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Ta có:
\(x^3-27-9\left(x-3\right)=\left(x-3\right)\left(x^2+3x+9\right)-9\left(x-3\right)\)
\(=\left(x-3\right)\left(x^2+3x+9-9\right)=\left(x-3\right)\left(x^2+3x\right)\)
\(=\left(x-3\right)\left(x+3\right)x=x\left(x^2-9\right)\)
(x^3-27)-9(x-3)=x(x^2-9)
<=>(x-3)(x^2+3x+9)-9(x-3)-x(x-3)(x+3)=0
<=>(x-3)(x^2+3x-x(x+3) )=0
<=>(x-3)(x^2+3x-x^2-3x)=0
<=>(x-3)=0
<=>x=3
![](https://rs.olm.vn/images/avt/0.png?1311)
Mình sẽ làm theo đề bài của mình nếu đúng thì ... nha
Biến đổi vế phải ta có :
( x + y) [ ( x - y)^2 + xy ] = ( x + y)( x^2 - 2xy + y^2 + xy)
= ( x+ y)( x^2 - xy+ y^2)
= x^3 + y^3
VẬy VT = VP đẳng thức được CM
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(ĐK:x\ne\dfrac{1}{2};x\ne1;x\ne\dfrac{3}{2};x\ne2;x\ne\dfrac{5}{2}\\ PT\Leftrightarrow\dfrac{1}{\left(2x-1\right)\left(x-1\right)}+\dfrac{1}{\left(x-1\right)\left(3x-2\right)}+\dfrac{1}{\left(3x-2\right)\left(x-2\right)}+\dfrac{1}{\left(x-2\right)\left(5x-2\right)}=\dfrac{4}{21}\\ \Leftrightarrow2\left[\dfrac{\dfrac{1}{2}}{\left(x-\dfrac{1}{2}\right)\left(x-1\right)}+\dfrac{\dfrac{1}{2}}{\left(x-1\right)\left(x-\dfrac{3}{2}\right)}+\dfrac{\dfrac{1}{2}}{\left(x-\dfrac{3}{2}\right)\left(x-2\right)}+\dfrac{\dfrac{1}{2}}{\left(x-2\right)\left(x-\dfrac{5}{2}\right)}\right]=\dfrac{4}{21}\)
\(\Leftrightarrow\dfrac{1}{x-1}-\dfrac{1}{x-\dfrac{1}{2}}+\dfrac{1}{x-\dfrac{3}{2}}-\dfrac{1}{x-1}+\dfrac{1}{x-2}-\dfrac{1}{x-\dfrac{3}{2}}+\dfrac{1}{x-\dfrac{5}{2}}-\dfrac{1}{x-2}=\dfrac{2}{21}\\ \Leftrightarrow\dfrac{1}{x-1}-\dfrac{1}{x-\dfrac{5}{2}}=\dfrac{2}{21}\\ \Leftrightarrow\dfrac{x-\dfrac{5}{2}-x+1}{\left(x-1\right)\left(x-\dfrac{5}{2}\right)}=\dfrac{2}{21}\\ \Leftrightarrow\dfrac{-\dfrac{3}{2}}{x^2-\dfrac{7}{2}x+\dfrac{5}{2}}=\dfrac{2}{21}\\ \Leftrightarrow x^2-\dfrac{7}{2}x+\dfrac{5}{2}=-\dfrac{63}{4}\\ \Leftrightarrow4x^2-14x+10=-63\\ \Leftrightarrow4x^2-14x+73=0\\ \Leftrightarrow x\in\varnothing\)
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a: \(=\dfrac{x+2}{x+2}=1\)
b: \(=\dfrac{2x+6}{x+3}=2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
-x^2+6x-11
=-(x^2-6x+11)
=-(x^2-6x+9+2)
=-(x-3)^2-2<=-2
Dấu = xảy ra khi x=3
![](https://rs.olm.vn/images/avt/0.png?1311)
40: Ta có: \(A=27x^3+8y^3-3x-2y\)
\(=\left(3x+2y\right)\left(9x^2-6xy+4y^2\right)-\left(3x+2y\right)\)
\(=\left(3x+2y\right)\left(9x^2-6xy+4y^2-1\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
từ trên ta có (x+2)/13+(2x+45)/15-(3x+8)/37-(4x+69)/9=0
(x+2)/13+1+(2x+45)/15-1-(3x+8)/37-1-(4x+69)/9+1=0
(x+15)/13+(2x+30)/15-((3x+8)/37+1)-((4x+69)/9-1)=0
(x+15)/13+2(x+15)/15-3(x+15)/37-4(x+15)/9=0
(x+15)(1/13+2/15-3/37-4/9)=0
suy ra x+15=0
x=-15
\(\frac{x+2}{13}+\frac{2x+45}{15}=\frac{3x+8}{37}+\frac{4x+69}{9}\)
<=> \(\left(\frac{x+2}{13}+1\right)+\left(\frac{2x+45}{15}-1\right)=\left(\frac{3x+8}{37}+1\right)+\left(\frac{4x+69}{9}-1\right)\)
<=> \(\frac{x+2+13}{13}+\frac{2x+45-15}{15}=\frac{3x+8+37}{37}+\frac{4x+69-9}{9}\)
<=> \(\frac{x+15}{13}+\frac{2\left(x+15\right)}{13}=\frac{3\left(x+15\right)}{37}+\frac{4\left(x+15\right)}{9}\)
<=> \(\frac{x+15}{13}+\frac{2\left(x+15\right)}{13}-\frac{3\left(x+15\right)}{37}-\frac{4\left(x+15\right)}{9}=0\)
<=> \(\left(x+15\right)\left(\frac{1}{13}+\frac{2}{13}-\frac{3}{37}-\frac{4}{9}\right)=0\)
Vì \(\frac{1}{13}+\frac{2}{13}-\frac{3}{37}-\frac{4}{9}\ne0\)
<=> x + 15 = 0
<=> x = -15
\(2x\left(x^2+1\right)-2x^2\left(x+1\right)=0\)
\(\Leftrightarrow2x^3+2x-2x^3-2x^2=0\)
\(\Leftrightarrow2x-2x^2=0\)
\(\Leftrightarrow2x\left(1-x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=0\\1-x=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
b,2x(x^2+1)-2x^2(x+1)=0
<=>2x^3+2x-2x^3-2x^2=0
<=>2x-2x^2=0
<=>2x(1-x)=0
<=>2x=0 hoặc 1-x=0
<=>x=0 hoặc x=1