Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) \(\sqrt{2x-1}=3\left(đk:x\ge\dfrac{1}{2}\right)\)
\(\Leftrightarrow2x-1=9\Leftrightarrow2x=10\Leftrightarrow x=5\)(thỏa đk)
b) \(\sqrt{1-3x}=\dfrac{1}{2}\left(đk:x\le\dfrac{1}{3}\right)\)
\(\Leftrightarrow1-3x=\dfrac{1}{4}\Leftrightarrow3x=\dfrac{3}{4}\Leftrightarrow x=\dfrac{1}{4}\)(thỏa đk)
c) \(\sqrt{\left(x-1\right)^2}=\dfrac{1}{2}\)
\(\Leftrightarrow\left|x-1\right|=\dfrac{1}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=\dfrac{1}{2}\\x-1=-\dfrac{1}{2}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{1}{2}\end{matrix}\right.\)
d) \(\sqrt{\left(1+2x\right)^2}=\dfrac{\sqrt{3}}{2}\)
\(\Leftrightarrow\left|1+2x\right|=\dfrac{\sqrt{3}}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}1+2x=\dfrac{\sqrt{3}}{2}\\1+2x=-\dfrac{\sqrt{3}}{2}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-2+\sqrt{3}}{4}\\x=-\dfrac{2+\sqrt{3}}{4}\end{matrix}\right.\)
e) \(\sqrt{\left(1-2x\right)^2}=\left|x-1\right|\)
\(\Leftrightarrow\left|1-2x\right|=\left|x-1\right|\)
\(\Leftrightarrow\left[{}\begin{matrix}1-2x=x-1\\1-2x=1-x\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=0\end{matrix}\right.\)
a: ĐKXĐ: x>=2/3
\(\dfrac{x-2}{\sqrt{3x-2}+2}=9\)
=>\(x-2=9\sqrt{3x-2}+18\)
=>\(9\sqrt{3x-2}=x-2-18=x-20\)
=>\(\Leftrightarrow\left\{{}\begin{matrix}x>=20\\81\left(3x-2\right)=x^2-40x+400\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x>=20\\x^2-40x+400-243x+162=0\end{matrix}\right.\)
=>x>=20 và x^2-283x+562=0
=>x=281(nhận) hoặc x=2(loại)
b: ĐKXĐ: x>=2/5
\(\sqrt{5x-2}=9\)
=>5x-2=81
=>5x=83
=>x=83/5
c: ĐKXĐ: x>=-1; x<>8
\(\dfrac{2x-16}{\sqrt{x+1}-3}=5\)
=>\(2x-16=5\sqrt{x+1}-15\)
=>\(\sqrt{25x+25}=2x-16+15=2x-1\)
=>\(\left\{{}\begin{matrix}x>=\dfrac{1}{2}\\4x^2-4x+1=25x+25\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>=\dfrac{1}{2}\\4x^2-29x-24=0\end{matrix}\right.\)
=>x=8(nhận) hoặc x=-3/4(loại)
\(a,\dfrac{x^2+x+2}{\sqrt{x^2+x+1}}=\dfrac{x^2+x+1+1}{\sqrt{x^2+x+1}}=\sqrt{x^2+x+1}+\dfrac{1}{\sqrt{x^2+x+1}}\left(1\right)\)
Áp dụng BĐT cosi: \(\left(1\right)\ge2\sqrt{\sqrt{x^2+x+1}\cdot\dfrac{1}{\sqrt{x^2+x+1}}}=2\)
Dấu \("="\Leftrightarrow x^2+x+1=1\Leftrightarrow x^2+x=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)
a) 1
b) \(2\sqrt{x-2}+\sqrt{x+2}\)
c)câu này để bạn tự làm nhé
Ta có : \(a\sqrt{32\left(b^2+c^2\right)}=2.2a\sqrt{2\left(b^2+c^2\right)}\le4a^2+2\left(b^2+c^2\right)\)
\(\left(b+c\right)^2\le2\left(b^2+c^2\right)\)
\(\Rightarrow12\le4\left(a^2+b^2+c^2\right)\Rightarrow a^2+b^2+c^2\ge3\)
Ngoài ra \(a\sqrt{\left(16+16\right)\left(b^2+b^2\right)}\ge a\left(4a+4b\right)\)
\(\left(b+c\right)^2\ge4bc\)
\(\Rightarrow ab+bc+ac\le3\)
\(VT=\frac{a^4}{ab+3a\sqrt{bc}}+\frac{b^4}{bc+3b\sqrt{ca}}+\frac{c^4}{ac+3c\sqrt{ba}}\)
\(\ge\frac{a^4}{ab+\frac{3}{2}\left(a^2+bc\right)}+\frac{b^4}{bc+\frac{3}{2}\left(b^2+ac\right)}+\frac{c^4}{ac+\frac{3}{2}\left(c^2+ab\right)}\)
\(\ge\frac{\left(a^2+b^2+c^2\right)^2}{\frac{3}{2}\left(a^2+b^2+c^2\right)+\frac{5}{2}\left(ab+bc+ac\right)}\)
\(\ge\frac{\left(a^2+b^2+c^2\right)^2}{\frac{3}{2}\left(a^2+b^2+c^2\right)+\frac{15}{2}}\)
Xét VT \(\ge\frac{3}{4}\)
\(\Leftrightarrow\left(a^2+b^2+c^2\right)^2\ge\frac{9}{8}\left(a^2+b^2+c^2\right)+\frac{45}{8}\)
\(\Leftrightarrow\left(a^2+b^2+c^2-3\right)+\left(a^2+b^2+c^2+\frac{15}{8}\right)\ge0\) ( luôn đúng với \(a^2+b^2+c^2\ge3\) )
\(\Rightarrowđpcm\)
Dấu " = " xảy ra khi a = b = c = 1
a) \(A=\sqrt{x-2}+\sqrt{6-x}\)
\(\Rightarrow A^2=x-2+6-x+2\sqrt{\left(x-2\right)\left(6-x\right)}\)
Ta có \(\sqrt{\left(x-2\right)\left(6-x\right)}\ge0,\forall x\)
Do đó \(A^2=4+2\sqrt{\left(x-2\right)\left(6-x\right)}\ge4\)
Mà A không âm \(\Leftrightarrow A\ge2\)
Dấu "=" \(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=6\end{matrix}\right.\)
Áp dụng BĐT Bunhiacopxky:
\(A^2=\left(\sqrt{x-2}+\sqrt{6-x}\right)^2\le\left(x-2+6-x\right)\left(1+1\right)=4\cdot2=8\)
\(\Leftrightarrow A\le\sqrt{8}\)
Dấu "=" \(\Leftrightarrow x-2=6-x\Leftrightarrow x=4\)
Mấy bài còn lại y chang nha
Tick hộ nha