Bài 2:

a: ĐKXĐ: \(x\notin\left\{0;-1;\dfrac{1}{2}\right\}\)

b: \(D=\left(\dfrac{x+2}{3x}+\dfrac{2}{x+1}-3\right):\dfrac{2-4x}{x+1}-\dfrac{3x-x^2+1}{3x}\)

\(=\dfrac{\left(x+2\right)\left(x+1\right)+6x-3\cdot3x\left(x+1\right)}{3x\left(x+1\right)}\cdot\dfrac{x+1}{2-4x}+\dfrac{x^2-3x-1}{3x}\)

\(=\dfrac{x^2+3x+2+6x-9x^2-9x}{3x}\cdot\dfrac{1}{2-4x}+\dfrac{x^2-3x-1}{3x}\)

\(=\dfrac{-8x^2+2}{3x}\cdot\dfrac{1}{-4x+2}+\dfrac{x^2-3x-1}{3x}\)

\(=\dfrac{-2\left(2x-1\right)\left(2x+1\right)}{3x\cdot\left(-2\right)\left(2x-1\right)}+\dfrac{x^2-3x-1}{3x}\)

\(=\dfrac{2x+1}{3x}+\dfrac{x^2-3x-1}{3x}\)

\(=\dfrac{2x+1+x^2-3x-1}{3x}=\dfrac{x^2-x}{3x}=\dfrac{x-1}{3}\)

c: Khi x=1 thì \(D=\dfrac{1-1}{3}=0\)

Bài 1:

a: \(2x^2-8x=0\)

=>\(x^2-4x=0\)

=>x(x-4)=0

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

b: \(\left(x+2\right)^2-x\left(x-1\right)=10\)

=>\(x^2+4x+4-x^2+x=10\)

=>5x+4=10

=>5x=6

=>\(x=\dfrac{6}{5}\)

c: \(x^3-6x^2+9x=0\)

=>\(x\left(x^2-6x+9\right)=0\)

=>\(x\left(x-3\right)^2=0\)

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

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

có a+b+a+c=10+25

2a+b+c=35

2a+25=35

2a=35-25

2a=10

a=10:2

a=5

mà a+b=10

=>b=10-5=5

mà b+c=25

=>c=25-5=20 bạn cho mình tk nhé

Ta thấy a,b,c,mỗi cái được lặp lại 2 lần nên tổng a+b+c là:

         (10+25+25):2=30

         K mình nha

2x[a+b+c]=10+25+25=60,[a+b+c]=60:2=30

a=30-25=5

b=10-5=5

c=25-5=20

-Co:a+b+a+c=10+25

      2a+b+c  =35

      2a+25    =35

      2a          =35 - 25

      2a          =10

        a          =10:2

        a          =5

        Ma a+b=10

=>b=10-5=5

    Ma b+c=25

=>c=25-5=20

20 chuẩn 100%

trong vio làm gì có 20

(a+b)+(a+c)+(b+c)=a+b+a+c+b+c=a+a+b+b+c+c=2a+2b+2c=2.(a+b+c)=10+25+25=60

Vậy a+b+c bằng:

60 chia 2=30

nhớ tích đúng nha bạn

a+b+a+c+b+c=10+25.2

=>2(a+b+c)=60

=>a+b+c=30

Vay :a+b+c=30

Cộng tất cả các điều kiện được : \(a+b+a+c+b+c=10+25\cdot2\)

\(\Rightarrow2(a+b+c)=10+50\)

\(\Rightarrow2(a+b+c)=60\)

\(\Rightarrow a+b+c=60:2=30\)

Vậy: ....

a: x-2b+c=1-2+3=2

b: x-2b+c=5-2*0+2=7

c: x-2b+c=3-2*2-4=-5