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TN
2 tháng 10 2016
a)A=x(x+1)(x+2)(x+3)
\(=\left(x^2+3x\right)\left(x^2+3x+2\right)\)
Đặt \(t=x^2+3x\) ta đc:
\(t\left(t+2\right)\)\(=t^2+2t+1-1\)
\(=\left(t+1\right)^2-1\ge-1\)
Dấu = khi \(t=-1\Rightarrow x^2+3x=-1\)\(\Rightarrow\)\(x=\frac{-3\pm\sqrt{5}}{2}\)
Vậy MinA=-1 khi \(x=\frac{-3\pm\sqrt{5}}{2}\)
b)\(B=\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)
Với a,b,c dương ta áp dụng Bđt Cô si 3 số:
\(a+b+c\ge3\sqrt[3]{abc}\)
\(\Rightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge3\sqrt[3]{\frac{1}{abc}}\)
\(\Rightarrow\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge9\)
Dấu = khi a=b=c
Vậy MinB=9 khi a=b=c
c)\(C=a^2+b^2+c^2\)
Áp dụng Bđt Bunhiacopski 3 cặp số ta có:
\(\left(1^2+1^2+1^2\right)\left(a^2+b^2+c^2\right)\ge\left(1a+1b+1c\right)^2=\left(\frac{3}{2}\right)^2=\frac{9}{4}\)
\(\Rightarrow3\left(a^2+b^2+c^2\right)\ge\frac{9}{4}\)
\(\Rightarrow a^2+b^2+c^2\ge\frac{3}{4}\)
\(\Rightarrow C\ge\frac{3}{4}\)
Dấu = khi \(a=b=c=\frac{1}{2}\)
Vậy MinC=\(\frac{3}{4}\) khi \(a=b=c=\frac{1}{2}\)
MT
2 tháng 2 2016
câu a) rút x theo y thế vào A rồi áp dụng HĐT
b)rút xy thế vào B
c)HĐT
d)rút x theo y thé vào C
rồi dùng BĐT cô-si
e)BĐT chưa dấu giá trị tuyệt đối
28 tháng 8 2021
a) \(\sqrt{3x+10}=4\left(đk:x\ge-\dfrac{10}{3}\right)\Leftrightarrow3x+10=16\Leftrightarrow x=2\)
b) \(\sqrt{9x^2-6x+1}=\sqrt{x^2+8x+16}\Leftrightarrow\sqrt{\left(3x-1\right)^2}=\sqrt{\left(x+4\right)^2}\Leftrightarrow3x-1=x+4\Leftrightarrow2x=5\Leftrightarrow x=\dfrac{5}{2}\)
c) \(\sqrt{2x+1}=3\left(đk:x\ge-\dfrac{1}{2}\right)\Leftrightarrow2x+1=9\Leftrightarrow x=4\)
d) \(\sqrt{2x+1}+1=x\left(đk:x\ge1\right)\Leftrightarrow\sqrt{2x+1}=x-1\Leftrightarrow2x+1=x^2-2x+1\Leftrightarrow x^2-4x=0\Leftrightarrow x\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)\(\Leftrightarrow x=4\)(do \(x\ge1\))
1 tháng 12 2021
a,ĐKXĐ:\(x\ge2\)
\(4\sqrt{x-2}+\sqrt{9x-18}-\sqrt{\dfrac{x-2}{4}}=26\\ \Leftrightarrow4\sqrt{x-2}+3\sqrt{x-2}-\dfrac{\sqrt{x-2}}{2}=26\\ \Leftrightarrow8\sqrt{x-2}+6\sqrt{x-2}-\sqrt{x-2}=52\\ \Leftrightarrow13\sqrt{x-2}=52\\ \Leftrightarrow\sqrt{x-2}=4\\ \Leftrightarrow x-2=16\\ \Leftrightarrow x=18\left(tm\right)\)
b,ĐKXĐ:\(x\in R\)
\(3x+\sqrt{4x^2-8x+4}=1\\ \Leftrightarrow2\sqrt{x^2-2x+1}=1-3x\\ \Leftrightarrow\left|x-1\right|=\dfrac{1-3x}{2}\\ \Leftrightarrow\left[{}\begin{matrix}x-1=\dfrac{1-3x}{2}\\x-1=\dfrac{3x-1}{2}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}2x-2=1-3x\\2x-2=3x-1\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{5}\left(tm\right)\\x=-1\left(tm\right)\end{matrix}\right.\)
c, ĐKXĐ:\(x\ge0\)
\(\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)=7\\ \Leftrightarrow\sqrt{x}\left(2\sqrt{x}+1\right)-2\left(2\sqrt{x}+1\right)=7\\ \Leftrightarrow2x+\sqrt{x}-4\sqrt{x}-2=7\\ \Leftrightarrow2x-3\sqrt{x}-9=0\\ \Leftrightarrow\left(2x+3\sqrt{x}\right)-\left(6\sqrt{x}+9\right)=0\\ \Leftrightarrow\sqrt{x}\left(2\sqrt{x}+3\right)-3\left(2\sqrt{x}+3\right)=0\\ \Leftrightarrow\left(\sqrt{x}-3\right)\left(2\sqrt{x}+3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=3\\2\sqrt{x}=-3\left(vô.lí\right)\end{matrix}\right.\\ \Leftrightarrow x=9\left(tm\right)\)
NV
Nguyễn Việt Lâm
Giáo viên
23 tháng 6 2019
a/ \(\left(x-2\right)^2=11+6\sqrt{2}\)
\(\Leftrightarrow\left(x-2\right)^2=\left(3+\sqrt{2}\right)^2\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=3+\sqrt{2}\\x-2=-3-\sqrt{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5+\sqrt{2}\\x=-1-\sqrt{2}\end{matrix}\right.\)
b/ \(x^2-10x+25=27-10\sqrt{2}\)
\(\Leftrightarrow\left(x-5\right)^2=\left(5-\sqrt{2}\right)^2\)
\(\Leftrightarrow\left[{}\begin{matrix}x-5=5-\sqrt{2}\\x-5=\sqrt{2}-5\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=10-\sqrt{2}\\x=\sqrt{2}\end{matrix}\right.\)
c/ \(4x^2+4x+1=28-10\sqrt{3}\)
\(\Leftrightarrow\left(2x+1\right)^2=\left(5-\sqrt{3}\right)^2\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+1=5-\sqrt{3}\\2x+1=\sqrt{3}-5\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{4-\sqrt{3}}{2}\\x=\frac{-6+\sqrt{3}}{2}\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
23 tháng 6 2019
d/ \(x^2+2\sqrt{5}x+5=21-4\sqrt{5}\)
\(\Leftrightarrow\left(x+\sqrt{5}\right)^2=\left(2\sqrt{5}-1\right)^2\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\sqrt{5}=2\sqrt{5}-1\\x+\sqrt{5}=1-2\sqrt{5}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{5}-1\\x=1-3\sqrt{5}\end{matrix}\right.\)
e/ \(x^2+2\sqrt{12}x+12=13-4\sqrt{3}\)
\(\Leftrightarrow\left(x+2\sqrt{3}\right)^2=\left(2\sqrt{3}-1\right)^2\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2\sqrt{3}=2\sqrt{3}-1\\x+2\sqrt{3}=1-2\sqrt{3}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=1-4\sqrt{3}\end{matrix}\right.\)
f/ \(4x^2-12\sqrt{2}x+18=51-10\sqrt{2}\)
\(\Leftrightarrow\left(2x-3\sqrt{2}\right)^2=\left(5\sqrt{2}-1\right)^2\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-5\sqrt{2}=5\sqrt{2}-1\\2x-2\sqrt{2}=1-5\sqrt{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{10\sqrt{2}-1}{2}\\x=\frac{1-3\sqrt{2}}{2}\end{matrix}\right.\)
e, \(E=\sqrt{x^2-2x+1}+\sqrt{x^2-6x+9}\)
\(=\sqrt{\left(x-1\right)^2}+\sqrt{\left(x-3\right)^2}\)
\(=\left|x-1\right|+\left|x-3\right|=\left|x-1\right|+\left|3-x\right|\)
Áp dụng bất đẳng thức \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\) có:
\(E\ge\left|x-1+3-x\right|=\left|2\right|=2\)
Dấu " = " khi \(\left\{{}\begin{matrix}x-1\ge0\\3-x\ge0\end{matrix}\right.\Rightarrow1\le x\le3\)
Vậy \(MIN_E=2\) khi \(1\le x\le3\)
f, \(F=\sqrt{x+9-6\sqrt{x}}+\sqrt{x+1-2\sqrt{x}}\)
\(=\sqrt{\left(\sqrt{x}-3\right)^2}+\sqrt{\left(\sqrt{x}-1\right)^2}\)
\(=\left|\sqrt{x}-3\right|+\left|\sqrt{x}-1\right|=\left|3-\sqrt{x}\right|+\left|\sqrt{x}-1\right|\)
Áp dụng bất đẳng thức \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\) có:
\(F\ge\left|3-\sqrt{x}+\sqrt{x}-1\right|=\left|2\right|=2\)
Dấu " = " khi \(\left\{{}\begin{matrix}3-\sqrt{x}\ge0\\\sqrt{x}-1\ge0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x\le\sqrt{3}\\x\ge1\end{matrix}\right.\)
Vậy \(MIN_F=2\) khi \(1\le x\le\sqrt{3}\)