=>36/18>x>8/15

=>2>x>8/15

mà x nguyên

nên x=1

c: Ta có: \(\left(x+1\right)^2\ge0\forall x\)

\(\left(y-\dfrac{1}{3}\right)^2\ge0\forall y\)

Do đó: \(\left(x+1\right)^2+\left(y-\dfrac{1}{3}\right)^2\ge0\forall x,y\)

\(\Leftrightarrow\left(x+1\right)^2+\left(y-\dfrac{1}{3}\right)^2-10\ge-10\forall x,y\)

Dấu '=' xảy ra khi x=-1 và \(y=\dfrac{1}{3}\)

a: \(A=\dfrac{2\cdot8^4\cdot27^2+44\cdot6^9}{2^7\cdot6^7+2^7\cdot40\cdot9^4}\)

\(=\dfrac{2\cdot2^{12}\cdot3^6+2^2\cdot11\cdot2^9\cdot3^9}{2^7\cdot3^7\cdot2^7+2^7\cdot2^3\cdot5\cdot3^8}\)

\(=\dfrac{2^{13}\cdot3^6+2^{11}\cdot3^9\cdot11}{2^{14}\cdot3^7+2^{10}\cdot5\cdot3^8}\)

\(=\dfrac{2^{11}\cdot3^6\left(2^2+3^3\cdot11\right)}{2^{10}\cdot3^7\left(2^4+5\cdot3\right)}\)

\(=\dfrac{2\cdot301}{3\cdot31}=\dfrac{602}{93}\)

\(4A-3x^2+7-6x=x^2+3A-4x-3\)

\(\Rightarrow4A-3A=\left(x^2+3x^2\right)-\left(4x-6x\right)-\left(3+7\right)\)

\(\Rightarrow A=4x^2-\left(-2x\right)-10\)

\(\Rightarrow A=4x^2+2x-10\)

\(a,\left|2x+\dfrac{1}{2}\right|=0\\ \Leftrightarrow2x+\dfrac{1}{2}=0\\ \Leftrightarrow2x=-\dfrac{1}{2}\\ \Leftrightarrow x=-\dfrac{1}{4}\\ b,\left|3x+\dfrac{3}{4}\right|=3\\ \Leftrightarrow\left[{}\begin{matrix}3x+\dfrac{3}{4}=3\\3x+\dfrac{3}{4}=-3\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}3x=\dfrac{9}{4}\\3x=-\dfrac{15}{4}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=-\dfrac{5}{4}\end{matrix}\right.\)

Sửa đề: -24x^2y^3

9x^4-24x^2y^3+16y^6

=9x^4-2*3x^2*4y^3+16y^6

=(3x^2-4y^3)^2

C

C

C

Giải pt à bạn

Bài 1:

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

\(=x^2+x+x^2-x-2x^2\)

\(=2x^2-2x^2\)

\(=0\)

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

\(=x^3-x^2+x+2x^2-2x+2-x^3-x^2-x+2x^2+2x+2\)

\(=\left(x^3-x^3\right)+\left(-x^2+2x^2-x^2+2x^2\right)+\left(x-2x-x+2x\right)+\left(2+2\right)\)

\(=2x^2+4\)

c) \(\left(3-x\right)^2+2\left(x-3\right)\left(x+7\right)+\left(x+7\right)^2\)

\(=\left(x-3\right)^2+2\left(x-3\right)\left(x+7\right)+\left(x+7\right)^2\)

\(=\left[\left(x-3\right)+\left(x+7\right)\right]^2\)

\(=\left(x-3+x+7\right)^2\)

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

Bài 2:

a) \(x^2-4-4\left(2-x\right)\)

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

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

\(=\left(x+2\right)\left(x-2\right)+4\left(x-2\right)\)

\(=\left(x-2\right)\left(x+2+4\right)\)

\(=\left(x-2\right)\left(x+6\right)\)

b) \(x^3+8\left(1-2x\right)^3\)

\(=x^3+\left[2\left(1-2x\right)\right]^3\)

\(=\left[x+2\left(1-2x\right)\right]\left[x^2-x\cdot2\left(1-2x\right)+2^2\left(1-2x\right)^2\right]\)

\(=\left(x+2-4x\right)\left[x^2-2x\left(1-2x\right)+4\left(1-4x+4x^2\right)\right]\)

\(=\left(2-3x\right)\left(x^2-2x+4x^2+4-16x+16x^2\right)\)

\(=\left(2-3x\right)\left(21x^2-18x+4\right)\)