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

`a)`

Có `x;y` TLN theo công thức `xy=10`

`=>x=10/y`

vậy hệ số tỉ lệ là `10`

`b)`

Thay `x=5` vào `x=10/y` ; ta đc :

`5=10/y`

`=>y=2`

Vậy nếu `x=5` thì `y=2`

Thay `x=-1` vào `x=10/y` ta đc :

`-1=10/y`

`y=-10`

Vậy nếu `x=-1` thì `y=-10`

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

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

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

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

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 \(x=4\)

(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

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

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

\(=x^2-x\)

\(\left(\dfrac{1}{15}+\dfrac{1}{35}+\dfrac{1}{63}\right)x=1\)

\(\Leftrightarrow\dfrac{1}{9}x=1\)

\(\Leftrightarrow x=1:\dfrac{1}{9}\)

\(\Leftrightarrow x=9\)

=>1/2(2/15+2/35+2/63)*x=1

=>1/2(1/3-1/5+1/5-1/7+1/7-1/9)*x=1

=>1/2*2/9*x=1

=>x*1/9=1

=>x=9

\(\left|x-3\right|-2x=1\left(đk:x\ge-\dfrac{1}{2}\right)\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x-3=1+2x\left(x\ge3\right)\\x-3=-1-2x\left(-\dfrac{1}{2}\le x< 3\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-2\left(loại\right)\\x=\dfrac{2}{3}\left(tm\right)\end{matrix}\right.\)