k) 3x(z−2)+5(−x−2)

=3xz−6x−5x−10

=3xz−x−10

=x(3z−1)−10

\(x^{10}+x^5+1=x^{10}+x^9+x^8-x^9-x^8-x^7+x^7+x^6+x^5-x^6-x^5-x^4\)

\(+x^5+x^4+x^3-x^3-x^2-x+x^2+x+1\)

\(=x^8\left(x^2+x+1\right)-x^7\left(x^2+x+1\right)+x^5\left(x^2+x+1\right)-x^4\left(x^2+x+1\right)\)

\(+x^3\left(x^2+x+1\right)-x\left(x^2+x+1\right)+x^2+x+1\)

\(=\left(x^2+x+1\right)\left(x^8-x^7+x^5-x^4+x^3-x+1\right)\)

\(x^2-408x+2015\)

\(=x^2-403x-5x+2015\)

\(=\left(x^2-403x\right)-\left(5x-2015\right)\)

\(=x\left(x-403\right)-5\left(x-403\right)\)

\(=\left(x-5\right)\left(x-403\right)\)

(x - 3)³ + 3 - x

= (x - 3)³ - (x - 3)

= (x - 3)[(x-3)²-1]

= (x-3)(x-3-1)(x-3+1)

= (x-3)(x-4)(x-2)

Bài 1:

\(P=3x^2+x-1\)

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

\(=3\left(x^2+2x.\frac{1}{6}+\frac{1}{36}-\frac{13}{36}\right)\)

\(=3\left(x+\frac{1}{6}\right)^2-\frac{13}{12}\ge\frac{-13}{12}\)\(\forall x\)

Dấu '' = '' xảy ra khi: \(\left(x+\frac{1}{6}\right)^2=0\Rightarrow x=\frac{-1}{6}\)

Vậy \(MinP=\frac{-13}{12}\) khi \(x=\frac{-1}{6}\)

Bài 2:

a) Không có điều kiện

b) Nghiệm vô tỉ

Bạn xem lại đề hai phần này nhé.

c) \(\left(x-2\right)^3-x^3+6x^2=14\)

\(\Rightarrow x^3-6x^2+12x-8-x^3+6x^2-14=0\)

\(\Rightarrow\left(x^3-x^3\right)+\left(-6x^2+6x^2\right)+12x+\left(-8-14\right)=0\)

\(\Rightarrow12x-22=0\)

\(\Rightarrow x=\frac{11}{6}\)

d) \(8x^2+30x+7=0\)

\(\Rightarrow8x^2+28x+2x+7=0\)

\(\Rightarrow\left(8x^2+28x\right)+\left(2x+7\right)=0\)

\(\Rightarrow4x\left(2x+7\right)+\left(2x+7\right)=0\)

\(\Rightarrow\left(4x+1\right)\left(2x+7\right)=0\)

\(\Rightarrow\orbr{\begin{cases}4x+1=0\\2x+7=0\end{cases}}\Rightarrow\orbr{\begin{cases}4x=-1\\2x=-7\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{-1}{4}\\x=\frac{-7}{2}\end{cases}}\)

a) Xét tam giác ADC và tam giác BCD, ta có:

AD = BC

\(\widehat{ADC}=\widehat{BCD}\)

CD chung

=> Tam giác ADC = tam giác BCD (c.g.c) 

b) Theo đề ra, ta có: AB song song CD

\(\Rightarrow\widehat{ACD}=\widehat{EAB}\) (Hai góc so le trong bằng nhau)

\(\Rightarrow\widehat{BDC}=\widehat{EBA}\) (Hai góc so le trong bằng nhau)

Theo phần a), ta có: \(\widehat{ACD}=\widehat{BDC}\)

\(\Rightarrow\widehat{EAB}=\widehat{EBA}\)

=> Tam giác EAB cân tại E

=> EA = EB

E D C B A