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a/ Áp dụng BĐT Bunhiacopxki :
\(5^2=\left(1.x+2.y\right)^2\le\left(1^2+2^2\right)\left(x^2+y^2\right)\Leftrightarrow5A\ge25\Leftrightarrow A\ge5\)
Đẳng thức xảy ra khi \(\begin{cases}x=\frac{y}{2}\\x+2y=5\end{cases}\) \(\Leftrightarrow\begin{cases}x=1\\y=2\end{cases}\)
Vậy MaxA = 5 <=> (x;y) = (1;2)
b/ Áp dụng BĐT Cauchy : \(5=x+2y\ge2\sqrt{2xy}\Rightarrow xy\le\frac{25}{8}\)
Đẳng thức xảy ra khi \(\begin{cases}x=2y\\x+2y=5\end{cases}\) \(\Leftrightarrow\begin{cases}x=\frac{5}{2}\\y=\frac{5}{4}\end{cases}\)
Vậy MaxA = 25/8 <=> (x;y) = (5/2;5/4)
f(x) = (x2- x + 1)2016 = a4032 . x4032 + a4031 . x4031 +.....+ a1 . x + a0
=>f(1)=\(\left(1^2-1+1\right)^{2016}=a_{4032}+a_{4031}+......+a_1+a_0\)=1
vậy tổng các hệ số bằng 1
ĐKXĐ: \(a;b;c\in Z\)
Xét hiệu: (a3 + b3 + c3) - (a + b + c)
= (a3 - a) + (b3 - b) + (c3 - c)
= a.(a2 - 1) + b.(b2 -1) + c.(c2 - 1)
= (a - 1).a.(a + 1) + (b - 1).b.(b + 1) + (c - 1).c.(c + 1)
Vì (a - 1).a.(a + 1); (b - 1).b.(b + 1) và (c - 1).c.(c + 1) đều là tích 3 số nguyên liên tiếp nên mỗi tích này chia hết cho 2 và 3
Do (2;3)=1 nên mỗi tích này chia hết cho 6
\(\Rightarrow\left(a-1\right).a.\left(a+1\right)+\left(b-1\right).b.\left(b+1\right)+\left(c-1\right).c.\left(c+1\right)⋮6\)
hay \(\left(a^3+b^3+c^3\right)-\left(a+b+c\right)⋮6\)
Mà \(a+b+c=2016^{2016}⋮6\) nên \(a^3+b^3+c^3⋮6\left(đpcm\right)\)
\(\left\{{}\begin{matrix}25=5^2\\16=4^2\\25\left(x+y\right)^2=\left[5\left(x+y\right)\right]^2\\16\left(x-y\right)^2=\left[4\left(x-y\right)\right]^2\end{matrix}\right.\)
\(A=\left[5\left(x+y\right)-4\left(x-y\right)\right]\left[5\left(x+y\right)+4\left(x-y\right)\right]\)
\(A=\left(x+9y\right)\left(9x+y\right)\)
a: \(\left(ax+1\right)\left(ax+b\right)=x^2+7\)
\(\Leftrightarrow a^2x^2+abx+ax+b=x^2+7\)
\(\Leftrightarrow a^2x^2+ax\left(b+1\right)+b=x^2+7\)
\(\Leftrightarrow\left\{{}\begin{matrix}a^2=1\\b=7\\a\left(b+1\right)=0\end{matrix}\right.\Leftrightarrow\left(a,b\right)\in\varnothing\)
b: \(\Leftrightarrow ax^3+acx^2+ax+x^2b+cxb+b=x^3-3x+2\)
\(\Leftrightarrow ax^3+x^2\left(ac+b\right)+x\left(a+bc\right)+b=x^3-3x+2\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=1\\ac+b=0\\a+bc=3\\b=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=1\\b=2\\c+2=0\\1+2\cdot\left(-2\right)=3\end{matrix}\right.\Leftrightarrow\left(a,b,c\right)\in\varnothing\)
Queen Material Giải:
\(25\left(x+y\right)^2-16\left(x-y\right)^2\)
\(=25\left(x^2+2xy+y^2\right)-16\left(x^2-2xy+y^2\right)\)
\(=25x^2+50xy+25y^2-16x^2+32xy-16y^2\)
\(=9x^2+82xy+9y^2\)
\(=x\left(9x+y\right)+9y\left(9x+y\right)\)
\(=\left(x+9y\right)\left(9x+y\right)\).
a)\(\left(x^2+x\right)^2+4\left(x^2+x\right)-12\)
\(=\left(x^2+x+4\right)\left(x^2+x\right)-12\)
Đặt \(t=x^2+x\) ta có:
\(\left(t+4\right)t-12=t^2+4t-12\)
\(=\left(t-2\right)\left(t+6\right)=\left(x^2+x-2\right)\left(x^2+x+6\right)\)
\(=\left(x-1\right)\left(x+2\right)\left(x^2+x+6\right)\)
b)\(x^8+x+1\)
\(=x^8-x^2+\left(x^2+x+1\right)\)
\(=x^2\left(x^6-1\right)+\left(x^2+x+1\right)\)
\(=x^2\left(x^3+1\right)\left(x^3-1\right)+\left(x^2+x+1\right)\)
\(=x^2\left(x^3+1\right)\left(x-1\right)\left(x^2+x+1\right)+\left(x^2+x+1\right)\)
\(=\left(x^2+x+1\right)\left[x^2\left(x^3+1\right)\left(x-1\right)+1\right]\)
\(x^4+2002x^2+2001x+2002\)
\(=x^4+x^2+1+2001x^2+2001x+2001\)
\(=\left(x^4+2x^2+1\right)-x^2+2001\left(x^2+x+1\right)\)
\(=\left(x^2+1-x\right)\left(x^2+1+x\right)+2001\left(x^2+x+1\right)\)
\(=\left(x^2+x+1\right)\left(x^2+1-x+2001\right)\)
\(=\left(x^2+x+1\right)\left(x^2-x+2002\right)\)
\(x^4+2007x^2-2006x+2007\)
\(=x^4+2x^2+1-x^2+2006\left(x^2-x+1\right)\)
\(=\left(x^2+1\right)^2-x^2+2006\left(x^2-x+1\right)\)
\(=\left(x^2+1+x\right)\left(x^2+1-x\right)+2006\left(x^2-x+1\right)\)
\(=\left(x^2-x+1\right)\left(x^2+x+1+2006\right)\)
\(=\left(x^2-x+1\right)\left(x^2+x+2007\right)\)