M = x - x mũ 2 phần 1 - x mũ 2 + 1 phần 1 +x . 3x - 1 phần x + 1 - 1
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13 tháng 4 2023
\(M=\left(\dfrac{x-x^2}{\left(x-1\right)^2}+\dfrac{1}{1+x}-\dfrac{x}{x-1}\right)\cdot\left(\dfrac{3x-1}{x}+\dfrac{1}{x+1}-1\right)\)
\(=\left(\dfrac{-x}{x-1}-\dfrac{x}{x-1}+\dfrac{1}{x+1}\right)\cdot\dfrac{\left(3x-1\right)\left(x+1\right)+x-x\left(x+1\right)}{x\left(x+1\right)}\)
\(=\dfrac{-2x\left(x+1\right)+x-1}{\left(x-1\right)\left(x+1\right)}\cdot\dfrac{3x^2+2x-1+x-x^2-x}{x\left(x+1\right)}\)
\(=\dfrac{-2x^2-2x+x-1}{\left(x-1\right)\left(x+1\right)}\cdot\dfrac{2x^2+2x-1}{x\left(x+1\right)}\)
\(=\dfrac{-2x^2-x-1}{\left(x-1\right)\left(x+1\right)}\cdot\dfrac{2x^2+2x-1}{x\left(x+1\right)}\)
\(=\dfrac{\left(-2x^2-x-1\right)\left(2x^2+2x-1\right)}{x\left(x+1\right)^2\cdot\left(x-1\right)}\)
MD
14 tháng 4 2023
\(\left(\dfrac{3x-1}{x+1}-1\right)\)bạn sửa lại đề bào thế này
22 tháng 10 2019
1. \(6x^3-8=40\\ 6x^3=48\\ x^3=8\\ \Rightarrow x=2\)Vậy x = 2
2. \(4x^5+15=47\\ 4x^5=32\\ x^5=8\\ \Rightarrow x\in\varnothing\left(\text{vì }x\in N\right)\)Vậy x ∈ ∅
3. \(2x^3-4=12\\ 2x^3=16\\ x^3=8\\ \Rightarrow x=2\)Vậy x = 2
4. \(5x^3-5=0\\ 5x^3=5\\ x^3=1\\ \Rightarrow x=1\)Vậy x = 1
5. \(\left(x-5\right)^{2016}=\left(x-5\right)^{2018}\\ \Rightarrow\left[{}\begin{matrix}x-5=0\\x-5=1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=5\\x=6\end{matrix}\right.\)Vậy \(x\in\left\{5;6\right\}\)
6. \(\left(3x-2\right)^{20}=\left(3x-1\right)^{20}\\ \Rightarrow3x-2=3x-1\\ 3x-3x=2-1\\ 0=1\left(\text{vô lí}\right)\)Vậy x ∈ ∅
7. \(\left(3x-1\right)^{10}=\left(3x-1\right)^{20}\\ \left(3x-1\right)^{10}=\left[\left(3x-1\right)^2\right]^{10}\\ \Rightarrow\left(3x-1\right)^2=3x-1\\ \left(3x-1\right)^2-\left(3x-1\right)=0\\ \left(3x-1\right)\left[\left(3x-1\right)-1\right]=0\\ \left(3x-1\right)\left(3x-2\right)=0\\ \Rightarrow\left[{}\begin{matrix}3x-1=0\\3x-2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}3x=1\\3x=2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{1}{3}\left(\text{loại vì }x\in N\right)\\x=\frac{2}{3}\left(\text{loại vì }x\in N\right)\end{matrix}\right.\)Vậy x ∈ ∅
8. \(\left(2x-1\right)^{50}=2x-1\\ \left(2x-1\right)^{50}-\left(2x-1\right)=0\\ \left(2x-1\right)\left[\left(2x-1\right)^{49}-1\right]=0\\ \Rightarrow\left[{}\begin{matrix}2x-1=0\\\left(2x-1\right)^{49}=1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=1\\2x-1=1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\frac{1}{2}\left(\text{loại vì }x\in N\right)\\x=1\left(t/m\right)\end{matrix}\right.\)Vậy x = 1
9. \(\left(\frac{x}{3}-5\right)^{2000}=\left(\frac{x}{3}-5\right)^{2008}\\ \left(\frac{x}{3}-5\right)^{2008}-\left(\frac{x}{3}-5\right)^{2000}=0\\ \left(\frac{x}{3}-5\right)^{2000}\left[\left(\frac{x}{3}-5\right)^8-1\right]=0\\ \Rightarrow\left[{}\begin{matrix}\left(\frac{x}{3}-5\right)^{2000}=0\\\left(\frac{x}{3}-5\right)^8=1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}\frac{x}{3}-5=0\\\frac{x}{3}-5=1\\\frac{x}{3}-5=-1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}\frac{x}{3}=5\\\frac{x}{3}=6\\\frac{x}{3}=4\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=5\cdot3=15\\x=6\cdot3=18\\x=4\cdot3=12\end{matrix}\right.\)Vậy \(x\in\left\{15;18;12\right\}\)
22 tháng 10 2019
\(1.6x^3-8=40\\ \Leftrightarrow6x^3=48\\ \Leftrightarrow x^3=8\Leftrightarrow x^3=2^3=\left(-2\right)^3\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)
Vậy \(x\in\left\{2;-2\right\}\)
\(2.4x^3+15=47\) (T nghĩ đề là mũ 3)
\(\Leftrightarrow4x^3=32\Leftrightarrow x^3=8=2^3=\left(-2\right)^3\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)
Vậy \(x\in\left\{2;-2\right\}\)
Câu 3, 4 tương tự nhé.
5 tháng 8 2021
,(3x-1) mũ 2=9/16
<=> (3x-1)^2 = ( ±3/4)^2
<=> l3x-1l = 3/4
Hoặc 3x-1 = 3/4
<=> 3x= 3/4 + 1
<=> x = 7/4 : 3
<=> x= 7/1
XO
23 tháng 7 2021
a) \(x^2-\frac{1}{49}=0\)
<=> \(\left(x-\frac{1}{7}\right)\left(x+\frac{1}{7}\right)=0\)
<=> \(\orbr{\begin{cases}x-\frac{1}{7}=0\\x+\frac{1}{7}=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{7}\\x=-\frac{1}{7}\end{cases}}\)
Vậy x = \(\pm\frac{1}{7}\)
b) \(64-\frac{1}{4}x^2=0\)
<=> \(\left(8-\frac{1}{2}x\right)\left(8+\frac{1}{2}x\right)=0\)
<=> \(\orbr{\begin{cases}8-\frac{1}{2}x=0\\8+\frac{1}{2}x=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=16\\x=-16\end{cases}}\)
Vậy \(x=\pm16\)
c) 9x2 + 12x + 4 = 0
<=> (3x + 2)2 = 0
<=> 3x + 2 = 0
<=> x = -2/3
Vậy x = -2/3
e) \(x^2+\frac{1}{4}=x\)
<=> \(x^2-x+\frac{1}{4}=0\)
<=> \(\left(x-\frac{1}{2}\right)^2=0\)
<=> \(x=\frac{1}{2}\)
Vậy \(x=\frac{1}{2}\)
23 tháng 7 2021
d, sửa đề : \(x^2+4=4x\Leftrightarrow x^2-4x+4=0\Leftrightarrow\left(x-2\right)^2=0\Leftrightarrow x=2\)
i, \(4-\frac{12}{x}+\frac{9}{x^2}=0\)ĐK : \(x\ne0\)
Vì \(x\ne0\)Nhân 2 vế với \(x^2\)phương trình có dạng
\(4x^2-12x+9=0\Leftrightarrow\left(2x-3\right)^2=0\Leftrightarrow x=\frac{3}{2}\)
25 tháng 2 2020
c/C=\(\frac{2x^2+2x}{1-x}-\frac{x}{x-1}=\frac{2x^2+2x+x}{1-x}=\frac{2x^2+3x}{1-x}\)
d/C thuộc Z thì C=\(\frac{\left(2x^2-2x\right)+\left(5x-5\right)+5}{1-x}=\frac{-2x\left(1-x\right)-5\left(1-x\right)+5}{1-x}=-2x-5+\frac{5}{1-x}\Rightarrow1-x\in\left(+-1,+-5\right)\Rightarrow\left\{{}\begin{matrix}x=0\\x=2\\x=-4\\x=6\end{matrix}\right.\)
25 tháng 2 2020
a/A đã rút gọn B=\(\frac{1-2x}{x^2-3x+2}+\frac{x+1}{x-2}=\frac{1-2x}{\left(x-1\right)\left(x-2\right)}+\frac{\left(x+1\right)\left(x-1\right)}{\left(x-1\right)\left(x-2\right)}=\frac{1-2x+x^2-1}{\left(x-1\right)\left(x-2\right)}=\frac{x\left(x-2\right)}{\left(x-1\right)\left(x-2\right)}=\frac{x}{x-1}\)b/\(\left|x-2\right|=3\Rightarrow\left\{{}\begin{matrix}x-2=3\\x-2=-3\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=5\\x=-1\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}B=\frac{2.5^2+2.5}{1-5}=-15\\B=\frac{2.\left(-1\right)^2+2\left(-1\right)}{1-\left(-1\right)}=0\end{matrix}\right.\)
\(M=\dfrac{x-x^2}{1-x^2}+\dfrac{1}{1+x}\cdot\dfrac{3x-1}{x+1}-1\)
\(=\dfrac{x\left(1-x\right)}{\left(1-x\right)\left(1+x\right)}+\dfrac{3x-1}{\left(x+1\right)^2}-1\)
\(=\dfrac{x^2+x+3x-1-x^2-2x-1}{\left(x+1\right)^2}\)
\(=\dfrac{2x-2}{\left(x+1\right)^2}\)
đề yêu cầu gì vậy bạn