(x+2)² +x(x-1)<2x²+1
x²+4x+4+x²-x<2x²+1
3x+4<1
x< -1

(x-1)^3-x(x-2)^2+1

= x^3-3x^2+3x-1-x(x^2-4x+4)+1

= x^3-3x^2+3x-1- x^3+4x^2-4x+1

= x^2-x 

= x(x-1)

HỌC TỐT!

@Zịt_siu_lừi

\(=x^3-3x^2+3x-1-x\left(x^2-4x+4\right)+1\)

\(=x^3-3x^2+3x-1-x^3+4x^2-4x+1\)

\(=x^2-x\)

1) A=1/8; 2) B=9472; 3) 0 và 10.

1: \(A=2x^3y^4-5x\cdot x^2y^4+xy^2\cdot x^2y^2=-2x^3y^4=-2\cdot\left(-1\right)^3\cdot\dfrac{1}{16}=\dfrac{1}{8}\)

2: \(B=9x^4y^6\cdot\left(-4xy\right)+19x^3y^5\cdot\left(-2\right)x^2y^2\)

\(=-36x^5y^7-38x^5y^7\)

\(=-74x^5y^7=-74\cdot\left(-1\right)^5\cdot2^7=9472\)

3: \(f\left(-1\right)=3\cdot\left(-1\right)^4+7\cdot\left(-1\right)^3+4\cdot\left(-1\right)^2-2\cdot\left(-1\right)-2=0\)

\(f\left(1\right)=3+7+4-2-2=10\)

1.

Đặt \(x-2=t\ne0\Rightarrow x=t+2\)

\(B=\dfrac{4\left(t+2\right)^2-6\left(t+2\right)+1}{t^2}=\dfrac{4t^2+10t+5}{t^2}=\dfrac{5}{t^2}+\dfrac{2}{t}+4=5\left(\dfrac{1}{t}+\dfrac{1}{5}\right)^2+\dfrac{19}{5}\ge\dfrac{19}{5}\)

\(B_{min}=\dfrac{19}{5}\) khi \(t=-5\) hay \(x=-3\)

2.

Đặt \(x-1=t\ne0\Rightarrow x=t+1\)

\(C=\dfrac{\left(t+1\right)^2+4\left(t+1\right)-14}{t^2}=\dfrac{t^2+6t-9}{t^2}=-\dfrac{9}{t^2}+\dfrac{6}{t}+1=-\left(\dfrac{3}{t}-1\right)^2+2\le2\)

\(C_{max}=2\) khi \(t=3\) hay \(x=4\)

\(\Leftrightarrow\left|3x-2\right|>x+1\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}3x-2>0\\x+1< 0\end{matrix}\right.\\\left\{{}\begin{matrix}x+1>0\\3x-2>x^2+2x+1\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>-1\\x^2+2x+1-3x+2< 0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x>-1\\x^2-x+3< 0\end{matrix}\right.\Leftrightarrow x\in\varnothing\)

a.x^3-1^3

b.x^3-5^3

c)(2x)^3+3^3

d)x^3+1/2^3

\(y=f\left(x\right)=4x^2-7.\)

\(y=1.\rightarrow f\left(1\right)=4.1^2-7=4-7=-3.\)

giúp mình với, đi mà mọi người

Xx4/5=1/2

       X=1/2:4/5

       X=5/8

X x 4/5 = 1/2

X          = 1/2 : 4/5

X          = 5/8