\(\left(x-3\right)^6=\left(x-3\right)^7\)
Help me guys,plssssssssss
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LD
11 tháng 6 2019
\(P=\frac{\left(x+\frac{1}{x}\right)^6-\left(x^6+\frac{1}{x^6}\right)-2}{\left(x+\frac{1}{x}\right)^3-\left(x^3+\frac{1}{x^3}\right)}\)
\(=\frac{\left(x+\frac{1}{x}\right)^6-\left[\left(x^3\right)^2+2x^3\cdot\frac{1}{x^3}+\left(\frac{1}{x^3}\right)^2\right]}{\left(x+\frac{1}{x}\right)^3-\left(x^3+\frac{1}{x^3}\right)}\)
\(=\frac{\left(x+\frac{1}{x}\right)^6-\left(x^3+\frac{1}{x^3}\right)^2}{\left(x+\frac{1}{x}\right)^3-\left(x^3+\frac{1}{x^3}\right)}\)
\(=\frac{\left[\left(x+\frac{1}{x}\right)^3-\left(x^3+\frac{1}{x^3}\right)\right]\left[\left(x+\frac{1}{x}\right)^3+\left(x^3+\frac{1}{x^3}\right)\right]}{\left(x+\frac{1}{x}\right)^3-\left(x^3+\frac{1}{x^3}\right)}\)
\(=\left(x+\frac{1}{x}\right)^3+\left(x^3+\frac{1}{x^3}\right)\ge\left(2\sqrt{x\cdot\frac{1}{x}}\right)^3+2\sqrt{x^3\cdot\frac{1}{x^3}}=8+2=10\)
Dấu "=" khi x = 1
NM
2
24 tháng 8 2023
\(\left(x-\dfrac{1}{5}\right):\left(x-1\dfrac{6}{7}\right)< 0\)
\(\Rightarrow\left(x-\dfrac{1}{5}\right):\left(x-\dfrac{13}{7}\right)< 0\)
\(TH1:\left\{{}\begin{matrix}x-\dfrac{1}{5}>0\\x-\dfrac{13}{7}< 0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x>\dfrac{1}{5}\\x< \dfrac{13}{7}\end{matrix}\right.\) \(\Leftrightarrow\dfrac{1}{5}< x< \dfrac{13}{7}\)
\(TH2:\left\{{}\begin{matrix}x-\dfrac{1}{5}< 0\\x-\dfrac{13}{7}>0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x< \dfrac{1}{5}\\x>\dfrac{13}{7}\end{matrix}\right.\) (vô lý nên loại)
Vậy \(\dfrac{1}{5}< x< \dfrac{13}{7}\) thỏa mãn đề bài
9 tháng 11 2017
Ta có: \(\left(8x^2-2x+7\right)\left(4x-6x^2-3\right)=\left(6x^2+3x+4\right)\left(9x-8x^2-6\right)\)
\(\Rightarrow\left(8x^2-2x+7\right)\left(4x-6x^2-3\right)-\left(6x^2+3x+4\right)\left(9x-8x^2-6\right)=0\)
\(\Rightarrow14x^3-33x^2+16x+3=0\) (Rút gọn vế đầu)
\(\Rightarrow14x^2\left(x-1\right)-19x\left(x-1\right)-3\left(x-1\right)=0\)
\(\Rightarrow\left(14x^2-19x-3\right)\left(x-1\right)=0\)
\(\Rightarrow\left[7x\left(2x-3\right)+\left(2x-3\right)\right]\left(x-1\right)=0\)
\(\Rightarrow\left(7x+1\right)\left(2x-3\right)\left(x-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{7}\\x=1\\x=\dfrac{3}{2}\end{matrix}\right.\).
Vậy \(x\in\left\{-\dfrac{1}{7};1;\dfrac{3}{2}\right\}.\)
25 tháng 7 2022
a: =>4x-6-9=5-3x-3
=>4x-15=-3x+2
=>7x=17
hay x=17/7
b: \(\Leftrightarrow\dfrac{2}{3x}-\dfrac{1}{4}=\dfrac{4}{5}-\dfrac{7}{x}+2\)
=>2/3x+21/3x=4/5+2+1/4=61/20
=>23/3x=61/20
=>3x=23:61/20=460/61
hay x=460/183
please
13 tháng 11 2017
\(A=\dfrac{1}{\left(x+1\right)\left(x+3\right)}+\dfrac{1}{\left(x+3\right)\left(x+5\right)}+\dfrac{1}{\left(x+5\right)\left(x+7\right)}+\dfrac{1}{\left(x+7\right)\left(x+9\right)}+\dfrac{1}{\left(x+9\right)\left(x+11\right)}\)\(A=\dfrac{1}{2}\left(\dfrac{1}{x+1}-\dfrac{1}{x+3}+\dfrac{1}{x+3}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+7}+\dfrac{1}{x+7}-\dfrac{1}{x+9}+\dfrac{1}{x+9}-\dfrac{1}{x+11}\right)\)
\(A=\dfrac{1}{2}\left(\dfrac{1}{x+1}-\dfrac{1}{x+11}\right)\)
\(A=\dfrac{1}{2}\left(\dfrac{x+11}{\left(x+1\right)\left(x+11\right)}-\dfrac{x+1}{\left(x+1\right)\left(x+11\right)}\right)\)
\(A=\dfrac{1}{2}\left(\dfrac{x+11-x-1}{\left(x+1\right)\left(x+11\right)}\right)=\dfrac{1}{2}.\dfrac{10}{\left(x+1\right)\left(x+11\right)}=\dfrac{10}{2\left(x+1\right)\left(x+11\right)}\)
28 tháng 3 2020
a) ĐKXĐ: x khác +2
\(\frac{x-2}{2+x}-\frac{3}{x-2}-\frac{2\left(x-11\right)}{x^2-4}\)
<=> \(\frac{x-2}{2+x}-\frac{3}{x-2}=\frac{2\left(x-11\right)}{\left(x-2\right)\left(x+2\right)}\)
<=> (x - 2)^2 - 3(2 + x) = 2(x - 11)
<=> x^2 - 4x + 4 - 6 - 3x = 2x - 22
<=> x^2 - 7x - 2 = 2x - 22
<=> x^2 - 7x - 2 - 2x + 22 = 0
<=> x^2 - 9x + 20 = 0
<=> (x - 4)(x - 5) = 0
<=> x - 4 = 0 hoặc x - 5 = 0
<=> x = 4 hoặc x = 5
làm nốt đi
NM
1
26 tháng 9 2023
`#3107`
`(x - 3)^5 = 4(x - 3)^3`
`=> (x - 3)^5 - 4(x - 3)^3 = 0`
`=> (x - 3)^3 * [ (x - 3)^2 - 4] = 0`
`=>`\(\left[{}\begin{matrix}\left(x-3\right)^3=0\\\left(x-3\right)^2-4=0\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x-3=0\\\left(x-3\right)^2=4\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=3\\\left(x-3\right)^2=\left(\pm2\right)^2\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=3\\x-3=2\\x-3=-2\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=3\\x=5\\x=1\end{matrix}\right.\)
Vậy, `x \in {1; 3; 5}.`
22 tháng 4 2021
|x|+|x+2|=3
=>x={3;-3} và x+2=3 hoặc x+2=-3
=>x={3;-3} và x=3-2=1 hoặc x=-3-2=-5
Vậy x={3;-3;1;-5}
CHẮC MÌNH KHÔNG ĐÚNG ĐÂU NÊN MONG BẠN THÔNG CẢM NHA
22 tháng 4 2021
Vì |x| luôn > hoặc = 0 với mọi x
| x+2l luôn > hoặc = 0 với mọi x
=>|x| + | x+2l luôn > hoặc = 0 với mọi x
=>x + x+2=3
<=> 2x + 2 = 3
<=> 2x = 1
<=> x= 1/2
Trả lời:
\(\left(x-3\right)^6=\left(x-3\right)^7\)
\(\Leftrightarrow\left(x-3\right)^6-\left(x-3\right)^7=0\)
\(\Leftrightarrow\left(x-3\right)^6\left(1-x+3\right)=0\)
\(\Leftrightarrow\left(x-3\right)^6\left(4-x\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}\left(x-3\right)^6=0\\4-x=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x-3=0\\4-x=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=3\\x=4\end{cases}}}\)
Vậy x = 3; x = 4
( x -3 )^6 = ( x -3 )^7
x = 3 , 4
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