Bài 1: Cm giá trị của biểu thức sau không phụ thuộc vào biến x:
a) \(\left(2x+3\right).\left(4x^2-6x+9\right)-2\left(4x^3-1\right)\)
b)\(\left(x+3\right)^3+\left(x+9\right).\left(x^2+27\right)\)
c)\(\left(x+y\right).\left(x^2-xy+y^2\right)+\left(x-y\right).\left(x^2+xy+y^2\right)-2x^3\)
Bài 2: Tìm x biết:
a) \(\left(x+2\right)^2-9=0\)
b) \(\left(x+2\right)^2-x^2+4=0\)
c) \(\left(x-3\right)^2-4=0\)
d) \(x^2-2x=24\)
e) \(\left(2x-1\right)^2+\left(x+3\right)^2-5\left(x-7\right)\left(x+7\right)=0\)
Bài 1:
a) \(\left(2x+3\right)\cdot\left(4x^2-6x+9\right)-2\left(4x^3-1\right)\)
\(=8x^3-12x^2+18x+12x^2-18x+27-8x^3-3=27-3=24\)
--> đpcm
b) Sửa đề: \(\left(x+3\right)^3-\left(x+9\right)\left(x^2+27\right)\)
\(=x^3+9x^2+27x+27-\left(x^3+27x+9x^2+243\right)\)
\(=x^3+9x^2+27x+27-x^3-27x-9x^2-243=27-243=-216\)
--> đpcm
c) \(\left(x+y\right)\left(x^2-xy+y^2\right)+\left(x-y\right)\left(x^2+xy+y^2\right)-2x^3\)
\(=x^3+y^3+x^3-y^3-2x^3=2x^3-2x^3=0\)
--> đpcm
B1: a) \(\left(2x+3\right)\left(4x^2-6x+9\right)-2\left(4x^3-1\right)\)
\(=8x^3-27-8x^3+2\)
\(=-25\)
b) c) Làm theo câu a áp dụng HĐT.
B2:
a) \(\left(x+2\right)^2-9=0\)
\(\Rightarrow\left(x+2+3\right)\left(x+2-3\right)=0\)
\(\Rightarrow\left(x+5\right)\left(x-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x+5=0\\x-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-5\\x=1\end{matrix}\right.\)
Vậy \(\left\{{}\begin{matrix}x=-5\\x=1\end{matrix}\right..\)
Mấy câu b,c,d bn chịu khó tạo HĐT nhé.
e) \(\Rightarrow4x^2-4x+1+x^2+6x+9-5x^2+245=0\)
\(\Rightarrow2x=-255\)
\(\Rightarrow x=-\dfrac{255}{2}\)
Vậy \(x=-\dfrac{255}{2}\)