Bài 1 : Tìm x .
a ) Ta có :
\(3x^3-7x^2+6x-14=0\)
\(\Leftrightarrow\left(3x^3-7x^2\right)+\left(6x-14\right)=0\)
\(\Leftrightarrow x^2\left(3x-7\right)+2\left(3x-7\right)=0\)
\(\Leftrightarrow\left(3x-7\right)\left(x^2+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-7=0\\x^2+2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{7}{3}\\loại\left(x^2+2>0\right)\end{matrix}\right.\)
Vậy \(x=\dfrac{7}{3}\)
Câu b :
\(6x^3+16x^2-150x-400=0\)
\(\Leftrightarrow\left(6x^3+16x^2\right)-\left(150x+400\right)=0\)
\(\Leftrightarrow x^2\left(6x+16\right)-25\left(6x+16\right)=0\)
\(\Leftrightarrow\left(6x+16\right)\left(x^2-25\right)=0\)
\(\Leftrightarrow\left(6x+16\right)\left(x-5\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}6x+16=0\\x-5=0\\x+5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-\dfrac{16}{6}\\x=5\\x=-5\end{matrix}\right.\)
Vậy \(x=-\dfrac{16}{6};x=5;x=-5\)
Bài 2 : Tính giá trị của biểu thức .
Ta có :
\(A=2x^3+x^2y-2xy-y^2\)
\(A=\left(2x^3+x^2y\right)-\left(2xy+y^2\right)\)
\(A=x^2\left(2x+y\right)-y\left(2x+y\right)\)
\(A=\left(2x+y\right)\left(x^2-y\right)\)
Thay \(x=25;y=125\) vào biểu thức vừa rút gọn ta có :
\(A=\left(2.25+125\right)\left(25^2-125\right)\)
\(A=175.500\)
\(A=87500\)
Bài 3 :Tính nhanh :
Ta có :
\(100^2-99^2+98^2-97^2+-...+2^2-1^2\)
\(=\left(100^2-99^2\right)+\left(98^2-97^2\right)+-.+\left(2^2-1^2\right)\)
\(=\left(100+99\right)\left(100-99\right)+\left(98+97\right)\left(98-97\right)+--.+\left(2+1\right)\left(2-1\right)\)
\(=100+99+98+97+-..+2+1\)
\(=\dfrac{100\left(100+1\right)}{2}\)
\(=5050\)
