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Xét ΔBAC có \(cosB=\frac{a^2+c^2-b^2}{2\cdot a\cdot c}\)
=>\(\left(4\sqrt2\right)^2+10^2-b^2=2\cdot4\sqrt2\cdot10\cdot cos45=8\sqrt2\cdot10\cdot\frac{\sqrt2}{2}=80\)
=>\(b^2=32+100-80=32+20=52\)
=>\(b=\sqrt{52}=2\sqrt{13}\)
Xét ΔABC có cos C=\(\frac{a^2+b^2-c^2}{2\cdot a\cdot b}\)
=>cosC=\(\frac{32+52-100}{2\cdot4\sqrt2\cdot2\sqrt{13}}=\frac{-16}{16\sqrt{26}}=-\frac{1}{\sqrt{26}}\)
=>\(\sin C=\sqrt{1-cos^2C}=\frac{5}{\sqrt{26}}\)
Diện tích tam giác CAB là:
\(S_{CAB}=\frac12\cdot CA\cdot CB\cdot\sin C\)
\(=\frac12\cdot\frac{5}{\sqrt{26}}\cdot2\sqrt{13}\cdot4\sqrt2=\frac{5\cdot2\cdot4}{2}=5\cdot4=20\)
Xét ΔABC có \(\frac{AB}{\sin C}=2R\)
=>\(2R=10:\frac{5}{\sqrt{26}}=\frac{10\sqrt{26}}{5}=2\sqrt{26}\)
=>\(R=\sqrt{26}\)
Ta có: \(S_{BCA}=\frac12\cdot AB\cdot AC\cdot\sin A\)
=>\(\frac12\cdot10\cdot2\sqrt{13}\cdot\sin A=20\)
=>\(\sin A=\frac{20}{10\sqrt{13}}=\frac{2}{\sqrt{13}}\)
\(S_{ACB}=\frac12\cdot BC\cdot h_{A}\)
=>\(\frac12\cdot4\sqrt2\cdot h_{A}=20\)
=>\(h_{A}=\frac{20}{2\sqrt2}=\frac{10}{\sqrt2}=5\sqrt2\)
Xét ΔBAC có \(cosB=\frac{a^2+c^2-b^2}{2\cdot a\cdot c}\)
=>\(\left(4\sqrt2\right)^2+10^2-b^2=2\cdot4\sqrt2\cdot10\cdot cos45=8\sqrt2\cdot10\cdot\frac{\sqrt2}{2}=80\)
=>\(b^2=32+100-80=32+20=52\)
=>\(b=\sqrt{52}=2\sqrt{13}\)
Xét ΔABC có cos C=\(\frac{a^2+b^2-c^2}{2\cdot a\cdot b}\)
=>cosC=\(\frac{32+52-100}{2\cdot4\sqrt2\cdot2\sqrt{13}}=\frac{-16}{16\sqrt{26}}=-\frac{1}{\sqrt{26}}\)
=>\(\sin C=\sqrt{1-cos^2C}=\frac{5}{\sqrt{26}}\)
Diện tích tam giác CAB là:
\(S_{CAB}=\frac12\cdot CA\cdot CB\cdot\sin C\)
\(=\frac12\cdot\frac{5}{\sqrt{26}}\cdot2\sqrt{13}\cdot4\sqrt2=\frac{5\cdot2\cdot4}{2}=5\cdot4=20\)
Xét ΔABC có \(\frac{AB}{\sin C}=2R\)
=>\(2R=10:\frac{5}{\sqrt{26}}=\frac{10\sqrt{26}}{5}=2\sqrt{26}\)
=>\(R=\sqrt{26}\)
Ta có: \(S_{BCA}=\frac12\cdot AB\cdot AC\cdot\sin A\)
=>\(\frac12\cdot10\cdot2\sqrt{13}\cdot\sin A=20\)
=>\(\sin A=\frac{20}{10\sqrt{13}}=\frac{2}{\sqrt{13}}\)
\(S_{ACB}=\frac12\cdot BC\cdot h_{A}\)
=>\(\frac12\cdot4\sqrt2\cdot h_{A}=20\)
=>\(h_{A}=\frac{20}{2\sqrt2}=\frac{10}{\sqrt2}=5\sqrt2\)
1.
\(\left|mx-3\right|=mx-3\Leftrightarrow mx-3\ge0\)
\(\Leftrightarrow mx\ge3\)
\(x^2-4=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\) \(\Rightarrow B=\left\{-2;2\right\}\)
\(B\backslash A=B\Leftrightarrow A\cap B=\varnothing\)
\(\Leftrightarrow\left\{{}\begin{matrix}-2m< 3\\2m< 3\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}m>-\frac{3}{2}\\m< \frac{3}{2}\end{matrix}\right.\)
\(\Rightarrow-\frac{3}{2}< m< \frac{3}{2}\)
2.
\(A=\left(-\infty;-3\right)\cup\left(\sqrt{6};+\infty\right)\)
À thôi nhìn tập \(C_RB\) thấy kì kì
Đề là \(\left(-5;2\right)\cup\left(\sqrt{3};\sqrt{11}\right)\) hay \(\left(-5;-2\right)\cup\left(\sqrt{3};\sqrt{11}\right)\) vậy bạn?
Vì đề như bạn ghi thì \(2>\sqrt{3}\) nên \(\left(-5;2\right)\cup\left(\sqrt{3};\sqrt{11}\right)=\left(-5;\sqrt{11}\right)\) luôn còn gì, người ta ghi dạng hợp 2 khoảng làm gì nữa?
Đề là (-5;2) \(\cup\) (\(\sqrt{3}\); \(\sqrt{11}\)) đó bạn!
f/
\(sin2A+sin2B+sin2C=2sin\left(A+B\right).cos\left(A-B\right)+2sinC.cosC\)
\(=2sinC.cos\left(A-B\right)+2sinC.cosC\)
\(=2sinC\left(cos\left(A-B\right)+cosC\right)\)
\(=2sinC\left[cos\left(A-B\right)-cos\left(A+B\right)\right]\)
\(=4sinC.sinA.sinB\)
g/
\(cos^2A+cos^2B+cos^2C=\frac{1}{2}+\frac{1}{2}cos2A+\frac{1}{2}+\frac{1}{2}cos2B+cos^2C\)
\(=1+\frac{1}{2}\left(cos2A+cos2B\right)+cos^2C\)
\(=1+cos\left(A+B\right).cos\left(A-B\right)+cos^2C\)
\(=1-cosC.cos\left(A-B\right)+cos^2C\)
\(=1-cosC\left(cos\left(A-B\right)-cosC\right)\)
\(=1-cosC\left[cos\left(A-B\right)+cos\left(A+B\right)\right]\)
\(=1-2cosC.cosA.cosB\)
d/ \(sinA+sinB+sinC=2sin\frac{A+B}{2}cos\frac{A-B}{2}+2sin\frac{C}{2}.cos\frac{C}{2}\)
\(=2cos\frac{C}{2}.cos\frac{A-B}{2}+2sin\frac{C}{2}.cos\frac{C}{2}\)
\(=2cos\frac{C}{2}\left(cos\frac{A-B}{2}+sin\frac{C}{2}\right)\)
\(=2cos\frac{C}{2}\left(cos\frac{A-B}{2}+cos\frac{A+B}{2}\right)\)
\(=4cos\frac{C}{2}.cos\frac{A}{2}.cos\frac{B}{2}\)
e/
\(cosA+cosB+cosC=2cos\frac{A+B}{2}cos\frac{A-B}{2}+1-2sin^2\frac{C}{2}\)
\(=1+2sin\frac{C}{2}.cos\frac{A-B}{2}-2sin^2\frac{C}{2}\)
\(=1+2sin\frac{C}{2}\left(cos\frac{A-B}{2}-sin\frac{C}{2}\right)\)
\(=1+2sin\frac{C}{2}\left(cos\frac{A-B}{2}-cos\frac{A+B}{2}\right)\)
\(=1+4sin\frac{C}{2}.sin\frac{A}{2}sin\frac{B}{2}\)
khó quá
anh