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a ) \(a+b+c=0\)
\(\Leftrightarrow\left(a+b+c\right)^2=0\)
\(\Leftrightarrow a^2+b^2+c^2+2\left(ab+bc+ca\right)=0\)
\(\Leftrightarrow a^2+b^2+c^2+2.0=0\)
\(\Leftrightarrow a^2+b^2+c^2=0\)
Do \(a^2\ge0;b^2\ge0;c^2\ge0\)
\(\Rightarrow a^2+b^2+c^2\ge0\)
Dấu " = " xảy ra \(\Leftrightarrow a=b=c=0\) ( * )
Thay * vào biểu thức M , ta được :
\(M=\left(0-1\right)^{1999}+0^{2000}+\left(0+1\right)^{2001}\)
\(=-1^{1999}+0+1^{2001}\)
\(=-1+0+1\)
\(=0\)
Vậy \(M=0\)
\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{1}{abc}\)
\(\Leftrightarrow\dfrac{bc}{abc}+\dfrac{ac}{abc}+\dfrac{ab}{abc}=\dfrac{1}{abc}\)
\(\Leftrightarrow\dfrac{bc+ac+ab-1}{abc}=0\)
\(\Leftrightarrow bc+ac+ab-1=0\)
\(\Leftrightarrow bc+ac+ab=1\)
Mà \(a^2+b^2+c^2=1\)
\(\Rightarrow bc+ac+ab=a^2+b^2+c^2\)
\(\Rightarrow2bc+2ac+2ab=2a^2+2b^2+2c^2\)
\(\Rightarrow2a^2+2b^2+2c^2-2bc-2ac-2ab=0\)
\(\Rightarrow\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(a^2-2ac+c^2\right)=0\)
\(\Rightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2=0\)
Do \(\left(a-b\right)^2\ge0;\left(b-c\right)^2\ge0;\left(a-c\right)^2\ge0\)
\(\Rightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2\ge0\)
Dấu " = " xảy ra \(\Leftrightarrow a=b=c\)
Mà \(P=\dfrac{a+b}{b+c}+\dfrac{b+c}{c+a}+\dfrac{c+a}{a+b}\)
\(\Rightarrow P=\dfrac{a+b}{a+b}+\dfrac{b+c}{b+c}+\dfrac{a+c}{a+c}\)
\(\Rightarrow P=1+1+1=3\)
Vậy \(P=3\)
Lời giải:
Kẻ \(NT\perp BC, CH\perp AD\) \(\Rightarrow NT\parallel CH\)
Hiển nhiên $ABCH$ là hình vuông\(\Rightarrow AH=AB=\frac{AD}{2}\Rightarrow HD=\frac{AD}{2}=HC\)
\(\Rightarrow \triangle HCD\) vuông cân tại $H$
\(\Rightarrow 45^0=\angle DCH=\angle TNC\), kéo theo tam giác \(NCT\) vuông cân tại $T$ \(\Rightarrow NT=CT\)
Xét thấy:
\(\left\{\begin{matrix} \angle BAM=\angle TMN(=90^0-\angle AMB)\\ \angle ABM=\angle MTN=90^0\end{matrix}\right.\Rightarrow \triangle ABM\sim \triangle MTN\)
\(\Rightarrow \frac{AB}{BM}=\frac{MT}{TN}\Leftrightarrow \frac{BC}{BM}=\frac{MT}{CT}\)
\(\Leftrightarrow BC.CT=MT.BM\Leftrightarrow (BM+MC)(MT-MC)=MT.BM\)
\(\Leftrightarrow MC.MT-BM.MC-MC^2=0\)
\(\Leftrightarrow MT-BM-MC=0\Leftrightarrow CT=BM\)
Khi đó, vì \(\triangle ABM\sim \triangle MTN\Rightarrow \frac{AM}{MN}=\frac{BM}{TN}=\frac{BM}{CT}=1\)
\(\Leftrightarrow AM=MN\) hay tam giác $AMN$ vuông cân .
\(e,\)
\(\left(\dfrac{1}{3}a^3b+\dfrac{1}{3}a^2b^2-\dfrac{1}{4}ab^3\right):5ab\)
\(=\dfrac{1}{15}a^2+\dfrac{1}{15}ab-\dfrac{1}{20}b^2\)
\(f,\)
\(\left(-\dfrac{2}{3}x^5y^2+\dfrac{3}{4}x^4y^3-\dfrac{4}{5}x^3y^4\right):6x^2y^2\)
\(=-\dfrac{1}{9}x^3+\dfrac{1}{8}x^2y-\dfrac{2}{15}xy^2\)
\(g,\)
\(\left(\dfrac{3}{4}a^6b^3+\dfrac{6}{5}a^3b^4-\dfrac{5}{10}ab^5\right):\left(\dfrac{3}{5}ab^3\right)\)
\(=\dfrac{5}{4}a^5+2a^2b-\dfrac{5}{6}b^2\)
A B C N D M
Giải
Ta có \(\dfrac{S_{BMN}}{S_{ABN}}=\dfrac{BM}{BA}\) (chung đường cao từ N)
mà \(\dfrac{AM}{AB}=\dfrac{1}{3}\)
Do đó: \(\dfrac{AB-AM}{AB}=\dfrac{3-1}{3}\) hay \(\dfrac{BM}{AB}=\dfrac{2}{3}\)
Nên \(\dfrac{S_{BMN}}{S_{ABN}}=\dfrac{2}{3}\)
Tương tự: \(\dfrac{S_{ABN}}{S_{ABC}}=\dfrac{BN}{BC}=\dfrac{1}{3}\) (chung đường cao từ A)
\(\Rightarrow\) \(\dfrac{S_{BMN}}{S_{ABN}}.\dfrac{S_{ABN}}{S_{ABC}}=\dfrac{2}{3}.\dfrac{1}{3}\)
\(\Rightarrow\) \(\dfrac{S_{BMN}}{S_{ABC}}=\dfrac{2}{9}\)
Tương tự: \(\dfrac{S_{DNC}}{S_{ABC}}=\dfrac{2}{9}\); \(\dfrac{S_{ADM}}{S_{ABC}}=\dfrac{2}{9}\)
Vậy SMND = SABC - SADM - SBMN - SDNC
= SABC - 3 . \(\dfrac{2}{9}\)SABC = \(\dfrac{1}{3}\)SABC = \(\dfrac{1}{3}\) . 30
= 10 (cm2)