\(\frac{a}{b}=\frac{c}{d}\Rightarrow ad=bc\)

a)\(\frac{a-b}{a+b}=\frac{c-d}{c+d}\)

\(\Leftrightarrow\left(a-b\right)\left(c+d\right)=\left(c-d\right)\left(a+b\right)\)

\(\Leftrightarrow ac-bc+ad-bd=ac-ad+bc-bd\)

\(\text{Thay }ad=bc\text{ vào}\Rightarrow ac-ad+ad-bd=ac-ad+ad-bd\)

\(\text{Đây là đẳng thức đúng }\Rightarrow\frac{a-b}{a+b}=\frac{c-d}{c+d}\text{ là đúng }\)

b)\(\text{Tương tự*}\)

a) \(\frac{a}{b}=\frac{c}{d}\Leftrightarrow\frac{a}{b}+1=\frac{c}{d}+1\Leftrightarrow\frac{a+b}{b}=\frac{c+d}{d}\Leftrightarrow\frac{b}{a+b}=\frac{d}{c+d}\)

\(\Leftrightarrow\frac{-2b}{a+b}+1=\frac{-2d}{c+d}+1\Leftrightarrow\frac{a-b}{a+b}=\frac{c-d}{c+d}\)

b) \(\frac{a}{b}=\frac{c}{d}\Leftrightarrow\frac{4a}{b}-5=\frac{4c}{d}-5\Leftrightarrow\frac{4a-5b}{b}=\frac{4c-5d}{d}\Leftrightarrow\frac{b}{4a-5b}=\frac{d}{4c-5d}\)

\(\Leftrightarrow\frac{11b}{4a-5b}+1=\frac{11d}{4c-5d}+1\Leftrightarrow\frac{4a+6b}{4a-5b}=\frac{4c+6d}{4c-5d}\Leftrightarrow\frac{2a+3b}{4a-5b}=\frac{2c+3d}{4c-5d}\)

\(\Leftrightarrow\frac{2a+3b}{2c+3d}=\frac{4a-5b}{4c-5d}\)

Bài 4: 

\(x=130^0\)

`@` `\text {Ans}`

`\downarrow`

`4,`

Vì `\text {MN // BC}`

`=>` $\widehat {B} = \widehat {BMN} = 114^0 (\text {2 góc đối đỉnh})$

Ta có: \(\left\{{}\begin{matrix}\widehat{\text{BMN}}+\widehat{\text{AMN}}=180^0\left(\text{2 góc kề bù}\right)\\\widehat{\text{CNM}}+\widehat{\text{ANM}}=180^0\left(\text{2 góc kề bù}\right)\end{matrix}\right.\)

`=>`\(\left\{{}\begin{matrix}\widehat{\text{AMN}}=180^0-114^0=66^0\\\widehat{\text{ANM}}=180^0-130^0=50^0\end{matrix}\right.\)

Xét `\Delta AMN`:

\(\widehat{\text{A}}+\widehat{\text{M}}+\widehat{\text{N}}=180^0\left(\text{định lý tổng 3 góc trong 1 tgiac}\right)\)

`=>`\(\widehat{\text{A}}+66^0+50^0=180^0\)

`=>`\(\widehat{\text{A}}=180^0-66^0-50^0=64^0\)

Mà \(\widehat{\text{A}}=\widehat{\text{x}}\)

`=>`\(\widehat{\text{x}}=64^0\)

Vậy, số đo của góc `x = 64^0.`

ngủ hết rùi 

à 1d

2b

3b

4d

mk nhìn đề bài mà choáng á!!!!

e: \(E=\dfrac{x^2-9-x^2+4-x^2+9}{\left(x+3\right)\left(x-2\right)}\)

\(=\dfrac{x+2}{x+3}\)

a: \(A=\dfrac{4x^2+x^2-2x+1+x^2+2x+1}{\left(x-1\right)\left(x+1\right)}\)

\(=\dfrac{6x^2+2}{\left(x-1\right)\left(x+1\right)}\)

\(A=\dfrac{-4x^2+x^2-2x+1-x^2-2x-1}{\left(1-x\right)\left(1+x\right)}=\dfrac{-4x\left(x+1\right)}{\left(1-x\right)\left(1+x\right)}=\dfrac{4x}{x-1}\\ C=\dfrac{-x^2-4x-4+x^2-4x+4-4x^2}{\left(x-2\right)\left(x+2\right)}=\dfrac{-4x\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{4x}{2-x}\\ E=\dfrac{x^2-9-x^2+4x-4-x^2+9}{\left(x-2\right)\left(x+3\right)}=\dfrac{-\left(x-2\right)^2}{\left(x-2\right)\left(x+3\right)}=\dfrac{2-x}{x+3}\)

a) (-26) + (-32)   b) 57 + 264   c) (-267) + (-473)   d) (-5) +8

=-(26 + 32)         =321             =-(267 + 473)         =8-5

=-58                                        =-740                     =3

e) 1000 + (-327)    f) (-5679) + 5679   g) (-2364) + (-175)

=1000-327             =0                          =- (2364 + 175)

=673                                                    =-2539

h) 136 + (-36)

=136 - 36

=100

27. unforgetable
câu 31 đổi thành ...if I can give him his guide....
câu 32 thiếu on trong turned on và đổi thành has been turned on vì câu gốc là HTHT
câu 33 rút gọn mệnh đề ok
câu 34 hơi phân vân nhưng mình thấy đúng