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d/ \(B=180^0-\left(A+C\right)=75^0\)
\(\Rightarrow b=c=4,5\)
\(\frac{a}{sinA}=\frac{b}{sinB}\Rightarrow a=\frac{b.sinA}{sinB}=\frac{9}{4}\left(\sqrt{6}-\sqrt{2}\right)\)
e/ \(cosA=\frac{b^2+c^2-a^2}{2bc}\Rightarrow a=\sqrt{b^2+c^2-2bc.cosA}\approx23\)
\(cosB=\frac{a^2+c^2-b^2}{2ac}=\frac{433}{460}\Rightarrow B\approx19^043'\)
\(\Rightarrow C=180^0-\left(A+B\right)=...\)
f/ \(cosA=\frac{b^2+c^2-a^2}{2bc}=\frac{11}{15}\Rightarrow A\approx42^050'\)
\(cosB=\frac{a^2+c^2-b^2}{2ac}=\frac{17}{35}\Rightarrow B\approx60^056'\)
\(C=180^0-\left(A+B\right)=...\)
a/ \(cosA=\frac{b^2+c^2-a^2}{2bc}=-\frac{1}{2}\Rightarrow A=120^0\)
\(cosB=\frac{a^2+c^2-b^2}{2ac}=\frac{\sqrt{2}}{2}\Rightarrow B=45^0\)
\(C=180^0-\left(A+B\right)=15^0\)
b/\(A=180^0-\left(B+C\right)=79^037'\)
\(\frac{a}{sinA}=\frac{b}{sinB}=\frac{c}{sinC}\Rightarrow\left\{{}\begin{matrix}b=\frac{sinB}{sinA}.a\approx61\\c=\frac{sinC}{sinA}.a\approx102\end{matrix}\right.\)
c/\(\frac{a}{sinA}=\frac{b}{sinB}\Rightarrow sinB=\frac{bsinA}{a}\approx0,6\Rightarrow B\approx36^052'\)
\(\Rightarrow C=180^0-\left(A+B\right)=75^045'\)
\(\frac{a}{sinA}=\frac{c}{sinC}\Rightarrow c=\frac{a.sinC}{sinA}\approx21\)
Ta luôn có \(\left(\dfrac{1}{\sqrt{a}}-\dfrac{1}{\sqrt{b}}\right)^2\ge0\forall a;b\)
\(\Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{2}{\sqrt{ab}}\)
\(\Leftrightarrow2\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\ge\dfrac{2}{\sqrt{ab}}+\dfrac{1}{a}+\dfrac{1}{b}\)
\(\Leftrightarrow\dfrac{2\left(a+b\right)}{ab}\ge\left(\dfrac{1}{\sqrt{a}}+\dfrac{1}{\sqrt{b}}\right)^2\)
\(\Leftrightarrow\sqrt{\dfrac{2\left(a+b\right)}{ab}}\ge\dfrac{1}{\sqrt{a}}+\dfrac{1}{\sqrt{b}}\)
Tương tự cho 2 BĐT còn lại rồi cộng theo vế:
\(\sqrt{2}\left(\sqrt{\dfrac{a+b}{ab}}+\sqrt{\dfrac{b+c}{bc}}+\sqrt{\dfrac{a+c}{ac}}\right)\ge2\left(\dfrac{1}{\sqrt{a}}+\dfrac{1}{\sqrt{b}}+\dfrac{1}{\sqrt{c}}\right)\)
\(\Leftrightarrow\sqrt{\dfrac{a+b}{ab}}+\sqrt{\dfrac{b+c}{bc}}+\sqrt{\dfrac{a+c}{ac}}\ge\sqrt{\dfrac{2}{a}}+\sqrt{\dfrac{2}{b}}+\sqrt{\dfrac{2}{c}}\)
\("="\Leftrightarrow a=b=c\)
\(P=\sqrt{a^2+\dfrac{1}{a^2}}+\sqrt{b^2+\dfrac{1}{b^2}}+\sqrt{c^2+\dfrac{1}{c^2}}\)
\(\Leftrightarrow\sqrt{\dfrac{97}{4}}P=\sqrt{4+\dfrac{81}{4}}\sqrt{a^2+\dfrac{1}{a^2}}+\sqrt{4+\dfrac{81}{4}}\sqrt{b^2+\dfrac{1}{b^2}}+\sqrt{4+\dfrac{81}{4}}\sqrt{c^2+\dfrac{1}{c^2}}\)
\(\ge\left(2a+\dfrac{9}{2a}\right)+\left(2b+\dfrac{9}{2b}\right)+\left(2c+\dfrac{9}{2c}\right)\)
\(=2\left(a+b+c\right)+\dfrac{9}{2}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)
\(\ge4+\dfrac{9}{2}.\dfrac{9}{a+b+c}=4+\dfrac{81}{4}=\dfrac{97}{4}\)
\(\Rightarrow P\ge\sqrt{\dfrac{97}{4}}\)
PS: Lần sau chép đề cẩn thận nhé bạn.
*, \(A< 1\Rightarrow\dfrac{\sqrt{a}-4}{\sqrt{a}-2}< 1\)
\(\Leftrightarrow\dfrac{\sqrt{a}-4}{\sqrt{a}-2}-1< 0\Leftrightarrow\dfrac{\sqrt{a}-4-1\left(\sqrt{a}-2\right)}{\sqrt{a}-2}< 0\Leftrightarrow\dfrac{-2}{\sqrt{a}-2}< 0\)Do -2<0 nên \(\sqrt{a}-2>0\Leftrightarrow a>4\)
Vậy \(a>4\) thì A<1. câu sau cmtt
\(B=\dfrac{\sqrt{a}-2}{\sqrt{a}+1}< 0\)
Đk do a trong căn\(\Rightarrow a\ge0\Rightarrow\sqrt{a}+1\ge1\)
do B<0 mà mẫu dương nên tử số phải âm hay\(\sqrt{a}-2>0\Leftrightarrow a>4\)
vậy a>4 thì B<0
Lớp 10 mà k làm dc cái này là phải xem lại nhé
\(a+b+c=2\sqrt{a}+2\sqrt{b}+2\sqrt{c}-3\)
\(\Leftrightarrow a+b+c-2\sqrt{a}-2\sqrt{b}-2\sqrt{c}+3=0\)
\(\Leftrightarrow\left(a-2\sqrt{a}+1\right)+\left(b-2\sqrt{b}+1\right)+\left(c-2\sqrt{c}+1\right)=0\)
\(\Leftrightarrow\left(\sqrt{a}-1\right)^2+\left(\sqrt{b}-1\right)^2+\left(\sqrt{c}-1\right)^2=0\)
Xảy ra khi \(a=b=c=1\)