sqrt{x^2+1}=a>0\\\sqrt{\dfrac{x}{2}}=b\ge0\end{matrix}\right.\) ta được:
\(a^2-ab-2b^2=0\)
\(\Leftrightarrow\left(a+b\right)\left(a-2b\right)=0\)
\(\Leftrightarrow a-2b=0\) (do \(a+b>0\))
\(\Leftrightarrow\sqrt{x^2+1}=2\sqrt{\dfrac{x}{2}}\)
\(\Leftrightarrow x^2+1=2x\)
\(\Leftrightarrow x=1\)
Đặt \(tan\left(x+\dfrac{\pi}{3}\right)=t\)
\(\Rightarrow t^2+\left(\sqrt{3}-1\right)t-\sqrt{3}=0\)
\(\Leftrightarrow t\left(t-1\right)+\sqrt{3}\left(t-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}t=1\\t=-\sqrt{3}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}tan\left(x+\dfrac{\pi}{3}\right)=1\\tan\left(x+\dfrac{\pi}{3}\right)=-\sqrt{3}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x+\dfrac{\pi}{3}=\dfrac{\pi}{4}+k\pi\\x+\dfrac{\pi}{3}=-\dfrac{\pi}{3}+k\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{12}+k\pi\\x=-\dfrac{2\pi}{3}+k\pi\end{matrix}\right.\)
ĐKXĐ: \(\left\{{}\begin{matrix}-1\le x\le3\\x\ne1\end{matrix}\right.\)
\(\dfrac{\sqrt{x+1}\left(\sqrt{x+1}+\sqrt{3-x}\right)}{2\left(x-1\right)}>x-\dfrac{1}{2}\)
\(\Leftrightarrow\dfrac{x+1+\sqrt{-x^2+2x+3}}{x-1}>2x-1\)
- TH1: Với \(x>1\) BPT tương đương:
\(x+1+\sqrt{-x^2+2x+3}>\left(2x-1\right)\left(x-1\right)\)
\(\Leftrightarrow\sqrt{-x^2+2x+3}>2x^2-4x\)
Đặt \(\sqrt{-x^2+2x+3}=t\ge0\Rightarrow2x^2-4x=-2t^2+6\)
BPt trở thành: \(t>-2t^2+6\Leftrightarrow2t^2+t-6>0\)
\(\Rightarrow t>\dfrac{3}{2}\Rightarrow-x^2+2x+3>\dfrac{9}{4}\Rightarrow1< x< \dfrac{2+\sqrt{7}}{2}\)
TH2: với \(x< 1\) BPT tương đương:
\(x+1+\sqrt{-x^2+2x+3}< \left(2x-1\right)\left(x-1\right)\)
\(\Leftrightarrow\sqrt{-x^2+2x+3}< 2x^2-4x\)
Tương tự như trên, đặt \(t=\sqrt{-x^2+2x+3}\ge0\) ta được \(0\le t< \dfrac{3}{2}\)
\(\Rightarrow-x^2+2x+3< \dfrac{9}{4}\) \(\Rightarrow-1\le x< \dfrac{2-\sqrt{7}}{2}\)
Vậy nghiệm của BPT là: \(\left[{}\begin{matrix}-1\le x< \dfrac{2-\sqrt{7}}{2}\\1< x< \dfrac{2+\sqrt{7}}{2}\end{matrix}\right.\)