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a)
\(\frac{x^2-16}{4x-x^2}=\frac{x^2-4^2}{x(4-x)}=\frac{(x-4)(x+4)}{x(4-x)}=\frac{x+4}{-x}\)
b) \(\frac{x^2+4x+3}{2x+6}=\frac{x^2+x+3x+3}{2(x+3)}=\frac{x(x+1)+3(x+1)}{2(x+3)}=\frac{(x+1)(x+3)}{2(x+3)}=\frac{x+1}{2}\)
c)
\(\frac{15x(x+y)^3}{5y(x+y)^2}=\frac{5.3.x(x+y)^2.(x+y)}{5y(x+y)^2}=\frac{3x(x+y)}{y}\)
d) \(\frac{5(x-y)-3(y-x)}{10(x-y)}=\frac{5(x-y)+3(x-y)}{10(x-y)}=\frac{8(x-y)}{10(x-y)}=\frac{8}{10}=\frac{4}{5}\)
e) \(\frac{2x+2y+5x+5y}{2x+2y-5x-5y}=\frac{7x+7y}{-3x-3y}=\frac{7(x+y)}{-3(x+y)}=\frac{-7}{3}\)
f) \(\frac{x^2-xy}{3xy-3y^2}=\frac{x(x-y)}{3y(x-y)}=\frac{x}{3y}\)
g) \(\frac{2ax^2-4ax+2a}{5b-5bx^2}=\frac{2a(x^2-2x+1)}{5b(1-x^2)}=\frac{2a(x-1)^2}{5b(1-x)(1+x)}\)
\(=\frac{2a(x-1)}{5b(-1)(x+1)}=\frac{2a(1-x)}{5b(x+1)}\)
\(BDT\Leftrightarrow\dfrac{x^4}{x^2y^2}+\dfrac{y^4}{x^2y^2}+\dfrac{4x^2y^2}{x^2y^2}\ge3\left(\dfrac{x^2}{xy}+\dfrac{y^2}{xy}\right)\)
\(\Leftrightarrow\dfrac{x^4+y^4-2x^2y^2+6x^2y^2}{x^2y^2}\ge\dfrac{3\left(x^2+y^2\right)}{xy}\)
\(\Leftrightarrow\dfrac{x^4+y^4-2x^2y^2}{x^2y^2}\ge\dfrac{3x^2+3y^2}{xy}-\dfrac{6xy}{xy}\)
\(\Leftrightarrow\dfrac{\left(x^2-y^2\right)^2}{x^2y^2}\ge\dfrac{3\left(x^2-2xy+y^2\right)}{xy}=\dfrac{3\left(x-y\right)^2}{xy}\)
\(\Leftrightarrow\left(x-y\right)^2\left[\dfrac{\left(x+y\right)^2-3xy}{x^2y^2}\right]\ge0\)
\(\Leftrightarrow\left(x-y\right)^2\left(\dfrac{x^2+y^2-xy}{x^2y^2}\right)\ge0\) (luôn đúng)
Vậy BĐT đã được chứng minh
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)
\(\dfrac{x^2}{y^2}+\dfrac{y^2}{x^2}+4\ge3\left(\dfrac{x}{y}+\dfrac{y}{x}\right)\)
\(\Leftrightarrow\dfrac{2x^2}{y^2}+\dfrac{2y^2}{x^2}+8\ge6\left(\dfrac{x}{y}+\dfrac{y}{x}\right)\)
\(\Leftrightarrow\left(\dfrac{x^2}{y^2}+2+\dfrac{y^2}{x^2}\right)-4\left(\dfrac{x}{y}+\dfrac{y}{x}\right)+4+\left(\dfrac{x^2}{y^2}-2.\dfrac{x}{y}+1\right)+\left(\dfrac{y^2}{x^2}-2.\dfrac{y}{x}+1\right)\ge0\)\(\Leftrightarrow\left(\dfrac{x}{y}+\dfrac{y}{x}-2\right)^2+\left(\dfrac{x}{y}-1\right)^2+\left(\dfrac{y}{x}-1\right)^2\ge0\) (đúng)
cách khác
đặt \(\dfrac{x}{y}+\dfrac{y}{x}=t\Rightarrow\left|t\right|\ge2\)
\(\Leftrightarrow t^2-3t+2\ge0\)
\(\Leftrightarrow\left(t-1\right)\left(t-2\right)\ge0\)
điều này luôn đúng với mọi |t| >=2 => dpcm
kết luận điều kiện đề hơi thừa
cái cần c/m đúng với mọi x,y khác 0
Ta có : \(\dfrac{\left(ax+by+cz\right)^2}{x^2+y^2+z^2}=a^2+b^2+c^2\)
\(\Leftrightarrow\left(ax+by+cz\right)^2=\left(a^2+b^2+c^2\right)\left(x^2+y^2+z^2\right)\)
\(\Leftrightarrow a^2x^2+b^2y^2+c^2z^2+2axby+2axcz+2bycz=a^2x^2+b^2x^2+c^2x^2+a^2y^2+b^2y^2+c^2y^2+a^2z^2+b^2z^2+c^2z^2\)
\(\Leftrightarrow2axby+2axvz+2bycz=a^2y^2+b^2x^2+a^2z^2+c^2x^2+b^2z^2+c^2y^2\)
\(\Leftrightarrow a^2y^2+b^2x^2+a^2z^2+c^2x^2+b^2z^2+c^2y^2-2axby-2azcx-2bycz=0\)
\(\Leftrightarrow\left(a^2y^2-2axby+b^2x^2\right)+\left(a^2z^2-2azcx+c^2x^2\right)+\left(b^2z^2-2bycz+c^2y^2\right)=0\)
\(\Leftrightarrow\left(ay-bx\right)^2+\left(az-cx\right)^2+\left(bz-cy\right)^2=0\)
Do \(\left(ay-bx\right)^2\ge0;\left(az-cx\right)^2\ge0;\left(bz-cy\right)^2\ge0\)
\(\Rightarrow\left\{{}\begin{matrix}ay-bx=0\\az-cx=0\\bz-cy=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}ay=bx\\az=cx\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{a}{x}=\dfrac{b}{y}\\\dfrac{c}{z}=\dfrac{a}{x}\end{matrix}\right.\)
\(\Rightarrow\dfrac{a}{x}=\dfrac{b}{y}=\dfrac{c}{z}\left(đpcm\right)\)
:D
Lời giải:
1.
\(\frac{a^3-4a^2-a+4}{a^3-7a^2+14a-8}=\frac{a^2(a-4)-(a-4)}{(a^3-8)-(7a^2-14a)}=\frac{(a-4)(a^2-1)}{(a-2)(a^2+2a+4)-7a(a-2)}\)
\(=\frac{(a-4)(a-1)(a+1)}{(a-2)(a^2-5a+4)}=\frac{(a-4)(a-1)(a+1)}{(a-2)(a-1)(a-4)}=\frac{a+1}{a-2}\)
2.
\(\frac{x^2y^2+1+(x^2-y)(1-y)}{x^2y^2+1+(x^2+y)(1+y)}=\frac{x^2y^2+1+x^2-x^2y-y+y^2}{x^2y^2+1+x^2+x^2y+y+y^2}\)
\(=\frac{(x^2y^2-x^2y+x^2)+(y^2-y+1)}{(x^2y^2+x^2y+x^2)+(y^2+y+1)}\)
\(=\frac{x^2(y^2-y+1)+(y^2-y+1)}{x^2(y^2+y+1)+(y^2+y+1)}=\frac{(x^2+1)(y^2-y+1)}{(x^2+1)(y^2+y+1)}=\frac{y^2-y+1}{y^2+y+1}\)
\(\dfrac{x^2}{y^2}+\dfrac{y^2}{x^2}+4\ge3\left(\dfrac{x}{y}+\dfrac{y}{x}\right)\)
\(\Leftrightarrow2\left(\dfrac{x^2}{y^2}+\dfrac{y^2}{x^2}+4\right)\ge6\left(\dfrac{x}{y}+\dfrac{y}{x}\right)\)
\(\Leftrightarrow\dfrac{2x^2}{y^2}+\dfrac{2y^2}{x^2}+8\ge\dfrac{6x}{y}+\dfrac{6y}{x}\)
\(\Leftrightarrow\left(\dfrac{x^2}{y^2}+2+\dfrac{y^2}{x^2}\right)-4\left(\dfrac{x}{y}+\dfrac{y}{x}\right)+4+\dfrac{x^2}{y^2}-2.\dfrac{x}{y}+1+\dfrac{y^2}{x^2}-2.\dfrac{y}{x}+1\ge0\)
\(\Leftrightarrow\left(\dfrac{x}{y}+\dfrac{y}{x}\right)^2-4.\left(\dfrac{x}{y}+\dfrac{y}{x}\right)+4+\left(\dfrac{x}{y}-1\right)^2+\left(\dfrac{y}{x}-1\right)^2\ge0\)
\(\Leftrightarrow\left(\dfrac{x}{y}+\dfrac{y}{x}-2\right)^2+\left(\dfrac{x}{y}-1\right)^2+\left(\dfrac{y}{x}-1\right)^2\ge0^{\left(1\right)}\)
\(^{\left(1\right)}\)đúng \(\Rightarrowđpcm\)
Áp dụng BĐT : x4 + y4 ≥ 2x2y2
=> \(\dfrac{x^2}{y^2}+\dfrac{y^2}{x^2}\) ≥ 2 ( x , y > 0 )
TT , \(\dfrac{x}{y}+\dfrac{y}{x}\) ≥ 2 ( x , y > 0 )
Ta có : \(\dfrac{x^2}{y^2}+\dfrac{y^2}{x^2}\) + 4 ≥ 6 ( 1 )
\(3\left(\dfrac{x}{y}+\dfrac{y}{x}\right)\) ≥ 6 ( 2 )
Từ ( 1 ; 2) => đpcm
Lời giải:
Ta có:
\(\frac{4x^2y^2}{(x^2+y^2)^2}+\frac{x^2}{y^2}+\frac{y^2}{x^2}\geq 3\)
\(\Leftrightarrow \frac{4x^2y^2}{(x^2+y^2)^2}-1+\frac{x^2}{y^2}+\frac{y^2}{x^2}-2\geq 0\)
\(\Leftrightarrow \frac{4x^2y^2-(x^2+y^2)^2}{(x^2+y^2)^2}+\left(\frac{x}{y}-\frac{y}{x}\right)^2\geq 0\)
\(\Leftrightarrow \frac{-(x^2-y^2)^2}{(x^2+y^2)^2}+\frac{(x^2-y^2)^2}{x^2y^2}\geq 0\)
\(\Leftrightarrow (x^2-y^2)^2\left(\frac{1}{x^2y^2}-\frac{1}{(x^2+y^2)^2}\right)\geq 0\)
\(\Leftrightarrow \frac{(x^2-y^2)^2(x^4+y^4+x^2y^2)}{x^2y^2(x^2+y^2)^2}\geq 0\) (luôn đúng)
Do đó ta có đpcm.
Dấu bằng xảy ra khi $x=y$
\(A=\dfrac{4x^2y^2}{\left(x^2+y^2\right)^2}+\dfrac{x^2}{y^2}+\dfrac{y^2}{x^2}\)
x,y khác 0
<=>\(A=\dfrac{4}{\left(\dfrac{x}{y}+\dfrac{y}{x}\right)^2}+\left(\dfrac{x}{y}\right)^2+\left(\dfrac{y}{x}\right)^2\)
\(A+2=\dfrac{4}{\left(\dfrac{x}{y}+\dfrac{y}{x}\right)^2}+\left(\dfrac{x}{y}+\dfrac{y}{x}\right)^2=m\)
\(\left(\dfrac{x}{y}+\dfrac{y}{x}\right)^2=t;t\ge4\)
\(m=\dfrac{4}{t}+t\Leftrightarrow t^2-mt+4=0\)
f(t) có nghiệm t>= 4<=>\(\left\{{}\begin{matrix}m^2-16\ge0\\\dfrac{m+\sqrt{m^2-16}}{2}\ge4\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\left|m\right|\ge4\\m^2-16\ge m^2-16m+64\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\left|m\right|\ge4\\m\ge5\end{matrix}\right.\) \(\Leftrightarrow A+2\ge5;A\ge3=>dpcm\)