A=-m+n-p-m+n+p                                                                                                                         A=(-m-m)+(n+n)+(p-p)                                                                                                                   A=-2m+2n

rút gọn các biểu thức

a.x+8-x-2

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

= 0 + 6 = 6

b.a-m+7-8+m

= a - (m-m) + (7-8)

= a -0 - 1 = a-1

a.x+8-x-2

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

= 0 + 6 = 6

b.a-m+7-8+m

= a - (m-m) + (7-8)

= a -0 - 1 = a-1

a)\(\sqrt{8+4\sqrt{3}}-\sqrt{8-4\sqrt{3}}=\sqrt{\dfrac{1}{2}\left(16+8\sqrt{3}\right)}-\sqrt{\dfrac{1}{2}\left(16-8\sqrt{3}\right)}\)

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

b)\(=\dfrac{\sqrt{16+2.4\sqrt{5}+5}}{4+\sqrt{5}}.\sqrt{\left(2-\sqrt{5}\right)^2}\)\(=\dfrac{\sqrt{\left(4+\sqrt{5}\right)^2}}{4+\sqrt{5}}\left|2-\sqrt{5}\right|=\sqrt{5}-2\)

a) Ta có: \(\sqrt{8+4\sqrt{3}}-\sqrt{8-4\sqrt{3}}\)

\(=\sqrt{6}+\sqrt{2}-\sqrt{6}+\sqrt{2}\)

\(=2\sqrt{2}\)

b) Ta có: \(\dfrac{\sqrt{21+8\sqrt{5}}}{4+\sqrt{5}}\cdot\sqrt{9-4\sqrt{5}}\)

\(=\left(4+\sqrt{5}\right)\left(4-\sqrt{5}\right)\)

=16-5=11

Ta có: (2x+3)(5x-1) - 5x(2x-7) = (2x+3)(5x-1) - 5x( 2x+3-10) 

= (2x+3)(5x-1) - 5x(2x+3) + 50x

= (2x+3) (5x - 1 - 5x) + 50x

= (2x+3) .-1 +50x

Thay x =0 => (2.0+3) . -1 +50.0 = -3

\(\sqrt{13-\sqrt{160}}-\sqrt{53+4\sqrt{60}}=\sqrt{\left(2\sqrt{2}-\sqrt{5}\right)^2}-\sqrt{\left(4\sqrt{3}+\sqrt{5}\right)^2}\)

\(=2\sqrt{2}-\sqrt{5}-4\sqrt{3}-\sqrt{5}\)

\(=2\sqrt{2}-4\sqrt{3}-2\sqrt{5}\)

\(\sqrt{\left(4+\sqrt{3}\right)\sqrt{19-8\sqrt{3}}+3}=\sqrt{\left(4+\sqrt{3}\right)\sqrt{\left(4-\sqrt{3}\right)^2}+3}\)

\(=\sqrt{\left(4+\sqrt{3}\right)\left(4-\sqrt{3}\right)+3}=\sqrt{4-3+3}=2\)

a) Ta có: \(\sqrt{13-\sqrt{160}}-\sqrt{53+4\sqrt{60}}\)

\(=2\sqrt{2}-\sqrt{5}-4\sqrt{3}+\sqrt{5}\)

\(=2\sqrt{2}-4\sqrt{3}\)

b) Ta có: \(\sqrt{\left(4+\sqrt{3}\right)\cdot\sqrt{19-8\sqrt{3}+3}}\)

\(=\sqrt{\left(4+\sqrt{3}\right)\left(4-\sqrt{3}\right)+3}\)

=4

\(b)\) Thay \(m=1234^3\)\(;\)\(n=-1\) và \(p=5678^3\) ta được : 

\(A=\left[3.1234^3+4.\left(-1\right)-5.5678^3\right]-\left[3.1234^3-4.\left(-1\right)-5.5678^3\right]\)

\(A=3.1234^3-4-5.5678^3-3.1234^3-4+5678^3\)

\(A=\left(3.1234^3-3.1234^3\right)+\left(-4-4\right)+\left(-5.5678^3+5.5678^3\right)\)

\(A=0+\left(-8\right)+0\)

\(A=-8\)

Vậy giá trị của biểu thức \(A=\left(3m+4n-5p\right)-\left(3m-4n-5p\right)\) tại \(m=1234^3\)\(;\)\(n=-1\) và \(p=5678^3\) là \(-8\)

Chúc bạn học tốt ~ 

\(a)\) \(A=\left(3m+4n-5p\right)-\left(3m-4n-5p\right)\)

\(A=3m+4n-5p-3m+4n+5p\)

\(A=\left(3m-3m\right)+\left(4n+4n\right)+\left(-5p+5p\right)\)

\(A=0+8n+0\)

\(A=8n\)

Vậy \(A=8n\)

Chúc bạn học tốt ~ 

Lời giải:

 Áp dụng HĐT: $(a-b)^3=a^3-b^3-3ab(a-b)$ cho cả hai bạn.

a. 

$M=x^3-1-3x(x-1)-3x(x-1)^2+3x^2(x-1)+x^3$
$=2x^3-1+3x(x-1)[-1-(x-1)+x]$

$=2x^3-1+3x(x-1).0=2x^3-1$

b.

$D=[(x-y)-x]^3=-y^3$

mai mk thi rùi cầu cho các bạn trai xinh gái đẹp giúp mk với huhu

a.Chứng tỏ rằng B = 1/2² + 1/3² + 1/4² + 1/5² + 1/6² + 1/7² +1/8² < 1

b.Cho S = 3/1.4 + 3/4.7 + 3/7.10 +......+3/40.43 + 3/43.46 hãy chứng tỏ rằng S < 1

Sửa đề: 1/3²=1/2³

Giải:

A=1+1/2+1/2²+1/2³+..1/2²⁰¹²

2A=2+1+1/2+1/2²+...+1/2²⁰¹¹

2A-A=(2+1+1/2+1/2²+...+1/2²⁰¹¹)-(1+1/2+1/2²+1/2³+...+1/2²⁰¹²)

A=2-2²⁰¹²

Chúc bạn học tốt!