\(x^3+27+\left(x+3\right)\left(x-9\right)=0\)

\(\Rightarrow\left(x^3+27\right)+\left(x+3\right)\left(x-9\right)=0\)

\(\Rightarrow\left(x+3\right)\left(x^2-3x+9\right)+\left(x+3\right)\left(x-9\right)=0\)

\(\Rightarrow\left(x+3\right)\left(x^2-3x+9+x-9\right)=0\)

\(\Rightarrow\left(x+3\right)\left(x^2-2x\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x+3=0\\x^2-2x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-3\\x\left(x-2\right)=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-3\\x=0\\x-2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-3\\x=0\\x=2\end{matrix}\right.\)

Vậy \(x\in\left\{-3;0;2\right\}\)

Theo đề ra, ta có:

\(x\in Z\Rightarrow2x+3;x-1\in Z\)

Mà: \(P\in Z\Rightarrow\dfrac{2x+3}{x-1}\in Z\Rightarrow2x+3⋮x-1\Rightarrow2x-2+5⋮x-1\Rightarrow2\left(x-1\right)+5⋮x-1\)

\(\Rightarrow5⋮x-1\Rightarrow x-1\inƯ\left(5\right)=\left\{\pm1;\pm5\right\}\Rightarrow x\in\left\{2;6;0;-4\right\}\)

\(\Rightarrow\dfrac{2x+3}{x-1}\in\left\{7;\pm3;1\right\}\Rightarrow P\in\left\{7;\pm3;1\right\}\)

Vậy giá trị lớn nhất của \(P=7\) khi \(x=2\)

\(x^2\left(x-3\right)^2-\left(x-3\right)^2-x^2+1\)

\(=\left(x^2-1\right)\left(x-3\right)^2-\left(x^2-1\right)\)

\(=\left(x^2-1\right)\left(x-3-1\right)\left(x-3+1\right)\)

\(=\left(x^2-1\right)\left(x-4\right)\left(x-2\right)\)

\(\left(x-2\right)^3-6x^2+12x-8-x^3+27+6x^2+12x+6=15\)

\(\Rightarrow x^3-6x^2+12x-8-x^3+27+6x^2+12x+6=15\)

\(\Rightarrow24x+25=15\)

\(\Rightarrow24x=-10\)

\(\Rightarrow x=\dfrac{-5}{12}\)

\(1.uncommon\)

\(2.donating\)

\(3.way\)

\(4.charitable\)

\(5.helping\)

\(6.disabled\)

\(7.how\)

\(8.feed\)

\(9.care\)

\(10.parents\)

Theo đề ra ta có: \(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}\) và \(4y-3x=12\)

\(\dfrac{y}{3}=\dfrac{4y}{12};\dfrac{3x}{6}\Rightarrow\dfrac{y}{3}=\dfrac{4y}{12}=\dfrac{z}{4}\) và \(4y-3z=12\)

Áp dụng tính chất của dãy tỉ số bằng nhau

\(\dfrac{4y-3x}{12-6}=\dfrac{12}{6}=2\)

\(\Rightarrow\dfrac{x}{2}=2\Rightarrow x=2.2=4\)

\(\Rightarrow\dfrac{y}{3}=2\Rightarrow y=3.2=6\)

\(\Rightarrow\dfrac{z}{4}=2\Rightarrow z=4.2=8\)

a) \(4x^3y^2-8x^2y+12xy^2=4xy.x^2y-4xy.2x+4xy.3y=4xy\left(x^2y-2x+3y\right)\)

b) \(3x^2-6xy-5x+10y=\left(3x^2-6xy\right)-\left(5x-10y\right)=3x\left(x-2y\right)-5\left(x-2y\right)=\left(x-2y\right)\left(3x-5\right)\)

c) \(x^2-49+4y^2-4xy=\left(x^2-4xy+4y^2\right)-49=\left(x-2y\right)^2-7^2=\left(x-2y-7\right)\left(x-2y+7\right)\)

d) \(x^2-6x-16=\left(x^2-8x\right)+\left(2x-16\right)=x\left(x-8\right)+2\left(x-8\right)=\left(x-8\right)\left(x+2\right)\)

\(10x^2-60x=10x.x-6.10x=10x\left(x-6\right)\)

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