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b)x^3 - 6x^2 +11x-6=0
<=>x^3 - x^2 - 5x^2 +5x + 6x - 6=0
<=>x^2(x - 1) - 5x(x - 1) +6(x - 1)=0
<=>(x-1).(x^2 - 5x + 6)=0
<=>(x - 1).(x^2 - 2x - 3x + 6)=0
<=>(x - 1).[(x(x-2)-3(x-2)]=0
<=>(x-1)(x-2)(x-3)=0
<=>x-1=0hoac x-2=0 hoac x-3=0
<=>x=1hoac x=2 hoac x=3
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a) \(-10x^2-17x+6\)
\(=-10x^2-20x+3x+6\)
\(=-10x\left(x+2\right)+3\left(x+2\right)\)
\(=\left(x+2\right)\left(3-10x\right)\)
b)\(x^2-10x+9\)
\(=x^2-x-9x+9\)
\(=x\left(x-1\right)-9\left(x-1\right)\)
\(=\left(x-1\right)\left(x-9\right)\)
c) \(x^2-10x+24\)
\(=x^2-4x-6x+24\)
\(=x\left(x-4\right)-6\left(x-4\right)\)
\(=\left(x-4\right)\left(x-6\right)\)
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\(x^2-10x+24=0\\ \Leftrightarrow x^2-10x+25-1=0\\ \Leftrightarrow\left(x-5\right)^2-1=0\\ \Leftrightarrow\left(x-6\right)\left(x-4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=6\\x=4\end{matrix}\right.\\ VậyS=\left\{6;4\right\}\)
\(x^2-10x+24=0\)
\(\Leftrightarrow x^2-10x+25-1=0\)
\(\Leftrightarrow\left(x-5\right)^2-1=0\)
\(\Leftrightarrow\left(x-5-1\right)\left(x-5+1\right)=0\)
\(\Leftrightarrow\left(x-6\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=6\\x=4\end{matrix}\right.\)
Vậy: \(S\in\left\{6;4\right\}\)
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a) Ta có: \(\dfrac{x^2-10x-29}{1971}+\dfrac{x^2-10x-27}{1973}=\dfrac{x^2-10x-1971}{29}+\dfrac{x^2-10x-1973}{27}\)
\(\Leftrightarrow\dfrac{x^2-10x-29}{1971}-1+\dfrac{x^2-10x-27}{1973}-1=\dfrac{x^2-10x-1971}{29}-1+\dfrac{x^2-10x-1973}{27}-1\)
\(\Leftrightarrow\dfrac{x^2-10x-2000}{1971}+\dfrac{x^2-10x-2000}{1973}=\dfrac{x^2-10x-1971}{29}+\dfrac{x^2-10x-1973}{27}\)
\(\Leftrightarrow\dfrac{x^2-10x-2000}{1971}+\dfrac{x^2-10x-2000}{1973}-\dfrac{x^2-10x-1971}{29}-\dfrac{x^2-10x-1973}{27}=0\)
\(\Leftrightarrow\left(x^2-10x-2000\right)\left(\dfrac{1}{1971}+\dfrac{1}{1973}-\dfrac{1}{29}-\dfrac{1}{27}\right)=0\)
mà \(\dfrac{1}{1971}+\dfrac{1}{1973}-\dfrac{1}{29}-\dfrac{1}{27}\ne0\)
nên \(x^2-10x-2000=0\)
\(\Leftrightarrow x^2+40x-50x-2000=0\)
\(\Leftrightarrow x\left(x+40\right)-50\left(x+40\right)=0\)
\(\Leftrightarrow\left(x+40\right)\left(x-50\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+40=0\\x-50=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-40\\x=50\end{matrix}\right.\)
Vậy: S={-40;50}
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đặt t= x2 +10x+20 ta có phương trình <=> (t-4)(t+4)+16=0
<=> t2-16+16=0
<=> t=0
<=>x2 +10x+20=0
giải nghiệm ra ta được x=-5+ căn 5 và x= -5- căn 5
mấy bạn chi mk ý kiến nha .thanks
\(\left(x^2+10x\right)\left(x^2+10x+24\right)+128\)
\(=x^4+10x^3+24x^2+10x^3+100x^2+240x+128\)
\(=x^4+20x^3+124x^2+240x+128\)