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

\(\Leftrightarrow3x-4=\dfrac{2}{3}\)

\(\Leftrightarrow3x=\dfrac{14}{3}\)

\(\Leftrightarrow x=\dfrac{14}{9}\)

Số thứ 2 là : 6948+193=7131

Tổng 3 số là 7483 x3 =22449

Số thứ 3 là :     22449-(6948 +7131 )=8370

vậy số thứ 3 là 8370

:))

\(3x^2-12x+1=3x^2-12x+12-11=3\left(x^2-4x+4\right)-11\)\(=3\left(x-2\right)^2-11\)

Ta thấy: \(3\left(x-2\right)^2\ge0\Rightarrow3\left(x-2\right)^2-11\ge-11\)

vậy GTNN của 3x²-12x+1 là -11 

tick cho mk nhé!

\(B=1+\dfrac{1}{2}\left(1+2\right)+\dfrac{1}{3}\left(1+2+3\right)+...+\dfrac{1}{20}\left(1+2+...+20\right)\)

\(B=1+\dfrac{1}{2}.2.3:2+\dfrac{1}{3}.3.4:2+...+\dfrac{1}{20}.20.21:2\)

\(B=\dfrac{2}{2}+\dfrac{3}{2}+...+\dfrac{21}{2}\)

\(B=\dfrac{2+3+...+21}{2}\)

\(B=\dfrac{230}{2}\)

\(\Rightarrow B=115\)

\(\dfrac{2011x}{xy+2011x+2011}+\dfrac{y}{yz+y+2011}+\dfrac{z}{xz+z+x}\)

\(=\dfrac{x^2yz}{xy+x^2yz+xyz}+\dfrac{y}{yz+y+xyz}+\dfrac{z}{xz+z+1}\)

\(=\dfrac{x^2yz}{xy\left(1+xz+z\right)}+\dfrac{y}{y\left(z+1+xz\right)}+\dfrac{z}{xz+z+1}\)

\(=\dfrac{xz}{1+xz+z}+\dfrac{1}{1+xz+z}+\dfrac{z}{1+xz+z}\)

\(=\dfrac{xz+1+z}{1+xz+z}\)

\(=1\) ( Đpcm )

\(A=\dfrac{\left(x^2y-y\right)\left(y+1\right)+x^2y^2-1}{\left(x^2+y\right)\left(y+1\right)+x^2y^2+1}\)

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

\(A=\dfrac{\left(x^2y^2+x^2y\right)-\left(y^2+y\right)+\left(x^2-1\right)}{\left(x^2y^2+x^2y\right)+\left(y^2+y\right)+\left(x^2+1\right)}\)

\(A=\dfrac{x^2y\left(y+1\right)-y\left(y+1\right)+\left(x^2-1\right)}{x^2y\left(y+1\right)+y\left(y+1\right)+\left(x^2+1\right)}\)

\(A=\dfrac{\left(x^2y-y\right)\left(y+1\right)+\left(x^2-1\right)}{\left(x^2y+y\right)\left(y+1\right)+\left(x^2+1\right)}\)

\(A=\dfrac{y\left(x^2-1\right)\left(y+1\right)+\left(x^2-1\right)}{y\left(x^2+1\right)\left(y+1\right)+\left(x^2+1\right)}\)

\(A=\dfrac{\left(x^2-1\right)\left[y\left(y+1\right)+1\right]}{\left(x^2+1\right)\left[y\left(y+1\right)+1\right]}\)

\(A=\dfrac{\left(x^2-1\right)\left(y^2+y+1\right)}{\left(x^2+1\right)\left(y^2+y+1\right)}\)

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

MTC: \(\left(x-y\right)^2\left(x+y\right)^2\)

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

\(=\dfrac{x^2\left(x+y\right)-2xy^2+y^2\left(x-y\right)}{\left(x-y\right)^2\left(x+y\right)^2}\)

\(=\dfrac{x^3+x^2y-2xy^2+y^2x-y^3}{\left(x-y\right)^2\left(x+y\right)^2}\)

\(=\dfrac{x^3+x^2y-xy^2-y^3}{\left(x-y\right)^2\left(x+y\right)^2}\)

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

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

\(=\dfrac{1}{x-y}\)

\(4x\left(x+y\right)\left(x+y+z\right)\left(x+z\right)+y^2z^2\)

\(= 4( x^2+ xz+xy+yz)( x^2+xy+xz) + y^2z^2 \)

Đặt \(x^2+xy+xz = a \) ta được:

\( 4(a+yz)a+y^2z^2 \)

\(= 4a^2+ 4ayz + y^2z^2 \)

\(= (2a+yz)^2 \)

\(=\left(2x^2+2xy+2xz+yz\right)^2\)