a) a² -6a +9

= a² - 2.a.3 + 3²

= (a-3)²

b) 1/4 x² + 2xy² + 4y⁴

= (1/2x)² + 2 . 1/2x . 2y² + (2y²)²

=(1/2 x + 2y²)² 

Trả lời:

a, \(a^2-6a+9=a^2-2.x.3+3^2=\left(a-3\right)^2\)

b, \(\frac{1}{4}x^2+2xy^2+4y^4=\left(\frac{1}{2}x\right)^2+2.\frac{1}{2}x.2y^2+\left(2y^2\right)^2=\left(\frac{1}{2}x+2y^2\right)^2\)

x³ - 5x² + 5 - x = 0 

<=> x²(x - 5) - (x - 5) = 0 

<=> (x² - 1)(x - 5) = 0 

<=> (x - 1)(x + 1)(x - 5) = 0

<=> x = 1 hoặc x = -1 hoặc x = 5

Vậy \(x\in\left\{1;-1;5\right\}\)

Trả lời:

\(x^3-5x^2+5-x=0\)

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

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

\(\Leftrightarrow\left(x-5\right)\left(x^2-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-5=0\\x^2-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=5\\x=\pm1\end{cases}}}\)

Vậy x = 5; x = 1; x = - 1 là nghiệm của pt.

X^3 - 2x^2 + x - 2 = 0

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

=>(x^2+1)(x-2)=0

=>x^2+1=0 or x-2=0

=> x^2=-1 ( vô lí ) or x=2

=>x=2

Trả lời:

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

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

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

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

\(\Leftrightarrow x-2=0\)( vì \(x^2+1\ge1>0\forall x\) )

\(\Leftrightarrow x=2\)

Vậy x = 2 là nghiệm của pt.

\(Q\left(2\right)=4a+2b+c\)

\(Q\left(-1\right)=a-b+c\)

\(Q\left(2\right)+Q\left(-1\right)=5a+b+2c=0\)

\(\Leftrightarrow Q\left(2\right)=-Q\left(-1\right)\)

\(Q\left(2\right).Q\left(-1\right)=-Q\left(-1\right)^2\le0\)

Q(-1) = a.(-1)2 + b.(-1) + c = a - b + c Q2) = a.22 + b.2 + c = 4a + 2b + c => Q(-1) + Q(2) = a - b + c + 4a + 2b + c = 5a + b + 2c = 0 hay Q(-1) + Q(2) = 0 => Q(-1) = -Q(2) Ta có :Q(-1).Q(2) = -Q(2).Q(2) = -Q2(2) ≤ 0 vì Q2(2) ≥ 0 Vậy ..... đpcm .

Xét hình thang ABCD (AB//CD) có:

AM=MD=12AD

BN=NC=12BC

⇒MN⇒MN là đường trung bình

⇒ \(\hept{\begin{cases}MN=(AB+CD)/2=3AB/2\\MN//AB//CD\end{cases}} \)

Xét △ABD có:

AM=MD=12AD

AP//AB

⇒AP=12AB       (1)

Xét △ABC có:

BN=NC=12BC

NQ//AB

⇒NQ=12AB(2)

Ta lại có:

MP+PQ+QN=MN

⇔PQ=MN−MP−NQ

⇔PQ=3AB2−12AB−12AB

⇔PQ=12AB(3)

Từ (1)(2)(3)⇒MP=PQ=QN

2x(x - 7) + 5x - 35 = 0

=> 2x(x - 7) + 5(x - 7) = 0

=> (x - 7)(2x + 5) = 0

=> x - 7 = 0 hoặc 2x + 5 = 0

=> x = 7 hoặc x = -5/2

Trả lời:

\(2x\left(x-7\right)+5x-35=0\)

\(\Leftrightarrow2x\left(x-7\right)+5\left(x-7\right)=0\)

\(\Leftrightarrow\left(x-7\right)\left(2x+5\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-7=0\\2x+5=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=7\\x=-\frac{5}{2}\end{cases}}}\)

Vậy x = 7; x = - 5/2 là nghiệm của pt.

\(C=3x^2+5x+3=3\left(x^2+\frac{5}{3}x+1\right)=3\left(x^2+2.\frac{5}{6}x+\frac{25}{36}+\frac{11}{36}\right)\)

\(=3\left(x+\frac{5}{6}\right)^2+\frac{11}{12}\ge\frac{11}{12}\)

Dấu \(=\)khi \(x=-\frac{5}{6}\)

\(D=7-4x^2-2x=-4\left(x^2+\frac{1}{2}x-\frac{7}{4}\right)=-4\left(x^2+2.\frac{1}{4}x+\frac{1}{16}-\frac{29}{16}\right)\)

\(=-4\left(x+\frac{1}{4}\right)^2+\frac{29}{4}\le\frac{29}{4}\)

Dấu \(=\)khi \(x=-\frac{1}{4}\).

Trả lời:

\(x^{10}-1=\left(x^5\right)^2-1^2=\left(x^5-1\right)\left(x^5+1\right)\)

...

\(x^{10}-1=\left(x^5-1\right)\left(x^5+1\right)\)

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