HOC24
Lớp học
Môn học
Chủ đề / Chương
Bài học
Ta có:
1) \(3\overrightarrow{IA}-2\overrightarrow{IB}+\overrightarrow{IC}=\overrightarrow{0}\)
\(\overrightarrow{IA}+\overrightarrow{IC}+2\left(\overrightarrow{IA}-\overrightarrow{IB}\right)=\overrightarrow{0}\)
\(2\overrightarrow{ID}=2\overrightarrow{AB}\) (D trung điểm AC)
\(\overrightarrow{ID}=\overrightarrow{AB}\)
=> I là điểm thứ tư của h.b.h ABDI
2) \(\overrightarrow{MN}=3\overrightarrow{MA}-2\overrightarrow{MB}+\overrightarrow{MC}\)
\(=3\left(\overrightarrow{MI}+\overrightarrow{IA}\right)-2\left(\overrightarrow{MI}+\overrightarrow{IB}\right)+\overrightarrow{MI}+\overrightarrow{IC}\)
\(=2\overrightarrow{MI}\) \(\left(3\overrightarrow{IA}-2\overrightarrow{IB}+\overrightarrow{IC}=\overrightarrow{0}\right)\)
=> MN đi qua điểm I cố định
\(A=\dfrac{bc}{8a^2}+\dfrac{ca}{b^2}+\dfrac{ab}{c^2}\)
\(=\dfrac{\left(bc\right)^3+8\left(ca\right)^3+8\left(ab\right)^3}{8\left(abc\right)^2}\)
\(=\dfrac{\left(bc\right)^3+\left(2ca\right)^3+\left(2ab\right)^3}{8\left(abc\right)^2}\)
\(=\dfrac{\left(bc\right)^3+\left(2ab+2ca\right)^3-3.2ca.2ab\left(2ab+2ca\right)}{8\left(abc\right)^2}\)
\(=\dfrac{\left(bc\right)^3+\left(-bc\right)^3-3.2ca.2ab.\left(-bc\right)}{8\left(abc\right)^2}\)
\(=\dfrac{12\left(abc\right)^2}{8\left(abc\right)^2}=\dfrac{12}{8}\)
\(x^2+1=x^2+xy+yz+zx\)
\(=x\left(x+y\right)+z\left(x+y\right)=\left(x+y\right)\left(x+z\right)\)
Tương tự:
\(\left\{{}\begin{matrix}y^2+1=\left(y+z\right)\left(y+x\right)\\z^2+1=\left(z+y\right)\left(z+x\right)\end{matrix}\right.\)
\(A=x\sqrt{\dfrac{\left(x+y\right)\left(y+z\right)\left(z+x\right)\left(y+z\right)}{\left(x+y\right)\left(z+x\right)}}+y\sqrt{\dfrac{\left(z+x\right)\left(y+z\right)\left(x+y\right)\left(z+x\right)}{\left(x+y\right)\left(y+z\right)}}+z\sqrt{\dfrac{\left(x+y\right)\left(z+x\right)\left(y+z\right)\left(x+y\right)}{\left(z+x\right)\left(y+z\right)}}\)
\(=x\left|y+z\right|+y\left|z+x\right|+z\left|x+y\right|\)
TH1: x,y,z <0
\(A=-x\left(y+z\right)-y\left(z+x\right)-z\left(x+y\right)=-2\)
TH2: x,y,z>0
\(A=x\left(y+z\right)+y\left(z+x\right)+z\left(x+y\right)=2\)
Gọi H là điểm đối xứng với A qua M
Xét tam giác AMB và tam giác HMC có:
\(\left\{{}\begin{matrix}HM=AM\\\widehat{AMB}=\widehat{HMC}\\MB=MC\end{matrix}\right.\)
\(\Rightarrow\Delta AMB=\Delta HMC\left(c.g.c\right)\)
\(\Rightarrow HC=AB=6cm\)
Xét tam giác HAC có:
\(AH^2+HC^2=10^2\left(8^2+6^2=10^2\right)\)
\(\Rightarrow\widehat{AHC}=90^o\)
Mà \(\Delta AMB=\Delta HMC\)
\(\Rightarrow\widehat{MAB}=\widehat{MHC}=90^o\left(đpcm\right)\)
Số lớn là:\(\left(225+25\right):2=125\)
Số bé là:
\(125-25=100\)
\(\widehat{A}=180^o-42^o-56^o=82^o\)
\(\dfrac{AC}{sinB}=\dfrac{AB}{sinC}=\dfrac{BC}{sinA}\)\(\Leftrightarrow\left\{{}\begin{matrix}AC=\dfrac{21.sin42}{sin56}\simeq16,95cm\\BC=\dfrac{21.sin82}{sin56}\simeq25,1cm\end{matrix}\right.\)
Câu 5:
\(A=\dfrac{2}{x}+\dfrac{6}{y}+\dfrac{9}{3x+y}\)
\(=\dfrac{2\left(3x+y\right)}{xy}+\dfrac{9}{3x+y}\)
\(=\dfrac{3x+y}{6}+\dfrac{9}{3x+y}\left(xy=12\right)\)
\(=\dfrac{3x+y}{16}+\dfrac{9}{3x+y}+\dfrac{5\left(3x+y\right)}{48}\)
\(\ge2\sqrt{\dfrac{3x+y}{16}.\dfrac{9}{3x+y}}+\dfrac{5.2\sqrt{3x.y}}{48}\)
\(=2\sqrt{\dfrac{9}{16}}+\dfrac{10\sqrt{3.12}}{48}=\dfrac{11}{4}\)
Dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=6\end{matrix}\right.\)
Chọn A
\(\sqrt{24+8\sqrt{9-x^2}}=x+2\sqrt{3-x}+4\) \(\left(Đk:-3\le x\le3\right)\)
\(\sqrt{4\left(x+3\right)+8\sqrt{9-x^2}+4\left(3-x\right)}=x+2\sqrt{3-x}+4\)
\(\sqrt{\left(2\sqrt{x+3}+2\sqrt{3-x}\right)^2}=x+2\sqrt{3-x}+4\)
\(2\sqrt{x+3}+2\sqrt{3-x}=x+2\sqrt{3-x}+4\)
\(2\sqrt{x+3}=x+4\)
\(4\left(x+3\right)=x^2+8x+14\)
\(x^2+4x+2=0\)
\(\Delta=16-8=8\)
\(\Delta>0\)=> phương trình có 2 nghiệm phân biệt
\(\left[{}\begin{matrix}x=\dfrac{-4+2\sqrt{2}}{2}=-2+\sqrt{2}\\x=\dfrac{-4-2\sqrt{2}}{2}=-2-\sqrt{2}\end{matrix}\right.\)