Toán Lớp 8: bài 9: Tính giá trị biểu thức:
d) Q= $x(x-y)^{2}$ – $y(x-y)^{2}$+ $xy^{2}$- $x^{2}y$ tại x- y= 7 và xy= 9
Bài 10: Tìm x, biết:
a) 2-x= $2(x-2)^{3}$
b) $8x^{3}$ – 72x =0
c) $(x-1,5)^{6}$ + $2 (1,5-x)^{2}$= 0
d) $2x^{3}$+ $3x^{2}$ +3+2x =0
e) $x^{2}$ (x+1) -x (x+1)+ x(x-1) =0
f) $x^{3}$ -4x- 14x (x-2) =0 $x^{2}$
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9)\\
d)\\
Q = x{\left( {x – y} \right)^2} – y{\left( {x – y} \right)^2} + x{y^2} – {x^2}y\\
= x{\left( {x – y} \right)^2} – y{\left( {x – y} \right)^2} + xy\left( {y – x} \right)\\
= {\left( {x – y} \right)^2}.\left( {x – y} \right) – xy.\left( {x – y} \right)\\
= {\left( {x – y} \right)^3} – xy\left( {x – y} \right)\\
= {7^3} – 9.7\\
= 280\\
B10)\\
a)2 – x = 2{\left( {x – 2} \right)^3}\\
\Leftrightarrow 2{\left( {x – 2} \right)^3} + \left( {x – 2} \right) = 0\\
\Leftrightarrow \left( {x – 2} \right).\left( {2{{\left( {x – 2} \right)}^2} + 1} \right) = 0\\
\Leftrightarrow x – 2 = 0\\
\Leftrightarrow x = 2\\
Vậy\,x = 2\\
b)8{x^3} – 72x = 0\\
\Leftrightarrow 8x\left( {{x^2} – 9} \right) = 0\\
\Leftrightarrow 8x\left( {x – 3} \right)\left( {x + 3} \right) = 0\\
\Leftrightarrow x = 0;x = 3;x = – 3\\
Vậy\,x = 0;x = 3;x = – 3\\
c){\left( {x – 1,5} \right)^6} + 2.{\left( {1,5 – x} \right)^2} = 0\\
\Leftrightarrow {\left( {x – 1,5} \right)^2}.\left[ {{{\left( {x – 1,5} \right)}^4} + 2} \right] = 0\\
\Leftrightarrow x – 1,5 = 0\\
\Leftrightarrow x = 1,5\\
Vậy\,x = 1,5\\
d)2{x^3} + 3{x^2} + 3 + 2x = 0\\
\Leftrightarrow \left( {2x + 3} \right).{x^2} + 2x + 3 = 0\\
\Leftrightarrow \left( {2x + 3} \right)\left( {{x^2} + 1} \right) = 0\\
\Leftrightarrow 2x + 3 = 0\\
\Leftrightarrow x = – \dfrac{3}{2}\\
Vậy\,x = \dfrac{{ – 3}}{2}\\
e){x^2}\left( {x + 1} \right) – x\left( {x + 1} \right) + x\left( {x – 1} \right) = 0\\
\Leftrightarrow x\left( {x + 1} \right)\left( {x – 1} \right) + x\left( {x – 1} \right) = 0\\
\Leftrightarrow x.\left( {x – 1} \right).\left( {x + 1 + 1} \right) = 0\\
\Leftrightarrow x\left( {x – 1} \right)\left( {x + 2} \right) = 0\\
\Leftrightarrow x = 0;x = 1;x = – 2\\
Vậy\,x = 0;x = 1;x = – 2\\
f){x^3} – 4x – 14x\left( {x – 2} \right) = 0\\
\Leftrightarrow x.\left( {{x^2} – 4 – 14\left( {x – 2} \right)} \right) = 0\\
\Leftrightarrow x\left( {x – 2} \right)\left( {x + 2 – 14} \right) = 0\\
\Leftrightarrow x\left( {x – 2} \right)\left( {x – 12} \right) = 0\\
\Leftrightarrow x = 0;x = 2;x = 12\\
Vậy\,x = 0;x = 2;x = 12
\end{array}$