Tinh
a) \(\left(a+5\right)^2\)
b) \(\left(4-x\right)^2\)
c) \(\left(3a-1\right)^2\)
d) \(\left(5-3b\right)^2\)
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a, Ta thấy : \(\left\{{}\begin{matrix}\left(2a+1\right)^2\ge0\\\left(b+3\right)^2\ge0\\\left(5c-6\right)^2\ge0\end{matrix}\right.\)\(\forall a,b,c\in R\)
\(\Rightarrow\left(2a+1\right)^2+\left(b+3\right)^2+\left(5c-6\right)^2\ge0\forall a,b,c\in R\)
Mà \(\left(2a+1\right)^2+\left(b+3\right)^2+\left(5c-6\right)^2\le0\)
Nên trường hợp chỉ xảy ra là : \(\left(2a+1\right)^2+\left(b+3\right)^2+\left(5c-6\right)^2=0\)
- Dấu " = " xảy ra \(\left\{{}\begin{matrix}2a+1=0\\b+3=0\\5c-6=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=-\dfrac{1}{2}\\b=-3\\c=\dfrac{6}{5}\end{matrix}\right.\)
Vậy ...
b,c,d tương tự câu a nha chỉ cần thay số vào là ra ;-;
\(a)\left( {a - 3b} \right)\left( {a + 3b} \right) = {a^2} - {\left( {3b} \right)^2} = {a^2} - 9{b^2}\)
\(b)\left( {2{\rm{x}} + 5} \right)\left( {2{\rm{x}} - 5} \right) = {\left( {2{\rm{x}}} \right)^2} - {5^2} = 4{{\rm{x}}^2} - 25\)
\(c)\left( {4y - 1} \right)\left( {4y + 1} \right) = {\left( {4y} \right)^2} - {1^2} = 16{y^2} - 1\)
a: \(=3x+5-3x+\dfrac{5}{3}-3x-1=3x+\dfrac{17}{3}\)
b: \(=\left(3a+2-3a+2\right)^2=4^2=16\)
1.
Áp dụng BĐT Cauchy-Schwarz:
\(\dfrac{a}{2a+a+b+c}=\dfrac{a}{25}.\dfrac{\left(2+3\right)^2}{2a+a+b+c}\le\dfrac{a}{25}\left(\dfrac{2^2}{2a}+\dfrac{3^2}{a+b+c}\right)=\dfrac{2}{25}+\dfrac{9}{25}.\dfrac{a}{a+b+c}\)
Tương tự:
\(\dfrac{b}{3b+a+c}\le\dfrac{2}{25}+\dfrac{9}{25}.\dfrac{b}{a+b+c}\)
\(\dfrac{c}{a+b+3c}\le\dfrac{2}{25}+\dfrac{9}{25}.\dfrac{c}{a+b+c}\)
Cộng vế:
\(VT\le\dfrac{6}{25}+\dfrac{9}{25}.\dfrac{a+b+c}{a+b+c}=\dfrac{3}{5}\)
Dấu "=" xảy ra khi \(a=b=c\)
2.
Đặt \(\dfrac{x}{x-1}=a;\dfrac{y}{y-1}=b;\dfrac{z}{z-1}=c\)
Ta có: \(\dfrac{x}{x-1}=a\Rightarrow x=ax-a\Rightarrow a=x\left(a-1\right)\Rightarrow x=\dfrac{a}{a-1}\)
Tương tự ta có: \(y=\dfrac{b}{b-1}\) ; \(z=\dfrac{c}{c-1}\)
Biến đổi giả thiết:
\(xyz=1\Rightarrow\dfrac{abc}{\left(a-1\right)\left(b-1\right)\left(c-1\right)}=1\)
\(\Rightarrow abc=\left(a-1\right)\left(b-1\right)\left(c-1\right)\)
\(\Rightarrow ab+bc+ca=a+b+c-1\)
BĐT cần chứng minh trở thành:
\(a^2+b^2+c^2\ge1\)
\(\Leftrightarrow\left(a+b+c\right)^2-2\left(ab+bc+ca\right)\ge1\)
\(\Leftrightarrow\left(a+b+c\right)^2-2\left(a+b+c-1\right)\ge1\)
\(\Leftrightarrow\left(a+b+c-1\right)^2\ge0\) (luôn đúng)
a: Ta có: \(\left(3x-1\right)^2-2\left(5x-2\right)^2-2\left(x^2+x-1\right)\left(x-1\right)\)
\(=9x^2-6x+1-2\left(25x^2-20x+4\right)-2\left(x^3-x^2+x^2-x-x+1\right)\)
\(=9x^2-6x+1-50x^2+40x-8-2\left(x^3-2x+1\right)\)
\(=-41x^2+34x-7-2x^3+4x-2\)
\(=-2x^3-41x^2+38x-9\)
b: Ta có: \(\left(3a+1\right)^2+2\left(9a^2-1\right)+\left(3a-1\right)^2\)
\(=\left(3a+1+3a-1\right)^2\)
\(=36a^2\)
a/ \(\left\{{}\begin{matrix}\left(x-1\right)^2\ge0\\\left(y-5\right)^2\ge0\\\left(x-y+4\right)^2\ge0\end{matrix}\right.\) \(\Rightarrow\left(x-1\right)^2+\left(y-5\right)^2+\left(x-y+4\right)^2\ge0\)
\(A_{min}=0\) khi \(\left\{{}\begin{matrix}x=1\\y=5\end{matrix}\right.\)
b/ \(B=x^2y^2-6xy+9+x^2+4x+4-16\)
\(B=\left(xy-3\right)^2+\left(x+2\right)^2-16\ge-16\)
\(B_{min}=-16\) khi \(\left\{{}\begin{matrix}x=-2\\y=-\frac{3}{2}\end{matrix}\right.\)
c/ \(C=x^2+\frac{y^2}{4}+16+xy+8x+4y+\frac{59}{4}y^2-3y+2001\)
\(C=\left(x+\frac{y}{2}+4\right)^2+\frac{59}{4}\left(y-\frac{6}{59}\right)^2+\frac{118050}{59}\ge\frac{118050}{59}\)
\(C_{min}=\frac{118050}{59}\)
d/ \(D=\left(x^2-2x\right)\left(y^2+6y\right)+12\left(x^2-2x\right)+3\left(y^2+6y\right)+36\)
\(=\left(x^2-2x\right)\left(y^2+6y+12\right)+3\left(y^2+6y+12\right)\)
\(=\left(x^2-2x+3\right)\left(y^2+6y+12\right)\)
\(=\left[\left(x-1\right)^2+2\right]\left[\left(y+3\right)^2+3\right]\ge2.3=6\)
\(D_{min}=6\)
e/ \(E=a^2+\frac{b^2}{4}+\frac{9}{4}+ab-3a-\frac{3b}{2}+\frac{3b^2}{4}-\frac{3b}{2}+2014-\frac{9}{4}\)
\(=\left(a+\frac{b}{2}-\frac{3}{2}\right)^2+\frac{3}{4}\left(y-1\right)^2+2011\ge2011\)
\(E_{min}=2011\)
a) (a+5)^2=a^2+2.a.5+5^2=a^2+10a+25
b) (4-x)^2=4^2-2.4.x+x^2=16-8x+x^2
c) (3a-1)^2=(3a)^2-2.3a.1+1^2=9.a^2-6a+1
d) (5-3b)^2=5^2-2.5.3b+(3b)^2=25-30b+9.b^2