Toán Lớp 8: TGTCBT
a)x(x^2-y)-x^2(x+y)+y(x^2-y)với x=1/2;y= -100
5x(x-4y)-4y(y-5)với x=-1/5 ; y=-1/2
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Toán Lớp 8: TGTCBT
a)x(x^2-y)-x^2(x+y)+y(x^2-y)với x=1/2;y= -100
5x(x-4y)-4y(y-5)với x=-1/5 ; y=-1/2
Comments ( 1 )
a)x\left( {{x^2} – y} \right) – {x^2}\left( {x + y} \right) + y\left( {{x^2} – y} \right)\\
= \left( {{x^2} – y} \right)\left( {x + y} \right) – {x^2}\left( {x + y} \right)\\
= \left( {{x^2} – y – {x^2}} \right)\left( {x + y} \right)\\
= – y.\left( {x + y} \right)\\
= – 100.\left( {\dfrac{1}{2} – 100} \right)\\
= – 100.\dfrac{{ – 199}}{2}\\
= 50.199\\
= 9950\\
b)\\
5x\left( {x – 4y} \right) – 4y\left( {y – 5} \right)\\
= 5{x^2} – 20xy – 4{y^2} + 20y\\
= 5.{\left( {\dfrac{{ – 1}}{5}} \right)^2} – 20.\dfrac{{ – 1}}{5}.\dfrac{{ – 1}}{2} – 4.{\left( {\dfrac{{ – 1}}{2}} \right)^2} + 20.\dfrac{{ – 1}}{2}\\
= \dfrac{1}{5} – 2 – 1 – 10\\
= \dfrac{1}{5} – 13\\
= – \dfrac{{64}}{5}
\end{array}$