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

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 .-.??

ui sai rồi

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 \(\left|2x-5\right|=t\ge0\)

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

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

Tham khảo bài này :

(3x+1)(7x+3)=(5x-7)(3x+1)

<=> (3x+1)(7x+3)-(5x-7)(3x+1)=0

<=> (3x+1)(7x+3-5x+7)=0

<=> (3x+1)(2x+10)=0

<=> 2(3x+1)(x+5)=0

=> 3x+1=0 hoặc x+5=0

=> x= -1/3 hoặc x=-5

Vậy x = -1/3 hoặc x = -5

\(a,x^2+10x+25-4x\left(x+5\right)=0.\)

\(\Leftrightarrow\left(x+5\right)^2-4x\left(x+5\right)=0\)

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

\(\Leftrightarrow\orbr{\begin{cases}x+5=0\\5-3x=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-5\\x=\frac{5}{3}\end{cases}}}\)

\(b,\left(4x-5\right)^2-2\left(16x^2-25\right)=0\)

\(\Leftrightarrow\left(4x-5\right)^2-2\left(4x+5\right)\left(4x-5\right)=0\)

\(\Leftrightarrow-\left(4x-5\right)\left(4x+15\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}4x-5=0\\4x+15=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{5}{4}\\x=-\frac{15}{4}\end{cases}}}\)