6x3-9x2
25x2-0,09
X2-x-y2-y
(x2+4)2-16x2
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câu 1 B
câu 2 D
câu 3 ko bt
câu 4 x=-1/2; x = -(căn bậc hai(3)*i-1)/4;x = (căn bậc hai(3)*i+1)/4;
câu 5 x=-5/3, x=0, x=1
Câu 1: x2 + 2 xy + y2 bằng:
A. x2 + y2 B.(x + y)2 C. y2 – x2 D. x2 – y2
Câu 2: (4x + 2)(4x – 2) bằng:
A. 4x2 + 4 B. 4x2 – 4 C. 16x2 + 4 D. 16x2 – 4
Câu 3: 25a2 + 9b2 - 30ab bằng:
A.(5a-9b)2 B.(5a – 3b)2 C.(5a+3b)2 D.(5a)2 – (3b)2
Câu 4: 8x3 +1 bằng
A.(2x+1).(4x2-2x+1) B. (2x-1).(4x2+2x+1) C.(2x+1)3 D.(2x)3-13
Câu 5:Thực hiện phép nhân x(3x2 + 2x - 5) ta được:
A.3x3 - 2x2 – 5x B. 3x3 + 2x2 – 5x C. 3x3 - 2x2 +5x D. 3x3 + 2x2 + 5x
a: \(=5x\left(xy^2+3x+6y^2\right)\)
b: \(=\left(x-2\right)\left(x+3\right)-\left(x-2\right)\left(x+2\right)=\left(x-2\right)\left(x+3-x-2\right)=\left(x-2\right)\)
c: \(=\left(x-3\right)\left(x-4\right)\)
d: \(=x\left(x^2-2xy+y^2-9\right)\)
=x(x-y-3)(x-y+3)
e: \(=\left(x+y\right)^2-25=\left(x+y+5\right)\left(x+y-5\right)\)
f: \(=\left(x-4\right)\left(x+3\right)\)
này mình có vài câu không làm được, xin lỗi bạn nha
\(b,16x^2-8x+1=\left(4x-1\right)^2\\ c,4x^2+12xy+9y^2=\left(2x+3y\right)^2\\ e,=x^2+2x+1+y^2+2y+1+2\left(x+1\right)\left(y+1\right)\\ =\left(x+1\right)^2+2\left(x+1\right)\left(y+1\right)+\left(y+1\right)^2\\ =\left[\left(x+1\right)+\left(y+1\right)\right]^2=\left(x+y+2\right)^2\\ g,=x^2-2x\left(y+2\right)+\left(x+2\right)^2=\left[x-\left(y+2\right)\right]^2=\left(x-y-2\right)^2\\ h,=\left[x+\left(y+1\right)\right]^2=\left(x+y+1\right)^2\)
\(M=\dfrac{\dfrac{1}{16}}{x^2}+\dfrac{\dfrac{1}{4}}{y^2}+\dfrac{1}{z^2}\ge\dfrac{\left(\dfrac{1}{4}+\dfrac{1}{2}+1\right)^2}{x^2+y^2+z^2}=\dfrac{7}{4}\)
\(M_{min}=\dfrac{7}{4}\) khi \(\left(x;y;z\right)=\left(\dfrac{1}{2};\dfrac{1}{\sqrt{2}};1\right)\)
\(\left(x-y\right)^2-4=\left(x-y-2\right)\left(x-y+2\right)\)
\(9-\left(x-y\right)^2=\left(3-x+y\right)\left(3+x-y\right)\)
\(\left(x^2+4\right)^2-16x^2=\left(x^2-4x+4\right)\left(x^2+4x+4\right)=\left(x-2\right)^2\left(x+2\right)^2\)
\(M=\dfrac{\dfrac{1}{16}}{x^2}+\dfrac{\dfrac{1}{4}}{y^2}+\dfrac{1}{z^2}\ge\dfrac{\left(\dfrac{1}{4}+\dfrac{1}{2}+1\right)^2}{x^2+y^2+z^2}=\dfrac{49}{16}\)
\(M_{min}=\dfrac{49}{16}\) khi \(\left(x;y;z\right)=\left(\dfrac{1}{\sqrt{7}};\dfrac{2}{\sqrt{14}};\dfrac{2}{\sqrt{7}}\right)\)
a) \(4x^2-1\)
\(=\left(2x\right)^2-1^2\)
\(=\left(2x-1\right)\left(2x+1\right)\)
b) \(x^2-3y^2\)
\(=x^2-\left(y\sqrt{3}\right)^2\)
\(=\left(x-y\sqrt{3}\right)\left(x+y\sqrt{3}\right)\)
c) \(9x^2-\dfrac{1}{4}\)
\(=\left(3x\right)^2-\left(\dfrac{1}{2}\right)^2\)
\(=\left(3x-\dfrac{1}{2}\right)\left(3x+\dfrac{1}{2}\right)\)
d) \(\left(x-y\right)^2-4\)
\(=\left(x-y\right)^2-2^2\)
\(=\left(x-y-2\right)\left(x-y+2\right)\)
e) \(9-\left(x-y\right)^2\)
\(=3^2-\left(x-y\right)^2\)
\(=\left(3+x-y\right)\left(3-x+y\right)\)
f) \(\left(x^2+4\right)^2-16x^2\)
\(=\left(x^2+4\right)^2-\left(4x\right)^2\)
\(=\left(x^2-4x+4\right)\left(x^2+4x+4\right)\)
\(=\left(x-2\right)^2\left(x+2\right)^2\)
Bài 1:
a: \(=2x^2-3xy+5y^2\)
b: \(=\dfrac{2x^3-10x^2-11x^2+55x+12x-60}{x-5}=2x^2-11x+12\)
c: \(=\dfrac{6x^3+3x^2-10x^2-5x+4x+2}{2x+1}=3x^2-5x+2\)
c: \(=\dfrac{\left(x+3\right)^2-y^2}{x+y+3}=x+3-y\)
\(6x^3-9x^2=3x^2\left(2x-3\right)\\ 25x^2-0,09=\left(5x-0,3\right)\left(5x+0,3\right)\\ x^2-x-y^2-y=\left(x-y\right)\left(x+y\right)-\left(x+y\right)=\left(x+y\right)\left(x-y-1\right)\\ \left(x^2+4\right)^2-16x^2=\left(x^2-4x+4\right)\left(x^2+4x+4\right)=\left(x-2\right)^2\left(x+2\right)^2\)