\(2x^3+5x^2+5x+3\)

CÂU 2

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Khách

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Câu 1 : \(12x^6+3\)

Câu 2:\(-5x^6+3\)

cái này là +- đa thức gì đó mk học mà quên rồi

5 tháng 12 2018

1/ \(\frac{x-3}{3xy}\)+\(\frac{5x+3}{3xy}\)\(\frac{6x}{3xy}\)=\(\frac{3}{y}\)

2/\(\frac{5x-7}{2x-3}\)+\(\frac{4-3x}{2x-3}\)=\(\frac{2x-3}{2x-3}\)=1

3/\(\frac{11x-7}{3-5x}\)-\(\frac{6x+4}{5x-3}\)=\(\frac{11x-7}{3-5x}\)+\(\frac{6x+4}{3-5x}\)=\(\frac{17x-3}{3-5x}\)

4/\(\frac{3}{2x+6}\)-\(\frac{x-6}{2x^2+6x}\)=\(\frac{3x}{x\left(2x+6\right)}\)-\(\frac{x-6}{x\left(2x+6\right)}\)=\(\frac{2x-6}{x\left(2x+6\right)}\)

5/\(\frac{1}{2x-10}\)+\(\frac{2x}{3x^2-15x}\)=\(\frac{1}{2\left(x-5\right)}\)+\(\frac{2x}{3x\left(x-5\right)}\)=\(\frac{3x}{6x \left(x-5\right)}\)+\(\frac{4x}{6x\left(x-5\right)}\)

=\(\frac{7x}{6x\left(x-5\right)}\)=\(\frac{7}{6\left(x-5\right)}\)

27 tháng 8 2017

a, x2 -6x+8 = (x-3)2-1=(x-3-1).(x-3+1)=(x-4).(x-2)

b, 4x2-7x+3 = (2x-7/4)2-1/16=4.(x-2).(x-3/4)

c, (3x-1)2-(2x-3)2=[ 3x-1-(2x-3)].[3x-1+2x-3]=(x+2).(5x-4)

d, 5x-3x-2=2x-2=2.(x-1)

6 tháng 7 2017

Câu 1:

\(2x^3+5x^2+5x+3=2x^3+3x^2+2x^2+3x+2x+3\)

\(=x^2.\left(2x+3\right)+x.\left(2x+3\right)+\left(2x+3\right)\)

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

Chúc bạn học tốt!!!

20 tháng 10 2018

help me!!!

26 tháng 10 2022

b: \(=x^4+x^2+36-2x^3+12x^2-12x+x^2-6x+9\)

\(=x^4-2x^3+14x^2-18x+45\)

\(=x^4+9x^2-2x^3-18x+5x^2+45\)

\(=\left(x^2+9\right)\left(x^2-2x+5\right)\)

d: \(=2x^4+2x^3+6x^2-x^3-x^2-3x+x^2+x+3\)

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

e: \(=3x^4-3x^3-3x^2-2x^3+2x^2+2x+2x^2-2x-2\)

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

27 tháng 8 2017

\(a,\)\(x^4-4x^3+4x^2=0\)

\(\Leftrightarrow x^2.\left(x^2-4x+4\right)=0\)

\(\Leftrightarrow x^2.\left(x^2-2.x.2+2^2\right)=0\)

\(\Leftrightarrow x^2.\left(x-2\right)^2=0\)

\(\Leftrightarrow\orbr{\begin{cases}x^2=0\\\left(x-2\right)^2=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=0\\x-2=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=2\end{cases}}\)

\(b,\)\(x^2+5x+4=0\)

\(\Leftrightarrow x^2+x+4x+4=0\)

\(\Leftrightarrow x.\left(x+1\right)+4.\left(x+1\right)=0\)

\(\Leftrightarrow\left(x+1\right).\left(x+4\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+1=0\\x+4=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=-1\\x=-4\end{cases}}\)

\(c,\)\(9x-6x^2-3=0\)

\(\Leftrightarrow-3.\left(2x^2-3x+1\right)=0\)

\(\Leftrightarrow2x^2-3x+1=0\)

\(\Leftrightarrow2x^2-2x-x+1=0\)

\(\Leftrightarrow2x.\left(x-1\right)-\left(x-1\right)\)

\(\Leftrightarrow\left(x-1\right).\left(2x-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\2x-1=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=1\\2x=1\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=1\\x=\frac{1}{2}\end{cases}}\)

\(d,\)\(2x^2+5x+2=0\)

\(\Leftrightarrow2x^2+4x+x+2=0\)

\(\Leftrightarrow2x.\left(x+2\right)+\left(x+2\right)=0\)

\(\Leftrightarrow\left(x+2\right).\left(2x+1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+2=0\\2x+1=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=-2\\2x=-1\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=-2\\x=-\frac{1}{2}\end{cases}}\)

Câu 1: 

\(\dfrac{x^2-10x+21}{x^3-7x^2+x-7}=\dfrac{\left(x-7\right)\left(x-3\right)}{\left(x-7\right)\left(x^2+1\right)}=\dfrac{x-3}{x^2+1}\)

\(\dfrac{2x^2-x-15}{2x^3+5x^2+2x+5}=\dfrac{2x^2-6x+5x-15}{\left(2x+5\right)\left(x^2+1\right)}=\dfrac{\left(2x+5\right)\left(x-3\right)}{\left(2x+5\right)\left(x^2+1\right)}=\dfrac{x-3}{x^2+1}\)

Do đó: \(\dfrac{x^2-10x+21}{x^3-7x^2+x-7}=\dfrac{2x^2-x-15}{2x^3+5x^2+2x+5}\)

19 tháng 6 2018

Bài 2:

\(A=\dfrac{5x^3+5x}{x^4-1}=\dfrac{5x\left(x^2+1\right)}{\left(x^2-1\right)\left(x^2+1\right)}\)

.....= \(\dfrac{5x}{x^2-1}\)

\(B=\dfrac{x^2+5x+6}{x^2+6x+9}=\dfrac{x^2+2x+3x+6}{\left(x+3\right)^2}\)

.....= \(\dfrac{x\left(x+2\right)+3\left(x+2\right)}{\left(x+3\right)^2}=\dfrac{\left(x+2\right)\left(x+3\right)}{\left(x+3\right)^2}\)

.....= \(\dfrac{x+2}{x+3}\)

19 tháng 6 2018

Câu 1:

B = \(\dfrac{32x-8x^2+2x^3}{x^3+64}\)

....= \(\dfrac{2x\left(x^2-4x+16\right)}{\left(x+4\right)\left(x^2-4x+16\right)}=\dfrac{2x}{x+4}\)

4: \(3x^3-5x^2+5x-2\)

\(=3x^3-2x^2-3x^2+2x+3x-2\)

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

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

5: \(5x^3-12x^2+14x-4\)

\(=5x^3-2x^2-10x^2+4x+10x-4\)

\(=\left(5x-2\right)\left(x^2-2x+2\right)\)