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\(1.\) Gỉa sử : \(\sqrt{25-16}< \sqrt{25}-\sqrt{16}\)
\(\Leftrightarrow3< 1\) ( Vô lý )
\(\Rightarrow\sqrt{25-16}>\sqrt{25}-\sqrt{16}\)
\(2.\sqrt{a}-\sqrt{b}< \sqrt{a-b}\)
\(\Leftrightarrow\left(\sqrt{a}-\sqrt{b}\right)^2< a-b\)
\(\Leftrightarrow a-2\sqrt{ab}+b< a-b\)
\(\Leftrightarrow2b-2\sqrt{ab}< 0\)
\(\Leftrightarrow2\left(b-\sqrt{ab}\right)< 0\)
Ta có :\(a>b\Leftrightarrow ab>b^2\Leftrightarrow\sqrt{ab}>b\)
\(\RightarrowĐpcm.\)
\(2a.\) Áp dụng BĐT Cauchy , ta có :
\(a+b\ge2\sqrt{ab}\left(a;b\ge0\right)\)
\(\Leftrightarrow\dfrac{a+b}{2}\ge\sqrt{ab}\)
\(b.\) Áp dụng BĐT Cauchy cho các số dương , ta có :
\(\dfrac{1}{x}+\dfrac{1}{y}\ge\dfrac{2}{\sqrt{xy}}\left(x,y>0\right)\left(1\right)\)
\(\dfrac{1}{y}+\dfrac{1}{z}\ge\dfrac{2}{\sqrt{yz}}\left(y,z>0\right)\left(2\right)\)
\(\dfrac{1}{x}+\dfrac{1}{z}\ge\dfrac{2}{\sqrt{xz}}\left(x,z>0\right)\left(3\right)\)
Cộng từng vế của ( 1 ; 2 ; 3 ) , ta được :
\(2\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\ge2\left(\dfrac{1}{\sqrt{xy}}+\dfrac{1}{\sqrt{yz}}+\dfrac{1}{\sqrt{xz}}\right)\)
\(\Leftrightarrow\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\ge\dfrac{1}{\sqrt{xy}}+\dfrac{1}{\sqrt{yz}}+\dfrac{1}{\sqrt{xz}}\)
\(3a.\sqrt{x-4}=a\left(a\in R\right)\left(x\ge4;a\ge0\right)\)
\(\Leftrightarrow x-4=a^2\)
\(\Leftrightarrow x=a^2+4\left(TM\right)\)
\(3b.\sqrt{x+4}=x+2\left(x\ge-2\right)\)
\(\Leftrightarrow x+4=x^2+4x+4\)
\(\Leftrightarrow x^2+3x=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(TM\right)\\x=-3\left(KTM\right)\end{matrix}\right.\)
KL....
Lời giải:
Đặt \(\sqrt[3]{4-\sqrt{15}}=m\)
Khi đó \(a=\frac{1}{m}+m\Rightarrow a^3-3a=\frac{1}{m^3}+\frac{3}{m}+3m+m^3-3(\frac{1}{m}+m)\)
\(=\frac{1}{m^3}+m^3=\frac{1}{4-\sqrt{15}}+4-\sqrt{15}=4+\sqrt{15}+4-\sqrt{15}=8(*)\)
Đặt \(\sqrt[3]{\frac{25+\sqrt{621}}{2}}=n; \sqrt[3]{\frac{25-\sqrt{621}}{2}}=p\)
\(\Rightarrow n^3+p^3=25; np=\sqrt[3]{\frac{25^2-621}{4}}=1\)
\(\Rightarrow (n+p)^3=n^3+p^3+3np(n+p)=25+3(n+p)\)
Do đó:
\(b^3-b^2=\frac{1}{27}(1-n-p)^3-\frac{1}{9}(1-n-p)^2\)
\(=\frac{1}{27}[1-3(n+p)+3(n+p)^2-(n+p)^3]-\frac{1}{9}[1-2(n+p)+(n+p)^2]\)
\(=\frac{-2}{27}+\frac{n+p}{9}-\frac{(n+p)^3}{27}\)
\(=\frac{-2}{27}+\frac{n+p}{9}-\frac{25+3(n+p)}{27}=-1(**)\)
Từ \((*);(**)\Rightarrow a^3+b^3-b^2-3a+100=8+(-1)+100=107\)
a, \(\sqrt{\dfrac{289}{225}}=\sqrt{\dfrac{17^2}{15^2}}=\dfrac{17}{15}\)
b, \(\sqrt{2\dfrac{14}{25}}=\sqrt{\dfrac{64}{25}}=\sqrt{\dfrac{8^2}{5^2}}=\dfrac{8}{5}\)
c, \(\sqrt{\dfrac{0,25}{9}}=\sqrt{\dfrac{0,5^2}{3^2}}=\dfrac{0,5}{3}\)
d, \(\sqrt{\dfrac{8,1}{1,6}}=\sqrt{\dfrac{0,1}{0,1}.\dfrac{81}{16}}=\sqrt{1.\dfrac{81}{16}}=\dfrac{9}{4}\)
Chúc bạn học tốt!!!
a) \(\sqrt{\dfrac{289}{225}}\)
\(=\dfrac{\sqrt{289}}{\sqrt{225}}\)
\(=\dfrac{\sqrt{17^2}}{\sqrt{15^2}}\)
\(=\dfrac{17}{15}\)
b) \(\sqrt{2\dfrac{14}{15}}\)
\(=\sqrt{\dfrac{44}{15}}\)
\(=\dfrac{\sqrt{44}}{\sqrt{15}}\)
\(=\dfrac{2\sqrt{11}}{\sqrt{15}}\)
\(=\dfrac{2\sqrt{165}}{15}\)
c) \(\sqrt{\dfrac{0,25}{9}}\)
\(=\sqrt{\dfrac{1}{\dfrac{4}{9}}}\)
\(=\dfrac{\dfrac{1}{2}}{3}\)
\(=\dfrac{1}{6}\)
d) \(\sqrt{\dfrac{8,1}{1,6}}\)
\(=\sqrt{5,0625}\)
\(=\sqrt{\dfrac{81}{16}}\)
\(=\dfrac{9}{4}\)
a/ ĐKXĐ: $x\leq 2$
Áp dụng BĐT AM-GM:
$\sqrt{2-x}\leq (2-x)+\frac{1}{4}=\frac{9}{4}-x$
$\Rightarrow B=x+\sqrt{2-x}\leq x+\frac{9}{4}-x=\frac{9}{4}$
Vậy $B_{\max}=\frac{9}{4}$
Giá trị này đạt tại $2-x=\frac{1}{4}\Leftrightarrow x=\frac{7}{4}$
b/ ĐKXĐ: $x\geq \frac{-3}{2}$
PT $\Leftrightarrow \sqrt{2x+3}=16-x$
$\Rightarrow 2x+3=(16-x)^2=x^2-32x+256$
$\Leftrightarrow x^2-34x+253=0$
$\Leftrightarrow (x-23)(x-11)=0$
$\Rightarrow x=23$ hoặc $x=11$
Thử lại thấy $x=11$ thỏa mãn
Vậy tập nghiệm của phương trình là $\left\{11\right\}$
a: \(=\sqrt{3a}:\sqrt{b}\)
b: \(=\sqrt{a}:\sqrt{xy}\)
a) \(\sqrt{\dfrac{25}{81}.\dfrac{16}{49}.\dfrac{196}{9}}=\sqrt{\dfrac{25}{81}}.\sqrt{\dfrac{16}{49}}.\sqrt{\dfrac{196}{9}}=\dfrac{5}{9}.\dfrac{4}{7}.\dfrac{14}{3}=\dfrac{40}{27}\)
b) \(\sqrt{3\dfrac{1}{16}.2\dfrac{14}{25}.2\dfrac{34}{81}}=\sqrt{\dfrac{49}{16}.\dfrac{64}{25}.\dfrac{196}{81}}=\sqrt{\dfrac{49}{16}}.\sqrt{\dfrac{64}{25}}.\sqrt{\dfrac{196}{81}}=\dfrac{7}{4}.\dfrac{8}{5}.\dfrac{14}{9}=\dfrac{196}{45}\)
c) \(\dfrac{\sqrt{640}.\sqrt{34,3}}{\sqrt{567}}=\sqrt{\dfrac{640.34,3}{567}}=\sqrt{\dfrac{64.49}{81}}=\dfrac{\sqrt{64}.\sqrt{49}}{\sqrt{81}}=\dfrac{8.7}{9}=\dfrac{56}{9}\)
d) \(\sqrt{21,6}.\sqrt{810}.\sqrt{11^2-5^2}=\sqrt{21,6.810.\left(11^2-5^2\right)}=\sqrt{216.81.\left(11+5\right)\left(11-5\right)}=\sqrt{36^2.9^2.4^2}=36.9.4=1296\)
\(P=\sqrt{\dfrac{64}{25}}=\dfrac{8}{5}\)
a+b=13