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a/ ĐKXĐ: \(x^2+5x+2\ge0\Rightarrow x...\left(casio\right)\)
\(x^2+5x-2-3\sqrt{x^2+5x+2}=0\)
Đặt \(\sqrt{x^2+5x+2}=a\ge0\)
\(\Rightarrow a^4-4-3a=0\Rightarrow\left[{}\begin{matrix}a=-1< 0\left(l\right)\\a=4\end{matrix}\right.\)
\(\Rightarrow\sqrt{x^2+5x+2}=4\Leftrightarrow x^2+5x-14=0\Rightarrow\left[{}\begin{matrix}x=2\\x=-7\end{matrix}\right.\)
b/ \(x^2-6x+9+3x-22-\sqrt{x^2-3x+7}=0\)
\(\Leftrightarrow x^2-3x+7-\sqrt{x^2-3x+7}-20=0\)
Đặt \(\sqrt{x^2-3x+7}=a>0\)
\(a^2-a-20=0\Rightarrow\left[{}\begin{matrix}a=5\\a=-4< 0\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{x^2-3x+7}=5\Leftrightarrow x^2-3x-18=0\Rightarrow\left[{}\begin{matrix}x=-3\\x=6\end{matrix}\right.\)
c/ĐKXĐ: \(\left[{}\begin{matrix}x\ge-1\\x\le-2\end{matrix}\right.\)
\(x^2+3x+2-\sqrt{x^2+3x+2}-6=0\)
Đặt \(\sqrt{x^2+3x+2}=a\ge0\)
\(a^2-a-6=0\Rightarrow\left[{}\begin{matrix}a=-2< 0\left(l\right)\\a=3\end{matrix}\right.\)
\(\Rightarrow\sqrt{x^2+3x+2}=3\Leftrightarrow x^2+3x-7=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{-3+\sqrt{37}}{2}\\x=\dfrac{-3-\sqrt{37}}{2}\end{matrix}\right.\)
Người đi hỏi có thể gợi ý câu mình hỏi cơ à, ngầu vậy :)
ĐKXĐ: \(\left[{}\begin{matrix}x\ge1\\x\le-1\end{matrix}\right.\)
Đặt \(\sqrt[3]{x+1}=a;\sqrt[3]{x-1}=b\)
Ta có hệ: \(\left\{{}\begin{matrix}a-b=\sqrt{ab}\\a^3-b^3=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a^2-3ab+b^2=0\\a^3-b^3=2\end{matrix}\right.\)
Quy về hệ đối xứng loại 1 rồi đó, S P mà giải
1.A sai đề ?
1.B : \(x^2+x+6+2x\sqrt{x+3}=4\left(x+\sqrt{x+3}\right)\)
\(\Leftrightarrow x^2+x+6+2x\sqrt{x+3}=4x+4\sqrt{x+3}\)
\(\Leftrightarrow x^2+x+6+2x\sqrt{x+3}-4x-4\sqrt{x+3}=0\)
\(\Leftrightarrow x^2-3x+6+2x\sqrt{x+3}-4\sqrt{x+3}=0\)
\(\Leftrightarrow x^2-3x+6+2\sqrt{x+3}\left(x-2\right)=0\)
\(\Leftrightarrow x+3+2\sqrt{x+3}\left(x-2\right)+\left(x-2\right)^2-1=0\)
\(\Leftrightarrow\left(\sqrt{x+3}+x-2\right)^2-1=0\)
\(\Leftrightarrow\left(\sqrt{x-3}+x-3\right)\left(\sqrt{x-3}+x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-3}+x-3=0\\\sqrt{x-3}+x-1=0\end{matrix}\right.\)
Đến đây dễ rồi
Đáp án : \(\left[{}\begin{matrix}x=3\\x=\varnothing\end{matrix}\right.\)
2.A đang nghĩ
2.B
Áp dụng bất đẳng thức Cô-si :
\(\frac{x}{\sqrt{4x-1}}+\frac{\sqrt{4x-1}}{x}\ge2\sqrt{\frac{x\left(\sqrt{4x-1}\right)}{\left(\sqrt{4x-1}x\right)}}=2\)
Dấu "=" xảy ra \(\Leftrightarrow\frac{x}{\sqrt{4x-1}}=\frac{\sqrt{4x-1}}{x}\)
\(\Leftrightarrow x^2=4x-1\)
\(\Leftrightarrow x^2-4x+1=0\)
\(\Leftrightarrow x=2\pm\sqrt{3}\)( thỏa )
Vậy....
lời giải
a)
\(\left(x+1\right)\left(2x-1\right)+x\le2x^2+3\)
\(\Leftrightarrow2x^2+x-1+x\le2x^2+3\)
\(\Leftrightarrow2x\le4\Rightarrow x\le2\)
\(\)b) \(\left(x+1\right)\left(x+2\right)\left(x+3\right)-x>x^3+6x^2-5\)
\(\left(x^2+3x+2\right)\left(x+3\right)-x>x^3+6x^2-5\)
\(x^3+3x^2+3x^2+9x+2x+6-x>x^3+6x^2-5\)
\(10x+6>-5\Rightarrow x>-\dfrac{11}{10}\)
c)Đkxđ: x≥0
x+√x>(2√x+3)(√x−1)
⇔x+√x>2x+√x−3
⇔x−3>0
⇔x>3. (tmđk).
a/ ĐKXĐ: \(-2\le x\le5\)
\(\sqrt{x+2}+\sqrt{5-x}+\sqrt{\left(x+2\right)\left(5-x\right)}-4=0\)
Đặt \(\sqrt{x+2}+\sqrt{5-x}=a>0\Rightarrow\sqrt{\left(x+2\right)\left(5-x\right)}=\frac{a^2-7}{2}\)
\(\Rightarrow a+\frac{a^2-7}{2}-4=0\)
\(\Leftrightarrow a^2+2a-15=0\Rightarrow\left[{}\begin{matrix}a=3\\a=-5\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{\left(x+2\right)\left(5-x\right)}=\frac{a^2-7}{2}=1\)
\(\Leftrightarrow-x^2+3x+10=1\)
\(\Leftrightarrow x^2-3x-9=0\)
b/ \(\Leftrightarrow\sqrt{x+1}-\sqrt{4-x}+2\left(5+2\sqrt{\left(x+1\right)\left(4-x\right)}\right)=17\)
Đặt \(\sqrt{x+1}-\sqrt{4-x}=a\Rightarrow\sqrt{\left(x+1\right)\left(4-x\right)}=\frac{5-a^2}{2}\)
\(a+2\left(5+5-a^2\right)=17\)
\(\Leftrightarrow-2a^2+a+3=0\Rightarrow\left[{}\begin{matrix}a=-1\\a=\frac{3}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\sqrt{x+1}-\sqrt{4-x}=-1\\\sqrt{x+1}-\sqrt{4-x}=\frac{3}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x+1}+1=\sqrt{4-x}\\2\sqrt{x+1}=2\sqrt{4-x}+3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2+2\sqrt{x+1}=4-x\\4x+4=25-4x+12\sqrt{4-x}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x+1}=1-x\left(x\le1\right)\\12\sqrt{4-x}=8x-21\left(x\ge\frac{21}{8}\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=\left(1-x\right)^2\\144\left(4-x\right)=\left(8x-21\right)^2\end{matrix}\right.\)
c/ ĐKXĐ: \(0\le x\le1\)
Đặt \(\sqrt{x}+\sqrt{1-x}=a>0\Rightarrow\sqrt{x-x^2}=\frac{a^2-1}{2}\)
\(a^2-1=3\left(a-1\right)\Leftrightarrow a^2-3a+2=0\Rightarrow\left[{}\begin{matrix}a=1\\a=2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\sqrt{x-x^2}=\frac{a^2-1}{2}=0\\\sqrt{x-x^2}=\frac{a^2-1}{2}=\frac{3}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-x^2=0\\x-x^2=\frac{9}{4}\left(vn\right)\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
d/ ĐKXĐ: ...
Đặt \(\left\{{}\begin{matrix}\sqrt{5+2x}=a\ge0\\\sqrt{5-2x}=b\ge0\end{matrix}\right.\) ta được:
\(\left\{{}\begin{matrix}\left(3a-1\right)\left(3b-1\right)=16\\a^2+b^2=10\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}3ab-\left(a+b\right)=5\\\left(a+b\right)^2-2ab=10\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a+b=3ab-5\\\left(a+b\right)^2-2ab=10\end{matrix}\right.\)
\(\Rightarrow\left(3ab-5\right)^2-2ab=10\)
\(\Leftrightarrow9\left(ab\right)^2-32ab+15=0\Rightarrow\left[{}\begin{matrix}ab=3\\ab=\frac{5}{9}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\left(ab\right)^2=9\\\left(ab\right)^2=\frac{25}{81}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}25-4x^2=9\\25-4x^2=\frac{25}{81}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x^2=4\\x^2=\frac{500}{81}\end{matrix}\right.\)
Đặt \(t=\sqrt{x+2}+\sqrt{5-x}\Rightarrow t^2=7+2\sqrt{\left(x+2\right)\left(5-x\right)}\)
=> \(\sqrt{\left(x+2\right)\left(5-x\right)}=\dfrac{t^2-7}{2}\); t2 \(\ge\)7
=> t + \(\dfrac{t^2-7}{2}=4\) <=> \(\dfrac{t^2+2t-15}{2}=0\Leftrightarrow\left[{}\begin{matrix}t=3\\t=-5\end{matrix}\right.\)
t = 3 <=> \(\sqrt{x+2}+\sqrt{5-x}=3\Rightarrow x+2+5-x+2\sqrt{\left(x+2\right)\left(5-x\right)=9}\)<=> \(\sqrt{\left(x+2\right)\left(5-x\right)}=1\Leftrightarrow\left(x+2\right)\left(5-x\right)=1\Leftrightarrow-x^2+3x+9=0\Leftrightarrow\left[{}\begin{matrix}\dfrac{3+3\sqrt{5}}{2}\\\dfrac{3-3\sqrt{5}}{2}\end{matrix}\right.\)
thanks y