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1) =\(-3x^4+9x^3+11x^3-33x^2-2x^2+6x-16x+48\)
=\(-3x^3\left(x-3\right)+11x^2\left(x-3\right)-2x\left(x-3\right)-16\left(x-3\right)\)
= \(\left(x-3\right)\left(-3x^3+11x^2-2x-16\right)\)
= \(\left(x-3\right)\left(-3x^3+6x^2+5x^2-10x+8x-16\right)\)
=\(\left(x-3\right)\left(-3x^2\left(x-2\right)+5x\left(x-2\right)+8\left(x-2\right)\right)\)
= \(\left(x-3\right)\left(x-2\right)\left(-3x^2+5x+8\right)\)
= \(\left(x-3\right)\left(x-2\right)\left(x-\frac{8}{3}\right)\left(x+1\right)\)
Ý b lm theo ý tưởng tương tự nha bn :D
a) \(x^3-16x=0\)
\(\Leftrightarrow x\left(x^2-16\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x^2-16=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\pm4\end{cases}}\)
Vậy tập nghiệm \(S=\left\{-4;0;4\right\}\)
b) \(x^4-2x^3+10x^2-20x=0\)
\(\Leftrightarrow x^3\left(x-2\right)+10x\left(x-2\right)=0\)
\(\Leftrightarrow\left(x^3+10x\right)\left(x-2\right)=0\)
\(\Leftrightarrow x\left(x^2+10\right)\left(x-2\right)=0\)
Mà \(x^2+10>0\)nên \(\orbr{\begin{cases}x=0\\x-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=2\end{cases}}\)
Vậy tập nghiệm S = { 0;2}
a: Ta có: \(-3x^4+20x^3-35x^2-10x+48\)
\(=-\left(3x^4-20x^3+35x^2+10x-48\right)\)
\(=-\left(3x^4-9x^3-11x^3+33x^2+2x^2-6x+16x-48\right)\)
\(=-\left(x-3\right)\left(3x^3-11x^2+2x+16\right)\)
\(=-\left(x-3\right)\left(3x^3-6x^2-5x^2+10x-8x+16\right)\)
\(=-\left(x-3\right)\left(x-2\right)\left(3x^2-5x-8\right)\)
\(=-\left(x-3\right)\left(x-2\right)\left(3x-8\right)\left(x+1\right)\)
b: Ta có: \(-\left(2x^4+7x^3+x^2-7x-3\right)\)
\(=-\left(2x^4-2x^3+9x^3-9x^2+10x^2-10x+3x-3\right)\)
\(=-\left(x-1\right)\left(2x^3+9x^2+10x+3\right)\)
\(=-\left(x-1\right)\left(2x^3+2x^2+7x^2+7x+3x+3\right)\)
\(=-\left(x-1\right)\left(x+1\right)\left(2x^2+7x+3\right)\)
\(=-\left(x-1\right)\left(x+1\right)\cdot\left(x+3\right)\left(2x+1\right)\)
b, \(15\left(x+3\right)+20x\left(x+8\right)=15x+45+20x^2+160x\)
\(=20x^2+175x+45=...\)
c, \(6\left(x-9\right)-3x\left(y-x\right)=6x-54-3xy+3x^2\)
d, \(2xy+10x^2-x\) không phân tích được nữa nhé
e, \(4ab^2-28a+16b\)không phân tích được nữa nhé
g, \(a\left(a+b\right)=ab\left(a+b\right)< =>\left(a+b\right)\left(a-ab\right)=0< =>\left(a+b\right)a\left(1-b\right)=0\)
h, \(30a^2+6a-6=\left(\sqrt{30}a\right)^2+2.\sqrt{30}.\frac{3}{\sqrt{30}}+\frac{3}{10}-\frac{63}{10}\)
\(=\left(\sqrt{30}a+\frac{3}{\sqrt{30}}\right)^2-\sqrt{\frac{63}{10}}^2=\left(\sqrt{30}a+\frac{3}{\sqrt{30}}-\sqrt{\frac{63}{10}}\right)\left(\sqrt{30}a+\frac{3}{\sqrt{30}}+\sqrt{\frac{63}{10}}\right)\)
\(3x^4-20x^3+35x^2+10x-48=0\)
\(\Leftrightarrow\left(x^2-2x-3\right)\left(3x^2-14x+16\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x-3\right)\left(3x-8\right)\left(x-2\right)=0\)
\(\Rightarrow x=\left\{-1;3;\frac{8}{3};2\right\}\)
cảm ơn bạn