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a.\(\left(3x-1\right)\left(9x^2+3x+1\right)+\left(1-3x\right)^3-3x\left(9x-3\right)-\left(x+2\right)^3+x\left(x^2+6x+12\right)\)\(=27x^3-1+1^3-9x+27x^2-27x^3-27x^2+9x-x^3-6x^2-12x-8+x^3+6x^2+12x\)\(=\left(27x^3+1^3-27x^3-x^3+x^3\right)+\left(27x^2-27x^2-6x^2+6x^2\right)+\left(-9x+9x-12x+12x\right)+\left(-1-8\right)\)\(=1-9=8\)
b.
\(\left(2x-3\right)\left(x-2\right)\left(x+2\right)-2\left(x+3\right)^3-\left(x-4\right)^3+\left(x-3\right)\left(x^2+3x+9\right)+9x^2+110x\)\(=\left(2x-3\right)\left(x^2-4\right)-2\left(x^3+9x^2+27x\right)-\left(x^3-12x^2+48x-64\right)+x^3-27+9x^2+110x\)\(=2x^3-8x-3x^2+1-2x^3-18x^2-54x-x^3+12x^2-48x+64+x^3-27+9x^2+110x\)\(=\left(2x^3-2x^3-x^3+x^3\right)+\left(-3x^2-18x^2+2x^2+9x^2\right)+\left(-8x-54x-48x+110x\right)+\left(1+64-27\right)\)\(=38\)
\(\frac{x^4-x^3-x+1}{x^4+x^3+3x^2+2x+2}\)
\(=\frac{x^3\left(x-1\right)-\left(x-1\right)}{x^4+x^3+x^2+2x^2+2x+2}\)
\(=\frac{\left(x-1\right)\left(x^3-1\right)}{x^2\left(x^2+x+1\right)+2\left(x^2+x+1\right)}\)
\(=\frac{\left(x-1\right)\left(x-1\right)\left(x^2+x+1\right)}{\left(x^2+x+1\right)\left(x^2+2\right)}\)
\(=\frac{\left(x-1\right)^2}{\left(x^2+2\right)}\)
\(\Leftrightarrow4\left(x^2+x-2\right)-\left(4x^2+11x-3\right)=2x-2\)
\(\Leftrightarrow4x^2+4x-8-4x^2-11x+3=2x-2\)
=>-7x-5=2x-2
=>-9x=3
hay x=-1/3
Bài 1 :
\(\frac{x^3-9x}{15-5x}=\frac{-x^2-3x}{5}\left(ĐKXĐ:x\ne3\right)\)
\(\Leftrightarrow5\left(x^3-9x\right)=-\left(x^2+3x\right)\left(15-5x\right)\)
\(\Leftrightarrow5x^3-45x=5x^3-45\) ( luôn đúng )
Do đó : \(\frac{x^3-9x}{15-5x}=\frac{-x^2-3x}{5}\left(x\ne3\right)\)
P/s : Bài này thì xét tích chéo của hai số thôi nhé @
\(x^4-y^4\)
\(=\left(x^2-y^2\right)\left(x^2+y^2\right)\)
\(=\left(x+y\right)\left(x-y\right)\left(x^2+y^2\right)\)
\(\left(a-b\right)^3-\left(a-b\right)^3\)
\(=\left(a-b\right)^2\left(a-b-a+b\right)\)
\(\left(a^2+2ab+b^2\right)+\left(a+b\right)^3\)
\(=\left(a+b\right)^2+\left(a+b\right)^3\)
\(=\left(a+b\right)^2\left(a+b+1\right)\)
......giải ....
a. \(\left(x^2-y^2\right)\left(x^2+y^2\right)=\left(x-y\right)\left(x+y\right)\left(x^2+y^2\right)\)
b ...ko cần làm .. =0
c.. =(a+b)^2 +(a+b)^3=(a+b)[ (a+b)+ (a+b)^2 ]
... check mk đó .. The end•••
Bạn cứ giải như bình thường thôi. Không việc gì phải đoán mò cả!
\(A=\frac{\left(x-1\right)^2}{x^2-4x+3}=\frac{\left(x-1\right)^2}{\left(x-1\right)\left(x-3\right)}< 1\)
\(\Leftrightarrow\left(x-1\right)^2< \left(x-1\right)\left(x-3\right)\)
\(\Leftrightarrow2\left(x-1\right)< 0\)
\(\Leftrightarrow x< 1\)
Vậy tập nghiệm của bất phương trình là \(S=\left\{x< 3\right\}\)
\(ĐKXĐ:x\ne1;x\ne3\)
để \(A< 1\) thì \(\frac{\left(x-1\right)^2}{x^2-4x+3}< 1\Leftrightarrow\frac{\left(x-1\right)^2}{\left(x-1\right)\left(x-3\right)}-1< 0\)
\(\Leftrightarrow\frac{x-1}{x-3}-\frac{x-3}{x-3}< 0\)
\(\Leftrightarrow\frac{x-1-x+3}{x-3}< 0\)
\(\Leftrightarrow\frac{2}{x-3}< 0\)
\(\Rightarrow x-3< 0\) vì \(2>0\)
\(\Rightarrow x< 3\)
kết hợp với \(ĐKXĐ:x\ne1;x\ne3\) ta có \(\hept{\begin{cases}x< 3\\x\ne1\end{cases}}\) thì \(A< 1\)
a)\(4\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\)
\(=\dfrac{1}{2}.\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\)
\(=\dfrac{1}{2}.\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\)
\(=\dfrac{1}{2}.\left(3^8-1\right)\left(3^8+1\right)\)
\(=\dfrac{1}{2}.\left(3^{16}-1\right)\)
\(=\dfrac{1}{2}3^{16}-\dfrac{1}{2}\)
b) \(48\left(5^2+1\right)\left(5^4+1\right).....\left(5^{32}+1\right)\)
\(=2.\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right).....\left(5^{32}+1\right)\)
\(=2.\left(5^4-1\right)\left(5^4+1\right).....\left(5^{32}+1\right)\)
\(=2.\left(5^8+1\right).....\left(5^{32}+1\right)\)
\(=2.\left(5^{32}-1\right)\)
\(=2.5^{32}-2\)
Tham khảo nhé~
3 chấm ở giữa để kia làm j z