Toán Lớp 8: giúp tui với mà mn tui đang cần gấp
tui hứu vote 5 sao
bài 3 tìm x
(3x-5)(5x+2)-x(15x-1)=-2
3x^3-27x=0
bài 2 phân tích đa thức thành nhân tử
20x-125y
x^2-14x+49
x^2-7x+2x-14
x^2+10x+25-9y^2
bài 1 thực hiện phép tính
-6x(x-3)
(3x-7)(2x+5)
(x-7)^2 -3x(5x-1)
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Comments ( 1 )
B3)\\
a)\left( {3x – 5} \right)\left( {5x + 2} \right) – x\left( {15x – 1} \right) = – 2\\
\Leftrightarrow 15{x^2} + 6x – 25x – 10 – 15{x^2} + x + 2 = 0\\
\Leftrightarrow – 18x – 8 = 0\\
\Leftrightarrow 18x = – 8\\
\Leftrightarrow x = – \dfrac{4}{9}\\
Vậy\,x = – \dfrac{4}{9}\\
b)3{x^3} – 27x = 0\\
\Leftrightarrow 3x\left( {{x^2} – 9} \right) = 0\\
\Leftrightarrow 3x\left( {x – 3} \right)\left( {x + 3} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x – 3 = 0\\
x + 3 = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x = 3\\
x = – 3
\end{array} \right.\\
Vậy\,x = 0;x = 3;x = – 3\\
B2)\\
a)20x – 125y = 5\left( {4x – 25y} \right)\\
b){x^2} – 14x + 49 = {\left( {x – 7} \right)^2}\\
c){x^2} – 7x + 2x – 14\\
= x\left( {x – 7} \right) + 2\left( {x – 7} \right)\\
= \left( {x – 7} \right)\left( {x + 2} \right)\\
d){x^2} + 10x + 25 – 9{y^2}\\
= {\left( {x + 5} \right)^2} – {\left( {3y} \right)^2}\\
= \left( {x + 5 + 3y} \right)\left( {x + 5 – 3y} \right)\\
B1)\\
a) – 6x\left( {x – 3} \right) = – 6{x^2} + 18x\\
b)\left( {3x – 7} \right)\left( {2x + 5} \right)\\
= 6{x^2} + 15x – 14x – 35\\
= 6{x^2} + x – 35\\
c){\left( {x – 7} \right)^2} – 3x\left( {5x – 1} \right)\\
= {x^2} – 14x + 49 – 15{x^2} + 3x\\
= – 14{x^2} – 11x + 49
\end{array}$