Theo định lí tứ giác thì: ∠A+ ∠B+∠C+ ∠D =360⁰

Mà ∠A + ∠B + ∠C= 216⁰ => ∠D = 360⁰-216⁰=144⁰

∠B + ∠C + ∠D = 324⁰ => ∠A = 360⁰ -324⁰= 36⁰

∠C+ ∠D + ∠A = 288⁰ => ∠ B = 360⁰ - 288⁰=72⁰

=> ∠C = 360⁰ - 144⁰- 36⁰ - 72⁰ =108⁰

Vậy ∠A = 36⁰ ; ∠B = 72⁰ ; ∠C = 108⁰ ; ∠D = 144⁰

a, \(\left(2x+1\right)\left(1-2x\right)+\left(1-2x\right)^2=0\)

\(\Leftrightarrow\left(1-2x\right)\left(2x+1+1-2x\right)=0\Leftrightarrow x=\frac{1}{2}\)

b, \(2\left(x+1\right)^2-\left(x-3\right)\left(x+3\right)-\left(x-4\right)^2=0\)

\(\Leftrightarrow2\left(x^2+2x+1\right)-\left(x^2-9\right)-\left(x^2-8x+16\right)=0\)

\(\Leftrightarrow2x^2+4x+2-x^2+9-x^2+8x-16=0\Leftrightarrow12x-5=0\Leftrightarrow x=\frac{5}{12}\)

c, \(\left(x-5\right)^2-x\left(x-4\right)=9\Leftrightarrow x^2-10x+25-x^2+4x=9\)

\(\Leftrightarrow-6x+16=0\Leftrightarrow x=\frac{8}{3}\)

d, \(\left(x-5\right)^2+\left(x-4\right)\left(1-x\right)=0\)

\(\Leftrightarrow x^2-10x+25+x-x^2-4+4x=0\)

\(\Leftrightarrow-5x+21=0\Leftrightarrow x=\frac{21}{5}\)

Ta có 4E = 4x² - 4xy + 12y² - 8x - 40y + 80 

= 4x² - 4xy + y² - 4(2x - y) + 4 + 11y² - 44y + 44 + 32 

= (2x - y)² - 4(2x - y) + 4 + 11(y² - 4y + 4) + 32 

 = (2x - y - 2)² + (y - 2)² + 32 \(\ge32\)

=> 4E \(\ge32\)

=> E \(\ge\)8

Dấu "=" xảy ra <=> \(\hept{\begin{cases}2x-y-2=0\\y-2=0\end{cases}}\Leftrightarrow x=y=2\)

Vậy Min E = 8 <=> x = y = 2 

x(a+b) - a-b

=x(a+b) - (a+b)

=(x-1)(a+b)

x( a - b ) - a + b = x( a - b ) - ( a - b ) = ( a - b )( x - 1 )

a( x - y ) - x + y = a( x - y ) - ( x - y ) = ( x - y )( a - 1 )

mấy ý kia không sửa được nữa nên nghỉ:)

\(\left(3x-5\right)^2-x\left(3x-5\right)=0\)

\(< =>\left(3x-5\right)\left(3x-5-x\right)=0\)

\(< =>\orbr{\begin{cases}3x-5=0\\2x-5=0\end{cases}< =>\orbr{\begin{cases}x=\frac{5}{3}\\x=\frac{5}{2}\end{cases}}}\)

Trả lời:

\(\left(3x-5\right)^2-x\left(3x-5\right)=0\)

\(\Leftrightarrow\left(3x-5\right)\left(3x-5-x\right)=0\)

\(\Leftrightarrow\left(3x-5\right)\left(2x-5\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}3x-5=0\\2x-5=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{5}{3}\\x=\frac{5}{2}\end{cases}}}\)

Vậy x = 5/3; x = 5/2 là nghiệm của pt.

Trả lời:

\(\left(x-1\right)^2-\left(2x\right)^2=0\)

\(\Leftrightarrow\left(x-1-2x\right)\left(x-1+2x\right)=0\)

\(\Leftrightarrow\left(-x-1\right)\left(3x-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}-x-1=0\\3x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-1\\x=\frac{1}{3}\end{cases}}}\)

Vậy x = - 1; x = 1/3 là nghiệm của pt.

\(\left(x-1\right)^2-\left(2x\right)^2=0\)

\(\Leftrightarrow\left(x-1-2x\right)\left(x-1+2x\right)=0\)

\(\Leftrightarrow\left(-x-1\right)\left(3x-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}-x-1=0\\3x-1=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=-1\\3x=1\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=-1\\x=\frac{1}{3}\end{cases}}\)

Vậy \(S=\left\{-1;\frac{1}{3}\right\}\)

\(\left(x-1\right)^2-\left(2x\right)^2=0\)

\(\Rightarrow\left(x-1-2x\right)\left(x-1+2x\right)=0\)

\(\Rightarrow\left(-x-1\right)\left(3x-1\right)=0\)

\(\Rightarrow\orbr{\begin{cases}-x-1=0\\3x-1=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}-x=1\\3x=1\end{cases}}\Rightarrow\orbr{\begin{cases}x=-1\\x=\frac{1}{3}\end{cases}}\)