3 + 2 ( 5 - x ) = 11 

      2 ( 5 - x ) = 11 - 3

      2 ( 5 - x ) = 8

            5 - x   = 8 ÷ 2

            5 - x   = 4

                 x   = 5 - 4

                 x    = 1 

Vậy x = 1

3 + 2( 5 - x ) = 11

2( 5 - x ) = 11 - 3

2(5 - x ) = 8

5 - x = 8 : 2

5 - x = 4

=> x = 5 - 4 = 1

thanks

a: Để C vô nghĩa thì x+2=0

hay x=-2

Để C có nghĩa thì x+2<>0

hay x<>-2

\(C=\dfrac{2\left(x^2-2x+4\right)}{\left(x+2\right)\left(x^2-2x+4\right)}=\dfrac{2}{x+2}\)

Để C=0 thì \(x\in\varnothing\)

Để C>0 thì x+2>0

hay x>-2

Để C<0 thì x+2<0

hay x<-2

b: \(C=\dfrac{2\left(x^2-2x+4\right)}{\left(x+2\right)\left(x^2-2x+4\right)}=\dfrac{2}{x+2}\)

Kết quả ra là 3 và 1 mà bạn .-.??

Đặt \(x^2-4x+10=a\ge6\)

\(\Leftrightarrow\frac{21}{a}-a+4=0\)

\(\Leftrightarrow-a^2+4a+21=0\) \(\Rightarrow\left[{}\begin{matrix}a=7\\a=-3\left(l\right)\end{matrix}\right.\)

\(\Rightarrow x^2-4x+10=7\Leftrightarrow x^2-4x+3=0\Rightarrow\left[{}\begin{matrix}x=1\\x=3\end{matrix}\right.\)

Đặt \(x^2-4x=t\)

ta có pt : \(\frac{21}{t+10}-\left(t+6\right)=0\)

\(\Leftrightarrow21-\left(t+6\right)\left(t+10\right)=0\)

\(\Leftrightarrow21-t^2-10t-6t-60=0\)

\(\Leftrightarrow t^2+16t+39=0\)

\(\left[{}\begin{matrix}t=-3\\t=-13\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x^2-4x+3=0\\x^2-4x+13=0\end{matrix}\right.\Leftrightarrow}\left[{}\begin{matrix}\left\{{}\begin{matrix}x=3\\x=1\end{matrix}\right.\\\left(x-2\right)^2+9>0\left(\text{vô nghiệm}\right)\end{matrix}\right.\)

vậy pt có nghiệm x = 1 và x = 3

x²+10x+25-4x(x+5)=0

⇔(x+5)²-4x(x+5)=0

⇔(x+5)(x+5-4x)=0

⇔(x+5)(5-3x)=0

⇔\(\left\{{}\begin{matrix}x+5=0\\5-3x=0\end{matrix}\right.\Leftrightarrow\left\{{} }\left\{{}\begin{matrix}x=-5\\x=\dfrac{5}{3}\end{matrix}\right.\)

\(\left(5+6\right)\times3:2=16,5\)

(5+6)x3:2= 15x3

a/ Đặt \(\left|x\right|=t\ge0\Rightarrow t^2-t-2=0\Rightarrow\left[{}\begin{matrix}t=-1\left(l\right)\\t=2\end{matrix}\right.\)

\(\Rightarrow\left|x\right|=2\Rightarrow x=\pm2\)

b/ \(\Leftrightarrow\left(x+1\right)^2+\left|x+1\right|-6=0\)

Đặt \(\left|x+1\right|=t\ge0\Rightarrow t^2+t-6=0\Rightarrow\left[{}\begin{matrix}t=-3\left(l\right)\\t=2\end{matrix}\right.\)

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

c/ \(\Leftrightarrow\left(x+1\right)^2-5\left|x+1\right|+4=0\)

Đặt \(\left|x+1\right|=t\ge0\Rightarrow t^2-5t+4=0\Rightarrow\left[{}\begin{matrix}t=1\\t=4\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}\left|x+1\right|=1\\\left|x+1\right|=4\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x+1=1\\x+1=-1\\x+1=4\\x+1=-4\end{matrix}\right.\)

d. \(\Leftrightarrow\left(x-1\right)^2+5\left|x-1\right|+4=0\)

Đặt \(\left|x+1\right|=t\ge0\Rightarrow t^2+5t+4=0\Rightarrow\left[{}\begin{matrix}t=-1\left(l\right)\\t=-4\left(l\right)\end{matrix}\right.\)

Vậy pt vô nghiệm

e. \(\Leftrightarrow\left(x-2\right)^2+2\left|x-2\right|-3=0\)

Đặt \(\left|x-2\right|=t\ge0\)

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

\(\Rightarrow\left|x-2\right|=1\Leftrightarrow\left[{}\begin{matrix}x-2=1\\x-2=-1\end{matrix}\right.\)

f. \(\Leftrightarrow\left(2x-5\right)^2+4\left|2x-5\right|-12=0\)

Đặt \(\left|2x-5\right|=t\ge0\)

\(\Rightarrow t^2+4t-12=0\Rightarrow\left[{}\begin{matrix}t=2\\t=-6\left(l\right)\end{matrix}\right.\)

\(\Rightarrow\left|2x-5\right|=2\Rightarrow\left[{}\begin{matrix}2x-5=2\\2x-5=-2\end{matrix}\right.\)

45x45x3,14=6358,5 nha

~HT~

K cho mình nha 

@@@@@@@@@@@@@@@@@

45 x 45 x 3,14 = 2025 x 3,14 = 6358,5

TL

45x45x3,14=6358,5

nha

_HT_