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Các câu hỏi dưới đây có thể giống với câu hỏi trên
2 tháng 4 2017
Vào đây nhé bạn: Câu hỏi của Công chúa Fine - Toán lớp 7 | Học trực tuyến
15 tháng 5 2022
a: Xét tứ giác ABDC có
M là trung điểm của AD
M là trung điểm của BC
Do đó; ABDC là hình bình hành
SUy ra: AB=DC
=>AC>DC
=>\(\widehat{ADC}>\widehat{DAC}\)
b: Xét ΔABC có AB<AC
mà HB là hình chiếu của AB trên BC
và HC là hình chiếu của AC trên BC
nên HB<HC
Xét ΔEBC có
HB là hình chiếu của EB trên BC
HC là hình chiếu của EC trên BC
mà HB<HC
nên EB<EC
NH
3 tháng 8 2017
- Theo dề bài ta có:
\(\left(a+b\right):\left(b+c\right):\left(c+a\right)=6:7:8\)
=> \(\dfrac{a+b}{6}=\dfrac{b+c}{7}=\dfrac{c+a}{8}\)
- Áp dụng tính chất của dãy tỉ só bằng nhau ta có:
\(\dfrac{a+b}{6}=\dfrac{b+c}{7}=\dfrac{c+a}{8}\)\(=\dfrac{a+b+b+c+c+a}{6+7+8}=\dfrac{\left(a+b+c\right).2}{21}=\dfrac{14.2}{21}=\dfrac{28}{21}=\dfrac{4}{3}\)
- Suy ra:
\(a+b=\dfrac{4}{3}.6=8\)
- Vì \(a+b+c=14\)
nên \(\Rightarrow c=14-8=6\)
- Vậy c = 6
MV
3 tháng 8 2017
\(\left(a+b\right):\left(b+c\right):\left(c+a\right)=6:7:8\\ \Rightarrow\dfrac{a+b}{6}=\dfrac{b+c}{7}=\dfrac{c+a}{8}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{a+b}{6}=\dfrac{b+c}{7}=\dfrac{c+a}{8}=\dfrac{a+b+b+c+c+a}{6+7+8}=\dfrac{2a+2b+2c}{21}=\dfrac{2\left(a+b+c\right)}{21}=\dfrac{2\cdot14}{21}=\dfrac{28}{21}=\dfrac{4}{3}\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{a+b}{6}=\dfrac{4}{3}\Rightarrow a+b=8\Rightarrow c=6\\\dfrac{b+c}{7}=\dfrac{4}{3}\Rightarrow b+c=9\dfrac{1}{3}\Rightarrow a=4\dfrac{2}{3}\\\dfrac{c+a}{8}=\dfrac{4}{3}\Rightarrow c+a=10\dfrac{2}{3}\Rightarrow b=3\dfrac{1}{3}\end{matrix}\right.\)
Vậy \(a=4\dfrac{2}{3};b=3\dfrac{1}{3};c=6\)
QA
18 tháng 6 2021
Trả lời:
Thay x = 7 vào P, ta được:
\(P=\left(x-4\right)^{\left(x-5\right)^{\left(x-6\right)^{\left(x+5\right)^{\left(x+6\right)}}}}\)
\(=\left(7-4\right)^{\left(7-5\right)^{\left(7-6\right)^{\left(7+5\right)^{\left(7+6\right)}}}}\)
\(=3^{2^{1^{12^{13}}}}\)
\(=9\)
4 tháng 6 2021
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17 tháng 5 2022
a: Xét ΔOAM vuông tại A và ΔOBM vuông tại B có
OM chung
\(\widehat{AOM}=\widehat{BOM}\)
Do đó; ΔOAM=ΔOBM
Suy ra: OA=OB
hay ΔOAB cân tại O
mà \(\widehat{AOB}=60^0\)
nên ΔOAB đều
b: Xét ΔBMF vuông tại B và ΔAME vuông tại A có
MB=MA
\(\widehat{BMF}=\widehat{AME}\)
Do đó;ΔBMF=ΔAME
Anhr mik tìm đc nha bn !!!
5 tháng 2 2022
Đề sai rồi bạn. Đã cho ΔABC rồi thì làm sao A,B,C thẳng hàng được?