tìm x biết
(2x+4/3)(x2+2021)=0
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Những câu hỏi liên quan
7 tháng 11 2021
\(x\left(5-6x\right)+\left(2x-1\right)\left(3x+\text{4}\right)=6\\ \Leftrightarrow5x-6x^2+6x^2+8x-3x-4=6\)
\(\Leftrightarrow10x-4=6\)
\(\Leftrightarrow10x=6+4\\ \Leftrightarrow10x=10\\ \Leftrightarrow x=\dfrac{10}{10}\)
\(\Leftrightarrow x=1\)
\(x^2\left(x-2021\right)-x+2021=0\)
\(\Leftrightarrow x^2\left(x-2021\right)-(x-2021)=0\)
\(\Leftrightarrow\left(x-2021\right)\left(x^2-1\right)=0\)
\(\Leftrightarrow\left(x-2021\right)\left(x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2021=0\\x-1=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2021\\x=1\\x=-1\end{matrix}\right.\)
MH
16 tháng 9 2021
a) \(\left(x-1\right)^3\)
\(=x^3-3x^2+3x-1\)
b) \(\left(2x-3y\right)^3\)
\(=\left(2x\right)^3-3\left(2x\right)^23y+3.2x\left(3y\right)^3+\left(3y\right)^3\)
\(=8x^3-36x^2y+54xy^2-27y^3\)
16 tháng 9 2021
Bài 3:
a: Ta có: \(\left(x-2\right)^3-x^2\left(x-6\right)=5\)
\(\Leftrightarrow x^3-6x^2+12x-8-x^3+6x^2=5\)
\(\Leftrightarrow12x=13\)
hay \(x=\dfrac{13}{12}\)
b: Ta có: \(\left(x-1\right)\left(x^2+x+1\right)-x\left(x+2\right)\left(x-2\right)=4\)
\(\Leftrightarrow x^3-1-x^3+4x=4\)
\(\Leftrightarrow4x=5\)
hay \(x=\dfrac{5}{4}\)
31 tháng 12 2021
\(a,\left(3x+1\right)\left(3x-1\right)-\left(18x^3+5x^2-2x\right):2x\\ =\left(9x^2-1\right)-\left(9x^2+\dfrac{5}{2}x-1\right)\\ =9x^2-1-9x^2-\dfrac{5}{2}x+1=\dfrac{5}{2}x\)
\(b,3x\left(x-2021\right)-x+2021=0\\ \Rightarrow b,3x\left(x-2021\right)-\left(x-2021\right)=0\\ \Rightarrow\left(x-2021\right)\left(3x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=2021\\x=\dfrac{1}{3}\end{matrix}\right.\)
CM
11 tháng 9 2019
a) x = 1; x = - 1 3 b) x = 2.
c) x = 3; x = -2. d) x = -3; x = 0; x = 2.
30 tháng 10 2021
\(a,\Rightarrow x^2+4x+4+x^2-2x+1+x^2-9-3x^2=-8\\ \Rightarrow2x=-4\Rightarrow x=-2\\ b,\Rightarrow\left(x-2021\right)\left(2022x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=2021\\x=\dfrac{1}{2022}\end{matrix}\right.\\ c,\Rightarrow\left(x^2-9\right)-\left(x-3\right)\left(2x+7\right)=0\\ \Rightarrow\left(x-3\right)\left(x+3\right)-\left(x-3\right)\left(2x+7\right)=0\\ \Rightarrow\left(x-3\right)\left(x+3-2x-7\right)=0\\ \Rightarrow\left(x-3\right)\left(-4-2x\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
NA
2
4 tháng 10 2021
a) \(5\left(x+3\right)-2x\left(3+x\right)=0\\ \Leftrightarrow\left(x+3\right)\left(5-2x\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x+3=0\\5-2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=\dfrac{5}{2}\end{matrix}\right.\)
b) \(4x\left(x-2021\right)-x+2021=0\\ \Leftrightarrow4x\left(x-2021\right)-\left(x-2021\right)=0\\ \Leftrightarrow\left(4x-1\right)\left(x-2021\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}4x-1=0\\x-2021=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{4}\\x=2021\end{matrix}\right.\)
Bạn tự kết luận cả 2 câu giúp mình nhé.
4 tháng 10 2021
a: \(5\left(x+3\right)-2x\left(x+3\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(5-2x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=\dfrac{5}{2}\end{matrix}\right.\)
b: Ta có: \(4x\left(x-2021\right)-x+2021=0\)
\(\Leftrightarrow\left(x-2021\right)\left(4x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2021\\x=\dfrac{1}{4}\end{matrix}\right.\)
\(\left(2x+\frac{4}{3}\right).\left(x^2+2021\right)=0\)
\(\Rightarrow\orbr{\begin{cases}2x+\frac{4}{3}=0\\x^2+2021=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}2x=-\frac{4}{3}\\x^2=-2021\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=-\frac{4}{6}\\x∈Ø\left(\text{ do }x^2≥0∀x\right)\end{cases}}\)