Cho tam giac ABC vs trọng tâm G.Gọi I la trung diem của doan AG va K la diem tren cạnh AB sao cho AK=1/5 AB
a/ hay phan tich AI,AK,CI,CK theo vecto a= vecto CA,veto b =vecto CB
b/ c/m 3 điểm C,I,K thang hàng
Gọi M là trung điểm BC
+) vecto AI=vecto IG=vecto GM
+) vecto AI=1/3vecto AM=1/3(vecto CM-vecto CA)=2/3vecto CB-1/3vecto CA
+) vecto AK=1/5vecto AB=1/5vecto CB-1/5vectoCA
+) vecto CK=vecto CA+vecto AK=vecto CA+1/5vecto AB
=vecto CA+1/5vecto CB-1/5vecto CA=1/5vecto CB+4/5vecto CA
+)vecto CI=vecto CA+vecto AI= vecto CA+1/3vecto AM
=vecto CA+1/3vecto AC+1/6vecto CB=2/3vecto CA+1/6vecto CB
b/
+) vecto CI =2/3vecto CA+1/6vecto CB=5(4/30vecto CA+1/30vecto CB)
+) vecto CK=6(4/30vecto CA+1/30vecto CB)
do đó 1/5vecto CI=1/6vecto CK
Nên C,I,K thẳng hàng.
1) Cho hai doan thang AC.BD cat nhau tai I. Phat bieu nao sau day?
A) Goc BAC va DIC dong vi B) Goc ADB va CBD so le trong
C) Goc BAC va ACD so le trong D) Goc CAD va ACB so le trong
2) Cho tam giac ABC co BAC=90 do. Qua B dung duong thang b vuong goc voi AB va qua C dung duong thang c vuong goc voi AC. Phat bieu nao sau day la sai?
A) b va c cat nhau B) AB//c C) AC//b D) Ca A,B,C deu sai
3) Cho hai duong thang song song a,b. Mot duong thang c vuong goc voi ca hai duong thang a va b. Goi d la mot duong thang bat ky. Phat bieu nao sau day la dung?
A) Neu d khong co diem chung voi a thi cung khong co diem chung voi b
B) Neu d cat a thi d cung cat b
C) Neu d vuong goc voi a thi d song song voi c
D) Ca ba paht bieu tren deu dung.
4) Cho ba diem M,N,P khong thang hang. Phat bieu nao sau day sai?
A) Chi co mot duong thang di qua M va song song voi NP B) Co mot tam giac MNP
C) Chi co mot duong thang di qua M va vuong goc voi NP D) Tat ca deu sai
5) Cho biet hai duong trung truc cua hai doan thang AB va CDD cat nhau tai O. Trong cac cau sau, cau nao dung:
A) AB//CD B) AB\(\perp\)CD C) AB va CD cat nhau D) Tat ca deu sai
Câu 1: B
Câu 2: D
Câu 3: D
Câu 4: D
Câu 5: D
1) Ba diem A,B,C tao thanh 1 tam giac vuong tai C, AC = 4cm, BC = 3cm va nam trong 1 dien truong deu. Vecto cuong do dien truong vecto E song song voi AC, huong tu A → C va co do lon E = 5000 V/m. Tinh:
a) UAC, UCB, UAB
b) Cong cua dien truong khi mot electron (e) di chuyen tu A den B
cho tam giac ABC co trong tam G , H la diem doi xung cua B qua G . chung minh rang
a, vecto AH= 2\3 vecto AC - 1\3 vecto AB
b, vecto CH =1\3(vecto AB +vecto AC )
c, goi M la trung diem BC . chung minh rang : vecto MH = 1\6 vecto AC + 5\6 vecto AB
Cho dung tron (o;r) va diem a nam.ngoai duong tron sao cho oa=3r .ve tiep tuyên ab cua duong tron(o) ve day cung bc vuong goc voi oa tai h .ke duong kinh cd (o) , ad cat(o) tai m .tip tuyên tai m (o) cat ab,ac lan luot la p,q .goi k la giao diem cua pq voi tiep tuyên tai d cua (o) cmr k,b,c thang hang
Bài toán nào quá khó mà hs giỏi ko giải đc thì đó là bài toán lừa đảo.
1) Cho hinh thang ABCD vuong tai A, D co AB = a, AD = 2a va CD = 3a. Goi M,N lan luot la trung diem cua cac canh AD va DC. Khi do /2 vecto AM + 1/2 vecto DC/ bang :
A. \(\dfrac{5a}{2}\) B. 5a C. 3a D.\(\dfrac{3a}{2}\)
Lời giải:
\(|2\overrightarrow{AM}+\frac{1}{2}\overrightarrow{DC}|=|\overrightarrow{AD}+\overrightarrow{DN}|=|\overrightarrow{AN}|=AN\)
Áp dụng định lý Pitago cho tam giác $ADN$ vuông tại $D$ ta có:
\(AN=\sqrt{AD^2+DN^2}=\sqrt{(2a)^2+(\frac{3a}{2})^2}=\frac{5}{2}a\)
Đáp án A
Cau 5: Trong cac khang dinh sau, khang dinh nao dung, khang dinh nao sai?
1.Ba diem phan biet la ba diem thang hang.
2.Voi ba diem phan biet luon co hai diem nam ve cung phia doi voi diem con lai.
3.Voi ba diem thang hang phan biet luon co hai diem nam khac phia voi diem con lai.
4.Voi ba diem A,B,C thang hang thi diem B luon nam giua hai diem A,C.
5.Co mot va chi mot duong thang di qua hai diem (phan biet) cho truoc.
6.Hai duong thang phan biet thi song song voi nhau.
7.Hai tia chung goc thi cung nam tren mot duong thang.
8.Hai tia chung goc thi doi nhau
9.Neu diem M nam giua diem A,B thi no la trung diem cua doan thang AB.
10.Neu MA=MB thi M la trung diem cua doan thang AB.
11.Neu MA+MB= AB va MA=MB thi M la trung diem cua doan thang AB.
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Bai 1
Cho tam giac nhon ABC , diem M thuoc canh BC , goi D la diem doi xung voi M qua AB , goi E la diem doi xung voi M qua AC. Goi I,K la giao diem cua DE voi AB va AC
a,CMR ; MA la tia fan giac cua goc IMK
b,Tim vi tri cua diem M de DE co do dai nho nhat
Bai 2
Cho hai diem A va B nam cung phia doi voi duong thang d.goi AH , BK la cac duong vuong goc ke tu A , B den d.Goi C la diem bat ki nam giua H va K
a,Ve diem A' doi xung voi A.cmr goc ACH = goc A'CH
b,Gia su goc ACH = goc BKC ,cmr khi do 3 diem A',C,B thang hang
c,Neu cach dung diem C nam giua H va K sao cho goc ACH = goc BKC
SACH CAC DANG TOAN VA PHUONG PHAP GIAI LOP 8
1) Cho luc giac deu ABCDEF tam O. So cac vecto khac vecto khong cung phuong voi vecto OC co diem dau va diem cuoi la cac dinh cua luc giac la:
A.4 B. 6 C.7 D.9
Bài 1: Cho năm điểm bất kì A, B, C, D, E. CMR:
Vecto AB + vecto DE - vecto DB + vecto BC = Vecto AC + BE
Bài 2: Chó sáu điểm bất kì A, B, C, D, E, F. CMR:
a) Vecto AD + vecto BE + vecto CF = Vecto AE + Vecto BF + vecto CD
b) Vecto AB + vecto CD = Vecto AD + vecto CB
c)Vecto AB - vecto CD = Vecto AB - vecto BD
Bài 3: Cho tam giác ABC nội tiếp trong đường tròn (O). Gọi H là trực tâm và I là trung điểm của BC. Vẽ đường kính AK. CMR: Vecto IH + vecto IB + vecto IK + vecto IC = Vecto 0
Bài 4: Cho hình bình hành ABCD với O là tâm. CMR:
a) Vecto CO - vecto OB = Vecto BA
b) Vecto AB - vecto BC = Vecto DB
c) Vecto DA - vecto DB = Vecto OD - vecto OC
d) Vecto DA - vecto DB + vecto DC = Vecto 0
Bài 4: Cho tam giác ABC vuông cân tại A, trọng tâm G. cạnh AB=a. Gọi I là trung điểm BC. Tính độ dài vecto sau:
a) Vecto a= vecto AB + vecto AC
b) Vecto b= vecto AB + vecto AC + vecto AG
c) Vecto c= vecto BA + vecto BC
d) Vecto d= vecto AB - vecto AC + vecto BI
1) Ta có:\(\overrightarrow{AB}+\overrightarrow{DE}-\overrightarrow{DB}+\overrightarrow{BC}=\overrightarrow{AE}+\overrightarrow{BC}=\overrightarrow{AC}+\overrightarrow{CE}+\overrightarrow{BE}+\overrightarrow{EC}\)
\(=\overrightarrow{AC}+\overrightarrow{BE}+\overrightarrow{CE}+\overrightarrow{EC}=\overrightarrow{AC}+\overrightarrow{BE}\left(đpcm\right)\)2) a) Ta có: \(\overrightarrow{AD}+\overrightarrow{BE}+\overrightarrow{CF}=\overrightarrow{AE}+\overrightarrow{ED}+\overrightarrow{BF}+\overrightarrow{FE}+\overrightarrow{CD}+\overrightarrow{DF}\)\(=\overrightarrow{AE}+\overrightarrow{BF}+\overrightarrow{CD}+\overrightarrow{ED}+\overrightarrow{DF}+\overrightarrow{FE}\)
\(=\overrightarrow{AE}+\overrightarrow{BF}+\overrightarrow{CD}\left(đpcm\right)\)
b) Ta có: \(\overrightarrow{AB}+\overrightarrow{CD}=\overrightarrow{AD}+\overrightarrow{DB}+\overrightarrow{CB}+\overrightarrow{BD}\)
\(=\overrightarrow{AD}+\overrightarrow{CB}+\overrightarrow{DB}+\overrightarrow{BD}=\overrightarrow{AD}+\overrightarrow{CB}\left(đpcm\right)\)c) \(\overrightarrow{AB}-\overrightarrow{CD}=\overrightarrow{AB}-\overrightarrow{BD}\)
\(\overrightarrow{AB}+\overrightarrow{DC}=\overrightarrow{AB}+\overrightarrow{DB}\)
Ta có: \(\overrightarrow{AB}+\overrightarrow{DC}=\overrightarrow{AB}+\overrightarrow{DB}+\overrightarrow{BC}\) ( đề bài bị lỗi gì à ?? :v ) hay do mình =))
