(x + 3)^2 – 25 x^3(x^2 – 1) – (x^2 – 1) (x – 4)2 + (x – 4) Giá trị của biểu thức A = x2 – y2 + 2y – 1 với x = 75; y = 26 Tìm x biết 2×2 – x – 1 = 0 Gi

Đáp án:

$\begin{array}{l}
a){\left( {x + 3} \right)^2} – 25\\
 = {\left( {x + 3} \right)^2} – {5^2}\\
 = \left( {x + 3 – 5} \right)\left( {x + 3 + 5} \right)\\
 = \left( {x – 2} \right)\left( {x + 8} \right)\\
b){x^3}\left( {{x^2} – 1} \right) – \left( {{x^2} – 1} \right)\\
 = \left( {{x^2} – 1} \right)\left( {{x^3} – 1} \right)\\
 = \left( {x – 1} \right)\left( {x + 1} \right)\left( {x – 1} \right)\left( {{x^2} + x + 1} \right)\\
 = \left( {x + 1} \right){\left( {x – 1} \right)^2}\left( {{x^2} + x + 1} \right)\\
c){\left( {x – 4} \right)^2} + \left( {x – 4} \right)\\
 = \left( {x – 4} \right)\left( {x – 4 + 1} \right)\\
 = \left( {x – 4} \right)\left( {x – 3} \right)\\
B2)\\
A = {x^2} – {y^2} + 2y – 1\\
 = {x^2} – \left( {{y^2} – 2y + 1} \right)\\
 = {x^2} – {\left( {y – 1} \right)^2}\\
 = \left( {x – y + 1} \right)\left( {x + y – 1} \right)\\
Thay\,x = 75;y = 26\,vào\,A\\
 \Rightarrow A = \left( {75 – 26 + 1} \right)\left( {75 + 26 – 1} \right)\\
 = 50.100\\
 = 5000\\
B3)2{x^2} – x – 1 = 0\\
 \Rightarrow 2{x^2} – 2x + x – 1 = 0\\
 \Rightarrow 2x\left( {x – 1} \right) + x – 1 = 0\\
 \Rightarrow \left( {x – 1} \right)\left( {2x + 1} \right) = 0\\
 \Rightarrow \left[ \begin{array}{l}
x – 1 = 0\\
2x + 1 = 0
\end{array} \right.\\
 \Rightarrow \left[ \begin{array}{l}
x = 1\\
x =  – \frac{1}{2}
\end{array} \right.\\
Vậy\,x = 1\,;x =  – \frac{1}{2}\\
B4)\\
4{\left( {x + y} \right)^2} – 9{\left( {x – y} \right)^2}\\
 = {\left( {2x + 2y} \right)^2} – {\left( {3x – 3y} \right)^2}\\
 = \left( {2x + 2y – 3x + 3y} \right)\left( {2x + 2y + 3x – 3y} \right)\\
 = \left( { – x + 5y} \right)\left( {5x – y} \right)\\
 = \left( { – 2 + 5.4} \right)\left( {5.2 – 4} \right)\left( {do:x = 2;y = 4} \right)\\
 = 18.6\\
 = 108\\
B50\\
a)49{\left( {y – 4} \right)^2} – 9{\left( {y + 2} \right)^2}\\
 = {\left( {7y – 28} \right)^2} – {\left( {3y + 6} \right)^2}\\
 = \left( {7y – 28 + 3y + 6} \right)\left( {7y – 28 – 3y – 6} \right)\\
 = \left( {10y – 22} \right)\left( {4y – 34} \right)\\
 = 4\left( {5y – 11} \right)\left( {2y – 17} \right)\\
b)9{x^6} + 24{x^3}{y^2} + 16{y^4}\\
 = {\left( {3{x^3} + 4{y^2}} \right)^2}\\
c)36 – 12x + {x^2}\\
 = {\left( {6 – x} \right)^2}
\end{array}$