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23 tháng 12 2023

Câu 5: B

Câu 6: 

a: ĐKXĐ: \(x-2\ne0\)

=>\(x\ne2\)

b: ĐKXĐ: \(x+1\ne0\)

=>\(x\ne-1\)

8:

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

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

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

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

7: 

\(\dfrac{8x^3yz}{24xy^2}\)

\(=\dfrac{8xy\cdot x^2z}{8xy\cdot3y}\)

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

2 tháng 10 2015

 

\(A=\left[\left(a+b\right)+\left(c+d\right)\right]^2+\left[\left(a+b\right)-\left(c+d\right)\right]^2+\left[\left(a-b\right)+\left(c-d\right)\right]^2+\left[\left(a-b\right)-\left(c-d\right)\right]^2\)

Ta có

\(\left[\left(a+b\right)+\left(c+d\right)\right]^2=\left(a+b\right)^2+2\left(a+b\right)\left(c+d\right)+\left(c+d\right)^2\)

\(\left[\left(a+b\right)-\left(c+d\right)\right]^2=\left(a+b\right)^2-2\left(a+b\right)\left(c+d\right)+\left(c+d\right)^2\)

\(\left[\left(a-b\right)+\left(c-d\right)\right]^2=\left(a-b\right)^2+2\left(a-b\right)\left(c-d\right)+\left(c-d\right)^2\)

\(\left[\left(a-b\right)-\left(c-d\right)\right]^2=\left(a-b\right)^2-2\left(a-b\right)\left(c-d\right)+\left(c-d\right)^2\)

\(A=2\left(a+b\right)^2+2\left(a-b\right)^2+2\left(c+d\right)^2+2\left(c-d\right)^2\)

\(A=2\left(a^2+2ab+b^2+a^2-2ab+b^2+c^2+2cd+d^2+c^2-2cd+d^2\right)\)

\(A=4\left(a^2+b^2+c^2+d^2\right)\)

 

25 tháng 11 2016

a ) \(\frac{\left(a+b\right)^2-c^2}{a+b+c}=\frac{\left(a+b+c\right)\left(a+b-c\right)}{a+b+c}=a+b-c\)

b ) \(\frac{a^2+b^2-c^2+2ab}{a^2-b^2+c^2+2ac}=\frac{a^2+2ab+b^2-c^2}{a^2+2ac+c^2-b^2}\)

\(=\frac{\left(a+b\right)^2-c^2}{\left(a+c\right)^2-b^2}=\frac{\left(a+b+c\right)\left(a+b-c\right)}{\left(a+c+b\right)\left(a+c-b\right)}=\frac{a+b-c}{a-b+c}\)

25 tháng 11 2016

a) \(\frac{\left(a+b\right)^2-c^2}{a+b+c}=\frac{\left(a+b+c\right)\left(a+b-c\right)}{a+b+c}=a+b-c\)

b) \(\frac{a^2+b^2-c^2+2ab}{a^2-b^2+c^2+2ac}=\frac{\left(a^2+2ab+b^2\right)-c^2}{\left(a^2+2ac+c^2\right)-b^2}=\frac{\left(a+b\right)^2-c^2}{\left(a+c\right)^2-b^2}=\frac{\left(a+b+c\right)\left(a+b-c\right)}{\left(a+c+b\right)\left(a+c-b\right)}=\frac{a+b-c}{a+c-b}\)

2 tháng 12 2018

\(\frac{\left(a-b\right)\left(c-d\right)}{\left(b^2-a^2\right)\left(d^2-c^2\right)}=\frac{\left(b-a\right)\left(d-c\right)}{\left(b-a\right)\left(b+a\right)\left(d-c\right)\left(d+c\right)}=\frac{1}{\left(a+b\right)\left(c+d\right)}\)

\(\frac{m^4-m}{2m^2+2m+2}=\frac{m\left(m^3-1\right)}{2m^2+2m+2}=\frac{m\left(m-1\right)\left(m^2+m+1\right)}{2\left(m^2+m+1\right)}=\frac{m\left(m-1\right)}{2}\)

24 tháng 10 2016

⇔2(a+b)2+2(c+d)2+2(a−b)2+2(d−c)2=2(2a2+2b2+2d2+2c2=4(∑a2) 

22 tháng 7 2023

a) \(\dfrac{3x^2y}{2xy^5}=\dfrac{3x}{2y^4}\)

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

c) \(\dfrac{ab^2-a^2b}{2a^2+a}=\dfrac{ab\left(b-a\right)}{a\left(2a+1\right)}=\dfrac{b\left(b-a\right)}{2a+1}=\dfrac{b^2-ab}{2a+1}\)

d) \(\dfrac{12\left(x^4-1\right)}{18\left(x^2-1\right)}=\dfrac{2\left(x^2-1\right)\left(x^2+1\right)}{3\left(x^2-1\right)}=\dfrac{2\left(x^2+1\right)}{3}\)

`a, (3x^2y)/(2xy^5)`

`= (3x)/(2y^4)`

`b, (3x^2-3x)/(x-1)`

`= (3x(x-1))/(x-1)`

`= 3x`

`c, (ab^2-a^2b)/(2a^2+a)`

`= (b(a-b))/((2a+1))`

`d, (12(x^4-1))/(18(x^2-1)) = (2(x^2+1))/3`.

14 tháng 7 2017

a) \(\cdot\left(m+n\right)^2-\left(m-n\right)^2+\left(m+n\right)\left(m-n\right)\)

\(=\left(m+n+m-n\right)\left(m+n-m+n\right)+\left(m+n\right)\left(m-n\right)\)

\(=\left(2m\cdot2n\right)+m^2-n^2\)

\(=4mn+m^2-n^2\)

b) \(\left(a+b\right)^2-\left(a-b\right)^2-2a^3\)

\(=\left(a+b+a-b\right)\left(a+b-a+b\right)-2a^3\)

\(=2ab-2a^3\)

c) \(\left(2x+1\right)^2+\left(2x-1\right)^2+2\left(4x^2-1\right)\)

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

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

\(=\left(4x\right)^2=16x^2\)

d) \(\left(a+b+c\right)^2-2\left(a+b+c\right)\left(b+c\right)+\left(b+c\right)^2\)

\(=\left(a+b+c-b-c\right)^2=a^2\)

14 tháng 7 2017

xin lỗi mk ghi sai đề ở bài :d) (a+b+c)^2-2(a+b+c)(b+c)+(b+c)^2