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19 tháng 9 2021
\(1,\\ a,=-35x^5y^4z\\ b,=6x^2-30x-6x^2-3x=-33x\\ c,=x^3-9x^2-2x^2+18x-x+9=x^3-11x^2+17x+9\\ 2,\\ A\left(x\right)+B\left(x\right)=10-2x+4x^3-5x^2-10x^3-5x+6x^2-20\\ =-6x^3+x^2-7x-10\\ A\left(x\right)-B\left(x\right)=10-2x+4x^3-5x^2+10x^3+5x-6x^2+20\\ =14x^3-11x^2+3x+30\\ 3,\\ a,M\left(x\right)=5x+20=0\\ \Leftrightarrow x=-4\\ b,N\left(x\right)=100x^2-49=0\\ \Leftrightarrow\left(10x-7\right)\left(10x+7\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{10}\\x=-\dfrac{7}{10}\end{matrix}\right.\\ c,P\left(x\right)=3x-15=0\\ \Leftrightarrow x=5\)
18 tháng 9 2021
Bài 1;
a)\(5x^3yz.\left(-7x^2y^3\right)=-35.x^5y^4z\)
b)\(6x\left(x-5\right)-x\left(6x+3\right)=6x^2-30x-6x^2-3x=-33x\)
c) \(\left(x-9\right)\left(x^2-2x-1\right)=x^3-2x^2-x-9x^2+18x+9=x^3-11x^2+17x+9\)
3 tháng 1 2022
\(=\dfrac{6x}{\left(x-3\right)\left(x+3\right)}-\dfrac{9+5x}{x-3}+\dfrac{x}{x+3}\)
\(=\dfrac{6x-\left(5x+9\right)\left(x+3\right)+x^2-3x}{\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{x^2+3x-5x^2-15x-9x-27}{\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{-4x^2-21x-27}{\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{-\left(4x^2+12x+9x+27\right)}{\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{-4x-9}{x-3}\)
28 tháng 7 2018
\(x\left(x-3\right)+x-3=0\)
\(\left(x-3\right)\left(x+1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-3=0\\x+1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=3\\x=-1\end{cases}}}\)
KL:......................
\(x^3-5x=0\)
\(x\left(x^2-5\right)=0\)
Làm tương tự như câu a
@_@ n...h..i......ề....u q...u.....................á!
28 tháng 12 2019
a) \(\frac{4x+3}{6x-4}+\frac{5x-9}{6x-4}\)
\(=\frac{4x+3+5x-9}{2\left(3x-2\right)}=\frac{9x-6}{2\left(3x-2\right)}\)
\(=\frac{3\left(3x-2\right)}{2\left(3x-2\right)}=\frac{3}{2}\)
b) \(\frac{2}{x-1}+\frac{3}{x+1}-\frac{4x-2}{x^2-1}\)
\(=\frac{2\left(x+1\right)+3\left(x-1\right)-4x+2}{\left(x-1\right)\left(x+1\right)}\)
\(=\frac{x+1}{\left(x-1\right)\left(x+1\right)}=\frac{1}{x-1}\)
VM
28 tháng 12 2019
a) \(\frac{4x+3}{6x-4}+\frac{5x-9}{6x-4}\)
\(=\frac{4x+3+5x-9}{6x-4}\)
\(=\frac{9x-6}{6x-4}\)
\(=\frac{3.\left(3x-2\right)}{2.\left(3x-2\right)}\)
\(=\frac{3}{2}.\)
b) \(\frac{2}{x-1}+\frac{3}{x+1}-\frac{4x-2}{x^2-1}\)
\(=\frac{2}{x-1}+\frac{3}{x+1}-\frac{4x-2}{\left(x-1\right).\left(x+1\right)}\)
\(=\frac{2.\left(x+1\right)}{\left(x-1\right).\left(x+1\right)}+\frac{3.\left(x-1\right)}{\left(x-1\right).\left(x+1\right)}-\frac{4x-2}{\left(x-1\right).\left(x+1\right)}\)
\(=\frac{2x+2}{\left(x-1\right).\left(x+1\right)}+\frac{3x-3}{\left(x-1\right).\left(x+1\right)}+\frac{-\left(4x-2\right)}{\left(x-1\right).\left(x+1\right)}\)
\(=\frac{2x+2+3x-3-4x+2}{\left(x-1\right).\left(x+1\right)}\)
\(=\frac{x+1}{\left(x-1\right).\left(x+1\right)}\)
\(=\frac{1}{x-1}.\)
Chúc bạn học tốt!
NV
Nguyễn Việt Lâm
Giáo viên
5 tháng 4 2019
Để ý rằng tất cả các biểu thức 2 vế của 4 bài đều không âm, cho nên ta bình phương 2 vế:
a/
\(\left(x^2-x+7\right)^2=\left(-5x+1\right)^2\)
\(\Leftrightarrow\left(x^2-x+7\right)^2-\left(-5x+1\right)^2=0\)
\(\Leftrightarrow\left(x^2-6x+8\right)\left(x^2+4x+6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-6x+8=0\\x^2+4x+6=0\left(vn\right)\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=2\\x=4\end{matrix}\right.\)
b/
\(\left(x^2+9\right)^2=\left(-6x+1\right)^2\)
\(\Leftrightarrow\left(x^2+9\right)^2-\left(-6x+1\right)^2=0\)
\(\Leftrightarrow\left(x^2-6x+10\right)\left(x^2+6x+8\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-6x+10=0\left(vn\right)\\x^2+6x+8=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=-2\\x=-4\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
6 tháng 4 2019
c/
\(\left(x^2+5x+7\right)^2-\left(3x+5\right)^2=0\)
\(\Leftrightarrow\left(x^2+2x+2\right)\left(x^2+8x+12\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+2x+2=0\left(vn\right)\\x^2+8x+12=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=-2\\x=-6\end{matrix}\right.\)
d/
\(\left(x^2+6x+9\right)^2-\left(2x+3\right)^2=0\)
\(\Leftrightarrow\left(x^2+4x+6\right)\left(x^2+8x+12\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+4x+6=0\left(vn\right)\\x^2+8x+12=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=-2\\x=-6\end{matrix}\right.\)
12 tháng 3 2020
\(\frac{6}{x^2-9}+\frac{5x}{x-3}+\frac{x}{x+3}\)
\(=\frac{6x}{\left(x-3\right)\left(x+3\right)}+\frac{5x}{x-3}+\frac{x}{x+3}\)
\(=\frac{6x}{\left(x-3\right)\left(x+3\right)}-\frac{5x\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}+\frac{x\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}\)
\(=\frac{6x+5x\left(x+3\right)+x\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}\)
\(=\frac{6x\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=\frac{6x}{x-3}\)
12 tháng 3 2020
\(\frac{6x}{x^2-9}+\frac{5x}{x-3}+\frac{x}{x+3}\left(x\ne\pm3\right)\)
\(=\frac{6x}{\left(x-3\right)\left(x+3\right)}+\frac{5x\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}+\frac{x\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}\)
\(=\frac{6x+5x^2+15x+x^2-3x}{\left(x-3\right)\left(x+3\right)}\)
\(=\frac{6x^2+18x}{\left(x-3\right)\left(x+3\right)}=\frac{6x\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=\frac{6x}{x-3}\)
\(5x\left(x-3\right)-x^2+6x-9\\ =5x\left(x-3\right)-\left(x^2-6x+9\right)\\ =5x\left(x-3\right)-\left(x-3\right)^2\\ =\left(x-3\right)\left(5x-x+3\right)\\ =\left(x-3\right)\left(4x+3\right)\)