I don't now 

sorry 

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

nha

b)  \(\left(3x-2\right)\left(x+1\right)^2\left(3x+8\right)=-16\)

\(\Leftrightarrow\)\(\left(3x-2\right)\left(3x+3\right)^2\left(3x+8\right)+144=0\)

Đặt:  \(3x+3=a\)pt trở thành:

\(\left(a-5\right)a^2\left(a+5\right)+144=0\)

\(\Leftrightarrow\)\(a^4-25a^2+144=0\)

\(\Leftrightarrow\)\(\left(a-4\right)\left(a-3\right)\left(a+3\right)\left(a+4\right)=0\)

đến đây bạn tìm a rồi tính x

c)  \(\left(4x-5\right)\left(2x-3\right)\left(x-1\right)=9\)

\(\Leftrightarrow\)\(\left(4x-5\right)\left(4x-6\right)\left(4x-4\right)-72=0\)

Đặt   \(4x-5=a\)pt trở thành:

\(a\left(a-1\right)\left(a+1\right)-72=0\)

\(\Leftrightarrow\)\(a^3-a-72=0\)

p/s: ktra lại đề

d)  \(\left(2x^2+x-2013\right)^2+4\left(x^2-5x-2012\right)^2=4\left(2x^2+x-2013\right)\left(x^2-5x-2012\right)\)

\(\Leftrightarrow\)\(\left(2x^2+x-2013\right)^2+4\left(x^2-5x-2012\right)^2-4\left(2x^2+x-2013\right)\left(x^2-5x-2012\right)=0\)

\(\Leftrightarrow\)\(\left[\left(2x^2+x-2013\right)-2\left(x^2-5x-2012\right)\right]^2=0\)

\(\Leftrightarrow\)\(\left(11x+2011\right)^2=0\)

đến đây làm nốt

a)  2x+1 là Ư(3x+2)

=>3x+2 chia hết cho 2x+1

<=>2(3x+2) chia hết cho 2x+1

<=>6x+4 chia hết cho 2x+1

<=>3(2x+1)+1 chia hết cho 2x+1

<=>1 chia hết cho 2x+1

=>2x+1 là Ư(1)

=>Ư(1)={-1;1}

Có:

TH1: 2x+1=-1

<=>2x=-2

<=>x=-1(t/m)

TH2: 2x+1=1

<=>2x=0

<=>x=0(t/m)

Vậy x thuộc {-1;0}

b)xy+x+y=2

<=>x(y+1)+y+1=3

<=>(y+1)(x+1)=3

=>y+1 và x+1 thuộc Ư(3)

=>Ư(3)={-1;1;-3;3}

Ta có bảng sau:

Vậy các cặp số (x;y) thỏa mãn là (0;2) và (2;0)

Shbh=a x h= 48 x (48 x \(\frac{1}{3}\) ) =768 (cm2 )

1. \(\left(3x-5\right)^{2010}+\left(y-1\right)^{2012}+\left(x-z\right)^{2014}=0\)

Vì \(\left(3x-5\right)^{2010}\ge0\forall x\); \(\left(y-1\right)^{2012}\ge0\forall y\); \(\left(x-z\right)^{2014}\ge0\forall x,z\)

\(\Rightarrow\left(3x-5\right)^{2010}+\left(y-1\right)^{2012}+\left(x-z\right)^{2014}\ge0\)

Dấu " = " xảy ra \(\Leftrightarrow\hept{\begin{cases}3x-5=0\\y-1=0\\x-z=0\end{cases}}\Leftrightarrow\hept{\begin{cases}3x=5\\y=1\\x=z\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\frac{5}{3}\\y=1\\z=\frac{5}{3}\end{cases}}\)

Vậy \(x=z=\frac{5}{3}\)và \(y=1\) 

\(1,\\ a,\dfrac{x^2}{x+1}+\dfrac{x}{x+1}=\dfrac{x^2+x}{x+1}=\dfrac{x\left(x+1\right)}{x+1}=x\)

\(b,\left(\dfrac{2xy}{x^2-y^2}+\dfrac{x-y}{2x+2y}\right):\dfrac{x+y}{2x}=\left(\dfrac{4xy}{2\left(x-y\right)\left(x+y\right)}+\dfrac{\left(x-y\right)^2}{2\left(x-y\right)\left(x+y\right)}\right).\dfrac{2x}{x+y}=\dfrac{4xy+x^2-2xy+y^2}{2\left(x-y\right)\left(x+y\right)}.\dfrac{2x}{x+y}=\dfrac{2x\left(x^2+2xy+y^2\right)}{2\left(x-y\right)\left(x+y\right)^2}=\dfrac{2x\left(x+y\right)^2}{2\left(x-y\right)\left(x+y\right)^2}=\dfrac{x}{x-y}\)

Câu 2: 

a: ĐKXĐ: \(x\ne1\)