Cho hình vẽ bên, biết a // b và  B ^...">
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16 tháng 7 2019

a) Vì B 2 ^ , A 1 ^  là cặp góc trong cùng phía nên ta có:

B 2 ^ + A 1 ^ = 180 0 ⇒ A 1 ^ = 180 0 − B 2 ^ = 180 0 − 45 0 = 135 0 .

b) Ta có B ^ 1 = A ^ 1 = 135 ∘  (hai góc đồng vị)

mà A ^ 3 = A ^ 1 = 135 ∘  (hai góc đối đỉnh)

Vậy  B ^ 1 = A ^ 3 = 135 ∘

c) Ta có A ^ 1 + A ^ 2 = 180 ∘ (hai góc kề bù) mà B ^ 1 = A ^ 1  (theo câu b)

Do đó  A ^ 2 + B ^ 1 = 180 ∘

NM
8 tháng 11 2021

a. ta có : \(\frac{5}{-3}=\frac{15}{-9}=-\frac{15}{9}\)

b.\(-\frac{1}{5}< 0< \frac{1}{100}\Rightarrow-\frac{1}{5}< \frac{1}{100}\)

c.\(\hept{\begin{cases}2^3=8\\3^2=9\end{cases}\Rightarrow2^3< 3^2}\)

26 tháng 8 2021

\(b^2=a.c\)\(=>\frac{a}{b}=\frac{b}{c}\)

Đặt : \(\frac{a}{b}=\frac{b}{c}=k\)

Ta có : \(a=b.k\)  

            \(b=c.k\)

\(=>\)\(\frac{a}{c}=\frac{b.k}{c}=\frac{c.k+k}{c}=k^2\left(1\right)\)

\(\left(\frac{a+2012b}{b+2012c}\right)^2=\left(\frac{bk+2012b}{ck+2012c}\right)^2=\left(\frac{b\left(k+2012\right)}{c\left(k+2012\right)}\right)^2=\left(\frac{b}{c}\right)^2=k^2\left(2\right)\)

Từ (1) và (2) \(=>\frac{a}{c}=\left(\frac{a+2012b}{b+2012c}\right)^2\left(đpcm\right)\)

Hok tốt~

Bài làm

a) Ta có:

\(P\left(x\right)=x^5-3x^2+7x^4-9x^3+x^2-\frac{1}{4}x\)

\(P\left(x\right)=x^5-2x^2+7x^4-9x^3-\frac{1}{4}x\)

\(P\left(x\right)=x^5+7x^4-9x^3-2x^2-\frac{1}{4}x\)

\(Q\left(x\right)=5x^4-x^5+x^2-2x^3+3x^2-\frac{1}{4}\)

\(Q\left(x\right)=5x^4-x^5-2x^3+4x^2-\frac{1}{4}\)

\(Q\left(x\right)=-x^5+5x^4-2x^3+4x^2-\frac{1}{4}\)

b) \(P\left(x\right)+Q\left(x\right)=x^5+7x^4-9x^3-2x^2-\frac{1}{4}x-x^5+5x^4-2x^3+4x^2-\frac{1}{4}\)

\(P\left(x\right)+Q\left(x\right)=12x^4-11x^3+2x^2-\frac{1}{4}x-\frac{1}{4}\)

Vậy \(P\left(x\right)+Q\left(x\right)=12x^4-11x^3+2x^2-\frac{1}{4}x-\frac{1}{4}\)

\(P\left(x\right)-Q\left(x\right)=x^5+7x^4-9x^3-2x^2-\frac{1}{4}x+x^5-5x^4+2x^3-4x^2+\frac{1}{4}\)

\(P\left(x\right)-Q\left(x\right)=2x^5-2x^4-7x^3-6x^2-\frac{1}{4}x-\frac{1}{4}\)

Vậy \(P\left(x\right)-Q\left(x\right)=2x^5-2x^4-7x^3-6x^2-\frac{1}{4}x-\frac{1}{4}\)

c) Ta có: 

\(P\left(1\right)=1^5+7.1^4-9.1^3-2.1^2-\frac{1}{4}.1\)

\(P\left(1\right)=-\frac{13}{4}\)

Vậy giá trị của biểu thức P = -13/4 khi x = 1

\(Q\left(0\right)=-0^5+5.0^4-2.0^3+4.0^2-\frac{1}{4}\)

\(Q\left(0\right)=-\frac{1}{4}\)

Vậy \(Q\left(0\right)=-\frac{1}{4}\)

14 tháng 5 2021

Cảm ơn bạn nha!

ĐỀ 3:Bài 1: Tính: a) ;......................................................................................... ......................................................................................... ......................................................................................... ......................................................................................... ............................................................................................
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ĐỀ 3:

Bài 1: Tính:

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c) ;

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Bài 2:  Tìm x:

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b) ;

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c) ;

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d) ;                  

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Bài 3: Tìm x, y biết:  ;

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Bài 4: Tìm 3 số x, y, z sao cho

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0
26 tháng 8 2021

A= 3x3 - (3x -2)x2  - 2x(x+1)

A= 3x3 - 3x3 + 2x2 - 2x2 -2x

A= -2x

Thay x =-20 vào A ta được:

A = -2.(-20) = 40

Vậy A= 40 khi x = -20 

b) C= x(2x+1) - x2(x+2) + x3 -x + 3

C= 2x2 + x - x3 - 2x2 + x3 -x +3

C= (2x2 - 2x2) + (x-x) - (x3 -x3) +3 

C = 3

Vậy C= 3

AH
Akai Haruma
Giáo viên
27 tháng 7 2024

Lời giải:

$b.b=ac\Rightarrow \frac{b}{c}=\frac{a}{b}$.
Đặt $\frac{b}{c}=\frac{a}{b}=k\Rightarrow b=ck; a=bk$.

Khi đó:

$\frac{a}{c}=\frac{bk}{c}=\frac{ck.k}{c}=k^2(1)$

Và:

$\frac{(a+2011b)^2}{(b+2011c)^2}=\frac{(bk+2011b)^2}{(ck+2011c)^2}$

$=\frac{b^2(k+2011)^2}{c^2(k+2011)^2}=\frac{b^2}{c^2}=\frac{(ck)^2}{c^2}=k^2(2)$

Từ $(1);(2)$ ta có đpcm.

 

20 tháng 4 2017

Vì a // b nên ta có:

a) ^B1 = ^A4 = 37° (2 góc so le trong)

Vậy ^B1 = 37°.

b) ^A1 = ^B4 (2 góc đồng vị).

c) ^B2 + ^A4 = 180° (2 góc trong cùng phía)

hay ^B2 + 37° =180°.

=> ^B2 = 180° - 37° = 143°.

Vậy ^B2 = 143°.

20 tháng 4 2017

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27 tháng 11 2021

a) Vì x và y là hai địa lượng tỉ lệ nghịch 

\(y=\frac{a}{x}=a=x.y\)

Thay \(a=2.4\)

Vậy \(a=8\)

b) \(x=\frac{a}{y}\)

c) Vì x là y là hai đại lượng tỉ lệ nghịch

\(x=\frac{a}{y}=x=\frac{a}{y}\)

Thay \(x=\frac{8}{-1}\); Thay \(x=\frac{8}{2}\)

\(\hept{\begin{cases}x=4\\x=8\end{cases}}\)