a: \(M=\dfrac{2\left(1-3x\right)\left(1+3x\right)}{3x\left(x+2\right)}\cdot\dfrac{3x}{2\left(1-3x\right)}=\dfrac{3x+1}{x+2}\)

a: Ta có: \(\left(x+3\right)\left(x+4\right)\left(x+5\right)\left(x+6\right)+1\)

\(=\left(x^2+9x+18\right)\left(x^2+9x+20\right)+1\)

\(=\left(x^2+9x\right)^2+38\left(x^2+9x\right)+360+1\)

\(=\left(x^2+9x\right)^2+2\cdot\left(x^2+9x\right)\cdot19+19^2\)

\(=\left(x^2+9x+19\right)^2\)

b. \(x^2+y^2+2x+2y+2\left(x+1\right)\left(y+1\right)+2\)

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

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

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

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

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

\(=\left(x-y-2\right)^2\)

d. \(x^2+2x\left(y+1\right)+y^2+2y+1\)

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

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

a,Ta có D= (1/3+2x+1/3-2x):1/3+2x

=2/3:1/3+2x

=2+2x

=2(x+1)

b, Từ câu a ta có

D=2(x+1)

Với x=3

=>2(x+1)

=2.4=8

KL

a,Ta có D= (1/3+2x+1/3-2x):1/3+2x

=2/3:1/3+2x

=2+2x

=2(x+1)

b, Từ câu a ta có

D=2(x+1)

Với x=3

=>2(x+1)

=2.4=8

\(=\frac{3x^2+9x-3}{x^2+x-2}-\frac{x+1}{x+2}-\frac{x-2}{x-1}\)

\(=\frac{3x^2+9x-3}{\left(x+2\right)\left(x-1\right)}-\frac{\left(x+1\right)\left(x-1\right)}{\left(x+2\right)\left(x-1\right)}-\frac{\left(x-2\right)\left(x+2\right)}{\left(x-1\right)\left(x+2\right)}\)

\(=\frac{3x^2+9x-3-\left(x^2-1\right)-\left(x^2-4\right)}{\left(x-1\right)\left(x+2\right)}\)

\(=\frac{3x^2+9x-3-x^2+1-x^2+4}{\left(x-1\right)\left(x+2\right)}\)

\(=\frac{x^2+9x+2}{\left(x-1\right)\left(x+2\right)}\)

hi bn 

bn ghi sai đề

a) x/-2=-4/y=2/4

*x/-2=2/4=>4x=(-2)x2=>x=-1

*-4/y=2/4=>(-4)x4=2y=>y=-8

b)2/x=y/-3

=>xy=-6(câu này đề hơi lạ)

c) x+1/2=8/x+1

=>16=(x+1)(x+1)=>x^2+2x+1=16=>(x+1)^2=16=>(x+1)^2=4^2=>x+1=4=>x=3

Đề nghị viết lại đề

a) \(\dfrac{x}{-2}=\dfrac{-4}{y}=\dfrac{2}{4}\) 

\(\Rightarrow\dfrac{x}{-2}=\dfrac{2}{4}\) 

\(\Rightarrow x=\dfrac{-2.2}{4}=-1\) 

\(\Rightarrow\dfrac{-4}{y}=\dfrac{2}{4}\) 

\(\Rightarrow x=\dfrac{-4.4}{2}=-8\) 

b) \(\dfrac{2}{x}=\dfrac{y}{-3}\) 

\(\Rightarrow x.y=2.-3\) 

\(\Rightarrow x.y=-6\) 

\(\Rightarrow x\) và \(y\inƯ\left(-6\right)=\left\{^+_-1;^+_-2;^+_-3;^+_-6\right\}\) 

\(\Rightarrow\left(x;y\right)=\left(-6;1\right);\left(1;-6\right);\left(6;-1\right);\left(-1;6\right);\left(-2;3\right);\left(3;-2\right);\left(2;-3\right);\left(-3;2\right)\) c) \(\dfrac{x+1}{2}=\dfrac{8}{x+1}\) 
\(\Rightarrow\left(x+1\right).\left(x+1\right)=2.8\) 

\(\Rightarrow\left(x+1\right)^2=16\) 

\(\Rightarrow\left(x+1\right)^2=4^2\) hoặc \(\left(x+1\right)^2=\left(-4\right)^2\) 

      \(x+1=4\) hoặc \(x+1=-4\) 

            \(x=3\) hoặc \(x=-5\)

=(2^2-1)*(2^2+1)*(2^4+1)*...*(2^64+1)
=(2^4-1)(2^2+1)*(2^4+1)*...*(2^64+1)
=(2^8-1)(2^8+1)*...*(2^64+1)
=...
=(2^64-1)(2^64+1)
=2^128-1

mk vua lm ra do!!!!!!

tks bạn nhưng mk làm ra lâu rồi sorry nha 

(mặc dù mk vẫn tk cho bạn)

\(=\)\(\left(2^2-1\right)\times\)\(\frac{2^4-1}{2^2-1}\)\(\times\frac{2^8-1}{2^4-1}\)\(\times\)...\(\times\)\(\frac{2^{128}-1}{2^{64}-1}\)

\(=\)\(2^{128}\)\(-1\)

Ta có: \(\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)

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

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

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

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

\(=\dfrac{\left(3^{16}-1\right)\left(3^{16}+1\right)}{2}\)

\(=\dfrac{3^{32}-1}{2}\)

Rút gọn: (3 + 1)(3² + 1)(3⁴ + 1)(3⁸ + 1)(3¹⁶ + 1)(3³² + 1)
A=(3 + 1)(3² + 1)(3⁴ + 1)(3⁸ + 1)(316 + 1)(3³² + 1)

A=(3-1)(3 + 1)(3² + 1)(3⁴ + 1)(3⁸ + 1)(3¹⁶ + 1)(3³² + 1)
A=(3²-1)(3² + 1)(3⁴ + 1)(3⁸ + 1)(3¹⁶ + 1)(3³² + 1)
A=(3⁴-1)(3⁴ + 1)(3⁸ + 1)(3¹⁶ + 1)(3³² + 1)
A=(3⁸-1)(3⁸ + 1)(3¹⁶ + 1)(3³² + 1)
A=(3¹⁶-1)(3¹⁶ + 1)(3³² + 1)
A=(3³² - 1)(3³² + 1)
A=3⁶⁴-1
=>A=(3⁶⁴-1) /2

Đặt \(A=\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\).Ta có : 

\(=>\left(3-1\right)A=\left(3-1\right)\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)

\(=>2A=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)

\(=>2A=\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)

...............................................................................

Cuối cùng \(=>2A=3^{64}-1\).

\(=>A=\frac{3^{64}-1}{2}\)

Đặt \(A=\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)

\(\Rightarrow2A=\left(3-1\right)\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)

\(=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)

\(=...........................................\)

\(=\left(3^{32}-1\right)\left(3^{32}+1\right)=3^{64}-1\)

\(\Rightarrow A=\frac{3^{64}-1}{2}\)

./ có nghĩa là phần đó