\(\dfrac{1}{8}\)

b)x2-6x+5

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Khách

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6 tháng 12 2019

a) \(x^3-\frac{1}{8}\)

=\(\left(x-\frac{1}{2}\right)\left(x^2+\frac{1}{2}x+\frac{1}{4}\right)\)

b)\(x^2-6x+5\)

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

=\(x\left(x-1\right)-5\left(x-1\right)\)

=\(\left(x-1\right)\left(x-5\right)\)

8 tháng 12 2017

Bài 1:

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

\(=36x^2+72x+1+36x^2-72x+1-2\left(36x^2-1\right)\)

\(=36x^2+72x+1+36x^2-72x+1-72x^2+2\)

\(=4\)

b) \(3\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

\(=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

\(=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

\(=\left(2^{16}-1\right)\left(2^{16}+1\right)\)

\(=2^{32}-1\)

c) \(x\left(2x^3-3\right)-x^2\left(5x+1\right)+x^2\)

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

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

d) \(3x\left(x-2\right)-5x\left(1-x\right)-8\left(x^2-3\right)\)

\(=3x^2-6x-5x+5x^2-8x^2+24\)

\(=-11x+24\)

8 tháng 12 2017

Bài 2:

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

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

\(=x^3+y^3\)

\(=\left(\dfrac{2}{3}\right)^3+\left(\dfrac{1}{3}\right)^3\)

\(=\dfrac{1}{3}\).

21 tháng 8 2017

\(e,\)

\(\left(\dfrac{1}{3}a^3b+\dfrac{1}{3}a^2b^2-\dfrac{1}{4}ab^3\right):5ab\)

\(=\dfrac{1}{15}a^2+\dfrac{1}{15}ab-\dfrac{1}{20}b^2\)

\(f,\)

\(\left(-\dfrac{2}{3}x^5y^2+\dfrac{3}{4}x^4y^3-\dfrac{4}{5}x^3y^4\right):6x^2y^2\)

\(=-\dfrac{1}{9}x^3+\dfrac{1}{8}x^2y-\dfrac{2}{15}xy^2\)

\(g,\)

\(\left(\dfrac{3}{4}a^6b^3+\dfrac{6}{5}a^3b^4-\dfrac{5}{10}ab^5\right):\left(\dfrac{3}{5}ab^3\right)\)

\(=\dfrac{5}{4}a^5+2a^2b-\dfrac{5}{6}b^2\)

21 tháng 8 2017

cam on

29 tháng 1 2019

a) \(\left(6x^3y^2-4x^2y^3-10x^2y^2\right):2xy\)

=\(\left(6x^3y^2:2xy\right)-\left(4x^2y^3:2xy\right)-\left(10x^2y^2:2xy\right)\)

\(=3x^2y-2xy^2-5xy\)

b) \(\dfrac{2y}{x-2}+\dfrac{5y}{x-2}\)

=\(\dfrac{2y+5y}{x-2}\)

=\(\dfrac{7y}{x-2}\)

c)\(\dfrac{xy}{3x-y}+\dfrac{3x^2}{y-3x}\)

\(=\dfrac{xy}{3x-y}-\dfrac{3x^2}{3x-y}\)

=\(\dfrac{x\left(y-3x\right)}{3x-y}\)

=\(\dfrac{-x\left(3x-y\right)}{3x-y}\)

=-x

d)\(\dfrac{x-1}{6x+12}.\dfrac{x+2}{x-1}\)

=\(\dfrac{\left(x-1\right)\left(x+2\right)}{6\left(x+2\right)\left(x-1\right)}\)

=\(\dfrac{1}{6}\)

26 tháng 10 2017

a) (x + 5)2 - (x - 3)2 = 2x - 7

(x + 5 - x + 3)(x + 5 + x - 3) = 2x - 7

8(2x + 2)= 2x - 7

16x + 16 = 2x - 7

16x - 2x = - 7 - 16

14x = - 23

x = - 23/14

b) (2x - 3)(4x2 + 6x + 9) = 98

(2x)3 - 33 = 98

8x3 - 27 = 98

8x3 = 125

x3 = 125/8

x3 = (5/2)3

x = 5/2

25 tháng 11 2022

Bài 2:

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

\(=3x^2+x-3+\dfrac{5x-5}{x^2+1}\)

Bài 3:

\(\dfrac{A}{B}=\dfrac{2x^3-x^2-x+1}{x^2-2x}\)

\(=\dfrac{2x^3-4x^2+3x^2-6x+5x+1}{x^2-2x}\)

\(=2x^2+3+\dfrac{5x+1}{x^2-2x}\)

=>\(2x^3-x^2-x+1=\left(x^2-2x\right)\left(2x^2+3\right)+5x+1\)

25 tháng 7 2019

\(9x^2-6x+2=9x^2-6x+1+1=\left(3x-1\right)^2+1>0\Rightarrowđpcm\)

\(x^2+x+1=x^2+x+\frac{1}{4}+\frac{3}{4}=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}>0\left(đpcm\right)\)

\(25x^2-20x+7=25x^2-20x+4+3=\left(5x-2\right)^2+3>0\left(đpcm\right)\)

\(9x^2-6xy+2y^2+1=\left(9x^2+6xy+y^2\right)+y^2+1=\left(3x+y\right)^2+y^2+1>0\left(đpcm\right)\)

\(\Leftrightarrow x^2+y^2\ge xy;x^2+y^2\ge2\sqrt{x^2y^2}=2\left|xy\right|\ge\left|xy\right|\ge xy\Rightarrowđpcm\)

25 tháng 7 2019

Cách khác câu e:

\(x^2-xy+y^2=x^2-2x.\frac{y}{2}+\frac{y^2}{4}+\frac{3y^2}{4}=\left(x+\frac{y}{2}\right)^2+\frac{3y^2}{4}\ge0\forall xy\) (đpcm)

12 tháng 3 2018

bài 1:

b,\(\dfrac{x+2}{x}=\dfrac{x^2+5x+4}{x^2+2x}+\dfrac{x}{x+2}\)(ĐKXĐ:x ≠0,x≠-2)

<=>\(\dfrac{\left(x+2\right)^2}{x\left(x+2\right)}=\dfrac{x^2+5x+4}{x\left(x+2\right)}+\dfrac{x^2}{x\left(x+2\right)}\)

=>\(x^2+4x+4=x^2+5x+4+x^2\)

<=>\(x^2-x^2-x^2+4x-5x+4-4=0\)

<=>\(-x^2-x=0< =>-x\left(x+1\right)=0< =>\left[{}\begin{matrix}x=0\left(loại\right)\\x+1=0< =>x=-1\left(nhận\right)\end{matrix}\right.\)

vậy...............

d,\(\left(x+3\right)^2-25=0< =>\left(x+3-5\right)\left(x+3+5\right)=0< =>\left(x-2\right)\left(x+8\right)=0< =>\left[{}\begin{matrix}x-2=0\\x+8=0\end{matrix}\right.< =>\left[{}\begin{matrix}x=2\\x=-8\end{matrix}\right.\)

vậy............

bài 3:

g,\(\dfrac{4}{x+1}-\dfrac{2}{x-2}=\dfrac{x+3}{x^2-x-2}\)(ĐKXĐ:x khác -1,x khác 2)

<=>\(\dfrac{4}{x+1}-\dfrac{2}{x-2}=\dfrac{x+3}{x^2-2x+x-2}\)

<=>\(\dfrac{4}{x+1}-\dfrac{2}{x-2}=\dfrac{x+3}{x\left(x-2\right)+\left(x-2\right)}\)

<=>\(\dfrac{4}{x+1}-\dfrac{2}{x-2}=\dfrac{x+3}{\left(x+1\right)\left(x-2\right)}\)

<=>\(\dfrac{4\left(x-2\right)}{\left(x+1\right)\left(x-2\right)}-\dfrac{2\left(x+1\right)}{\left(x+1\right)\left(x-2\right)}=\dfrac{x+3}{\left(x+1\right)\left(x-2\right)}\)

=>\(4x-8-2x-2=x+3\)

<=>\(x=13\)

vậy..............

mấy ý khác bạn làm tương tụ nhé

chúc bạn học tốt ^ ^

14 tháng 2 2018

a.

\(2\left(x+5\right)-x^2-5x=0\)

\(\Leftrightarrow2x+10-x^2-5x=0\)

\(\Leftrightarrow-x^2-3x+10=0\)

\(\Leftrightarrow x^2+3x-10=0\)

\(\Leftrightarrow x^2+5x-2x-10=0\)

\(\Leftrightarrow\left(x^2+5x\right)-\left(2x+10\right)=0\)

\(\Leftrightarrow x\left(x+5\right)-2\left(x+5\right)=0\)

\(\Leftrightarrow\left(x+5\right)\left(x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=2\end{matrix}\right.\)

b.

\(2x^2+3x-5=0\)

\(\Leftrightarrow2x^2-2x+5x-5=0\)

\(\Leftrightarrow\left(2x^2-2x\right)+\left(5x-5\right)=0\)

\(\Leftrightarrow2x\left(x-1\right)+5\left(x-1\right)=0\)

\(\Leftrightarrow\left(2x+5\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+5=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-5}{2}\\x=1\end{matrix}\right.\)

14 tháng 2 2018

bài 2:

ĐKXĐ: x khác -1

\(\dfrac{1-x}{x+1}+3=\dfrac{2x+3}{x+1}\)

\(\Leftrightarrow\dfrac{1-x+3\left(x+1\right)}{x+1}=\dfrac{2x+3}{x+1}\)

\(\Leftrightarrow1-x+3x+3=2x+3\)

\(\Leftrightarrow0x=-1\)

\(\Leftrightarrow x\in\varnothing\)

Suy ra pt vô nghiệm

b.

ĐKXĐ: x khác \(\dfrac{3}{2}\)

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

\(\Leftrightarrow\dfrac{x^2+4x+4}{2x-3}-\dfrac{2x-3}{2x-3}=\dfrac{x^2+10}{2x-3}\)

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

\(\Leftrightarrow2x-3=0\)

\(\Leftrightarrow x=\dfrac{3}{2}\) ( loại)

25 tháng 3 2018

a) ĐKXĐ: x khác 0

\(x+\dfrac{5}{x}>0\)

\(\Leftrightarrow x^2+5>0\) ( luôn đúng)

Vậy bất pt vô số nghiệm ( loại x = 0)

d)

\(\dfrac{x+1}{12}-\dfrac{x-1}{6}>\dfrac{x-2}{8}-\dfrac{x+3}{8}\)

\(\Leftrightarrow\dfrac{x+1}{12}-\dfrac{x-1}{6}>\dfrac{x-2-x-3}{8}\)

\(\Leftrightarrow\dfrac{x+1}{12}-\dfrac{x-1}{6}>\dfrac{-5}{8}\)

\(\Leftrightarrow2x+2-4x+4>-15\)

\(\Leftrightarrow-2x>-21\)

\(\Leftrightarrow x< \dfrac{21}{2}\)

Vậy....................

25 tháng 3 2018

a)\(x+\dfrac{5}{x}>0\left(ĐKXĐ:x\ne0\right)\)

\(\Leftrightarrow\dfrac{x^2+5}{x}>0\)

\(x^2+5>0\)

\(\Rightarrow x>0\)

d)\(\dfrac{x+1}{12}-\dfrac{x-1}{6}>\dfrac{x-2}{8}-\dfrac{x+3}{8}\)

\(\Leftrightarrow\dfrac{x+1}{12}-\dfrac{2x-2}{12}>\dfrac{-5}{8}\)

\(\Leftrightarrow\dfrac{-x+3}{12}>\dfrac{-5}{8}\)

\(\Leftrightarrow-x+3>-\dfrac{15}{2}\)

\(\Leftrightarrow-x>-\dfrac{21}{2}\)

\(\Leftrightarrow x< \dfrac{21}{2}\)