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a: Xét ΔAMB và ΔCMD có

MA=MC

\(\widehat{AMB}=\widehat{CMD}\)(hai góc đối đỉnh)

MB=MD

Do đó: ΔAMB=ΔCMD

b: ΔAMB=ΔCMD

=>AB=CD
 mà AB=AC

nên CD=CA

=>ΔCDA cân tại C

c: ΔABC cân tại A

mà AH là đường cao

nên H là trung điểm của BC

Xét ΔABC có

AH,BM là các đường trung tuyến

AH cắt BM tại I

Do đó: I là trọng tâm của ΔABC

Xét ΔIBC có

IH là đường cao

IH là đường trung tuyến

Do đó: ΔIBC cân tại I

=>IB=IC

Xét ΔABC có

BM là đường trung tuyến

I là trọng tâm

Do đó: \(BI=\dfrac{2}{3}BM=\dfrac{2}{3}\cdot\dfrac{1}{2}\cdot BD=\dfrac{1}{3}BD\)

=>BD=3BI

Xét ΔABC có

I là trọng tâm

CI cắt AB tại N

Do đó: N là trung điểm của AB; IN=1/2IC

=>\(IN=\dfrac{1}{2}IB\)

\(\dfrac{IN}{BD}=\dfrac{BI}{2}:3BI=\dfrac{BI}{2\cdot3BI}=\dfrac{1}{6}\)

13 tháng 5

BẠN KẾT BẠN VỚI MÌNH NHÉ XIN BẠN ĐÓ

1 tháng 10 2021

máy tính hay tv đấy 

1 tháng 10 2021

Máy tính như hacker í

27 tháng 12 2018

24 - 16(x - 1/2) = 23

=> 16(x - 1/2) = 24 - 23

=> 16(x - 1/2) = 1

=> x - 1/2 = 1/16

=> x = 1/16 + 1/2

=> x = 9/16

27 tháng 12 2018

\(24-16(x-\frac{1}{2})=23\)

\(16(x-\frac{1}{2})=24-23\)

\(16(x-\frac{1}{2})=1\)

\(x-\frac{1}{2}=\frac{1}{16}\)

\(x=\frac{1}{16}+\frac{1}{2}\)

\(x=\frac{9}{16}\)

Vậy số thực x cần tìm là \(\frac{9}{16}\)

Chúc bạn hok tốt ~

saiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiii

3 tháng 7 2021

B1: a)Dấu hiệu: Điểm ktra môn Toán của 1 nhóm hs

b)Điểm(x)    |  7  |  8  |  9  |  10  |

   Tần số(n) |  5  |  7  |  5  |     3  |  N=20

-Nhận xét: +Có 3 bạn đạt điểm cao nhất là 10 điểm

                  +Có 5 bạn điểm thấp là 7 điểm

                  +Có 20 bạn tham gia làm bài

c)AD CT tính số TBC:

     \(\dfrac{x_1.n_1+x_2.n_2+...+x_4.n_4}{N}\)

=\(\dfrac{7.5+8.7+9.5+10.3}{20}\)

=8,3

-Mo=8

 

Bài 4: 

a) Xét ΔCAE vuông tại C và ΔDAE vuông tại D có 

BE chung

AC=AD(gt)

Do đó: ΔCAE=ΔDAE(Cạnh huyền-cạnh góc vuông)

Suy ra: \(\widehat{CAE}=\widehat{DAE}\)(hai góc tương ứng)

mà tia AE nằm giữa hai tia AC,AB

nên AE là tia phân giác của \(\widehat{CAB}\)

b) Ta có: ΔCAE=ΔDAE(cmt)

nên EC=ED(hai cạnh tương ứng)

Ta có: BC=BD(gt)

nên B nằm trên đường trung trực của CD(Tính chất đường trung trực của một đoạn thẳng)(1)

Ta có: EC=ED(cmt)

nên E nằm trên đường trung trực của CD(Tính chất đường trung trực của một đoạn thẳng)(2)

Từ (1) và (2) suy ra BE là đường trung trực của CD(đpcm)

31 tháng 5 2020

*\(M+\left(5x^2-2xy\right)=6x^2+9xy-y^2\)

\(M=6x^2+9xy-y^2-\left(5x^2-2xy\right)\)

\(M=6x^2+9xy-y^2-5x^2+2xy\)

\(M=\left(6-5\right)x^2+\left(9+2\right)xy-y^2\)

\(M=x^2+11xy-y^2\)

\(\left(2x-5\right)^{2018}+\left(3y+4\right)^{2020}\le0\)

Ta có : \(\hept{\begin{cases}\left(2x-5\right)^{2018}\ge0\forall x\\\left(3y+4\right)^{2020}\ge0\forall y\end{cases}\Rightarrow}\left(2x-5\right)^{2018}+\left(3y+4\right)^{2020}\ge0\forall x,y\)

Mà đề cho \(\left(2x-5\right)^{2018}+\left(3y+4\right)^{2020}\le0\)

=> \(\left(2x-5\right)^{2018}+\left(3y+4\right)^{2020}=0\)

=> \(\hept{\begin{cases}2x-5=0\\3y+4=0\end{cases}\Rightarrow}\hept{\begin{cases}x=\frac{5}{2}\\y=-\frac{4}{3}\end{cases}}\)

Thay x = 5/2 ; y = -4/3 vào M ta được :

\(M=\left(\frac{5}{2}\right)^2+11\cdot\frac{5}{2}\cdot\left(-\frac{4}{3}\right)-\left(-\frac{4}{3}\right)^2\)

\(M=\frac{25}{4}+\frac{-110}{3}-\frac{16}{9}\)

\(M=\frac{-1159}{36}\)

Vậy giá trị của M = -1159/36 khi x = 5/2 ; y = -4/3

Không chắc nha 

Vì \(\hept{\begin{cases}\widehat{A}+\widehat{B}=50^o+30^o=180^o\\\widehat{C}+\widehat{B}=40^o+140^o=180^o\end{cases}}\)mà \(\hept{\begin{cases}\widehat{A}\text{ và }\widehat{B}\text{ là 2 góc trong cùng phía}\\\widehat{C}\text{ và }\widehat{B}\text{ là 2 góc trong cùng phía}\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}AD//BE\\CD//BE\end{cases}}\Rightarrow AD//CD\)

19 tháng 9 2016

Ta có :

\(xy=x:y\)

\(\Rightarrow y^2=1\)

\(\Rightarrow\left[\begin{array}{nghiempt}y=1\\y=-1\end{array}\right.\)

(+) y = 1

\(\Rightarrow x+1=x\) ( vô lý )

(+) \(y=-1\)

\(\Rightarrow x=\frac{1}{2}\) ( Nhận )

Vậy \(\left(x;y\right)=\left(\frac{1}{2};-1\right)\)

19 tháng 9 2016

thanks bạn nhìu nhìu nhayeu