a) \(2x^2+2x+1=0\)

\(\Rightarrow2x^2+2x=-1\)

\(\Rightarrow2x\left(x+1\right)=-1\)

⇒ Pt vô nghiệm

a: \(2x^2+2x+1=0\)

\(\text{Δ}=2^2-4\cdot2\cdot1=4-8=-4< 0\)

Vì Δ<0 nên phương trình vô nghiệm

a: Ta có: \(y\left(x^2-y^2\right)\cdot\left(x^2+y^2\right)-y\left(x^4-y^4\right)\)

\(=y\left(x^4-y^4\right)-y\left(x^4-y^4\right)\)

=0

b: Ta có: \(\left(2x+\dfrac{1}{3}\right)\left(4x^2-\dfrac{2}{3}x+\dfrac{1}{9}\right)-\left(8x^3-\dfrac{1}{27}\right)\)

\(=8x^3+\dfrac{1}{27}-8x^3+\dfrac{1}{27}\)

\(=\dfrac{2}{27}\)

c: Ta có: \(\left(x-1\right)^3-\left(x-1\right)\left(x^2+x+1\right)-3x\left(1-x\right)\)

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

=0

Ta có

(I): 4 x 2   +   4 x   –   9 y 2   +   1   =   ( 4 x 2   +   4 x   +   1 )   –   9 y 2   =   ( 2 x   +   1 ) 2   –   ( 3 y ) 2

= (2x + 1 + 3y)(2x + 1 – 3y) nên (I) đúng

Và

(II):

5 x 2   –   10 x y   +   5 y 2   –   20 z 2   =   5 ( x 2   –   2 x y   +   y 2   –   4 z 2 )     =   5 [ ( x   –   y ) 2   –   ( 2 z ) 2 ]  

= 5(x – y – 2z)(x – y + 2z) nên (II) sai

Đáp án cần chọn là: A

BÀi 1:

a: \(\left(a+b\right)^2-\left(a-b\right)^2\)

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

\(=2b\cdot2a=4ab\)

b: \(\left(a+2\right)^2-\left(a+2\right)\left(a-2\right)\)

\(=a^2+4a+4-\left(a^2-4\right)\)

\(=a^2+4a+4-a^2+4=4a+8\)

c: \(\left(2x+3\right)^2-4\left(x-1\right)\left(x+1\right)=49\)

=>\(4x^2+12x+9-4\left(x^2-1\right)=49\)

=>\(4x^2+12x+9-4x^2+4=49\)

=>12x+13=49

=>12x=36

=>x=3

d: \(Q=\left(x+3\right)^2+\left(x+3\right)\left(x-3\right)-2\left(x+2\right)\left(x-4\right)\)

\(=x^2+6x+9+x^2-9-2\left(x^2-4x+2x-8\right)\)

\(=2x^2+6x-2\left(x^2-2x-8\right)=2x^2+6x-2x^2+4x+16=10x+16\)

Khi x=1/2 thì Q=10*1/2+16=5+16=21

Bài 2:

a: \(A=\left(4x^2+y^2\right)\left(2x+y\right)\left(2x-y\right)\)

\(=\left(4x^2+y^2\right)\left(4x^2-y^2\right)=16x^4-y^4\)

b: \(\left(7x+1\right)^2-\left(x+7\right)^2\)

\(=49x^2+14x+1-\left(x^2+14x+49\right)\)

\(=49x^2+14x+1-x^2-14x-49=48x^2-48=48\left(x^2-1\right)\)

c: \(16x^2-\left(4x-5\right)^2=15\)

=>\(16x^2-\left(16x^2-40x+25\right)=15\)

=>\(16x^2-16x^2+40x-25=15\)

=>40x=40

=>x=1

d: \(A=-x^2+2x+3\)

\(=-x^2+2x-1+4\)

\(=-\left(x-1\right)^2+4\le4\forall x\)

Dấu '=' xảy ra khi x-1=0

=>x=1

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

\(=2x^2-2x-3x^2-12x+x^2+2x\) 

\(=-12x\) 

\(b.\left(2x-3\right)\left(3x+5\right)-\left(x-1\right)\left(6x+2\right)+3-5x\) 

\(=6x+10x-9x^2-15-6x^2-2x-6x-2+3-5x\) 

\(=-15x^2+3x-14\) 

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

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

\(=y^3+x^2y\) 

\(VT=6\left(x^2+y^2+z^2\right)+10\left(xy+yz+xz\right)+2\left(\frac{1}{2x+y+z}+\frac{1}{x+2y+z}+\frac{1}{x+y+2z}\right)\)

\(=6\left(x+y+z\right)^2-2\left(xy+yz+xz\right)+2\frac{9}{2x+y+z+x+2y+z+x+y+2z}\)

\(\ge6\left(x+y+z\right)^2-2\frac{\left(x+y+z\right)^2}{3}+2\frac{9}{4\left(x+y+z\right)}\)

\(=\: 6\cdot\left(\frac{3}{4}\right)^2-2\cdot\frac{\left(\frac{3}{4}\right)^2}{3}+2\cdot\frac{9}{4\cdot\frac{3}{4}}=9\)

Đề bài yêu cầu gì vậy em.

Ta có phương trình hoành độ giao điểm là

\(\dfrac{-1}{2}x^2=x-4\)

⇒\(\left[{}\begin{matrix}x=2\\x=-4\end{matrix}\right.\)

Ta có : a(2;y₁); b(-4;y₂). Do hai điểm a và b cùng thuộc đường thẳng d nên ta có:

\(\left\{{}\begin{matrix}y_1=x_1-4=2-4=-2\\y_2=x_2-4=-4-4=-8\end{matrix}\right.\)

Khi đó ta có:

y₁+y₂ -5(x₁+x₂)=-2-8-5(2-4)=0 ⇒đpcm

VẬY..............