Câu 5:
a) Do \(\widehat{HOG}=45^o\Rightarrow\widehat{BOG}+\widehat{DOH}=180^o-45^o=135^o\)
Xét tam giác DOH có \(\widehat{ODH}=45^o\Rightarrow\widehat{DHO}+\widehat{DOH}=135^o\)
Vậy nên \(\widehat{DHO}=\widehat{BOG}\)
Xét tam giác HOD và OGB có :
\(\widehat{ODH}=\widehat{GBO}=45^o\)
\(\widehat{DHO}=\widehat{BOG}\left(cmt\right)\)
Vậy nên \(\Delta HOD\sim\Delta OGB\left(g-g\right)\)
b) Do \(\Delta HOD\sim\Delta OGB\Rightarrow\frac{HD}{OB}=\frac{OD}{GB}\)
Đặt MB = x thì AD = 2x. Vậy \(OB=OD=x\sqrt{2}\)
Vậy thì \(HD.GB=OB.OD=x\sqrt{2}.x\sqrt{2}=2x^2=AD.BM\)
Từ đó suy ra \(HD.GB=AD.BM\Rightarrow\frac{HD}{MB}=\frac{AD}{GB}\)
Vậy ta có ngay \(\Delta ADH\sim\Delta GBM\left(c-g-c\right)\Rightarrow\widehat{AHD}=\widehat{GMB}\Rightarrow\widehat{HAB}=\widehat{GMB}\)
Chúng lại ở vị trí đồng vị nên MG // AH.
Bài 6:
Trên tia đối của tia DA, lấy điểm I sao cho \(\widehat{DCI}=\widehat{BAD}=\widehat{DAC}\)
Vậy thì \(\Delta BAD\sim\Delta ICD\left(g-g\right)\Rightarrow\widehat{ABD}=\widehat{CID};\frac{BD}{ID}=\frac{AD}{CD}\Rightarrow BD.CD=AD.ID\)
Từ đó ta cũng có \(\Delta BAD\sim\Delta IAC\left(g-g\right)\Rightarrow\frac{BA}{IA}=\frac{AD}{AC}\Rightarrow AB.AC=AI.AD\)
Từ đây \(AB.AC-BD.BC=AD.AI-AD.ID=AD\left(AI-ID\right)=AD^2\)
Vậy \(AD=\sqrt{AB.AC-BD.DC}\left(đpcm\right)\)
Bài 4:
a) Ta có a + b + c = 2.
+ Nếu cả ba cạnh lớn hơn 1 thì chu vị lớn hơn 3 (Loại)
+ Nếu chỉ tồn tại hai cạnh lớn hơn 1 thì chu vi lớn hơn 2 (Loại)
+ Nếu chỉ tồn tại một cạnh lớn hơn 1 thì tổng hai cạnh còn lại sẽ phải nhỏ hơn 1. Điều này vô lí vì theo bất đẳng thức trong tam giác, độ dài một cạnh phải nhỏ hơn tổng độ dài hai cạnh còn lại.
Tóm lại cả a, b và c đều phải nhỏ hơn 1.
b) Đề bài tương đương: Cho \(0< a,b,c< 1\). Chứng minh rằng \(a^2+b^2+c^2< 2\left(1-abc\right)\)
Ta có \(S=\sqrt{p\left(p-a\right)\left(p-b\right)\left(p-c\right)};p=2:2=1\)
\(\Rightarrow S^2=\left(1-a\right)\left(1-b\right)\left(1-c\right)=1-\left(a+b+c\right)+\left(ab+bc+ca\right)-abc\)
\(=-1+\left(ab+bc+ca\right)-abc>0\)
\(\Rightarrow abc+1< ab+bc+ca\)
Mà \(ab+bc+ca=\frac{\left(a+b+c\right)^2-\left(a^2+b^2+c^2\right)}{2}=\frac{4-\left(a^2+b^2+c^2\right)}{2}\)
\(\Rightarrow abc+1< \frac{4-\left(a^2+b^2+c^2\right)}{2}\Rightarrow a^2+b^2+c^2< 2-2abc=2\left(1-abc\right)\left(đpcm\right)\)
a/ Ta có:
\(a+a< a+b+c=2\)
\(\Leftrightarrow a< 1\)
Tương tự ta có \(\hept{\begin{cases}b< 1\\c< 1\end{cases}}\)
b/ Ta có: \(\left(1-a\right)\left(1-b\right)\left(1-c\right)>0\)
\(\Leftrightarrow1-\left(a+b+c\right)+\left(ab+bc+ca\right)-abc>0\)
\(\Leftrightarrow-1+\left(ab+bc+ca\right)-abc>0\)
\(\Leftrightarrow-2+2\left(ab+bc+ca\right)-2abc>0\)
\(\Leftrightarrow-2+a^2+b^2+c^2+2\left(ab+bc+ca\right)-2abc>a^2+b^2+c^2\)
\(\Leftrightarrow a^2+b^2+c^2< 2-2abc=2\left(1-abc\right)\)
7/ Ta có:
Thay \(x=9\) vào ta được
\(f\left(\sqrt{9}-1\right)+f\left(2\right)=9^2\)
\(\Leftrightarrow f\left(2\right)+f\left(2\right)=81\)
\(\Leftrightarrow f\left(2\right)=\frac{81}{2}\)
Thay \(x=\left(\sqrt{2}+1\right)^2\) vào ta được
\(f\left(\sqrt{\left(\sqrt{2}+1\right)^2}-1\right)+f\left(2\right)=\left(\sqrt{2}+1\right)^4\)
\(\Leftrightarrow f\left(\sqrt{2}\right)+f\left(2\right)=\left(\sqrt{2}+1\right)^4\)
\(\Leftrightarrow f\left(\sqrt{2}\right)=\left(\sqrt{2}+1\right)^4-f\left(2\right)=\left(\sqrt{2}+1\right)^4-\frac{81}{2}\)
cô nhầm hay sao ạ
BĐT tam giác là tổng 2 cạnh ;ớn hơn hoặc bằng cạnh còn lại chứ ạ
a+b\(\ge c\)suy ra c\(\le1\)
a+c\(\ge\)b suy ra b\(\le1\)
b+c\(\ge a\)suy ra a \(\le1\)
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