1: \(100-x^2=\left(10-x\right)\left(10+x\right)\)

2: \(b^2-a^2=\left(b-a\right)\left(b+a\right)\)

3: \(\left(3y\right)^2-\left(4x\right)^2=\left(3y-4x\right)\left(3y+4x\right)\)

4. (x - 2)(x + 2) = x² - 4

5. (2x - y)(2x + y) = 4x² - y²

6. \(\left(\dfrac{1}{2}x+y\right)\left(\dfrac{1}{2}x-y\right)=\dfrac{1}{4}x^2-y^2\)

( 3x - 6 ) . 3 = 18

3x - 6 = 18 : 3 = 6

3x = 6 + 6 = 12

x= 4

6x - 5 = 613

6x = 613 + 5 =618

x=103

x-36:18=12

x - 2 = 12

x= 12 + 2

x= 14

\(\left(3x-6\right).3=18\)

\(\Rightarrow3x-6=19:3=6\)

\(\Rightarrow3x=6+6=12\)

\(\Rightarrow x=12:3=4\)

\(6.x-5=613\)

\(\Rightarrow6x=613+ 5=618\)

\(\Rightarrow x=618:6=103\)

\(x-36:18=12\)

\(\Rightarrow x-2=12\)

\(\Rightarrow x=12+2=14\)

Ủng hộ nha m.n ^_^

\(\left(3x-6\right)\cdot3=18\)

\(\Rightarrow3x-6=18:3=6\)

\(\Rightarrow3x=6+6=12\)

\(\Rightarrow x=12:3=4\)

\(6\cdot x-5=613\)

\(\Rightarrow6\cdot x=613+5=618\)

\(\Rightarrow x=61=618:6=103\)

\(x-36:18=12\)

\(=x-2=12\)

\(\Rightarrow x=12+2=14\)

Ủng hộ mk nka!!!^_^^_^^_^

Bài 5:

Ta có : \(\widehat{A_1}+\widehat{A_3}=180^o\) (kề bù)

            \(100^o+\widehat{A_3}=180^o\)

            \(\widehat{A_3}=80^o\)

Ta có: \(\widehat{A_3}=\widehat{B_1}=80^o\)

            \(\widehat{A_3}\) và \(\widehat{B_1}\) ở vị trí đồng vị 

\(\Rightarrow AC//BD\)

\(\Rightarrow\widehat{C}_1=\widehat{D_1}=135^o\) (đồng vị)

\(x=135^o\)

b)

Ta có: \(\widehat{G_1}+\widehat{B_1}=180^o\left(120^o+60^o=180^o\right)\)

               \(\widehat{G_1}\) và \(\widehat{B_1}\) ở vị trí trong cùng phía

\(\Rightarrow QH//BK\)

\(\Rightarrow\widehat{H_1}=\widehat{K_1}=90^o\)(so le)

\(x=90^o\)

Bài 1: 

a: Xét ΔABI và ΔACI có

AB=AC

AI chung

BI=CI

Do đó: ΔABI=ΔACI

Bài 1: 

a: 23,4+3,5=26,9

b: 2,3+345,78=348,08

Xvvsdush

Hêuh