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đề là gì bạn có phải như mình làm ko
\(x^2+6x+9=\left(x+3\right)^2\)
\(x^2+8x+16=\left(x+4\right)^2\)
\(x^2+10x+25=\left(x+5\right)^2\)
\(x^2-12+36=\left(x-6\right)^2\)
\(x^2-14x+49=\left(x-7\right)^2\)
a) \(\frac{9x^2}{11y^2}:\frac{6x}{11y}=\frac{9x^2}{11y^2}\cdot\frac{11y}{6x}=\frac{3xy}{2}\)
b) \(\frac{x^2-49}{x-7}+x-2=\frac{\left(x-7\right)\left(x+7\right)}{x-7}+x-2=x+7+x-2=2x+5\)
c) \(\frac{3}{x+3}+\frac{1}{x-3}-\frac{18}{9-x^2}\)
= \(\frac{3\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}+\frac{1\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\frac{18}{\left(3-x\right)\left(x+3\right)}\)
= \(\frac{3x-9}{\left(x-3\right)\left(x+3\right)}+\frac{x+3}{\left(x-3\right)\left(x+3\right)}+\frac{18}{\left(x-3\right)\left(x+3\right)}\)
= \(\frac{3x-9+x+3+18}{\left(x-3\right)\left(x+3\right)}\)
= \(\frac{4x+12}{\left(x-3\right)\left(x+3\right)}\)
= \(\frac{4\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=\frac{4}{x-3}\)(đk: \(x-3\ne0\)=> \(x\ne3\))
a ) \(\frac{4}{x+2}+\frac{2}{x-2}+\frac{5x-6}{4-x^2}=\frac{4\left(x-2\right)+2\left(x+2\right)}{\left(x+2\right)\left(x-2\right)}+\frac{6-5x}{\left(x+2\right)\left(x-2\right)}=\frac{6x-4+6-5x}{\left(x+2\right)\left(x-2\right)}\)
\(=\frac{x+2}{\left(x+2\right)\left(x-2\right)}=\frac{1}{x+2}\)
b ) \(\frac{1-3x}{2x}+\frac{3x-2}{2x-1}+\frac{3x-2}{2x-4x^2}=\frac{\left(1-3x\right)\left(2x-1\right)+2x\left(3x-2\right)+2-3x}{2x\left(2x-1\right)}\)
\(=\frac{-6x^2+5x-1+6x^2-4x+2-3x}{2x\left(2x-1\right)}=\frac{-2x+1}{2x\left(2x-1\right)}=\frac{-1}{2x}\)
c ) \(\frac{1}{x^2+6x+9}+\frac{1}{6x-x^2-9}+\frac{x}{x^2-9}=\frac{1}{\left(x+3\right)^2}+\frac{1}{-\left(x-3\right)^2}+\frac{x}{\left(x+3\right)\left(x-3\right)}\)
\(=\frac{\left(x-3\right)^2-\left(x+3\right)^2+x\left(x+3\right)\left(x-3\right)}{\left(x+3\right)^2\left(x-3\right)^2}=\frac{-12x+x^3-9x}{\left(x+3\right)^2\left(x-3\right)^2}=\frac{x^3-21x}{x^4-18x^2+81}\)
d ) \(\frac{x^2+2}{x^3-1}+\frac{2}{x^2+x+1}+\frac{1}{1-x}=\frac{x^2+2+2\left(x-1\right)-\left(x^2+x+1\right)}{x^3-1}=\frac{x-1}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\frac{1}{x^2+x+1}\)
e ) \(\frac{x}{x-2y}+\frac{x}{x+2y}+\frac{4xy}{4y^2-x^2}=\frac{x\left(x+2y\right)+x\left(x-2y\right)-4xy}{\left(x-2y\right)\left(x+2y\right)}=\frac{2x\left(x-2y\right)}{\left(x-2y\right)\left(x+2y\right)}\)
\(=\frac{2x}{x+2y}\)
\(a,\frac{7}{x}-\frac{x}{x+6}+\frac{36}{x^2-6x}\)
\(=\frac{7}{x}-\frac{x}{x+6}+\frac{36}{x\left(x-6\right)}\)
\(=\frac{7\left(x-6\right)\left(x+6\right)-x\left(x-6\right)+36\left(x+6\right)}{x\left(x-6\right)\left(x+6\right)}\)
\(=\frac{7\left(x^2-6\right)-x^2+6x+36x+216}{x\left(x^2-6\right)}\)
\(=\frac{7x^2-42-x^2+6x+36x+216}{x\left(x^2-6\right)}\)
\(=\frac{6x^2+42x+216}{x\left(x^2-6\right)}\)
\(=\frac{6\left(x^2+7x+36\right)}{x\left(x^2-6\right)}\)
Đề sai nhé, phải là như này nè :
\(b,\frac{1}{x^2-x+1}-\frac{1}{x^2+x+1}-\frac{2x}{x^4-x^2+1}+\frac{4x^3}{x^8-x^4+1}\)
\(=\frac{x^2+x+1-\left(x^2-x+1\right)}{\left(x^2-x+1\right)\left(x^2+x+1\right)}\)\(-\frac{2x}{x^4-x^2+1}+\frac{4x^3}{x^8-x^4+1}\)
\(=\frac{x^2+x+1-x^2+x-1}{x^4+x^2+1}\)\(-\frac{2x}{x^4-x^2+1}+\frac{4x^3}{x^8-x^4+1}\)
\(=\frac{2x}{x^4+x^2+1}-\frac{2x}{x^4-x^2+1}+\frac{4x^3}{x^8-x^4+1}\)
\(=\frac{2x\left(x^4-x^2+1\right)-2x\left(x^4+x^2+1\right)}{\left(x^4+x^2+1\right)\left(x^4-x^2+1\right)}+\frac{4x^3}{x^8-x^4+1}\)
\(=\frac{2x^5-2x^3+2x-2x^5-2x^3-2x}{x^8-x^4+1}+\frac{4x^3}{x^8-x^4+1}\)
\(=-\frac{4x^3}{x^8-x^4+1}+\frac{4x^3}{x^8-x^4+1}=0\)
a) Áp dụng hằng đẳng thức số 3 bạn nhé
b) (2x + 3)(4x^2 - 6x +9) = 8x^3 + 9
Thay x= 120:2 = 60 vào biểu thức.
8* 60^3 + 9 = 1728009
c) = (2x + 1)^3
Thay x= -0,5 vào biểu thức
[2*(-0,5)+1]^3 = 0
d) = x^2 - 49 - x^2 - 2x - 1 = -50 - 2x
Thay x=49 vào biểu thức.
-50 - 2* 49 = -148
\(\frac{x^2+6x}{x-7}\div\frac{x^2-36}{x^2-14x+49}\)
\(=\frac{x\left(x+6\right)}{x-7}\times\frac{\left(x-7\right)^2}{\left(x-6\right)\left(x+6\right)}\)
\(=\frac{x\left(x-7\right)}{x-6}\)
\(\frac{x^2+6x}{x-7}\div\frac{x^2-36}{x^2-14x+49}\)
\(=\frac{x^2+6x}{x-7}.\frac{x^2-14x+49}{x^2-36}\)
\(=\frac{x\left(x+6\right)\left(x-7\right)^2}{\left(x-7\right)\left(x-6\right)\left(x+6\right)}\)
\(=\frac{x\left(x-7\right)}{x-6}\)