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a) \(22-x\left(1-4x\right)=\left(2x+3\right)^3\)
\(\Leftrightarrow22-x+4x^2=8x^3+36x^2+54x+27\)
\(\Leftrightarrow-x-54x+4x^2-36x^2-8x^3=-22+27\)
\(\Leftrightarrow-8x^3-32x^2-55x=5\Leftrightarrow-8x^3-32x^2-55x-5=0\)
Bn tự làm tiếp nhé
b) \(\frac{2x}{3}+\frac{2x-1}{6}=\frac{4-x}{3}\Leftrightarrow\frac{2.2x}{6}+\frac{2x-1}{6}=\frac{2\left(4-x\right)}{6}\)
\(\Leftrightarrow2.2x+2x-1=2\left(4-x\right)\Leftrightarrow4x+2x-1=8-2x\)
\(\Leftrightarrow6x-1=8-2x\Leftrightarrow8x=9\Leftrightarrow x=\frac{9}{8}\)
Vậy phương trình có tập nghiệm S ={9/8}
c) \(\frac{x-1}{2019}+\frac{x-2}{2018}=\frac{x-3}{2017}+\frac{x-4}{2016}\)
\(\Leftrightarrow\left(\frac{x-1}{2019}-1\right)+\left(\frac{x-2}{2018}-1\right)=\left(\frac{x-3}{2017}-1\right)+\left(\frac{x-4}{2016}-1\right)\)
\(\Leftrightarrow\frac{x-2020}{2019}+\frac{x-2020}{2018}-\frac{x-2020}{2017}-\frac{x-2020}{2016}=0\)
\(\Leftrightarrow\left(x-2020\right)\left(\frac{1}{2019}+\frac{1}{2018}+\frac{1}{2017}+\frac{1}{2016}\right)=0\)
Do \(\frac{1}{2019}+\frac{1}{2018}+\frac{1}{2017}+\frac{1}{2016}>0\)
Nên \(x-2020=0\Leftrightarrow x=2020\)
b. \(\left(2x+1\right)+\left(4x+3\right)+\left(6x+5\right)+...+\left(100x+99\right)=7600\)
\(\rightarrow\left(2x+4x+6x+...+100x\right)+\left(1+3+5+...+99\right)=7600\)
\(\rightarrow\frac{\left(2x+100x\right).50}{2}+\frac{\left(1+99\right).50}{2}=7600\)
\(\rightarrow51x.50+50.50=7600\)
\(\rightarrow51x.50+2500=7600\)
\(\rightarrow51x.50=7600-2500\)
\(\rightarrow51x.50=5100\)
\(\rightarrow50x=100\)
\(\rightarrow x=\frac{100}{50}=2\)
Vậy x = 2
câu 1:
x3-1+3x2-3x =(x-1)(x^2+x+1)+3x(x-1)=(x-1)(x^2+x+1+3x)=(x-1)(x^2+4x=1)
Câu 2 :
a) \(\left(x^4-2x^3+2x-1\right):\left(x^2-1\right)\)
\(=\left(x^4-x^2-2x^3+2x+x^2-1\right):\left(x^2-1\right)\)
\(=\left[x^2\left(x^2-1\right)-2x\left(x^2-1\right)+\left(x^2-1\right)\right]:\left(x^2-1\right)\)
\(=\left(x^2-1\right)\left(x^2-2x+1\right):\left(x^2-1\right)\)
\(=x^2-2x+1\)
b) \(\left(x^6-2x^5+2x^4+6x^3-4x^2\right):6x^2\)
\(=\frac{1}{6}x^4-\frac{1}{3}x^3+\frac{1}{3}x^2+x-\frac{2}{3}\)
Câu 3 :
Sửa đề :
\(\frac{3x^2+6x+12}{x^3-8}=\frac{3\left(x^2+2x+4\right)}{\left(x-2\right)\left(x^2+2x+4\right)}=\frac{3}{x-2}\)
a/
Nhận thấy ngay phương trình có 2 nghiệm \(\left[{}\begin{matrix}x=2019\\x=2018\end{matrix}\right.\)
- Với \(x>2019\Rightarrow\left\{{}\begin{matrix}x-2018>1\\x-2019>0\end{matrix}\right.\) \(\Rightarrow\left|x-2018\right|^{2019}+\left|x-2019\right|^{2018}>1\Rightarrow\) pt vô nghiệm
- Với \(x< 2018\Rightarrow\left\{{}\begin{matrix}x-2018< 0\\x-2019< -1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\left|x-2018\right|>0\\\left|x-2019\right|>1\end{matrix}\right.\)
\(\Rightarrow\left|x-2018\right|^{2019}+\left|x-2019\right|^{2018}>1\Rightarrow\) pt vô nghiệm
- Với \(2018< x< 2019\) viết lại pt:
\(\left|x-2018\right|^{2019}+\left|2019-x\right|^{2018}=1\)
Ta có: \(\left\{{}\begin{matrix}0< x-2018< 1\\0< 2019-x< 1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\left|x-2018\right|^{2019}< x-2018\\\left|2019-x\right|^{2018}< 2019-x\end{matrix}\right.\)
\(\Rightarrow\left|x-2018\right|^{2019}+\left|2019-x\right|^{2018}< x-2018+2019-x=1\)
\(\Rightarrow\) pt vô nghiệm
Vậy pt có đúng 2 nghiệm: \(\left[{}\begin{matrix}x=2018\\x=2019\end{matrix}\right.\)
b/
Thay \(x=0\) vào pt thấy không phải là nghiệm, chia cả tử và mẫu của các hạng tử vế trái cho x:
\(\frac{2}{x+\frac{1}{x}-1}-\frac{1}{x+\frac{1}{x}+1}=\frac{5}{3}\)
Đặt \(x+\frac{1}{x}=a\) phương trình trở thành:
\(\frac{2}{a-1}-\frac{1}{a+1}=\frac{5}{3}\)
\(\Leftrightarrow2\left(a+1\right)-\left(a-1\right)=\frac{5}{3}\left(a^2-1\right)\)
\(\Leftrightarrow5a^2-3a-14=0\) \(\Rightarrow\left[{}\begin{matrix}a=2\\a=-\frac{7}{5}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x+\frac{1}{x}=2\\x+\frac{1}{x}=-\frac{7}{5}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x^2-2x+1=0\\5x^2+7x+5=0\left(vn\right)\end{matrix}\right.\) \(\Rightarrow x=1\)
a) ĐKXĐ: \(x\ne\pm2\)
Ta có: \(\dfrac{x}{x+2}=\dfrac{x^2+4}{x^2-4}\)
\(\Leftrightarrow\dfrac{x}{x+2}=\dfrac{x^2+4}{\left(x+2\right)\left(x-2\right)}\)
\(\Leftrightarrow\dfrac{x\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}=\dfrac{x^2+4}{\left(x+2\right)\left(x-2\right)}\)
\(\Rightarrow x\left(x-2\right)=x^2+4\)
\(\Leftrightarrow x^2-2x=x^2+4\)
\(\Leftrightarrow-2x=4\Leftrightarrow x=-2\)(KTMĐK)
Vậy phương trình vô nghiệm
b) ĐKXĐ: \(x\ne3;x\ne-1\)
Ta có: \(\dfrac{x}{2x-6}+\dfrac{x}{2x+2}+\dfrac{2x}{\left(x+1\right)\left(3-x\right)}=0\)
\(\Leftrightarrow\dfrac{x}{2\left(x-3\right)}+\dfrac{x}{2\left(x+1\right)}-\dfrac{2x}{\left(x+1\right)\left(x-3\right)}=0\)
\(\Leftrightarrow\dfrac{x\left(x+1\right)}{2\left(x-3\right)\left(x+1\right)}+\dfrac{x\left(x-3\right)}{2\left(x+1\right)\left(x-3\right)}-\dfrac{2.2x}{2\left(x+1\right)\left(x-3\right)}=0\)
\(\Rightarrow x\left(x+1\right)+x\left(x-3\right)-2.2x=0\)
\(\Leftrightarrow x^2+x+x^2-3x-4x=0\)
\(\Leftrightarrow2x^2-6x=0\)
\(\Leftrightarrow2x\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(TMĐK\right)\\x=3\left(KTMĐK\right)\end{matrix}\right.\)
Vậy phương trình có nghiệm là \(x=0\)
\(B=\left(2x-1\right)^2+2.\left(2x-1\right)\left(2x-3\right)+\left(2x-3\right)^2+2019\)
\(=\left(2x-1+2x-3\right)^2+2019\)
\(=\left(4x-4\right)^2+2019\)
\(=\left(4.2018-4\right)^2+2019\)
\(=\left(8072-4\right)^2+2019\)
\(=8068^2+2019=65092624+2019=65094643\)