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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
\(a,=5\left(x^3+8y\right)\\ b,=\left(4x+y\right)^2-16=\left(4x+y-4\right)\left(4x+y+4\right)\\ c,=3\left(x^2+2\cdot\dfrac{7}{3}x-5\right)\\ =3\left(x^2+2\cdot\dfrac{7}{3}x+\dfrac{49}{9}-\dfrac{94}{9}\right)\\ =3\left(x+\dfrac{7}{3}-\dfrac{\sqrt{94}}{3}\right)\left(x+\dfrac{7}{3}+\dfrac{\sqrt{94}}{3}\right)\)
a: \(5x^3+40y=5\left(x^3+8y\right)\)
b: \(16x^2+8xy+y^2-16\)
\(=\left(4x+y\right)^2-16\)
\(=\left(4x+y-4\right)\left(4x+y+4\right)\)
này mình có vài câu không làm được, xin lỗi bạn nha
\(b,16x^2-8x+1=\left(4x-1\right)^2\\ c,4x^2+12xy+9y^2=\left(2x+3y\right)^2\\ e,=x^2+2x+1+y^2+2y+1+2\left(x+1\right)\left(y+1\right)\\ =\left(x+1\right)^2+2\left(x+1\right)\left(y+1\right)+\left(y+1\right)^2\\ =\left[\left(x+1\right)+\left(y+1\right)\right]^2=\left(x+y+2\right)^2\\ g,=x^2-2x\left(y+2\right)+\left(x+2\right)^2=\left[x-\left(y+2\right)\right]^2=\left(x-y-2\right)^2\\ h,=\left[x+\left(y+1\right)\right]^2=\left(x+y+1\right)^2\)
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: \(=\left(x-4\right)\left(x+3\right)\)
\(d,15x^2y+20xy^2+25xy=5xy\left(3x+4y+5\right)\\ e,\left(x+y\right)^2-25=\left(x+y\right)^2-5^2=\left(x+y-5\right)\left(x+y+5\right)\\ f,1-2y+y^2=\left(1-y\right)^2\\ h,4x^2+8xy-3x-6y=\left(4x^2+8xy\right)-\left(3x+6y\right)=4x\left(x+2y\right)-3\left(x+2y\right)=\left(x+2y\right)\left(4x-3\right)\)
\(16x^2-8xy+y^2+1=\left(4x-y\right)^2+1\ge1\)
Dấu \("="\Leftrightarrow4x=y\)
\(-4x^2+2x-1=-\left(4x^2-2\cdot2\cdot\dfrac{1}{2}x+\dfrac{1}{16}\right)-\dfrac{15}{16}=-\left(2x-\dfrac{1}{4}\right)^2-\dfrac{15}{16}\le-\dfrac{15}{16}\)
Dấu \("="\Leftrightarrow2x=\dfrac{1}{4}\Leftrightarrow x=\dfrac{1}{8}\)