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Bài 8:
Cho các số thực a,b,c,x,y thỏa mãn ax−by=√3ax−by=3.
Tìm GTNN của F=a2+b2+x2+y2+bx+ayF=a2+b2+x2+y2+bx+ay
Lời giải:
Sử dụng giả thiết ax−by=√3ax−by=3 ta có:
(a2+b2)(x2+y2)=(ax+by)2+(ax−by)2=(ax+by)2+3(a2+b2)(x2+y2)=(ax+by)2+(ax−by)2=(ax+by)2+3
Áp dụng bất đẳng thức CauchyCauchy , suy ra:
a2+b2=x2+y2=(a2+b2)+(x2+y2)≥2√(a2+b2)(x2+y2)=2√(ax+by)2+3a2+b2=x2+y2=(a2+b2)+(x2+y2)≥2(a2+b2)(x2+y2)=2(ax+by)2+3
Do đó, ta đưa về bài toán tìm GTNN của: 2√x2+3+x2x2+3+x trong đó x=ax+byx=ax+by
Ta có:
(2√x2+3+x)2=4(x2+3)+4x√x2+3+x2=(x2+3)+4x√x2+3+4x2+9=(√x2+3+2x)2+9≥9(2x2+3+x)2=4(x2+3)+4xx2+3+x2=(x2+3)+4xx2+3+4x2+9=(x2+3+2x)2+9≥9
⇒2√x2+3+x≥3⇒2x2+3+x≥3
Vậy MinT=3MinT=3
Bài 11:Cho các số a,b,c không âm không đồng thời bằng không. Chứng minh rằng;
∑2a2−bcb2−bc+c2≥3∑2a2−bcb2−bc+c2≥3
Không mất tính tổng quát, ta có thể giả sử bb là số nằm giữa aa và cc
BĐT đã cho tương đương với
∑2a2+(b−c)2b2−bc+c2≥6∑2a2+(b−c)2b2−bc+c2≥6
Áp dụng BĐT Cauchy-Schwarz, ta có
∑2a2b2−bc+c2≥2(a2+b2+c2)2∑a2(b2−bc+c2)=2(a2+b2+c2)22∑a2b2−abc∑a∑2a2b2−bc+c2≥2(a2+b2+c2)2∑a2(b2−bc+c2)=2(a2+b2+c2)22∑a2b2−abc∑a
∑(b−c)2b2−bc+c2≥[a(b−c)+b(a−c)+c(a−b)]22∑a2b2−abc∑a=4b2(a−c)22∑a2b2−abc∑a∑(b−c)2b2−bc+c2≥[a(b−c)+b(a−c)+c(a−b)]22∑a2b2−abc∑a=4b2(a−c)22∑a2b2−abc∑a
Do đó ta chỉ cần chứng minh
(a2+b2+c2)2+2b2(a−c)2≥6∑a2b2−3abc∑a(1)(a2+b2+c2)2+2b2(a−c)2≥6∑a2b2−3abc∑a(1)
Ta có
b2(a−c)2=[a(b−c)+c(a−b)]2=a2(b−c)2+c2(a−b)2+2ac(a−b)(b−c)b2(a−c)2=[a(b−c)+c(a−b)]2=a2(b−c)2+c2(a−b)2+2ac(a−b)(b−c)
≥a2(b−c)2+c2(a−b)2≥a2(b−c)2+c2(a−b)2
Suy ra
2b2(a−c)2≥a2(b−c)2+b2(c−a)2+c2(a−b)22b2(a−c)2≥a2(b−c)2+b2(c−a)2+c2(a−b)2
⇒VT(1)≥(∑a2)2+2∑a2b2−2abc∑a⇒VT(1)≥(∑a2)2+2∑a2b2−2abc∑a
Do đó ta chỉ còn phải chứng minh
(∑a2)2+2∑a2b2−2abc∑a≥6∑a2b2−3abc∑a(∑a2)2+2∑a2b2−2abc∑a≥6∑a2b2−3abc∑a
⇔∑a4+abc∑a≥2∑a2b2⇔∑a4+abc∑a≥2∑a2b2
BĐT này hiển nhiên đúng theo BĐT Schur
∑a4+abc∑a≥∑ab(a2+b2)∑a4+abc∑a≥∑ab(a2+b2)
Và BĐT AM-GM
∑ab(a2+b2)≥2∑a2b2∑ab(a2+b2)≥2∑a2b2
Kết thúc chứng minh
Đẳng thức xảy ra khi a=b=ca=b=c hoặc a=ba=b, c=0c=0 và các hoán vị.
\(a,\dfrac{2x+2y}{a^2+2ab+b^2}.\dfrac{ax-ay+bx-by}{2x^2-2y^2}\)
\(=\dfrac{2\left(x+y\right)}{\left(a+b\right)^2}.\dfrac{a\left(x-y\right)+b\left(x-y\right)}{2\left(x^2-y^2\right)}\)
\(=\dfrac{2\left(x+y\right)}{\left(a+b\right)^2}.\dfrac{\left(x-y\right)\left(a+b\right)}{2\left(x-y\right)\left(x+y\right)}\)
\(=\dfrac{1}{a+b}\)
\(b,\dfrac{a+b-c}{a^2+2ab+b^2-c^2}.\dfrac{a^2+2ab+b^2+ac+bc}{a^2-b^2}\)
\(=\dfrac{a+b-c}{\left(a+b\right)^2-c^2}.\dfrac{\left(a+b\right)^2+c\left(a+b\right)}{\left(a-b\right)\left(a+b\right)}\)
\(=\dfrac{a+b-c}{\left(a+b-c\right)\left(a+b+c\right)}.\dfrac{\left(a+b\right)\left(a+b+c\right)}{\left(a-b\right)\left(a+b\right)}\)
\(=\dfrac{1}{a-b}\)
\(c,\dfrac{x^3+1}{x^2+2x+1}.\dfrac{x^2-1}{2x^2-2x+2}\)
\(=\dfrac{\left(x+1\right)\left(x^2-x+1\right)}{\left(x+1\right)^2}.\dfrac{\left(x-1\right)\left(x+1\right)}{2\left(x^2-x+1\right)}\) \(=\dfrac{x-1}{2}\) \(d,\dfrac{x^8-1}{x+1}.\dfrac{1}{\left(x^2+1\right)\left(x^4+1\right)}\) \(=\dfrac{\left(x^4\right)^2-1}{x+1}.\dfrac{1}{\left(x^2+1\right)\left(x^4+1\right)}\) \(=\dfrac{\left(x^4-1\right)\left(x^4+1\right)}{x+1}.\dfrac{1}{\left(x^2+1\right)\left(x^4+1\right)}\) \(=\dfrac{\left(x^2+1\right)\left(x^2-1\right)}{x+1}.\dfrac{1}{x^2+1}\) \(=\dfrac{\left(x-1\right)\left(x+1\right)}{x+1}\) \(=x-1\) \(e,\dfrac{x-y}{xy+y^2}-\dfrac{3x+y}{x^2-xy}.\dfrac{y-x}{x+y}\) \(=\dfrac{x-y}{y\left(x+y\right)}-\dfrac{3x+y}{x\left(x-y\right)}.\dfrac{-\left(x-y\right)}{x+y}\) \(=\dfrac{x-y}{y\left(x+y\right)}-\dfrac{3x+y}{x}.\dfrac{-1}{x+y}\) \(=\dfrac{x-y}{y\left(x+y\right)}-\dfrac{-3x-y}{x\left(x+y\right)}\) \(=\dfrac{x\left(x-y\right)+y\left(3x+y\right)}{xy\left(x+y\right)}\) \(=\dfrac{x^2-xy+3xy+y^2}{xy\left(x+y\right)}\) \(=\dfrac{x^2+2xy+y^2}{xy\left(x+y\right)}\) \(=\dfrac{\left(x+y\right)^2}{xy\left(x+y\right)}=\dfrac{x+y}{xy}\)tìm giá trị của m để pt 2x-m=1-x nhận giá trị x=-2 là nghiệm
giải hộ e với :)
Ta có: \(\left(a^2+b^2\right)\left(x^2+y^2\right)=\left(ax+by\right)^2\)
\(\Leftrightarrow a^2x^2+a^2y^2+b^2x^2+b^2y^2=a^2x^2+b^2y^2+2axby\)
\(\Leftrightarrow a^2y^2-2axby+b^2x^2=0\)
\(\Leftrightarrow\left(ay-bx\right)^2=0\)
\(\Leftrightarrow ay=bx\)
hay \(\dfrac{a}{x}=\dfrac{b}{y}\)
Ta có : \(\left(a^2+b^2\right)\left(x^2+y^2\right)=\left(ax+by\right)^2\)
\(\Leftrightarrow a^2x^2+a^2y^2+b^2x^2+b^2y^2=a^2x^2+2abxy+b^2y^2\)
\(\Leftrightarrow a^2y^2-2abxy+b^2x^2=0\)
\(\Leftrightarrow\left(ay-bx\right)^2=0\)
\(\Leftrightarrow ay-bx=0\)
\(\Leftrightarrow ay=bx\Leftrightarrow\dfrac{a}{b}=\dfrac{x}{y}\)
đặt x/a=y/b=z/c=k
=>x=a.k,
y=b.k
z=c.k
=>(a^2k^2+b^2k^2+c^2k^2)(a^2+b^2+c^2)=k^2.(a^2+b^2+c^2)^2(1)
(ax+by+cz)^2=(a.a.k+b.b.k+c.c.k)^2=(a^2.k+b^2.k+c^2.k)^2
=k^2(a^2+b^2+c^2)(2)
từ (1)(2)=> nếu x/a=y/b=z/c thì (x2 + y2 + z2) (a2 + b2 + c2) = (ax + by + cz)2
=>
\(a,=\left(3x-5\right)\left(3x+3\right)=3\left(x+1\right)\left(3x-5\right)\\ b,=\left(5x-4-7x\right)\left(5x-4+7x\right)=\left(-2x-4\right)\left(12x-4\right)\\ =-8\left(x+2\right)\left(x-3\right)\\ c,=\left(2x+5-x+9\right)\left(2x+5+x-9\right)\\ =\left(x+14\right)\left(3x-4\right)\\ d,=\left(3x+1-2x+4\right)\left(3x+1+2x-4\right)\\ =\left(x+5\right)\left(5x-3\right)\\ e,=\left(6x+9-2x-2\right)\left(6x+9+2x+2\right)\\ =\left(4x+7\right)\left(8x+11\right)\\ f,=\left(2bc-b^2-c^2+a^2\right)\left(2bc+b^2+c^2-a^2\right)\\ =\left[a^2-\left(b-c\right)^2\right]\left[\left(b+c\right)^2-a^2\right]\\ =\left(a-b+c\right)\left(a+b-c\right)\left(b+c-a\right)\left(b+c+a\right)\\ g,=\left(ax+by-ay-bx\right)\left(ax+by+ay+bx\right)\\ =\left(a-b\right)\left(x-y\right)\left(a+b\right)\left(x+y\right)\)
\(h,=\left(a^2+b^2-5-2ab-4\right)\left(a^2+b^2-5+2ab+4\right)\\ =\left[\left(a-b\right)^2-9\right]\left[\left(a+b\right)^2-1\right]\\ =\left(a-b-3\right)\left(a-b+3\right)\left(a+b-1\right)\left(a+b+1\right)\)
a: \(\left(3x-1\right)^2-16\)
\(=\left(3x-1-4\right)\left(3x-1+4\right)\)
\(=\left(3x+3\right)\left(3x-5\right)\)
\(=3\left(x+1\right)\left(3x-5\right)\)
b: \(\left(5x-4\right)^2-49x^2\)
\(=\left(5x-4-7x\right)\left(5x-4+7x\right)\)
\(=\left(-2x-4\right)\left(12x-4\right)\)
\(=-8\left(x+2\right)\left(3x-1\right)\)
Bài 4:
Ta có: \(\left(x^3-x^2\right)-4x^2+8x-4=0\)
\(\Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-2\right)^2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
a: \(x^2-9-x^2\left(x^2-9\right)\)
\(=\left(x^2-9\right)-x^2\left(x^2-9\right)\)
\(=\left(x^2-9\right)\left(1-x^2\right)\)
\(=\left(1-x\right)\left(1+x\right)\left(x-3\right)\left(x+3\right)\)
b: \(x^2\left(x-y\right)+y^2\left(y-x\right)\)
\(=x^2\left(x-y\right)-y^2\left(x-y\right)\)
\(=\left(x-y\right)\left(x^2-y^2\right)\)
\(=\left(x-y\right)\left(x-y\right)\left(x+y\right)=\left(x-y\right)^2\cdot\left(x+y\right)\)
c: \(x^3+27+\left(x+3\right)\left(x-9\right)\)
\(=\left(x+3\right)\left(x^2-3x+9\right)+\left(x+3\right)\left(x-9\right)\)
\(=\left(x+3\right)\left(x^2-3x+9+x-9\right)\)
\(=\left(x+3\right)\left(x^2-2x\right)=x\left(x-2\right)\left(x+3\right)\)
d: \(x^2+5x+6\)
\(=x^2+2x+3x+6\)
\(=x\left(x+2\right)+3\left(x+2\right)=\left(x+2\right)\left(x+3\right)\)
e: \(3x^2-4x-4\)
\(=3x^2-6x+2x-4\)
\(=3x\left(x-2\right)+2\left(x-2\right)\)
\(=\left(x-2\right)\left(3x+2\right)\)
g: \(x^4+64y^4\)
\(=x^4+16x^2y^2+64y^4-16x^2y^2\)
\(=\left(x^2+8y^2\right)^2-\left(4xy\right)^2\)
\(=\left(x^2+8y^2-4xy\right)\left(x^2+8y^2+4xy\right)\)
h: \(a^2+b^2+2a-2b-2ab\)
\(=a^2-2ab+b^2+2a-2b\)
\(=\left(a-b\right)^2+2\left(a-b\right)=\left(a-b\right)\left(a-b+2\right)\)
i: \(\left(x+1\right)^2-2\left(x+1\right)\left(y-3\right)+\left(y-3\right)^2\)
\(=\left(x+1-y+3\right)^2\)
\(=\left(x-y+4\right)^2\)
k: \(x^2\left(x+1\right)-2x\left(x+1\right)+\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2-2x+1\right)\)
\(=\left(x+1\right)\left(x-1\right)^2\)
anh jaki kìa các fan của jaki!!!
a jaki:D