Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Bạn tham khảo:
Cho bất phương trình x2-6x +2(m+2)|x-3| +m2 +4m +12 >0có bao nhiêu giá trị nguyên của m ϵ [-10;10] để bất phương tình... - Hoc24
Câu a bạn coi lại đề
b. ĐKXĐ: \(x\ge0;x\ne1\)
\(\Leftrightarrow\dfrac{\sqrt{2x+1}+\sqrt{3x}}{1-x}=\dfrac{\sqrt{3x+2}}{1-x}\)
\(\Leftrightarrow\sqrt{2x+1}+\sqrt{3x}=\sqrt{3x+2}\)
\(\Leftrightarrow5x+1+2\sqrt{3x\left(2x+1\right)}=3x+2\)
\(\Leftrightarrow2\sqrt{6x^2+3x}=1-2x\) (\(x\le\dfrac{1}{2}\) )
\(\Leftrightarrow4\left(6x^2+3x\right)=4x^2-4x+1\)
\(\Leftrightarrow20x^2+16x-1=0\)
\(\Rightarrow x=\dfrac{-4+\sqrt{21}}{10}\)
\(\Leftrightarrow\dfrac{b^2+c^2-a^2}{2abc}+\dfrac{a^2+c^2-b^2}{2abc}+\dfrac{a^2+b^2-c^2}{2abc}=\dfrac{a}{bc}\)
\(\Leftrightarrow\dfrac{a^2+b^2+c^2}{2abc}=\dfrac{a}{bc}\)
\(\Leftrightarrow a^2+b^2+c^2=2a^2\)
\(\Leftrightarrow a^2=b^2+c^2\)
\(\Rightarrow\) Tam giác vuông tại A theo Pitago đảo
\(\Leftrightarrow\left\{{}\begin{matrix}x^3-x^2y-7\left(x-y\right)=x^2+y^2+2xy+4\\3x^2+y^2-8\left(x-y\right)+4=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x^2-7\right)\left(x-y\right)-x^2-2xy=y^2+4\\3x^2-8\left(x-y\right)=-y^2-4\end{matrix}\right.\)
Cộng vế:
\(\left(x^2-7\right)\left(x-y\right)-8\left(x-y\right)+2x^2-2xy=0\)
\(\Leftrightarrow\left(x^2-15\right)\left(x-y\right)+2x\left(x-y\right)=0\)
\(\Leftrightarrow\left(x-y\right)\left(x^2+2x-15\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=y\\x^2+2x-15=0\end{matrix}\right.\)
\(\Leftrightarrow...\)
\(f\left(x\right)=\left(m+1\right)x^2+mx+m\)
TH1: \(m+1=0\Leftrightarrow m=-1\Rightarrow f\left(x\right)>0,\forall x\in R\)
TH2: \(m+1\ne0\Leftrightarrow m\ne-1\)
Yêu cầu bài toán thỏa mãn khi \(\left\{{}\begin{matrix}\Delta=-3m^2-4m< 0\\m+1< 0\end{matrix}\right.\Leftrightarrow m< -\frac{4}{3}\)
Đ/s: \(m< -\frac{4}{3};m=-1\)
1a) \(\frac{x\sqrt{x}+y\sqrt{y}}{\sqrt{x}+\sqrt{y}}-\left(\sqrt{x}-\sqrt{y}\right)^2=\frac{\left(\sqrt{x}+\sqrt{y}\right)\left(x-\sqrt{xy}+y\right)}{\sqrt{x}+\sqrt{y}}-x+2\sqrt{xy}-y\)
\(=x-\sqrt{xy}+y-x+2\sqrt{xy}-y=\sqrt{xy}\)
b) \(\sqrt{\frac{x-2\sqrt{x}+1}{x+2\sqrt{x}+1}}=\sqrt{\frac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}+1\right)^2}}=\frac{\sqrt{x}-1}{\sqrt{x}+1}\)
2. a) ĐK: x \(\ge\)4
Ta có: \(\sqrt{16x-64}=2\) <=> \(4\sqrt{x-4}=2\) <=> \(\sqrt{x-4}=\frac{1}{2}\)
<=> \(x-4=\frac{1}{4}\) <=> \(x=\frac{17}{4}\left(tm\right)\)
b) \(16x^2-\left(1+\sqrt{2}\right)^2=0\) <=> \(\left(4x-1-\sqrt{2}\right)\left(4x+1+\sqrt{2}\right)=0\)
<=> \(\orbr{\begin{cases}x=\frac{1+\sqrt{2}}{4}\\x=\frac{-1-\sqrt{2}}{4}\end{cases}}\)
c)Đk: x \(\ge\)0
\(x-2\sqrt{3x}+3=4\) <=> \(\left(\sqrt{x}-\sqrt{3}\right)^2=4\)
<=> \(\left(\sqrt{x}-\sqrt{3}-2\right)\left(\sqrt{x}-\sqrt{3}+2\right)=0\)
<=> \(\orbr{\begin{cases}\sqrt{x}=\sqrt{3}+2\\\sqrt{x}=\sqrt{3}-2\left(loại\right)\end{cases}}\)
<=> \(x=5+4\sqrt{3}\)
3a) \(\frac{10+2\sqrt{10}}{\sqrt{5}+\sqrt{2}}+\frac{8}{1-\sqrt{5}}=\frac{\sqrt{20}.\left(\sqrt{5}+\sqrt{2}\right)}{\sqrt{5}+\sqrt{2}}+\frac{8\left(1+\sqrt{5}\right)}{\left(1-\sqrt{5}\right)\left(1+\sqrt{5}\right)}\)
\(=4\sqrt{5}-\frac{8\left(1+\sqrt{5}\right)}{5-1}=4\sqrt{5}-2\left(1+\sqrt{5}\right)=4\sqrt{5}-2-2\sqrt{5}=2\sqrt{5}-2\)
b) \(\frac{2\sqrt{8}-\sqrt{12}}{\sqrt{18}-\sqrt{48}}-\frac{\sqrt{5}+\sqrt{27}}{\sqrt{30}+\sqrt{162}}=\frac{4\sqrt{2}-2\sqrt{3}}{3\sqrt{2}-4\sqrt{3}}-\frac{\sqrt{5}+3\sqrt{3}}{\sqrt{30}+9\sqrt{2}}\)
\(=\frac{2\left(2\sqrt{2}-3\right)}{\sqrt{6}\left(\sqrt{3}-2\sqrt{2}\right)}-\frac{\sqrt{5}+3\sqrt{3}}{\sqrt{6}\left(\sqrt{5}+3\sqrt{3}\right)}=-\frac{2}{\sqrt{6}}-\frac{1}{\sqrt{6}}=-\frac{3}{\sqrt{6}}=-\frac{\sqrt{6}}{2}\)