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\(A=\left(x+2\right)\left(x^2-2x+4\right)-\left(x^3-2\right)\)
\(\Rightarrow A=\left(x^3+8\right)-\left(x^3-2\right)\)
\(\Rightarrow A=x^3+8-x^3+2\)
\(\Rightarrow A=\left(x^3-x^3\right)+\left(8+2\right)\)
\(\Rightarrow A=10\)
\(A=\left(x+2\right)\left(x^2-2x+4\right)-\left(x^3-2\right)\)
\(=x^3+8-x^3+2\)
\(=10\)
\(B=\left(x+2\right)\left(x-2\right)\left(x^2+2x+4\right)\left(x^2-2x+4\right)\)
\(=\left(x+2\right)\left(x^2-2x+4\right)\left(x-2\right)\left(x^2+2x+4\right)\)
\(=\left(x^3+8\right)\left(x^3-8\right)\)
\(=x^6-64\)
\(C=\left(x^2+3x+1\right)^2+\left(3x-1\right)^2-2\left(x^2+3x+1\right)\left(3x-1\right)\)
\(=\left(x^2+3x+1\right)^2-2\left(x^2+3x+1\right)\left(3x-1\right)+\left(3x-1\right)^2\)
\(=\left(x^2+3x+1-3x+1\right)^2\)
\(=\left(x^2+2\right)^2\)
\(D=\left(3x^3+3x+1\right)\left(3x^3-3x+1\right)-\left(3x^3+1\right)^2\)
\(=\left(3x^3+1+3x\right)\left(3x^3+1-3x\right)-\left(3x^3+1\right)^2\)
\(=\left(3x^3+1\right)^2-9x^2-\left(3x^3+1\right)^2\)
\(=-9x^2\)
\(E=\left(2x^2+2x+1\right)\left(2x^2-2x+1\right)-\left(2x^2+1\right)^2\)
\(=\left(2x^2+1+2x\right)\left(2x^2+1-2x\right)-\left(2x^2+1\right)^2\)
\(=\left(2x^2+1\right)^2-4x^2-\left(2x^2+1\right)^2\)
\(=-4x^2\)
a) \(\left(4x-1\right)^2-\left(3x+2\right)\left(3x-2\right)=\left(7x-1\right)\left(x+2\right)+\left(2x+1\right)^2-\left(4x^2+7\right)\)(1)
\(\Leftrightarrow\left(16x^2-8x+1\right)-\left(9x^2-4\right)=\left(7x^2+14x-x-2\right)+\left(4x^2+4x+1\right)-\left(4x^2+7\right)\)
\(\Leftrightarrow16x^2-8x+1-9x^2+4=7x^2+13x-2+4x^2+4x+1-4x^2-7\)
\(\Leftrightarrow7x^2-8x+5=7x^2+17x-8\)
\(\Leftrightarrow7x^2-8x-7x^2-17x=-8-5\)
\(\Leftrightarrow-25x=-13\)
\(\Leftrightarrow x=\dfrac{13}{25}\)
Vậy tập nghiệm phương trình (1) là \(S=\left\{\dfrac{13}{25}\right\}\)
\(\left(x+1\right)\left(x^2-x-x^2+x-1\right)=-\left(x+1\right)\)
\(\left(2a^2+1\right)^2-4a^2-\left(2a^2+1\right)^2=-4a^2\)
\(\left(a^2+b^2+c^2+a^2-b^2-c^2\right)\left(a^2+b^2+c^2-a^2+b^2+c^2\right)=2a^2\left(2b^2+2c^2\right)=4a^2b^2+4a^2c^2\)
\(\left(a-5\right)^2\left(a+5\right)^2=\left(a^2-25\right)^2\)
\(\left(3a^3+1\right)^2-9a^2-\left(3a^3+1\right)^2=-9a^2\)
a)Ta có:
\(\left(x-2\right)^2-\left(x-3\right)\left(x-1\right)\\ =x^2-4x+4-x^2+4x-3\\ =1\)
Vậy biểu thức \(\left(x-2\right)^2-\left(x-3\right)\left(x-1\right)\)không phụ thuộc vào biến
b) Ta có:
\(\left(x-1\right)^3-\left(x+1\right)^3+6\left(x+1\right)\left(x-1\right)\\ =x^3-3x^2+3x-1-x^3-3x^2-3x-1+6x^2-6\\ =-8\)
Vậy.....
c) Ta có:
\(\left(x-3\right)\left(x+3\right)\left(x^2+9\right)-\left(x^2-2\right)\left(x^2+2\right)\\ =\left(x^2-9\right)\left(x^2+9\right)-x^4+4\\ =x^4-81-x^4+4=-77\)
Vậy....
d) Ta có: \(\left(3x+1\right)^2-2\left(3x+1\right)\left(3x-5\right)+\left(3x-5\right)^2\\ =\left(3x+1-3x+5\right)^2\\ =6^2=36\)
Vậy....
Bài 2: a) \(3x^3-3x=0\Leftrightarrow3x\left(x^2-1\right)=0\Leftrightarrow\orbr{\begin{cases}x=0\\x=\pm1\end{cases}}\)
b) \(x^2-x+\frac{1}{4}=0\Leftrightarrow x^2-2.\frac{1}{2}+\left(\frac{1}{2}\right)^2=0\Leftrightarrow\left(x-\frac{1}{2}\right)^2=0\)
\(\Leftrightarrow x-\frac{1}{2}=0\Leftrightarrow x=\frac{1}{2}\)
`a, (3x-1)^3-(3x+1)^3`
`= (3x-1-3x-1)(9x^2-6x+1+9x^2-1+9x^2+6x+1`
`= (-2)(27x^2 +1)`
`= -54x^2-2`.
`b, (1+3x)^3 - (1-3x)^3`
`= 1+ 9x + 27x^2 + 27x^3 - 1 + 9x - 27x^2 + 27x^3`
`= 54x^3 + 18x`.
`c, = 54x^3 + 18x -1 +9x^2`.
a: =27x^3-27x^2+9x-1-27x^3-27x^2-9x-1
=-54x^2-2
b: =27x^3+27x^2+9x+1-27x^3+27x^2-9x+1
=54x^2+2
c: =54x^2+2+(3x-1)(3x+1)
=54x^2+2+9x^2-1
=63x^2+1