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\(a,\left(x+3\right)^3=x^3+9x^2+27x+27\\ b,\left(\dfrac{1}{2}-x\right)^3=\dfrac{1}{8}-\dfrac{3}{4}x+\dfrac{3}{2}x^2-x^3\\ c,\left(2x-y^2\right)^3=8x^3-12x^2y^2+6xy^4-y^6\\ d,\left(2x-\dfrac{y^2}{x}\right)^3=8x^3-\dfrac{12x^2y^2}{x}+\dfrac{6xy^4}{x^2}-\dfrac{y^6}{x^2}\\ =8x^3-12xy^2+6y^4-\dfrac{y^6}{x^2}\)
\(2x\left(x^2-7x-3\right)=2x^3-14x-6x\)
\(4xy^2\left(-2x^3+y^2-7xy\right)=-8x^4y^2+4xy^5-28x^2y^3\)
\(\)a: \(\left(x-2y\right)^3\)
\(=x^3-3\cdot x^2\cdot2y+3\cdot x\cdot\left(2y\right)^2-\left(2y\right)^3\)
\(=x^3-6x^2y+12xy^2-8y^3\)
b: \(\left(2x+y\right)^3=\left(2x\right)^3+3\cdot\left(2x\right)^2\cdot y+3\cdot2x\cdot y^2+y^3\)
\(=8x^3+12x^2y+6xy^2+y^3\)
c: \(\left(\dfrac{1}{3}x-1\right)^3=\left(\dfrac{1}{3}x\right)^3-3\cdot\left(\dfrac{1}{3}x\right)^2\cdot1+3\cdot\dfrac{1}{3}x\cdot1^2-1^3\)
\(=\dfrac{1}{27}x^3-\dfrac{1}{3}x^2+x-1\)
d: \(\left(x+\dfrac{1}{3}y\right)^3\)
\(=x^3+3\cdot x^2\cdot\dfrac{1}{3}y+3\cdot x\cdot\left(\dfrac{1}{3}y\right)^2+\left(\dfrac{1}{3}y\right)^3\)
\(=x^3+x^2y+\dfrac{1}{3}xy^2+\dfrac{1}{27}y^3\)
e: (2x-3y)3
\(=\left(2x\right)^3-3\cdot\left(2x\right)^2\cdot3y+3\cdot2x\cdot\left(3y\right)^2-\left(3y\right)^3\)
\(=8x^3-36x^2y+54xy^2-27y^3\)
f: \(\left(x^2-2y\right)^3\)
\(=\left(x^2\right)^3-3\cdot\left(x^2\right)^2\cdot2y+3\cdot x^2\cdot\left(2y\right)^2-\left(2y\right)^3\)
\(=x^6-6x^4y+12x^2y^2-8y^3\)
g: \(\left(\dfrac{1}{2}x-y\right)^3=\left(\dfrac{1}{2}x\right)^3-3\cdot\left(\dfrac{1}{2}x\right)^2\cdot y+3\cdot\dfrac{1}{2}x\cdot y^2-y^3\)
\(=\dfrac{1}{8}x^3-\dfrac{3}{4}x^2y+\dfrac{3}{2}xy^2-y^3\)
2: \(N=a^2-6ab+9b^2-a^2-6ab-9b^2-ab+2a+b-2\)
\(=-13ab+2a+b-2\)
\(=-13\cdot\dfrac{1}{2}\cdot\left(-3\right)+2\cdot\dfrac{1}{2}+\left(-3\right)-2\)
\(=\dfrac{39}{2}+1-3-2=\dfrac{39}{2}-4=\dfrac{31}{2}\)
3: \(P=4x^2-25-4x^2-4x-1=-4x-26\)
=-8020-26=-8046
4: \(Q=\left(y^2-9\right)\left(y^2+9\right)-\left(y^2+2\right)\left(y^2-2\right)\)
\(=y^4-81-y^4+4=-77\)
chuyển về dạng nguyên thể rồi tính thể chất khối lượng sau đó quay về đang tìm mũ của nhiều số làm ra rồi thì dễ lắm bạn ạ k minh nha
a)\(\left(x^2-2\right)\left(x^2+2x+2\right)\)
b)\(\left(x-1\right)\left(2x+1\right)\left(3x+7\right)\)
c)\(-2\left(x-4\right)\left(2x+1\right)\)
d)\(\left(x-5\right)\left(4x+1\right)\)
e)\(3\left(x-2\right)\left(3x-2\right)\)
g)\(2\left(a-b\right)^2\)
h)\(\left(xy-3\right)\left(5y^2-2z\right)\)
i)\(\left(4x+1\right)\left(2x-y\right)\)
l)\(abc^2\left(b-a\right)\left(b+c\right)\)
m)\(\left(x-y\right)\left(y-z\right)\left(x-z\right)\)
a. \(y=\frac{2}{2x+3}\in Z\)
\(\Rightarrow2x+3\in\left\{-2;-1;1;2\right\}\)
\(\Rightarrow2x\in\left\{-5;-4;-2;-1\right\}\). Vì x thuộc Z
\(\Rightarrow x\in\left\{-2;-1\right\}\)
b. \(y=\frac{2x-1}{2x-3}=\frac{2x-3+2}{2x-3}=1+\frac{2}{2x-3}\)
Vì y thuộc Z nên 2 / 2x - 3 thuộc Z
\(\Rightarrow2x-3\in\left\{-2;-1;1;2\right\}\)
\(\Rightarrow2x\in\left\{1;2;4;5\right\}\). Vì x thuộc Z
\(\Rightarrow x\in\left\{1;2\right\}\)
c. \(y=\frac{2x^2-1}{2x-3}=\frac{x\left(2x-3\right)+2x-3-x+2}{2x-3}=x+1-\frac{x+2}{2x-3}\)
Vì y thuộc Z nên x thuộc Z ; x + 2 / 2x - 3 thuộc Z
=> 2x + 4 / 2x - 3 thuộc Z
=> 2x - 3 + 7 / 2x - 3 thuộc Z
=> 7 / 2x - 3 thuộc Z
\(\Rightarrow2x-3\in\left\{-7;-1;1;7\right\}\)
\(\Rightarrow2x\in\left\{-4;2;4;10\right\}\)
\(\Rightarrow x\in\left\{-2;1;2;5\right\}\) ( tm x thuộc Z )
d,e tương tự
lm hết hộ mik