1. x(x + 1) - x² + 1 = 0

<=> x(x + 1) - (x² - 1) = 0

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

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

<=> x + 1 = 0\

<=> x = -1

2. 4x(x - 2) - 6 + 3x = 0

<=> 4x(x - 2) - (3x - 6) = 0

<=> 4x(x - 2) - 3(x - 2) = 0

<=> (4x - 3)(x - 2) = 0

<=> \(\left[{}\begin{matrix}4x-3=0\\x-2=0\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=2\end{matrix}\right.\)

3. x(x + 2) - 3(x + 2) = 0

<=> (x - 3)(x + 2) = 0

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

<=> \(\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)

Ko thấy j hết á bạn

1)

\(3\left(x-2\right)+4\left(x-1\right)=25\)

\(3x-6+4x-4=25\)

\(7x-10=25\\ 7x=35\\ x=5\)

2)

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

\(\left(5x-3\right)\left(x-2\right)-\left(x-1\right)\left(x-2\right)=0\)

\(\left(x-2\right)\left(5x-3-x+1\right)=0\)

\(\left(x-2\right)\left(4x-2\right)=0\)

\(=>\left[{}\begin{matrix}x-2=0\\4x-2=0\end{matrix}\right.=>\left[{}\begin{matrix}x=2\\x=\dfrac{1}{2}\end{matrix}\right.\)

3)

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

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

\(x^2-4x+4=4x^2-8x+4\)

\(x^2-4x+4-4x^2+8x-4=0\)

\(-3x^2+4x=0\)

\(x\left(-3x+4\right)=0\)

\(=>\left[{}\begin{matrix}x=0\\-3x+4=0\end{matrix}\right.=>\left[{}\begin{matrix}x=0\\x=\dfrac{4}{3}\end{matrix}\right.\)

\(1,x^2+4x+4=0\\ \Rightarrow\left(x+2\right)^2=0\\ \Rightarrow x+2=0\\ \Rightarrow x=-2\\ 2,x^2+4x+4=0\\ \Rightarrow\left(x+2\right)^2=0\\ \Rightarrow x+2=0\\ \Rightarrow x=-2\\ 3,\left(x+1\right)^2+2\left(x+1\right)=0\\ \Rightarrow\left(x+1\right)\left(x+1+2\right)=0\\ \Rightarrow\left(x+1\right)\left(x+3\right)=0\\ \Rightarrow\left[{}\begin{matrix}x+1=0\\x+3=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-1\\x=-3\end{matrix}\right.\)

x²+4x+4=0
(x+2)²=0
x+2=0
x=+-2
câu 1 giống câu 2
(x+1)²+2(x+1)=0
(x+1+2)(x+1)=0
Th1: x+3=0           Th2: x+1=0
            x=-3                      x=-1
vậy ...

1)
\(4x^2-4x+1-4x^2-16x-16=9\)
\(-20x-15=9\)
-20x=24
x=-1,2

3)
(2x+1)²=5²
2x+1=5
2x=4
x=2
 

\(1,\Rightarrow4x^2-4x+1-4x^2-16x-16=9\\ \Rightarrow-20x=23\Rightarrow x=-\dfrac{23}{20}\\ 2,\Rightarrow9x^2-6x+1+2x+6+11-11x^2=15\\ \Rightarrow2x^2+4x-3=0\\ \Rightarrow2\left(x^2+2x+1\right)-5=0\\ \Rightarrow2\left(x+1\right)^2-5=0\\ \Rightarrow\left[\sqrt{2}\left(x+1\right)-\sqrt{5}\right]\left[\sqrt{2}\left(x+1\right)+\sqrt{5}\right]=0\\ \Rightarrow\left[{}\begin{matrix}\sqrt{2}\left(x+1\right)=\sqrt{5}\\\sqrt{2}\left(x+1\right)=-\sqrt{5}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x+1=\sqrt{\dfrac{5}{2}}=\dfrac{\sqrt{10}}{2}\\x+1=-\sqrt{\dfrac{5}{2}}=\dfrac{-\sqrt{10}}{2}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{\sqrt{10}-2}{2}\\x=\dfrac{-\sqrt{10}-2}{2}\end{matrix}\right.\)

\(3,\Rightarrow\left(2x+1\right)^2-25=0\Rightarrow\left(2x+1-5\right)\left(2x+1+5\right)=0\\ \Rightarrow\left[{}\begin{matrix}2x=4\\2x=-6\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\)

\(4,\Rightarrow x^3+3x^2+3x+1-x^3-2x^2-2x+1-x^2=15\\ \Rightarrow x+2=15\Rightarrow x=13\)

`x xx1/4+x xx1/3=2`

`=> x xx (1/4+1/3)=2`

`=> x xx (3/12+4/12)=2`

`=> x xx 7/12=2`

`=> x=2:7/12`

`=>x=2xx12/7`

`=>x=24/7`

X x (1/4 +1/3)=2

X x 7/12=2

x= 2 : 7/12

x=24/7

\(x.\dfrac{1}{4}+x.\dfrac{1}{3}=2\)

\(x.\left(\dfrac{1}{4}+\dfrac{1}{3}\right)=2\)

\(x.\dfrac{7}{12}=2\)

\(x=\dfrac{24}{7}\)

`(3x-1)(x-3)-2(x-3)=9`

`-> 3x(x-3)-1(x-3)-2x+6=9`

`-> 3x^2-9x-x+3-2x+6=9`

`-> 3x^2-12x+9=9`

`-> 3x^2-12x=0`

`-> x(3x-12)=0`

`->`\(\left[{}\begin{matrix}x=0\\3x-12=0\end{matrix}\right.\)

`->`\(\left[{}\begin{matrix}x=0\\3x=12\end{matrix}\right.\)

`->`\(\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)

Vậy, `x={0 ; 4}`.

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

Lời giải:

$x+(x+1)+(x+2)+....+(x+21)=231$

$\underbrace{x+x+....+x}_{22}+(1+2+3+...+21)=231$

$22x+231=231$

$22x=0$

$x=0$