Bài 209 : đăng tách ra cho mn cùng làm nhé 

a,sửa đề :  \(A=\left(3x+1\right)^2-2\left(3x+1\right)\left(3x+5\right)+\left(3x+5\right)^2\)

\(=\left(3x+1-3x-5\right)^2=\left(-4\right)^2=16\)

b, \(B=\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{32}+1\right)\)

\(2B=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{32}+1\right)=\left(3^{32}-1\right)\left(3^{32}+1\right)\)

\(2B=3^{64}-1\Rightarrow B=\frac{3^{64}-1}{2}\)

c, \(C=\left(a+b-c\right)^2+\left(a-b+c\right)^2-2\left(b-c\right)^2\)

\(=2\left(a-b+c\right)^2-2\left(b-c\right)^2=2\left[\left(a-b+c\right)^2-\left(b-c\right)^2\right]\)

\(=2\left(a-b+c-b+c\right)\left(a-b+c+b-c\right)=2a\left(a-2b+2c\right)\)

Trả lời:

a) \(\frac{1}{4}x^2y+5x^3-x^2y^2=x^2\left(\frac{1}{4}y+5x-y^2\right)\)

 b) 5x ( x - 1 ) - 3y ( 1 - x ) = 5x ( x - 1 ) + 3y ( x - 1 ) = ( x - 1 )( 5x + 3y )

 c) 4x² - 25 = ( 2x )² - 5² = ( 2x - 5 )( 2x + 5 )

 d) 6x- 9x² = 3x ( 2 - 3x )

a) Áp dụng hằng đẳng thức thứ hai ta có

\(x^2 - 2xy + y^2\) \(<=>\) \((x +y )^2\)

b) Ta có :

\(x^2 - 2xy - 4z^2 + y^2 <=> (x^2 - 2xy + y^2) - (2z)^2 <=> ( x-y)^2 - (2z)^2 <=> ( x-y+2z) (x-y-2z)\)

Vậy (x^4 - x^3 - 3x^2 + x + 2) = (x^2 - x - 1)(x^2 - 1) + 1

\(P=\left(x-y\right)^2+\left(x+y\right)^2-2\left(x+y\right)\left(x-y\right)-4x^2=\left(x-y-x-y\right)^2-\left(2x\right)^2=\left(-2y\right)^2-\left(2x\right)^2\)

\(=\left(2y-2x\right)\left(2y+2x\right)=2\left(y-x\right)2\left(y+x\right)=4\left(x+y\right)\left(y-x\right)\)

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

\(x^3-2x^2-4xy^2+x=x\left(x^2-2x+1-4y^2\right)=x\left[\left(x-1\right)^2-\left(2y\right)^2\right]=x\left(x+2y-1\right)\left(x-2y-1\right)\)

\(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-8=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-8\)

Đặt \(x^2+7x+10=t\), ta có:

\(t\left(t+2\right)-8=t^2+2t-8=t^2-2t+4t-8=t\left(t-2\right)+4\left(t-2\right)=\left(t-2\right)\left(t+4\right)\)

\(=\left(x^2+7x+10+4\right)\left(x^2+7x+10-2\right)=\left(x^2+7x+14\right)\left(x^2+7x-8\right)\)

Trả lời:

a, \(-xy.\left(x^2+2xy-3\right)=-x^3y-2x^2y^2+3xy\)

b, \(\left(12x^6y^5-3x^3y^4+4x^2y\right):6x^2y\)

\(=12x^6y^5:6x^2y^2-3x^3y^4:6x^2y+4x^2y+6x^2y\)

\(=2x^4y^3-\frac{1}{2}xy^3+\frac{2}{3}\)

a.\(\left(-xy\right)\left(x^2+2xy-3\right)=-x^3y-2x^2y^2+6xy\)

b.\(\left(12x^6y^5-3x^3y^4+4x^2y\right):6x^2y=2x^4y^4-\frac{1}{2}xy^3+\frac{2}{3}\)

Bài 3:

a) ta có: \(A=x^2+4x+9\)

\(=x^2+4x+4+5=\left(x+2\right)^2+5\)

Ta có: \(\left(x+2\right)^2\ge0\forall x\)

\(\Rightarrow\left(x+2\right)^2+5\ge5\forall x\)

Dấu '=' xảy ra khi

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

Vậy: GTNN của đa thức \(A=x^2+4x+9\) là 5 khi x=-2

b) Ta có: \(B=2x^2-20x+53\)

\(=2\left(x^2-10x+\frac{53}{2}\right)\)

\(=2\left(x^2-10x+25+\frac{3}{2}\right)\)

\(=2\left[\left(x-5\right)^2+\frac{3}{2}\right]\)

\(=2\left(x-5\right)^2+2\cdot\frac{3}{2}\)

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

Ta có: \(\left(x-5\right)^2\ge0\forall x\)

\(\Rightarrow2\left(x-5\right)^2\ge0\forall x\)

\(\Rightarrow2\left(x-5\right)^2+3\ge3\forall x\)

Dấu '=' xảy ra khi

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

Vậy: GTNN của đa thức \(B=2x^2-20x+53\) là 3 khi x=5

c) Ta có : \(M=1+6x-x^2\)

\(=-x^2+6x+1\)

\(=-\left(x^2-6x-1\right)\)

\(=-\left(x^2-6x+9-10\right)\)

\(=-\left[\left(x-3\right)^2-10\right]\)

\(=-\left(x-3\right)^2+10\)

Ta có: \(\left(x-3\right)^2\ge0\forall x\)

\(\Rightarrow-\left(x-3\right)^2\le0\forall x\)

\(\Rightarrow-\left(x-3\right)^2+10\le10\forall x\)

Dấu '=' xảy ra khi

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

Vậy: GTLN của đa thức \(M=1+6x-x^2\) là 10 khi x=3

Bài 2:

a) \(\left(x+y\right)^2+\left(x^2-y^2\right)\)

\(=\left(x+y\right)^2+\left(x-y\right).\left(x+y\right)\)

\(=\left(x+y\right).\left(x+y+x-y\right)\)

\(=\left(x+y\right).2x\)

c) \(x^2-2xy+y^2-z^2+2zt-t^2\)

\(=\left(x^2-2xy+y^2\right)-\left(z^2-2zt+t^2\right)\)

\(=\left(x-y\right)^2-\left(z-t\right)^2\)

\(=\left[x-y-\left(z-t\right)\right].\left(x-y+z-t\right)\)

\(=\left(x-y-z+t\right).\left(x-y+z-t\right)\)

Chúc bạn học tốt!

Ta có A = 4a²(5a - 3b) - 5a²(4a + b) 

= 20a³ - 12a²b - 20a³ - 5a²b

= - 17a²b

Thay a = -2 ; b = -3 vào A ta được

A = -17.(-2)²(-3) = 204

Trả lời:

A = 4a² ( 5a - 3b ) - 5a² ( 4a + b ) 

= 20a³ - 12a²b - 20a³ - 5a²b

= - 17a²b

Thay a = - 2; b = - 3 vào A, ta có:

A = - 17 . ( - 2 )² . ( - 3 ) = 204

a) \(73^2-27^2=\left(73+27\right)\left(73-27\right)=100.46=4600\)

b) \(55^2+20^2-25^2+40.45=\left(55^2-25^2\right)+\left(20^2+40.45\right)\)

\(=\left(55-25\right)\left(55+25\right)+\left(40.10+40.45\right)=30.80+40.55\)
\(=40\left(60+55\right)=40.115=4600\)

a, ĐKXĐ : \(x-1\ne0\)

=> \(x\ne1\)

TH₁ : \(x-2\ge0\left(x\ge2\right)\)

=> \(\left|x-2\right|=x-2=1\)

=> \(x=3\left(TM\right)\)

- Thay x = 3 vào biểu thức P ta được :

\(P=\frac{3+2}{3-1}=\frac{5}{2}\)

TH₂ : \(x-2< 0\left(x< 2\right)\)

=> \(\left|x-2\right|=2-x=1\)

=> \(x=1\left(KTM\right)\)

Vậy giá trị của P là \(\frac{5}{2}\) .

a) \(P=\frac{x+2}{x-1}\) \(\left(ĐKXĐ:x\ne1\right)\)

Ta có: \(\left|x-2\right|=1\text{⇔}\left[{}\begin{matrix}x-2=1\\x-2=-1\end{matrix}\right.\text{⇔}\left[{}\begin{matrix}x=3\\x=1\end{matrix}\right.\) (loại x = 1 vì x ≠ 1)

Thay \(x=3\) vào P, ta có:

\(P=\frac{3+2}{3-2}=\frac{5}{1}=5\)

Vậy P = 5 tại x = 3.

b) \(Q=\frac{x-1}{x}+\frac{2x+1}{x^2+x}=\frac{x-1}{x}+\frac{2x+1}{x\left(x+1\right)}=\frac{x^2-1}{x\left(x+1\right)}+\frac{2x+1}{x\left(x+1\right)}\) (ĐKXĐ: x ≠ 0, x ≠ -1)

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