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a)Ta có : \(\dfrac{x+1}{1-x}\)( giữ nguyên )
\(\dfrac{x^2-2}{1-x}\)( giữ nguyên )
\(\dfrac{2x^2-x}{x-1}=\dfrac{x-2x^2}{1-x}\)
b)Ta có : \(\dfrac{1}{x-1}=\dfrac{x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{x^2+x+1}{x^3-1}\)
\(\dfrac{2x}{x^2+x+1}=\dfrac{2x\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{2x^2-2x}{x^3-1}\)
\(\dfrac{2x-3x^2}{x^3-1}\)(giữ nguyên )
c) MTC = ( x+ 2)2(x - 2)2
Do đó , ta có : \(\dfrac{1}{x^2+4x+4}=\dfrac{1}{\left(x+2\right)^2}=\dfrac{\left(x-2\right)^2}{\left(x+2\right)^2\left(x-2\right)^2}\)
\(\dfrac{1}{x^2-4x+4}=\dfrac{1}{\left(x-2\right)^2}=\dfrac{\left(x+2\right)^2}{\left(x-2\right)^2\left(x+2\right)^2}\)
\(\dfrac{x}{x^2-4}=\dfrac{x}{\left(x+2\right)\left(x-2\right)}=\dfrac{x\left(x^2-2^2\right)}{\left(x+2\right)^2\left(x-2\right)^2}=\dfrac{x^3-4x}{\left(x+2\right)^2\left(x-2\right)^2}\)
d) MTC = xyz( x - y)( y - z)( x - z)
Do đó , ta có : \(\dfrac{1}{x\left(x-y\right)\left(x-z\right)}=\dfrac{yz\left(y-z\right)}{xyz\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)
\(\dfrac{1}{y\left(y-x\right)\left(y-z\right)}=\dfrac{-xz\left(x-z\right)}{xyz\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)
\(\dfrac{1}{z\left(z-x\right)\left(z-y\right)}=\dfrac{xy\left(x-y\right)}{xyz\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)
Cộng các phân thức lại ta có :
\(\dfrac{yz\left(y-z\right)}{xyz\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)+\(\dfrac{-xz\left(x-z\right)}{xyz\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)+\(\dfrac{xy\left(x-y\right)}{xyz\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)
= \(\dfrac{yz\left(y-z\right)-xz\left(x-z\right)+xy\left(x-y\right)}{xyz\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)
a, Vì x2 ≥ 0 , 2y2 ≥ 0 với mọi x,y
=>x2+2y2+ 1 ≥ 1
=>Phân thức trên luôn có nghĩa
Ta có :
\(VT=\left(\dfrac{1}{2}xy-\dfrac{1}{3}y\right)\left(\dfrac{1}{4}x^2y^2+\dfrac{1}{6}xy^2+\dfrac{1}{9}y^2\right)\)
\(=\dfrac{1}{8}x^3y^3+\dfrac{1}{12}x^2y^3+\dfrac{1}{18}xy^3-\dfrac{1}{12}x^2y^3-\dfrac{1}{18}xy^3-\dfrac{1}{27}y^3\)
\(=\dfrac{1}{8}x^3y^3-\dfrac{1}{27}y^3=VT\)
\(\Rightarrow dpcm\)
Vậy : ..............
Phương Ann Nhã Doanh Đinh Đức Hùng Mashiro Shiina
Nguyễn Thanh Hằng Nguyễn Huy Tú Lightning Farron
Akai Haruma Võ Đông Anh Tuấn
mấy anh chị cm cho e thêm cái : \(\dfrac{ay+bx}{c}=\dfrac{bz+cy}{a}=\dfrac{cx+az}{b}\)
a)(x-1)(x+1)(x+2)
=(x2-1)(x+2)
=x3-x+2x2-2
b)\(\dfrac{1}{2}\)x2y(2x+y)(2x-y)
=\(\dfrac{1}{2}\)x2y(4x2-y2)
=2x4y-\(\dfrac{1}{2}\)x2y3
c)(x-\(\dfrac{1}{2}\))(x+\(\dfrac{1}{2}\))(4x-1)
=(x2-\(\dfrac{1}{4}\))(4x-1)
=4x3-x2-x+\(\dfrac{1}{4}\)
Bài 1:
Áp dụng BĐT Cauchy-Schwarz ta có:
\(C=\dfrac{1}{yz}+\dfrac{1}{xz}\ge\dfrac{\left(1+1\right)^2}{xz+yz}=\dfrac{4}{xz+yz}\)
Từ \(x+y+z=3\Rightarrow x+y=3-z\)
\(\Rightarrow C\ge\dfrac{4}{xz+yz}=\dfrac{4}{z\left(x+y\right)}=\dfrac{4}{z\left(3-z\right)}=\dfrac{4}{-z^2+3z}\)
Lại có: \(-z^2+3z=\dfrac{9}{4}-\left(z-\dfrac{3}{2}\right)^2\le\dfrac{9}{4}\)
\(\Rightarrow C\ge\dfrac{4}{-z^2+3z}\ge\dfrac{4}{\dfrac{9}{4}}=\dfrac{16}{9}\)
Đẳng thức xảy ra khi \(x=y=\dfrac{3}{4};z=\dfrac{3}{2}\)
Bài 2:
Từ \(5x^2-5xy+y^2+\dfrac{4}{x^2}=0\)
\(\Leftrightarrow\left(4x^2-4xy+y^2\right)+\left(x^2+\dfrac{4}{x^2}-4\right)+4=xy\)
\(\Leftrightarrow\left(2x-y\right)^2+\left(x-\dfrac{2}{x}\right)^2+4\ge xy\)
Dễ thấy: \(VT\ge4\forall x;y\)\(\Rightarrow VP\ge4\forall x;y\)
Đẳng thức xảy ra khi \(\left(x;y\right)=\left(\sqrt{2};2\sqrt{2}\right);\left(-\sqrt{2};-2\sqrt{2}\right)\)
Bài 3:
Từ \(a^2+b^2=4a+2b+540\)
\(\Leftrightarrow\left(a^2-4a+4\right)+\left(b^2-2b+1\right)=545\)
\(\Leftrightarrow\left(a-2\right)^2+\left(b-1\right)^2=545\)
Áp dụng BĐT Cauchy-Schwarz ta có:
\(\left (P-2063 \right )^2=\left [23(a-2)+4(b-1) \right ]^2\)
\(\leq (23^2+4^2)\left [ (a-2)^2+(b-1)^2 \right ]\)
\(\Rightarrow P\le545+2063=2608\)
Đẳng thức xảy ra khi \(a=25;b=5\)
mình cảm ơn nha