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PTĐTTNT?
1.Đặt \(a^2+a=t\)
\(\Rightarrow\left(a^2+a\right)\left(a^2+a+1\right)-2\)
\(=t\left(t+1\right)-2\)
\(=t^2+t-2\)
\(=t^2+2t-\left(t+2\right)\)
\(=t\left(t+2\right)-\left(t+2\right)\)
\(=\left(t+2\right)\left(t-1\right)\)
Sửa đề:
\(x^4+2011x^2+2010x+2011\)
\(=\left(x^4-x\right)+2011x^2+2011x+2011\)
\(=x\left(x^3-1\right)+2011\left(x^2+x+1\right)\)
\(=x\left(x-1\right)\left(x^2+x+1\right)+2011\left(x^2+x+1\right)\)
\(=\left(x^2+x+1\right)\left(x^2-x+2011\right)\)
3. \(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-120\)
\(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)-120\)
Đặt \(x^2+5x+4=t\)
\(\Rightarrow\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-120\)
\(=t\left(t+2\right)-120\)
\(=t^2+2t+1-121\)
\(=\left(t+1\right)^2-11^2\)
\(=\left(t+1-11\right)\left(t+1+11\right)\)
\(=\left(t-10\right)\left(t+12\right)\)
\(=\left(x^2+5x-6\right)\left(x^2+5x+16\right)\)
\(=\left[\left(x^2-x\right)+\left(6x-6\right)\right]\left(x^2+5x+16\right)\)
\(=\left[x.\left(x-1\right)+6\left(x-1\right)\right]\left(x^2+5x+16\right)\)
\(=\left(x-1\right)\left(x+6\right)\left(x^2+5x+16\right)\)
4. \(\left(x^2+x+4\right)^2+8x\left(x^2+x+1\right)+15x^2\)
\(=\left(x^2+x+4\right)^2+2.\left(x^2+x+1\right).4x+\left(4x\right)^2-x^2\)
\(=\left(x^2+x+4+4x\right)^2-x^2\)
\(=\left(x^2+4+5x-x\right)\left(x^2+5x+x+4\right)\)
\(=\left(x^2+4x+4\right)\left(x^2+6x+4\right)\)
\(=\left(x+2\right)^2\left[\left(x^2+2.x.3+3^2\right)-\left(\sqrt{5}\right)^2\right]\)
\(=\left(x+2\right)^2\left[\left(x+3\right)^2-\left(\sqrt{5}\right)^2\right]\)
\(=\left(x+2\right)^2\left(x+3-\sqrt{5}\right)\left(x+3+\sqrt{5}\right)\)
2.Tim x
a,(2x+1)2-4(x+2)2=9
<=> (4x2+4x+1)-4(x2+4x+4)=9
<=> -12x-15=9
<=> -12x=24
<=> x=-2
\(1a,\)\(\left(x^2-0,1\right)=\left(x-\sqrt{0,1}\right)\left(x+\sqrt{0,1}\right)\)
\(1b,\)\(\left(2a^2+b^2\right)^2=\left(2a^2\right)^2+2.2a^2.b^2+\left(b^2\right)^2=4a^4+4a^2b^2+b^4\)
\(1c,\)\(\left(a^2+5\right)\left(5-a^2\right)=\left(5+a^2\right)\left(5-a^2\right)=25-x^4\)
\(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-120=\left(x^2+5x+4\right)\left(x^2+5x+6\right)-120\)
Đặt: x2+5x+4=t
Ta có:
\(t\left(t+2\right)-120=t^2+2t-120=t^2+12t-10t-120=t\left(t+12\right)-10\left(t+12\right)\)
\(=\left(t+12\right)\left(t-10\right)=\left(x^2+5x+16\right)\left(x^2+5x-6\right)\)
\(A=\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)
\(\Rightarrow2A=8.\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)
\(=\left(3-1\right)\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)
\(=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)
.....
\(=\left(3^{64}-1\right)\left(3^{64}+1\right)\)
\(=3^{128}-1\)
\(\Rightarrow A=\frac{3^{128}-1}{2}\)
Ta có:
\(A=\left(3x-5\right)\left(2x+11\right)-\left(2x+3\right)\left(3x+7\right)\)
\(=\left[3x\left(2x+11\right)-5\left(2x+11\right)\right]-\left[2x\left(3x+7\right)+3\left(3x+7\right)\right]\)
\(=\left[\left(6x^2+33x\right)-\left(10x+55\right)\right]-\left[\left(6x^2+14x\right)+\left(9x+21\right)\right]\)
\(=\left[6x^2+23x-55\right]-\left[6x^2+23x+21\right]\)
\(=-55-21=-76\)
Vậy biểu thức A không phụ thuộc vào biến x, y.
1. F = x2 + 2y2 + 2xy - 4x - 10y + 15
F = (x2 + 2xy + y2) - 4(x + y) + 4 + (y2 - 6y + 9) + 2
F = (x + y)2 - 4(x + y) + 4 + (y - 3)2 + 2
F = (x + y - 2)2 + (y - 3)2 + 2\(\ge\)2 với mọi x;y
Dấu "=" xảy ra <=> \(\hept{\begin{cases}x+y-2=0\\y-3=0\end{cases}}\) <=> \(\hept{\begin{cases}x=2-y\\y=3\end{cases}}\) <=> \(\hept{\begin{cases}x=-1\\y=3\end{cases}}\)
Vậy Mìn = 2 khi x = -1 và y = 3