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a: \(=x^3-3x^2+3x-1-x^3+1-3x\left(1-x\right)\)
\(=-3x^2+3x-3x+3x^2=0\)
b: \(=8x^3+\dfrac{1}{27}-8x^3+\dfrac{1}{27}=\dfrac{2}{27}\)
a: Ta có: \(y\left(x^2-y^2\right)\cdot\left(x^2+y^2\right)-y\left(x^4-y^4\right)\)
\(=y\left(x^4-y^4\right)-y\left(x^4-y^4\right)\)
=0
b: Ta có: \(\left(2x+\dfrac{1}{3}\right)\left(4x^2-\dfrac{2}{3}x+\dfrac{1}{9}\right)-\left(8x^3-\dfrac{1}{27}\right)\)
\(=8x^3+\dfrac{1}{27}-8x^3+\dfrac{1}{27}\)
\(=\dfrac{2}{27}\)
c: Ta có: \(\left(x-1\right)^3-\left(x-1\right)\left(x^2+x+1\right)-3x\left(1-x\right)\)
\(=x^3-3x^2+3x-1-x^3+1-3x+3x^2\)
=0
1: Ta có: \(\left(x+3\right)\left(x^2-3x+9\right)-\left(x^3+54\right)\)
\(=x^3+27-x^3-54\)
=-27
2: Ta có: \(\left(2x+y\right)\left(4x^2-2xy+y^2\right)-\left(2x-y\right)\left(4x^2+2xy+y^2\right)\)
\(=8x^3+y^3-8x^3+y^3\)
\(=2y^3\)
\(1,=x^3+270-x^3-54=-27\\ 2,=8x^3+y^3-8x^3+y^3=2y^3\\ 3,=x^3-3x^2+3x-1-x^3-8+3x^2-48=3x-57\\ 4,=x^3-x-x^3-1=-x-1\\ 5,=8x^3-5\left(8x^3+1\right)=-32x^3-5\\ 6,=27+x^3-27=x^3\\ 7,làm.ở.câu.3\\ 8,=x^3-6x^2+12x-8+6x^2-12x+6-x^3-1+3x\\ =3x-3\)
có phải M=\(\dfrac{x+3}{3x}+\dfrac{2}{x+1}-3:\dfrac{2-4x}{x+1}-3x-x^2+\dfrac{1}{3x}\)
ko bạn
a) Ta có: \(\left(x-1\right)\left(x-2\right)\left(x^2+x+1\right)\left(x^2+2x+4\right)-x^6+9x^3\)
\(=\left(x-1\right)\left(x^2+x+1\right)\left(x-2\right)\left(x^2+2x+4\right)-x^6+9x^3\)
\(=\left(x^3-1\right)\left(x^3-8\right)-x^6+9x^3\)
\(=x^6-9x^3+8-x^6+9x^3=8\)
b) Ta có: \(\left(\dfrac{1}{3}+2x\right)\left(\dfrac{1}{9}-\dfrac{2}{3}x+4x^2\right)-\left(2x-\dfrac{1}{3}\right)\left(4x^2+\dfrac{2}{3}x+\dfrac{1}{4}\right)\)
\(=\dfrac{1}{27}+8x^3-8x^3+\dfrac{1}{27}\)
\(=\dfrac{2}{27}\)
c) Ta có: \(\left(x-1\right)^3-\left(x-1\right)\left(x^2+x+1\right)-3x\left(1-x\right)\)
\(=x^3-3x^2+3x-1-x^3+1-3x+3x^2\)
=0
d) Ta có: \(\left(x^2-y^2\right)\left(x^2+xy+y^2\right)\left(x^2-xy+y^2\right)-x^6+y^6\)
\(=\left(x-y\right)\left(x^2+xy+y^2\right)\left(x+y\right)\left(x^2-xy+y^2\right)-x^6+y^6\)
\(=\left(x^3-y^3\right)\left(x^3+y^3\right)-x^6+y^6\)
\(=x^6-y^6-x^6+y^6=0\)
1: \(\Leftrightarrow\left(x-3\right)\left(x+3\right)-\left(x-3\right)\left(5x+2\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(-4x+1\right)=0\)
hay \(x\in\left\{3;\dfrac{1}{4}\right\}\)
2: \(\Leftrightarrow\left(x-1\right)\left(x^2+x+1\right)-\left(x-1\right)\left(x^2-2x+16\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^2+x+1-x^2+2x-16\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(3x-15\right)=0\)
hay \(x\in\left\{1;5\right\}\)
3: \(\Leftrightarrow\left(x-1\right)\left(4x^2-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(2x-1\right)\left(2x+1\right)=0\)
hay \(x\in\left\{1;\dfrac{1}{2};-\dfrac{1}{2}\right\}\)
4: \(\Leftrightarrow x^2\left(x+4\right)-9\left(x+4\right)=0\)
\(\Leftrightarrow\left(x+4\right)\left(x-3\right)\left(x+3\right)=0\)
hay \(x\in\left\{-4;3;-3\right\}\)
5: \(\Leftrightarrow\left[{}\begin{matrix}3x+5=x-1\\3x+5=1-x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=-6\\4x=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-1\end{matrix}\right.\)
6: \(\Leftrightarrow\left(6x+3\right)^2-\left(2x-10\right)^2=0\)
\(\Leftrightarrow\left(6x+3-2x+10\right)\left(6x+3+2x-10\right)=0\)
\(\Leftrightarrow\left(4x+13\right)\left(8x-7\right)=0\)
hay \(x\in\left\{-\dfrac{13}{4};\dfrac{7}{8}\right\}\)
1.
\(\Leftrightarrow\left(x-3\right)\left(x+3\right)=\left(x-3\right)\left(5x-2\right)\)
\(\Leftrightarrow x+3=5x-2\)
\(\Leftrightarrow4x=5\Leftrightarrow x=\dfrac{5}{4}\)
2.
\(\Leftrightarrow\left(x-1\right)\left(x^2+x+1\right)=\left(x-1\right)\left(x^2-2x+16\right)\)
\(\Leftrightarrow x^2+x+1=x^2-2x+16\)
\(\Leftrightarrow3x=15\Leftrightarrow x=5\)
3.
\(\Leftrightarrow4x^2\left(x-1\right)-\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(4x^2-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{2};x=-\dfrac{1}{2}\end{matrix}\right.\)
Lời giải:
$P=(x+1)^3-(x+1)^3-[(x-1)^2+(x+1)^2]$
$=-[(x-1)^2+(x+1)^2]=-[(x^2-2x+1)+(x^2+2x+1)]=-2(x^2+1)$ phụ thuộc vào giá trị của biến nhé. Bạn xem lại đề.
$Q=(2x)^3-y^3+(2x)^3+y^3-16x^3$
$=8x^3-y^3+8x^3+y^3-16x^3=(8x^3+8x^3-16x^3)+(-y^3+y^3)=0+0=0$ không phụ thuộc vào giá trị của biến (đpcm)
bạn chỉ cần tính như nhân đa thức với đa thức sau đó rút gọn,kết quả ra là số thì bn gọi là ko phù hợp vào biến
nếu cần mik giải thì báo nha