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23 tháng 8 2016
a) \(\frac{x^2-y^2}{\left(x+y\right)\left(ay-\text{ax}\right)}=\frac{\left(x+y\right)\left(x-y\right)}{-a\left(x+y\right)\left(x-y\right)}=\frac{-1}{a}\)
b) \(\frac{2ax-2x-3y+3ay}{4ax+\text{4x}+6y+6ay}=\frac{2x\left(a-1\right)+3y\left(a-1\right)}{\text{4x}\left(a+1\right)+6y\left(a+1\right)}\)
\(=\frac{\left(a-1\right)\left(2x+3y\right)}{2\left(a+1\right)\left(2x+3y\right)}=\frac{a-1}{2\left(a+1\right)}\)
PN
1 tháng 8 2020
\(F=-3\left(x-8\right)\left(2x+1\right)-\left(x+5\right)\left(2-3x\right)-4x\left(x-6\right)\)
\(=-3\left(-3-8\right)\left(-6+1\right)-\left(5-3\right)\left(2+9\right)+12\left(-9\right)\)
\(=-3\left(-11\right)\left(-5\right)-\left(-2\right)11-12.9\)
\(=-165+22-108=22-273=-251\)
\(G=\left(5x-4\right)\left(5-2x\right)-7x\left(x^2-4x+3\right)+\left(x^2-4x\right)\left(7x-2\right)\)
\(=\left(5-4\right)\left(5-2\right)-7\left(1-4+3\right)+\left(1-4\right)\left(7-2\right)\)
\(=3-7.0+5.\left(-3\right)=3-15=-12\)
\(H=\left(-3x+5\right)\left(x-6\right)-\left(x-1\right)\left(x^2-2x+3\right)+\left(x+2\right)\left(x^2-3\right)\)
\(=\left(3+5\right)\left(-1-6\right)-\left(-1-1\right)\left(1+2+3\right)+\left(-1+2\right)\left(1-3\right)\)
\(=8\left(-7\right)-\left(-2\right)6+1\left(-2\right)=-56+12-2=-46\)
PN
1 tháng 8 2020
\(L=5x\left(x-1\right)\left(2x+3\right)-10x\left(x^2-4x+5\right)-\left(x-1\right)\left(x-4\right)\)
\(=-\frac{5}{3}\left(-\frac{4}{3}\right)\left(-\frac{2}{3}+3\right)+\frac{10}{3}\left(\frac{1}{9}+\frac{4}{3}+5\right)-\left(-\frac{4}{3}\right)\left(-\frac{1}{3}-4\right)\)
\(=\frac{20}{9}\left(\frac{7}{3}\right)+\frac{10}{3}\left(\frac{13}{9}+5\right)+\frac{4}{3}\left(-\frac{13}{3}\right)\)
\(=\frac{140}{27}+\frac{10}{3}.\frac{58}{9}-\frac{52}{9}\)
\(=\frac{140}{27}+\frac{580}{27}-\frac{156}{27}=\frac{140+580-156}{27}=\frac{720-156}{27}=\frac{564}{27}\)
\(M=-7x\left(x-5\right)-\left(x-1\right)\left(x^2-x-2\right)+x^2\left(x-3\right)-5x\left(x-8\right)\)
\(=\frac{-7}{2}\left(\frac{1}{2}-5\right)+\frac{\left(\frac{1}{4}-\frac{1}{2}-2\right)}{2}+\frac{1}{4}\left(\frac{1}{2}-3\right)-\frac{5}{2}\left(\frac{1}{2}-8\right)\)
\(=\frac{7}{2}.\frac{9}{2}-\frac{9}{8}-\frac{1}{4}.\frac{5}{2}+\frac{5}{2}.\frac{15}{2}\)
\(=\frac{63}{4}-\frac{9}{8}-\frac{5}{8}+\frac{75}{4}=\frac{138}{4}-\frac{7}{4}=\frac{131}{4}\)
25 tháng 8 2016
a) 5x2 ( 3x2 -7x+2)-15x(x-3)
=15x4-35x3+10x2-15x2+45x
=15x4-35x3-5x2+45x
c) (x+3)(x-3)(x-2)(x+1)
=(x2-9)(x2+x-2x-2)
=(x2-9)(x2-x-2)
=x4-x3-2x2-9x2+9x+18
=x4-x3-11x2+9x+18
d)(2x+1)2+(4x-1)2+2(2x+1)(4x+1)
=2x2+4x+1-16x2-8x+1
=2x2+4x+1-16x2-8x+1+16x2-4x+8x-2
=2x2+7
e) (2x2-3x)(5x2-2x+1)-10x2(x+3)
=10x4 -4x3+2x2-15x3+6x2-3 -10x2-30x
=10x4-19x3-2x2-30x-3
C
2
5 tháng 1 2021
Đề là biểu thức hay phân thức ( nếu là biểu thức thi :)
a, \(x^2-10x+25=\left(x-5\right)^2\)
\(x^2-3x-10=x^2+2x-5x-10=\left(x-5\right)\left(x+2\right)\)
\(4x+8=4\left(x+2\right)\)
5 tháng 1 2021
Nếu là phân thức thì =) p/s : viết đề hẳn hoi đi :v
a, \(\frac{x^2-10x+25}{x^2-3x-10}=\frac{\left(x-5\right)^2}{\left(x-5\right)\left(x+2\right)}=\frac{x-5}{x+2}\)
b, chả hiểu
LM
3
TL
7 tháng 12 2019
a) \(\frac{x^2+2x+4}{4x^3-32}=\frac{x^2+2x+4}{4\left(x^3-8\right)}=\frac{x^2+2x+4}{4\left(x-2\right)\left(x^2+2x+4\right)}=\frac{1}{4\left(x-2\right)}.\)
b) \(\frac{10x-15}{4x^2-9}=\frac{5\left(2x-3\right)}{\left(2x\right)^2-3^2}=\frac{5\left(2x-3\right)}{\left(2x-3\right)\left(2x+3\right)}=\frac{5}{2x+3}.\)
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
HAND!!!!
H
7 tháng 12 2019
\(\frac{x^2+2x+4}{4x^3-32}=\frac{\left(x+2\right)^2}{4\left(x^3-8\right)}=\frac{\left(x+2\right)^2}{4\left(x-2\right)\left(x^2+2x+4\right)}=\frac{x+2}{4\left(x^2+2x+4\right)}.\)
\(\frac{10x-15}{4x^2-9}=\frac{5\left(2x-3\right)}{\left(2x\right)^2-3^2}=\frac{5\left(2x-3\right)}{\left(2x-3\right)\left(2x+3\right)}=\frac{5}{2x+3}\)
23 tháng 7 2017
a) Ta có
Biến đổi tử phân số A
x^3-x^2-10x-8=(x^3-4x^2)+(3x^2-12x)+(2x-8)
=x^2(x-4)+3x(x-4)+2(x-4)=(x^2+3x+2)(x-4)
=(x^2+x+2x+2)(x-4)=[x(x+1)+2(x+1)](x-4)
=(x+1)(x+2)(x+4) (1)
Biến đổi mẫu của phân số A:
x^3-4x^2+5x-20=x^2(x-4)+5(x-4)=(x^2+5)(x-4) (2)
Từ (1) và (2) suy ra:
A=(x+1)(x+2)/x^2+5
23 tháng 1 2018
\(A=\dfrac{x^3-x^2-10x-8}{x^3-4x^2+5x-20}\\ ĐKXĐ:x\ne4\)
a) Với \(x\ne4\)
\(\text{Ta có : }A=\dfrac{x^3-x^2-10x-8}{x^3-4x^2+5x-20}\\ =\dfrac{x^3+x^2-2x^2-2x-8x-8}{\left(x^3-4x^2\right)+\left(5x-20\right)}\\ =\dfrac{\left(x^3+x^2\right)-\left(2x^2+2x\right)-\left(8x+8\right)}{x^2\left(x-4\right)+5\left(x-4\right)}\\ =\dfrac{x^2\left(x+1\right)-2x\left(x+1\right)-8\left(x+1\right)}{\left(x^2+5\right)\left(x-4\right)}\\ =\dfrac{\left(x^2-2x-8\right)\left(x+1\right)}{\left(x^2+5\right)\left(x-4\right)}\\ = \dfrac{\left(x^2-4x+2x-8\right)\left(x+1\right)}{\left(x^2+5\right)\left(x-4\right)}\\ =\dfrac{\left[\left(x^2-4x\right)+\left(2x-8\right)\right]\left(x+1\right)}{\left(x^2+5\right)\left(x-4\right)}\\ =\dfrac{\left[x\left(x-4\right)+2\left(x-4\right)\right]\left(x+1\right)}{\left(x^2+5\right)\left(x-4\right)}\\ =\dfrac{\left(x+2\right)\left(x-4\right)\left(x+1\right)}{\left(x^2+5\right)\left(x-4\right)}\\ =\dfrac{\left(x+2\right)\left(x+1\right)}{x^2+5}\)
Vậy \(A=\dfrac{\left(x+2\right)\left(x+1\right)}{x^2+5}\) với \(x\ne4\)
b) Với \(x\ne4\)
Để \(A\ge0\) thì \(\Rightarrow\dfrac{\left(x+2\right)\left(x+1\right)}{x^2+5}\ge0\) \(\Rightarrow\left(x+2\right)\left(x+1\right)\ge0\left(\text{Vì }x^2+5>0\right)\) Lập bảng xét dấu: x+2 x+1 (x+1)(x+2) (x+1)(x+2) x -2 -1 0 0 0 0 _ + + _ _ + + _ + \(\Rightarrow\left[{}\begin{matrix}x\le-2\\x\ge-1\end{matrix}\right.\) Vậy để \(A\ge0\) thì \(x\le-2;x\ge-1\)
\(\dfrac{x^3-x^2-10x-8}{x^3-4x^2+5x-20}\\ =\dfrac{\left(x^3+x^2\right)-\left(2x^2+2x\right)-\left(8x+8\right)}{x^2\left(x-4\right)+5\left(x-4\right)}\\ =\dfrac{x^2\left(x+1\right)-2x\left(x+1\right)-8\left(x+1\right)}{\left(x^2+5\right)\left(x-4\right)}\\ =\dfrac{\left(x+1\right)\left(x^2-2x-8\right)}{\left(x^2+5\right)\left(x-4\right)}\\ =\dfrac{\left(x+1\right)\left[\left(x^2-4x\right)+\left(2x-8\right)\right]}{\left(x^2+5\right)\left(x-4\right)}\)
\(=\dfrac{\left(x+1\right)\left[x\left(x-4\right)+2\left(x-4\right)\right]}{\left(x^2+5\right)\left(x-4\right)}\\ =\dfrac{\left(x+1\right)\left(x-4\right)\left(x+2\right)}{\left(x^2+5\right)\left(x-4\right)}\\ =\dfrac{\left(x+1\right)\left(x+2\right)}{x^2+5}\left(x\ne4\right)\)
cám oeen:)))