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\(B=\sqrt{\left(7x-\frac{11}{7}\right)^2+\left(\frac{8\sqrt{5}}{7}\right)^2}+\sqrt{\left(7x+\frac{11}{7}\right)^2+\left(\frac{8\sqrt{5}}{7}\right)^2}\)
\(B=\sqrt{\left(\frac{11}{7}-7x\right)^2+\left(\frac{8\sqrt{5}}{7}\right)^2}+\sqrt{\left(7x+\frac{11}{7}\right)^2+\left(\frac{8\sqrt{5}}{7}\right)^2}\)
dùng Bất đẳng thức Bunyakovsky
\(B\ge\sqrt{\left(\frac{22}{7}\right)^2+\left(\frac{16\sqrt{5}}{7}\right)^2}\)
\(B\ge6\)
dấu "=" khi x=0
\(49x^2-22x+9=\left(7x\right)^2-2.7.\dfrac{11}{7}x+\dfrac{121}{49}+\dfrac{320}{49}\)
\(=\left(7x-\dfrac{11}{7}\right)^2+\dfrac{320}{49}\ge\dfrac{320}{49}\) dấu"=" xảy ra<=>\(x=\dfrac{11}{49}\)
\(=>\sqrt{49x^2-22x+9}\ge\)\(\sqrt{\dfrac{320}{49}}=\dfrac{8\sqrt{5}}{7}\)
\(=>B\ge\dfrac{8\sqrt{5}}{7}+8\sqrt{38}\)
Câu 1:
\(\sqrt{x^2-2x+1}+\sqrt{x^2-4x+4}=3\)
\(\Leftrightarrow\left|x-1\right|+\left|x-2\right|=3\)(1)
Trường hợp 1: x<1
(1) trở thành 1-x+2-x=3
=>3-2x=3
=>x=0(nhận)
Trường hợp 2: 1<=x<2
(1) trở thành x-1+2-x=3
=>1=3(loại)
Trường hợp 3: x>=2
(1) trở thành x-1+x-2=3
=>2x-3=3
=>2x=6
hay x=3(nhận)
Bài 2:Áp dụng BĐT AM-GM ta có:
\(\frac{1}{x}+\frac{1}{y}\ge2\sqrt{\frac{1}{xy}}\)
\(\frac{1}{y}+\frac{1}{z}\ge2\sqrt{\frac{1}{yz}}\)
\(\frac{1}{x}+\frac{1}{z}\ge2\sqrt{\frac{1}{xz}}\)
CỘng theo vế 3 BĐT trên có:
\(2\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\ge2\left(\frac{1}{\sqrt{xy}}+\frac{1}{\sqrt{yz}}+\frac{1}{\sqrt{xz}}\right)\)
Khi x=y=z
Ta có: \(\frac{1}{\sqrt{1}}>\frac{1}{\sqrt{100}}\)
\(\frac{1}{\sqrt{2}}>\frac{1}{\sqrt{100}}\)
\(\frac{1}{\sqrt{3}}>\frac{1}{\sqrt{100}}\)
\(..........................\)
\(\frac{1}{\sqrt{99}}>\frac{1}{\sqrt{100}}\)
\(\frac{1}{\sqrt{100}}=\frac{1}{\sqrt{100}}\)
Cộng theo vế ta có:
\(\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+...+\frac{1}{\sqrt{100}}>\frac{1}{10}+\frac{1}{10}+...+\frac{1}{10}=\frac{100}{10}=10\)
a) \(6\sqrt{x-1}-\dfrac{1}{3}\cdot\sqrt{9x-9}+\dfrac{7}{2}\sqrt{4x-4}=24\) (ĐK: \(x\ge1\))
\(\Leftrightarrow6\sqrt{x-1}-\dfrac{1}{3}\cdot\sqrt{9\left(x-1\right)}+\dfrac{7}{2}\sqrt{4\left(x-1\right)}=24\)
\(\Leftrightarrow6\sqrt{x-1}-\dfrac{1}{3}\cdot3\sqrt{x-1}+\dfrac{7}{2}\cdot2\sqrt{x-1}=24\)
\(\Leftrightarrow6\sqrt{x-1}-\sqrt{x-1}+7\sqrt{x-1}=24\)
\(\Leftrightarrow12\sqrt{x-1}=24\)
\(\Leftrightarrow\sqrt{x-1}=\dfrac{24}{12}\)
\(\Leftrightarrow\sqrt{x-1}=2\)
\(\Leftrightarrow x-1=4\)
\(\Leftrightarrow x=4+1\)
\(\Leftrightarrow x=5\left(tm\right)\)
b) \(\dfrac{1}{2}\sqrt{4x+8}-2\sqrt{x+2}-\dfrac{3}{7}\sqrt{49x+98}=-8\) (ĐK: \(x\ge-2\))
\(\Leftrightarrow\dfrac{1}{2}\cdot2\sqrt{x+2}-2\sqrt{x+2}-\dfrac{3}{7}\cdot7\sqrt{x+2}=-8\)
\(\Leftrightarrow\sqrt{x+2}-2\sqrt{x+2}-3\sqrt{x+2}=-8\)
\(\Leftrightarrow-4\sqrt{x+2}=-8\)
\(\Leftrightarrow\sqrt{x+2}=\dfrac{-8}{-4}\)
\(\Leftrightarrow\sqrt{x+2}=2\)
\(\Leftrightarrow x+2=4\)
\(\Leftrightarrow x=4-2\)
\(\Leftrightarrow x=2\left(tm\right)\)
\(\sqrt{x^2-6x+9}+\sqrt{x^2-10x+25}+\sqrt{x^2-22x+121}\)
\(=\sqrt{\left(x-3\right)^2}+\sqrt{\left(x-5\right)^2}+\sqrt{\left(x-11\right)^2}\)
\(=\left|x-3\right|+\left|x-5\right|+\left|x-11\right|\)