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

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 \(x=4\)

a. 9x² - 6x - 3 = 0

<=> 3(3x² - 2x - 1) = 0

<=> 3(3x² - 3x + x - 1) = 0

<=> \(3\left[3x\left(x-1\right)+\left(x-1\right)\right]=0\)

<=> 3(3x + 1)(x - 1) = 0

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

b. (2x + 1)² - 4(x + 2)² = 9

<=> (2x + 1)² - \(\left[2\left(x+2\right)\right]^2=9\)

<=> (2x + 1 - 2x - 4)(2x + 1 + 2x + 4) = 9

<=> -3(4x + 5) = 9

<=> 4x + 5 = -3

<=> 5 + 3 = -4x

<=> -4x = 8

<=> -x = 2

<=> x = -2

a) \(\Leftrightarrow\left(9x^2-6x+1\right)-4=0\)

\(\Leftrightarrow\left(3x-1\right)^2-4=0\)

\(\Leftrightarrow3\left(x-1\right)\left(3x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{1}{3}\end{matrix}\right.\)

b) \(\Leftrightarrow4x^2+4x+1-4x^2-16x-16=9\)

\(\Leftrightarrow12x=-24\Leftrightarrow x=-2\)

c) \(\Leftrightarrow3x^2-6x+3-3x^2+15x=21\)

\(\Leftrightarrow9x=18\Leftrightarrow x=2\)

d) \(\Leftrightarrow x^2+6x+9-x^2-4x+32=1\)

\(\Leftrightarrow2x=-40\Leftrightarrow x=-20\)

(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

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

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

\(=x^2-x\)

f(1)=-1

=>m+3=-1

hay m=-4

b) Thay x=-1 vào biểu thức \(B=\dfrac{2x^2+5x+4}{x^2-4x+3}\), ta được:

\(B=\dfrac{2\cdot\left(-1\right)^2+5\cdot\left(-1\right)+4}{\left(-1\right)^2-4\cdot\left(-1\right)+3}=\dfrac{2\cdot1-5+4}{1+4+3}=\dfrac{1}{8}\)

Vậy: Khi x=-1 thì \(B=\dfrac{1}{8}\)

Ta có:

|x| = \(\dfrac{1}{3}\)

\(\Rightarrow x=\dfrac{1}{3};x=-\dfrac{1}{3}\)

`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`