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1,
a,\(2x\left(3x^2-5x+3\right)\)
\(=6x^3-10x^2+6x\)
b,\(-2x\left(x^2+5x-3\right)\)
\(=-2x^3-10x^2+6x\)
c,\(-\dfrac{1}{2}x\left(2x^3-4x+3\right)\)
\(=-x^4+2x^2-\dfrac{3}{2}x\)
Bài 2:
a) \(\left(2x-1\right)\left(x^2-5-4\right)\)
\(=\left(2x-1\right)\left(x^2-9\right)\)
\(=2x^3-18x-x^2+9\)
b) \(-\left(5x-4\right)\left(2x+3\right)\)
\(=-\left(10x^2+15x-8x-12\right)\)
\(=-10x^2-7x+12\)
c) \(\left(2x-y\right)\left(4x^2-2xy+y^2\right)\)
\(=8x^3-y^3\)
a) \(\left(3x-2\right)^2-\left(3x-5\right)\left(3x+2\right)=11\)
\(\Leftrightarrow\left(9x^2-12x+4\right)-\left(9x^2+6x-15x-10\right)=11\)
\(\Leftrightarrow9x^2-12x+4-9x^2-6x+15x+10=11\)
\(\Leftrightarrow-3x+3=0\)
\(\Leftrightarrow-3x=-3\)
\(\Leftrightarrow x=1\)
Vậy \(S=\left\{1\right\}\)
b) \(\left(4x-3\right)^2-\left(4x-5\right)\left(4x+5\right)=32\)
\(\Leftrightarrow\left(16x^2-24x+9\right)-\left(16x^2-25\right)=32\)
\(\Leftrightarrow16x^2-24x+9-16x^2+25=32\)
\(\Leftrightarrow-24x+2=0\)
\(\Leftrightarrow-24x=-2\)
\(\Leftrightarrow x=\dfrac{1}{12}\)
Vậy \(S=\left\{\dfrac{1}{12}\right\}\)
c) \(\left(5x-2\right)^2-\left(5x+3\right)\left(5x-5\right)=1\)
\(\Leftrightarrow\left(25x^2-20x+4\right)-\left(25x^2-25x+15x-15\right)=1\)
\(\Leftrightarrow25x^2-20x+4-25x^2+25x-15x+15=1\)
\(\Leftrightarrow-10x+18=0\)
\(\Leftrightarrow-10x=-18\)
\(\Leftrightarrow x=\dfrac{9}{5}\)
Vậy \(S=\left\{\dfrac{9}{5}\right\}\)
d) \(\left(x-4\right)^2-\left(x-7\right)\left(2x-3\right)=5-x^2\)
\(\Leftrightarrow\left(x^2-8x+16\right)-\left(2x^2-3x-14x+21\right)=5-x^2\)
\(\Leftrightarrow x^2-8x+16-2x^2+3x+14x-21=5-x^2\)
\(\Leftrightarrow x^2-8x+16-2x^2+3x+14x-21-5+x^2=0\)
\(\Leftrightarrow9x-10=0\)
\(\Leftrightarrow9x=10\)
\(\Leftrightarrow x=\dfrac{10}{9}\)
Vậy \(S=\left\{\dfrac{10}{9}\right\}\)
Cho mk hỏi vs ! Câu a bn rút gọn hay bn lm kiểu j mak tự nhiên 11 lại lôi đâu ra số 0 vậy ? Gt hộ mk vs, mk vẫn chưa hiểu cách bn lm ở câu a cho lắm !
a)(x - 4)2 - 25= 0
<--> ( x - 4)2 - 52 = 0
<--> ( x - 4 - 5 )( x - 4 + 5 ) = 0
<--> ( x - 4 - 5 ) = 0 <--> x - 9 = 0 <--> x = 9
hoặc
<--> ( x - 4 + 5 ) = 0 <--> x + 1 = 0 <--> x = -1
b)bài này tương tự bài a
\(a,\left(x-4\right)^2-25=0\)
\(\Rightarrow\left(x-4-5\right)\left(x-4+5\right)=0\)
\(\Rightarrow\left(x-9\right)\left(x+1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-9=0\\x+1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=9\\x=-1\end{matrix}\right.\)
\(b,\left(x-3\right)^2-\left(x+1\right)^2=0\)
\(\Rightarrow\left(x-3-x-1\right)\left(x-3+x+1\right)=0\)
\(\Rightarrow-4\left(2x-2\right)=0\)
\(\Rightarrow2\left(x-1\right)=0\)
\(\Rightarrow x-1=0\)
\(\Rightarrow x=1\)
\(c,\left(x^2-4\right)\left(2x+3\right)=\left(x^2-4\right)\left(x-1\right)\)
\(\Rightarrow2x+3=x-1\)
\(\Rightarrow2x-x=-1-3\)
\(\Rightarrow x=-4\)
\(d,\left(3x-7\right)^2-4\left(x+1\right)^2=0\)
\(\Rightarrow\left(3x-7\right)-\left[2\left(x+1\right)\right]^2=0\)
\(\Rightarrow\left(3x-7\right)^2-\left(2x+2\right)^2=0\)
\(\Rightarrow\left(3x-7-2x-2\right)\left(3x-7+2x+2\right)=0\)
\(\Rightarrow\left(x-9\right)\left(5x-5\right)=0\)
\(\Rightarrow5\left(x-9\right)\left(x-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-9=0\\x-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=9\\x=1\end{matrix}\right.\)
a: \(=6x^2-9x+14x-21-4x^2+20x-25-2x\left(x+6\right)+5-31x\)
\(=2x^2-6x-41-2x^2-12x\)
=-18x-41
b: \(=2x^2-6x-2x^2+6x+14=14\)
c: \(=x^3+1-x^3+1=2\)
\(\left(x-3\right)^2-\left(x^2-3x\right)=0\)
\(\left(x-3\right).\left(x-3\right)-x.\left(x-3\right)=0\)
\(\left(x-3\right).\left(x-3-x\right)=0\)
\(\left(x-3\right).3=0\)
\(x-3=0=>x=3\)
b) \(\left(x+2\right)\left(x+3\right)-\left(x-2\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x+3\right)+\left(x+2\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x+3+x+5\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(2x+8\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+2=0\\2x+8=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-2\\x=-4\end{matrix}\right.\)