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vuongnhatbac
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giangdzvodich
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Sayu
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nhi lê
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Nguyễn Lê Phước Thịnh
5 tháng 11 2021 lúc 23:34

Bài 5: 

a: Xét ΔBEC và ΔADC có 

\(\widehat{C}\) chung

\(\widehat{EBC}=\widehat{DAC}\)

Do đó: ΔBEC\(\sim\)ΔADC

 

Huy Lý
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Đào Thu Ngân
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Mạnh Lê
17 tháng 3 2017 lúc 20:43

Ta có sơ đồ :

Anh hiện nay :                          |----------|----------|----------|----------|

Anh trước đây ( em hiện nay ) : |----------|----------|----------|

Em trước đây :                        |----------|----------|

Tổng số phần bằng nhau là :

        4 + 3 = 7 ( phần )

1 phần có số tuổi là :

        56 : 7 = 8 ( tuổi )

Tuổi em hiện nay là :

        8 x 3 = 24 ( tuổi )

                      Đ/S : 24 tuổi           

nguyen ngoc quynh anh
17 tháng 3 2017 lúc 20:35

tuoi em = 24

Đào Thu Ngân
29 tháng 4 2017 lúc 11:25

cảm ơn các bạn nha

các bạn đúng rồi

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LÂM 29
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trinhthikhanhvy
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Nguyễn Hà Giang
30 tháng 11 2021 lúc 7:38

TK

Nội dung chính : tình yêu thương của anh em trong một gia đình là vô cùng quan trọng, trên thuận dưới hòa cả nhà sẽ hạnh phúc.

Nguyễn Rita
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Akai Haruma
4 tháng 8 2021 lúc 18:29

1a.

$x^2-5x+6=x^2-2x-(3x-6)=x(x-2)-3(x-2)=(x-2)(x-3)$

1b.

$3x^2+9x-30=3(x^2+3x-10)=3(x^2-2x+5x-10)$

$=3[x(x-2)+5(x-2)]=3(x-2)(x+5)$

1c.
$x^2-3x+2=(x^2-x)-(2x-2)=x(x-1)-2(x-1)=(x-1)(x-2)$

1d.

$x^2-9x+18=x^2-3x-(6x-18)=x(x-3)-6(x-3)=(x-3)(x-6)$

1e.

$x^2-6x+8=x^2-2x-(4x-8)=x(x-2)-4(x-2)=(x-2)(x-4)$

Akai Haruma
4 tháng 8 2021 lúc 18:31

1f.
$x^2-5x-14=x^2-7x+2x-14=x(x-7)+2(x-7)=(x+2)(x-7)$

1g.

$x^2+6x+5=(x^2+x)+(5x+5)=x(x+1)+5(x+1)=(x+1)(x+5)$

1h.

$x^2-7x+12=x^2-3x-(4x-12)=x(x-3)-4(x-3)=(x-3)(x-4)$

1i.

$x^2-7x+10=(x^2-2x)-(5x-10)=x(x-2)-5(x-2)=(x-2)(x-5)$

Akai Haruma
4 tháng 8 2021 lúc 18:35

2a.

$3x^2-5x-2=(3x^2-6x)+(x-2)=3x(x-2)+(x-2)$

$=(x-2)(3x+1)$

2b. 

$2x^2+x-6=(2x^2+4x)-(3x+6)=2x(x+2)-3(x+2)$

$=(2x-3)(x+2)$

2c.

$7x^2+50x+7=(7x^2+49x)+(x+7)$

$=7x(x+7)+(x+7)=(7x+1)(x+7)$

2d. 

$12x^2+7x-12=(12x^2-9x)+(16x-12)$

$=3x(4x-3)+4(4x-3)=(4x-3)(3x+4)$

2e.

$15x^2+7x-2=15x^2-3x+10x-2$

$=3x(5x-1)+2(5x-1)=(5x-1)(3x+2)$

Nguyễn Anh Kiệt
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Nguyễn Thái Thịnh
6 tháng 2 2022 lúc 19:22

Tất cả các điểm đã cho đều thuộc đồ thị hàm số \(y=x^2+2x+1\).

Thay \(x=0;y=1\) vào đồ thị hàm số \(y=x^2+2x+1\) ta có:

\(1=0^2+2.0+1\Rightarrow1=1\) (luôn đúng)

Tương tự với các điểm còn lại.