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\(x^3-7x-6=x^3-x-6x-6=x\left(x^2-1\right)-6\left(x-1\right)=x\left(x-1\right)\left(x+1\right)-6\left(x-1\right)\)

\(=\left(x-1\right)\left(x^2+x-6\right)=\left(x-1\right)\left(x^2+3x-2x-6\right)=\left(x-1\right)\left(x\left(x+3\right)-2\left(x+3\right)\right)\)

\(=\left(x-1\right)\left(x+3\right)\left(x-2\right)\)

2 tháng 7 2019

\(\left[\frac{2}{3x}-\frac{2}{x+1}\left(\frac{x+1}{3x}-x-1\right)\right]:\frac{x-1}{x}\)

\(=\left[\frac{2}{3x}-\frac{2\left(x+1\right)}{\left(x+1\right).3x}-\frac{2\left(-x-1\right)}{x+1}\right]:\frac{x-1}{x}\)

\(=\)\(\left[\frac{2}{3x}-\frac{2\left(x+1\right)}{\left(x+1\right).3x}+\frac{2\left(x+1\right)}{x+1}\right]:\frac{x-1}{x}\)

\(=\left[\frac{2}{3x}-\frac{2}{3x}+2\right]:\frac{x-1}{x}\)

\(=2.\frac{x}{x-1}=\frac{2x}{x-1}\)\(\left(đpcm\right)\)

2 tháng 7 2019

P=x^2(b-c)+b^2(c-x)+c^2(x-b)
P=X^2(b-c)+b^2[-(b-c)-(x-b)]+c^2(x-b)
P=x^2(b-c)-b^2(b-c)-b^2(x-b)+c^2(x-b)
P=(x^2-b^2)(b-c)- (b^2-c^2)(x-b)
P=(x+b)(x-b)(b-c)-(b+c)(b-c)(x-b)
P=(x-b)(b-c)(x+b-b-c)
P=(x-b)(b-c)(x-c)

2 tháng 7 2019

#)Giải :

Ta có : 

\(\hept{\begin{cases}\frac{ab}{b+c+a+b}\le\frac{ab}{4}\left(\frac{1}{a+c}+\frac{1}{b+c}\right)\\\frac{bc}{a+b+a+c}\le\frac{bc}{4}\left(\frac{1}{a+b}+\frac{1}{a+c}\right)\\\frac{ac}{b+c+a+b}\le\frac{ac}{4}\left(\frac{1}{b+c}+\frac{1}{a+b}\right)\end{cases}}\)

\(\Rightarrow VT\le\frac{1}{a+b}.\left(\frac{bc}{4}+\frac{ac}{4}\right)+\frac{1}{a+c}.\left(\frac{bc}{4}+\frac{ab}{4}\right)+\frac{1}{b+c}.\left(\frac{ac}{4}+\frac{ab}{4}\right)\)

\(=\frac{1}{a+b}.\frac{c\left(a+b\right)}{4}+\frac{1}{a+c}.\frac{b\left(a+c\right)}{4}+\frac{1}{b+c}.\frac{a\left(b+c\right)}{4}\)

\(=\frac{c}{4}+\frac{b}{4}+\frac{a}{4}\)

\(\Rightarrow\frac{a+b+c}{4}\)

\(\Rightarrowđpcm\)