-5/8=x/16
Y/10=(-4)/8
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từ giả thiết suy ra x-2y =1
=> (x+2y)(x2+4y2)(x4+16y4)=(x - 2y ) (x+2y)(x2+4y2)(x4+16y4)=
sd hằng đẳngthức để làm tiếp nhé
\(A=x^2-4y^4=\left(x-2y^2\right)\left(x+2y^2\right)\)
\(B=8x^3+1=\left(2x+1\right)\left(4x^2-2x+1\right)\)
\(C=54x^3-16y^3=2\left(27x^3-8y^3\right)=2\left(3x-2y\right)\left(9x^2+6xy+4y^2\right)\)
\(D=x^2-6x+8=\left(x^2-6x+9\right)-1=\left(x-3\right)^2-1=\left(x-3-1\right)\left(x-3+1\right)=\left(x-4\right)\left(x-2\right)\)
\(E=2x^2-5x+2=\left(2x^2-4x\right)-\left(x-2\right)=2x\left(x-2\right)-\left(x-2\right)=\left(x-2\right)\left(2x-1\right)\)
\(G=x^4+2x^2-3=\left(x^4+3x^2\right)-\left(x^2+3\right)=x^2\left(x^2+3\right)-\left(x^2+3\right)=\left(x^2+3\right)\left(x^2-1\right)=\left(x^2+3\right)\left(x-1\right)\left(x+1\right)\)
1) \(\left(x+1\right)^2-16y^2\)
\(=\left(x+1\right)^2-\left(4y\right)^2\)
\(=\left(x+1-4y\right).\left(x+1+4y\right)\)
2) \(2x^2-8\)
\(=\left[\left(\sqrt{2x}\right)^2\right]^2-\left(\sqrt{8}\right)^2\)
\(=\left[\left(\sqrt{2x}\right)^2-\sqrt{8}\right].\left[\left(\sqrt{2x}\right)^2+\sqrt{8}\right]\)
6) \(x^2+1-y^2+2x\)
\(=x^2+2x+1-y^2\)
\(=\left(x^2+2x+1\right)-y^2\)
\(=\left(x+1\right)^2-y^2\)
\(=\left(x+1-y\right).\left(x+1+y\right)\)
Chúc bạn học tốt!
2.\(\left(x^2-16y^2\right)-3x+12y=\left(x-4y\right)\left(x+4y\right)-3\left(x-4y\right)=\left(x-4y\right)\left(x+4y-3\right)\)
4.\(x^3+6x^2+12x+8=\left(x+2\right)^3\)
Áp dụng BĐT phụ \(a^2+b^2\ge\dfrac{1}{2}\left(a+b\right)^2\Leftrightarrow\left(a-b\right)^2\ge0\)
\(A\ge\dfrac{1}{2}\left(x+y+\dfrac{1}{x}+\dfrac{1}{y}\right)^2\ge\dfrac{1}{2}\left(x+y+\dfrac{4}{x+y}\right)^2=\dfrac{1}{2}\left(1+\dfrac{4}{1}\right)^2=\dfrac{25}{2}\)
Dấu "=" \(x=y=\dfrac{1}{2}\)
Đáp án A
y ' = - 2 sin 4 x , y ' ' = - 8 cos 4 x , y ' ' ' = 32 sin 4 x
⇒ y ' ' ' + y ' ' + 16 y ' + 16 y - 8 = 0
-5/8 = x / 16
x = 16 : -5/8
x= -25,6
Y/10 = (-4)/8
Y = (-4)/8 x 10
Y = -5
\(\frac{-5}{8}=\frac{x}{16}\)
\(=>8x=-5.16\)
\(=>8x=-80=>x=-10\)
\(\frac{y}{10}=\frac{-4}{8}\)
\(=>\frac{y}{10}=\frac{-1}{2}\)
\(=>2y=-10\)
\(=>x=-5\)