a ) 

Ta có : 

\(x^2+xy+x=x\left(x+y+1\right)\)

Thay \(x=77;y=22\)vào b/t , ta được : 

\(77\left(77+22+1\right)=77.100=7700\)

Vậy \(x^2+xy+x=7700\)tại \(x=77;y=22\)

b ) 

Ta có : 

\(x\left(x-y\right)+y\left(y-x\right)\)

\(=x\left(x-y\right)-y\left(x-y\right)\)

\(=\left(x-y\right)\left(x-y\right)\)

\(=\left(x-y\right)^2\)

Thay \(x=53;y=3\)vào b/t , ta được : 

\(\left(53-3\right)^2=50^2=2500\)

Vậy \(x\left(x-y\right)+y\left(y-x\right)=2500\) tại \(x=53;y=3\)

a.

\(x^2+xy+x=x\left(x+y+1\right)\)

Tại \(x=77;y=22\Rightarrow x\left(x+y+1\right)=77\left(77+22+1\right)=77.100=7700\)

b.

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

\(=\left(53-3\right)^2=50^2=2500\)

c.

\(x\left(x-1\right)-y\left(1-x\right)=x\left(x-1\right)+y\left(x-1\right)=\left(x+y\right)\left(x-1\right)\)

\(=\left(2001+1999\right)\left(2001-1\right)=4000.2000=8000000\)

Ta có:  x 2  + xy + x = x(x + y + 1)

Thay x = 77, y = 22 vào biểu thức, ta được:

x(x + y + 1) = 77.(77 + 22 + 1) = 77.100 = 7700

a.\(x=0;y=-1\)

\(\Rightarrow2.0-\dfrac{-1\left(0^2-2\right)}{0.-1-1}=0-\dfrac{2}{-1}=2\)

b.\(x=2\)

\(\Rightarrow4.2^2-3\left|2\right|-2=16-6-2=8\)

\(x=-3\)

\(\Rightarrow4.\left(-3\right)^2-3\left|-3\right|-2=36-9-2=25\)

c.\(x=-\dfrac{1}{5};y=-\dfrac{3}{7}\)

\(\Rightarrow5.\left(-\dfrac{1}{5}\right)^2-7.\left(-\dfrac{3}{7}\right)+6=5.\dfrac{1}{25}+3+6=\dfrac{1}{5}+3+6=\dfrac{46}{5}\)

thay x=2 và biểu thức A ta đc

\(A=4.2^2-3.\left|2\right|-2=4.4-6-2=16-6-2=8\)

thay x=-3  biểu thức A ta đc

\(A=4.\left(-3\right)^2-3.\left|-3\right|-2=4.9-9-2=36-9-2=25\)

thay x=-1/5 ; y=-3/7  biểu thức B ta đc

\(B=5.\left(-\dfrac{1}{5}\right)^2-7.\left(-\dfrac{3}{7}\right)+6\)

\(B=5\cdot\dfrac{1}{25}+3+6\)

\(B=\dfrac{1}{5}+3+6=\dfrac{46}{5}\)

a) \(x^2+xy+x\)

\(\Leftrightarrow x\left(x+y+1\right)\)

Tại x=77 và y=22 có:

\(\Leftrightarrow77\left(77+22+1\right)\)

\(=7700\)

b) \(x\left(x-y\right)+y\left(y-x\right)\)

\(\Leftrightarrow\left(x-y\right)\left(x+y\right)\)

\(\Leftrightarrow x^2-y^2\)

Tại x=53 và y=3, ta có:

\(53^2-3^2=2800\)

\(2,\\ a,=-3x^3y^3z^4\\ b,=\dfrac{1}{4}xy^2\cdot\dfrac{1}{4}x^4y^4\cdot\left(-\dfrac{4}{5}yz^2\right)=-\dfrac{1}{20}x^5y^7z^2\\ c,=-\dfrac{15}{14}x^6y^{11}z^{10}\\ 3,\\ a,=9\left(-1\right)\left(-\dfrac{1}{27}\right)=\dfrac{1}{3}\\ b,=-\dfrac{1}{5}\left(-8\right)=\dfrac{8}{5}\\ c,=\dfrac{4}{9}a\cdot36\left(-1\right)=-16a\)

a) \(P=x\left(x-y\right)+y\left(x-y\right)=\left(x-y\right)\left(x+y\right)=x^2-y^2=5^2-4^2=9\)

b) \(Q=x\left(x^2-y\right)-x^2\left(x+y\right)+y\left(x^2-x\right)=x^3-xy-x^3-x^2y+x^2y-xy=0\)

Ta có: x(x – y) + y(y – x) = x(x – y) – y(x – y) = (x – y)(x – y) = x - y 2

Thay x = 53, y = 3 vào biểu thức ta được:

x - y 2  = 53 - 3 2  = 50 2  = 2500