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`S=1+2+3+4+5+...+100`
\(=\left(1+100\right)\times100:2\\ =101\times50=5050\)
\(\left(2x-1\right)^4=16=\left(\pm2\right)^4\\ =>\left[{}\begin{matrix}2x-1=2\\2x-1=-2\end{matrix}\right.\\ =>\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{1}{2}\end{matrix}\right.\)
\(\left(2x+1\right)^3=125=5^3\\ =>2x+1=1\\ =>x=2\)
\(\left(b\right):\left(x-7\right)\left(x-5\right)+\left(1-x^2\right)=8\\ =>x^2-7x-5x+35+1-x^2=8\\ =>-12x+36=8\\ =>-12x=-28\\ =>x=\dfrac{28}{12}=\dfrac{7}{3}\)
\(\left(c\right):\left(\dfrac{1}{2}x-1\right)\left(3-2x\right)-x\left(x-5\right)=2\\ =>\dfrac{3}{2}x-3-x^2+2x-x^2+5x=2\\ =>-2x^2+\dfrac{17}{2}x-3=2\\ =>-2x^2+\dfrac{17}{2}x-5=0\\ =>4x^2-17x+10=0\\ =>\left(2x-\dfrac{17}{4}\right)^2=\dfrac{129}{16}\\=>\left[{}\begin{matrix}2x-\dfrac{17}{4}=\dfrac{\sqrt{129}}{4}\\2x-\dfrac{17}{4}=-\dfrac{\sqrt{129}}{4}\end{matrix}\right.\)
\(=>x=\dfrac{17\pm\sqrt{129}}{8}\)
\(\left(a\right):2x-7\sqrt{x}+3=0\left(x\ge0\right)\\ < =>\left(2x-6\sqrt{x}\right)-\left(\sqrt{x}-3\right)=0\\ < =>2\sqrt{x}\left(\sqrt{x}-3\right)-\left(\sqrt{x}-3\right)=0\\ < =>\left(2\sqrt{x}-1\right)\left(\sqrt{x}-3\right)=0\\ =>\left[{}\begin{matrix}2\sqrt{x}-1=0\\\sqrt{x}-3=0\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=\dfrac{1}{4}\left(TM\right)\\x=9\left(TM\right)\end{matrix}\right.\)
\(\left(b\right):3\sqrt{x}+5< 6\\ < =>3\sqrt{x}< 1\\ < =>\sqrt{x}< \dfrac{1}{3}\\ < =>0\le x< \dfrac{1}{9}\)
\(\left(c\right):x-3\sqrt{x}-10< 0\\ < =>\left(x-5\sqrt{x}\right)+\left(2\sqrt{x}-10\right)< 0\\ < =>\sqrt{x}\left(\sqrt{x}-5\right)+2\left(\sqrt{x}-5\right)< 0\\ < =>\left(\sqrt{x}-5\right)\left(\sqrt{x}+2\right)< 0\\ =>\left\{{}\begin{matrix}\sqrt{x}-5< 0\\\sqrt{x}+2>0\end{matrix}\right.\\ < =>\left\{{}\begin{matrix}0\le x< 25\\x\ge0\end{matrix}\right.< =>0\le x< 25\)
\(\left(d\right):x-5\sqrt{x}+6=0\left(x\ge0\right)\\ < =>\left(x-2\sqrt{x}\right)-\left(3\sqrt{x}-6\right)=0\\ < =>\sqrt{x}\left(\sqrt{x}-2\right)-3\left(\sqrt{x}-2\right)=0\\ < =>\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)=0\\ =>\left[{}\begin{matrix}\sqrt{x}-3=0\\\sqrt{x}-2=0\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=9\\x=4\end{matrix}\right.\left(TM\right)\)
\(\left(e\right):x+5\sqrt{x}-14< 0\\ < =>\left(x+7\sqrt{x}\right)-\left(2\sqrt{x}+14\right)< 0\\ < =>\sqrt{x}\left(\sqrt{x}+7\right)-2\left(\sqrt{x}+7\right)< 0\\ < =>\left(\sqrt{x}-2\right)\left(\sqrt{x}+7\right)< 0\\ =>\left\{{}\begin{matrix}\sqrt{x}+7>0\\\sqrt{x}-2< 0\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x\ge0\\0\le x< 4\end{matrix}\right.< =>0\le x< 4\)
Với `k=0` :
\(x=2.0.\left(0+2\right)=0\left(TM\right)\)
Với k = 1 :
\(x=2.1.\left(1+2\right)=6\left(TM\right)\)
Tương tự với `k=2,3`
\(=>Q=\left\{0;6;16;30\right\}\)
\(\left\{{}\begin{matrix}\\\end{matrix}\right.\) Này là dấu và nên phải thỏa mãn cả 2 đk trên và dưới => x = -3 vừa thỏa mãn trên vừa thỏa mãn dưới nha bạn
Cho \(K\left(x\right)=0\)
\(=>\left(x+3\right)^2+\left(x^2-9\right)^2=0\\ =>\left\{{}\begin{matrix}x+3=0\\x^2-9=0\end{matrix}\right.\\ =>\left\{{}\begin{matrix}x=-3\\x^2=9\end{matrix}\right.\\ =>\left\{{}\begin{matrix}x=-3\\x=\pm3\end{matrix}\right.=>x=-3\)
Vậy `x=-3` là nghiệm đa thức
Chiều rộng HCN là :
\(63:3=21\left(m\right)\)
Chu vi HCN là :
\(\left(63+21\right)\times2=168\left(m\right)\)
Vậy chu vi HV cũng là `168m`
Số cọc cần đóng là :
\(168:2+1=85\) (cọc)
Sửa đề : `P(x)=x^{4}+3x^{2}+4033`
Ta thấy : `x^{4},3x^{2}\ge0` với mọi `x`
`=>x^{4}+3x^{2}\ge0`
`=>P(x)=x^{4}+3x^{2}+4033\ge 4033>0`
Vậy `P(x)` vô nghiệm ( Do không có giá trị x thỏa mãn để `P(x)=0` )
\(\dfrac{5}{4}-\dfrac{1}{2}.x=\dfrac{2}{3}\\ =>\dfrac{x}{2}=\dfrac{5}{4}-\dfrac{2}{3}\\ =>\dfrac{x}{2}=\dfrac{7}{12}\\ =>x=\dfrac{7}{12}.2=\dfrac{7}{6}\)