`#3107.101107`

`x^2 + 5x - 6`

`= x^2 + 6x - x - 6`

`= (x^2 + 6x) - (x + 6)`

`= x(x + 6) - (x + 6)`

`= (x - 1)(x + 6)`

`#3107.101107`

`a)`

Vì ABCD là hình thang cân:

\(\Rightarrow\widehat{\text{A}}=\widehat{\text{B}};\text{ }\widehat{\text{C}}=\widehat{\text{D}};\text{AD = BC}\) (dấu hiệu nhận biết)

Xét `\Delta ABC` và `\Delta ABD:`

\(\text{AD = BC}\\ \widehat{\text{BAD}}=\widehat{\text{ABC}}\\ \text{AB chung}\)

\(\Rightarrow\) `\Delta ABC = \Delta ABD` (c - g - c)

\(\Rightarrow\widehat{\text{ABD}}=\widehat{\text{BAC}}\left(\text{2 góc tương ứng}\right)\)

Xét `\Delta OAB:`

\(\widehat{\text{OAB}}=\widehat{\text{OBA}}\)

\(\Rightarrow\) `\Delta OAB` là `\Delta` cân

`\Rightarrow OA = OB`

`b)`

Xét `\Delta ADC` và `\Delta BCD:`

\(\text{AD = BC (CMT)}\\ \widehat{\text{ADC}}=\widehat{\text{BCD}}\left(\text{CMT}\right)\\ \text{DC chung}\)

`\Rightarrow \Delta ADC = \Delta BCD` (c - g - c)

`\Rightarrow`\(\widehat{\text{ACD}}=\widehat{\text{BDC}}\left(\text{2 góc tương ứng}\right)\)

Xét `\Delta OCD:`

\(\widehat{\text{ACD}}=\widehat{\text{BDC}}\)

`\Rightarrow \Delta OCD` là `\Delta` cân

`\Rightarrow OC = OD.`

`#3107.101107`

`6.`

`a)`

\(\dfrac{2^{12}\cdot3^5-4^6\cdot9^2}{\left(2^2\cdot3\right)^6+8^4\cdot3^5}-\dfrac{5^{10}\cdot7^3-25^5\cdot49^2}{\left(125\cdot7\right)^3+5^9\cdot14^3}\\ =\dfrac{2^{12}\cdot3^5-\left(2^2\right)^6\cdot\left(3^2\right)^2}{2^{12}\cdot3^6+\left(2^3\right)^4\cdot3^5}-\dfrac{5^{10}\cdot7^3-\left(5^2\right)^5\cdot\left(7^2\right)^2}{125^3\cdot7^3+5^9\cdot2^3\cdot7^3}\\ =\dfrac{2^{12}\cdot3^5-2^{12}\cdot3^4}{2^{12}\cdot3^6+2^{12}\cdot3^5}-\dfrac{5^{10}\cdot7^3-5^{10}\cdot7^4}{\left(5^3\right)^3\cdot7^3+5^9\cdot2^3\cdot7^3}\\ =\dfrac{2^{12}\cdot3^4\left(3-1\right)}{2^{12}\cdot3^5\left(3+1\right)}-\dfrac{5^{10}\cdot7^3\left(1-7\right)}{5^9\cdot7^3\left(1+2^3\right)}\\ =\dfrac{2}{3\cdot4}-\dfrac{5\cdot\left(-6\right)}{1+8}\\ =\dfrac{2}{12}-\dfrac{-30}{9}\\ =\dfrac{7}{2}\)

`#3107.101107`

`#3107.101107`

\(\left(-\dfrac{1}{2}\right)^3\div\left(-\dfrac{1}{2}\right)^6\\ =\left(-\dfrac{1}{2}\right)^{3-6}\\ =\left(-\dfrac{1}{2}\right)^{-3}\\ =\left(-2\right)^3\\ =-8\)

`#3107.101107`

`1.`

`a)`

Ta có:

\(\widehat{\text{B}}=\widehat{\text{C}}=75^0\)

Mà `2` góc này nằm ở vị trí sole trong

\(\Rightarrow\text{AB // CD}\) (tính chất)

\(\Rightarrow\) Tứ giác ABCD là hình thang

`b)`

- Để ABCD là hình thang cân

\(\Rightarrow\widehat{\text{D}}=\widehat{\text{C}}\); \(\widehat{\text{A}}=\widehat{\text{B}}\) (dấu hiệu nbiet)

\(\Rightarrow\) `x` có số đo là `75^0`

`2.`

`a)`

Ta có:

\(\widehat{\text{D}}=\widehat{\text{C}}=65^0\)

Mà `2` góc này là đáy của hình thang

\(\Rightarrow\) ABCD là hình thang cân

`b)`

Vì ABCD là hình thang

\(\Rightarrow\widehat{\text{A}}+\widehat{\text{D}}=180^0\left(\text{AB // CD}\right)\)

\(\Rightarrow\widehat{\text{A}}=180^0-65^0=115^0\)

Vì ABCD là hình thang cân

\(\Rightarrow\widehat{\text{A}}=\widehat{\text{B}}=115^0.\)

`#3107.101107`

\(\dfrac{328}{435}\times\dfrac{468}{432}\times\dfrac{435}{164}\times\dfrac{432}{984}\times\dfrac{164}{468}\\ =\left(\dfrac{328}{435}\times\dfrac{435}{164}\times\dfrac{164}{468}\right)\times\left(\dfrac{468}{432}\times\dfrac{432}{984}\right)\\ =\dfrac{328}{468}\times\dfrac{468}{984}\\ =\dfrac{328}{984}\\ =\dfrac{1}{3}\)

Không đăng lại CH nhé.