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hihi

thik nhất cái ảnh cuổi tuy k đẹp>.< nhưng chả hỉu s t vẫn thik haha

<3

số chia : 97 

số bị chia :315

\(a,\left(x-2\right)\left(x^2+2x+4\right)-\left(x-1\right)^3+7\)

\(=x^3-8-x^3+3x^2-3x+1+7\)

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

\(b,x\left(x+2\right)\left(2-x\right)+\left(x+3\right)\left(x^2-3x+9\right)\)

\(=x\left(4-x^2\right)+x^3+27\)

\(=4x-x^3+x^3+27=4x-27\)

\(1,5x^2-4\left(x^2-2x+1\right)+20=0\)

\(\Leftrightarrow5x^2-4x^2+8x-4+20=0\)

\(\Leftrightarrow x^2+8x+16=0\)

\(\Leftrightarrow\left(x+4\right)^2=0\)

\(\Rightarrow x+4=0\Rightarrow x=-4\)

\(2,x\left(x-2\right)-5x+10=0\)

\(\Leftrightarrow x\left(x-2\right)-5\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-5\right)\left(x-2\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x-2=0\\x-5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\\x=5\end{matrix}\right.\)

\(a,\dfrac{x}{2}+\dfrac{x}{3}+\dfrac{x}{4}+....+\dfrac{x}{1000}=0\)

\(\Leftrightarrow x\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+...+\dfrac{1}{1000}\right)=0\)

Vì \(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+...+\dfrac{1}{1000}\ne0\Rightarrow x=0\)

Vậy pt có tập nghiệm \(S=\left\{0\right\}\)

\(b,3x+2+5=3x+7\)

\(\Leftrightarrow3x-3x=7-2-5\)

\(\Leftrightarrow0=0\)

Vậy x bằng mọi giá trị

\(a,-x^4\left(yx\right)^2\left(-x\right)^2\left(-y\right)^3=x^8y^5\)

\(\dfrac{1}{2}ax^3\left(-xy\right)\left(-y\right)^2=\dfrac{1}{2}ax^4y^2\)

\(-\dfrac{4}{5}y\left(\dfrac{3}{2}x^2y\right)^4=-\dfrac{81}{20}x^8y^5\)

\(3x^2+x=6\)

\(\Leftrightarrow3x^2+x-6=0\)

\(\Leftrightarrow3\left(x^2+\dfrac{1}{3}x+\dfrac{1}{36}\right)-\dfrac{73}{12}=0\)

\(\Leftrightarrow3\left(x+\dfrac{1}{6}\right)^3=\dfrac{73}{12}\)

\(\Rightarrow\left(x+\dfrac{1}{6}\right)^2=\dfrac{73}{36}\)

\(\Rightarrow\left[{}\begin{matrix}x+\dfrac{1}{6}=\sqrt{\dfrac{73}{36}}\\x+\dfrac{1}{6}=-\sqrt{\dfrac{73}{36}}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\sqrt{\dfrac{73}{36}}-\dfrac{1}{6}\\x=-\sqrt{\dfrac{73}{36}}-\dfrac{1}{6}\end{matrix}\right.\)

\(\left(x^2+3x+1\right)\left(x^2+3x+2\right)-6\)

Đặt \(x^2+3x+1=t\) ta được:

\(t\left(t+1\right)-6\)

\(=t^2+t-6\)

\(=t^2-2t+3t-6\)

\(=t\left(t-2\right)+3\left(t-2\right)\)

\(=\left(t+3\right)\left(t-2\right)\)

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

\(=\left(x^2+3x+4\right)\left(x^2+3x-1\right)\)