Bài làm:

đk: \(x\ne-3;x\ne1\)

Ta có: \(\frac{x^2+6x+9}{1-x}\cdot\frac{\left(x-1\right)^3}{2\left(x+3\right)^3}\)

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

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

\(=-\frac{x^2-2x+1}{2x+6}\)

\(ĐKXĐ:\hept{\begin{cases}x\ne-3\\x\ne1\end{cases}}\)

\(\frac{x^2+6x+9}{1-x}.\frac{\left(x-1\right)^3}{2\left(x+3\right)^3}=\frac{-\left(x+3\right)^2}{x-1}.\frac{\left(x-1\right)^3}{2\left(x+3\right)^3}=\frac{-\left(x-1\right)^2}{2\left(x+3\right)}\)

                          A B O C D

Vì ABCD là hình thang \(\Rightarrow AB//CD\)\(\Rightarrow\widehat{OAB}=\widehat{OCD}\); \(\widehat{OBA}=\widehat{ODC}\)( so le trong )

Xét \(\Delta AOB\)và \(\Delta COD\)ta có:

+) \(\widehat{AOB}=\widehat{COD}\)( đối đỉnh )

+) \(\widehat{OAB}=\widehat{OCD}\)( chứng minh trên )

+) \(\widehat{OBA}=\widehat{OCD}\)( chứng minh trên )

\(\Rightarrow\Delta AOB~\Delta COD\)( \(g.g.g\) ) ( đpcm ) 

Bài làm:

đk: \(x\le\frac{5}{2}\)

Ta có: \(\left|5-2x\right|=1-x\)

\(\Leftrightarrow\orbr{\begin{cases}5-2x=1-x\\5-2x=x-1\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=4\\3x=6\end{cases}}\Rightarrow\orbr{\begin{cases}x=4\left(ktm\right)\\x=2\end{cases}}\)

Vậy x = 2

\(ĐKXĐ:x\le1\)

\(\left|5-2x\right|=1-x\)\(\Leftrightarrow\orbr{\begin{cases}5-2x=-\left(1-x\right)\\5-2x=1-x\end{cases}}\Leftrightarrow\orbr{\begin{cases}5-2x=-1+x\\x=4\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}3x=6\\x=4\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=2\\x=4\end{cases}}\)( cả 2 đều không thỏa mãn )

Vậy phương trình vô nghiệm

Bài làm:

Ta có: \(3x^2-10x+7=0\)

\(\Leftrightarrow\left(3x^2-3x\right)-\left(7x-7\right)=0\)

\(\Leftrightarrow3x\left(x-1\right)-7\left(x-1\right)=0\)

\(\Leftrightarrow\left(3x-7\right)\left(x-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}3x-7=0\\x-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{7}{3}\\x=1\end{cases}}\)

\(3x^2-10x+7=0\)

\(\Leftrightarrow\left(3x^2-3x\right)-\left(7x-7\right)=0\)

\(\Leftrightarrow3x\left(x-1\right)-7\left(x-1\right)=0\)

\(\Leftrightarrow\left(3x-7\right)\left(x-1\right)=0\Leftrightarrow\orbr{\begin{cases}3x-7=0\\x-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{7}{3}\\x=1\end{cases}}\)

Vậy x= 7/3 hoặc x = 1

Ta có x³ - 4x² + 4x = 0

=> \(x^3-2x^2-2x^2+4x=0\)

=> x²(x - 2) - 2x(x - 2) = 0

=> (x² - 2x)(x - 2) = 0

=> x(x - 2)(x - 2) = 0

=> x(x - 2)² = 0

=> \(\orbr{\begin{cases}x=0\\x-2=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=2\end{cases}}\)

Vậy \(x\in\left\{0;2\right\}\)

Bài làm:

Ta có: \(x^3-4x^2+4x=0\)

\(\Leftrightarrow x\left(x^2-4x+4\right)=0\)

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

\(\Leftrightarrow\orbr{\begin{cases}x=0\\\left(x-2\right)^2=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=0\\x=2\end{cases}}\)

5x(x-3)-2x+6=0

5x(x - 3) - 2(x-3) = 0
(X-3)(5X-2)=0

x-3=0

5x-2=0
=> X=3 HOẶC X= 2/5

A = x² - 10x + 12

= ( x² - 10x + 25 ) - 13

= ( x - 5 )² - 13

( x - 5 )² ≥ 0 ∀ x => ( x - 5 )² - 13 ≥ -13

Đẳng thức xảy ra <=> x - 5 = 0 => x = 5

=> MinA = -13 <=> x = 5

B = 6y² + 4y - 1

= 6( y² + 2/3y + 1/9 ) - 5/3

= 6( y + 1/3 )² - 5/3

6( y + 1/3 )² ≥ 0 ∀ x => 6( y + 1/3 )² - 5/3 ≥ -5/3

Đẳng thức xảy ra <=> y + 1/3 = 0 => y = -1/3

=> MinB = -5/3 <=> y = -1/3

C = x² + y² - 2x - 6y - 1

= ( x² - 2x + 1 ) + ( y² - 6y + 9 ) - 11

= ( x - 1 )² + ( y - 3 )² - 11

\(\hept{\begin{cases}\left(x-1\right)^2\ge0\forall x\\\left(y-3\right)^2\ge0\forall y\end{cases}\Rightarrow}\left(x-1\right)^2+\left(y-3\right)^2-11\ge-11\)

Đẳng thức xảy ra <=> \(\hept{\begin{cases}x-1=0\\y-3=0\end{cases}}\Rightarrow\hept{\begin{cases}x=1\\y=3\end{cases}}\)

=> MinC = -11 <=> x = 1 ; y = 3

D = 2x² + 3y² - x - 3y + 5

= 2( x² - 1/2x + 1/16 ) + 3( y² - y + 1/4 ) + 33/8

= 2( x - 1/4 )² + 3( y - 1/2 )² + 33/8

\(\hept{\begin{cases}2\left(x-\frac{1}{4}\right)^2\ge0\forall x\\3\left(y-\frac{1}{2}\right)^2\ge0\forall y\end{cases}}\Rightarrow2\left(x-\frac{1}{4}\right)^2+3\left(y-\frac{1}{2}\right)^2+\frac{33}{8}\ge\frac{33}{8}\)

Đẳng thức xảy ra <=> \(\hept{\begin{cases}x-\frac{1}{4}=0\\y-\frac{1}{2}=0\end{cases}}\Rightarrow\hept{\begin{cases}x=\frac{1}{4}\\y=\frac{1}{2}\end{cases}}\)

=> MinD = 33/8 <=> x = 1/4 ; y = 1/2

\(\text{Z là Oxi =}>NTK_Z=NTK_O=16\left(\text{đvC}\right)\)

\(=>NTK_Y=1,5NTK_Z=1,5.16=24\left(\text{đvC}\right)\)

\(=>NTK_X=\frac{1}{2}NTK_Y=\frac{1}{2}.24=12\left(\text{đvC}\right)\)

          \(\text{Vậy NTK của X là 12 đvC.}\)