Rút gọn A: \(\frac{x-1}{x+2}-\frac{x+1}{2-x}-\frac{x^2-2x+4}{x^2-4}\)
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\(\frac{x^2+2}{2xy^3}-\frac{2x+2}{2xy^3}=\frac{x^2+2-2x-2}{2xy^3}=\frac{x^2-2x}{2xy^3}=\frac{x\left(x-2\right)}{2xy^3}=\frac{x-2}{2y^3}\)
\(\frac{4}{x-5}-\frac{1}{x+5}+\frac{13x-x^2}{25-x^2}=\frac{4}{x-5}-\frac{1}{x+5}+\frac{x^2-13x}{x^2-25}\)
\(=\frac{4\left(x+5\right)}{\left(x-5\right)\left(x+5\right)}-\frac{x-5}{\left(x-5\right)\left(x+5\right)}+\frac{x^2-13x}{\left(x-5\right)\left(x+5\right)}\)
\(=\frac{4x+20-x+5+x^2-13x}{\left(x-5\right)\left(x+5\right)}\)
\(=\frac{x^2-10x+25}{\left(x-5\right)\left(x+5\right)}=\frac{\left(x-5\right)^2}{\left(x-5\right)\left(x+5\right)}=\frac{x-5}{x+5}\)
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a) x(x - 3) + 5x = x2 - 8
=> x2 - 3x + 5x - x2 + 8 = 0
=> 2x + 8 = 0
=> 2x = -8
=> x = -4
b) 3(x + 4) - x2 - 4x = 0
=> 3(x + 4) - x(x + 4) = 0
=> (3 - x)(x + 4) = 0
=> \(\orbr{\begin{cases}3-x=0\\x+4=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=3\\x=-4\end{cases}}\)
Vậy \(x\in\left\{3;-4\right\}\)là giá trị cần tìm
c) 7x3 + 12x2 - 4x = 0
=> x(7x2 + 12x - 4) = 0
=> x(7x2 + 14x - 2x - 4) = 0
=> x[7x(x + 2) - 2(x + 2)] = 0
=> x(x + 2)(7x - 2) = 0
=> x = 0 hoặc x + 2 = 0 hoặc 7x - 2 = 0
=> x = 0 hoặc x = -2 hoặc x = 2/7
Vậy \(x\in\left\{0;-2;\frac{2}{7}\right\}\)là giá trị cần tìm
x( x - 3 ) + 5x = x2 - 8
⇔ x2 - 3x + 5x - x2 + 8 = 0
⇔ 2x + 8 = 0
⇔ 2x = -8
⇔ x = -4
3( x + 4 ) - x2 - 4x = 0
⇔ 3( x + 4 ) - x( x + 4 ) = 0
⇔ ( x + 4 )( 3 - x ) = 0
⇔ x = -4 hoặc x = 3
7x3 + 12x2 - 4x = 0
⇔ x( 7x2 + 12x - 4 ) = 0
⇔ x( 7x2 + 14x2 - 2x - 4 ) = 0
⇔ x[ 7x( x + 2 ) - 2( x + 2 ) ] = 0
⇔ x( x + 2 )( 7x - 2 ) = 0
⇔ x = 0 hoặc x = -2 hoặc x= 2/7
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a, \(5x^2y+10xy=5xy\left(x+2\right)\)
b, \(x^2-2xy+y^2-25=\left(x-y\right)^2-5^2=\left(x-y-5\right)\left(x-y+5\right)\)
c, \(x^3-8+2x\left(x-2\right)=\left(x-2\right)\left(x^2+2x+4\right)+2x\left(x-2\right)\)
\(=\left(x-2\right)\left[\left(x^2+2x+4\right)+2x\right]=\left(x-2\right)\left(x+2\right)^2\)
d, \(x^4+x^2y^2+y^4\):<
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bđt <=> \(\frac{abc}{a^3\left(b+c\right)}+\frac{abc}{b^3\left(c+a\right)}+\frac{abc}{c^3\left(a+b\right)}\)
\(=\frac{bc}{a^2\left(b+c\right)}+\frac{ac}{b^2\left(c+a\right)}+\frac{ab}{c^2\left(a+b\right)}\)
\(=\frac{1}{a^2\left(\frac{1}{c}+\frac{1}{b}\right)}+\frac{1}{b^2\left(\frac{1}{a}+\frac{1}{c}\right)}+\frac{1}{c^2\left(\frac{1}{b}+\frac{1}{a}\right)}\)
\(=\frac{\frac{1}{a^2}}{\frac{1}{c}+\frac{1}{b}}+\frac{\frac{1}{b^2}}{\frac{1}{a}+\frac{1}{c}}+\frac{\frac{1}{c^2}}{\frac{1}{b}+\frac{1}{a}}\)(1)
Đặt 1/a = x ; 1/b = y ; 1/c = z
(1) được viết lại thành \(\frac{x^2}{y+z}+\frac{y^2}{x+z}+\frac{z^2}{y+x}\)
Theo Bunyakovsky dạng phân thức ta có :
\(\frac{x^2}{y+z}+\frac{y^2}{x+z}+\frac{z^2}{y+x}\ge\frac{\left(x+y+z\right)^2}{2\left(x+y+z\right)}=\frac{x+y+z}{2}\)
Lại có \(x+y+z\ge3\sqrt{xyz}=3\sqrt{\frac{1}{a}\cdot\frac{1}{b}\cdot\frac{1}{c}}=3\sqrt{\frac{1}{abc}}=3\)( Cauchy )
=> \(\frac{x^2}{y+z}+\frac{y^2}{x+z}+\frac{z^2}{y+x}\ge\frac{x+y+z}{2}\ge\frac{3}{2}\)
Hay \(\frac{abc}{a^3\left(b+c\right)}+\frac{abc}{b^3\left(c+a\right)}+\frac{abc}{c^3\left(a+b\right)}\ge\frac{3}{2}\)( đpcm )
Đẳng thức xảy ra <=> x = y = z = 1
hay a = b = c = 1
cho a^3 +b^3+c^3=3abc và a+b+c khác 0 tính giá trị của biểu thức M=a^2020+b^2020+c^2020/(a+b+c)^2020
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Ta có : a3 + b3 + c3 = 3abc
=> (a + b)(a2 - ab + b2) + c3 - 3abc = 0
=> (a + b)3 - 3ab(a + b) + c3 - 3abc = 0
=> [(a + b)3 + c3] - [(3ab(a + b) + 3abc] = 0
=> (a + b + c)(a2 + b2 + 2ab - ac - bc + c2) - 3ab(a + b + c) = 0
=> (a + b + c)(a2 + b2 + c2 - ab - ac - bc) = 0
=> a2 + b2 + c2 - ab- ac - bc = 0
=> 2(a2 + b2 + c2 - ab- ac - bc) = 0
=> 2a2 + 2b2 + 2c2 - 2ab - 2ac - 2bc = 0
=> (a2 - 2ab + b2) + (b2 - 2bc + c2) + (a2 - 2ac + c2) = 0
=> (a - b)2 + (b - c)2 + (a - c)2 = 0
=> \(\hept{\begin{cases}a-b=0\\b-c=0\\c-a=0\end{cases}}\Rightarrow a=b=c\)
Khi đó M = \(\frac{a^{2020}+b^{2020}+c^{2020}}{\left(a+b+c\right)^{2020}}=\frac{3.c^{2020}}{\left(3c\right)^{2020}}+\frac{3c^{2020}}{3^{2020}.c^{2020}}=\frac{1}{3^{2019}}\)
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\(\frac{x-1}{x+2}-\frac{x+1}{2-x}-\frac{x^2-2x+4}{x^2-4}\)
= \(\frac{x-1}{x+2}+\frac{x+1}{x-2}-\frac{x^2-2x+4}{x^2-4}\)
= \(\frac{\left(x-1\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}+\frac{\left(x+2\right)\left(x+1\right)}{\left(x-2\right)\left(x+2\right)}-\frac{x^2-2x+4}{\left(x-2\right)\left(x+2\right)}\)
= \(\frac{x^2-3x+2+x^2+3x+2-x^2+2x-4}{\left(x-2\right)\left(x+2\right)}=\frac{x^2+2x}{\left(x-2\right)\left(x+2\right)}=\frac{x\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}=\frac{x}{x-2}\)
im,em moi lop 3m :)