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HB
1
3 tháng 9 2018
\(\frac{149-x}{25}+\frac{170-x}{23}+\frac{187-x}{21}+\frac{200-x}{19}=10\)
\(\Rightarrow\frac{149-x}{25}-1+\frac{170-x}{23}-2+\frac{187-x}{21}-3+\frac{200-x}{19}-4=0\)
\(\Rightarrow\frac{124-x}{25}+\frac{124-x}{23}+\frac{124-x}{21}+\frac{124-x}{19}=0\)
\(\Rightarrow\left(124-x\right)\left(\frac{1}{25}+\frac{1}{23}+\frac{1}{21}+\frac{1}{19}\right)=0\)
Mà \(\frac{1}{25}+\frac{1}{23}+\frac{1}{21}+\frac{1}{19}>0\Rightarrow x-124=0\Rightarrow x=124\)
H
2
20 tháng 4 2018
Ta có : \(\frac{149-x}{25}+\frac{170-x}{23}+\frac{187-x}{21}+\frac{200-x}{19}=10\)
\(\Leftrightarrow\frac{149-x}{25}-1+\frac{170-x}{23}-2+\frac{187-x}{21}-3+\frac{200-x}{19}-4=0\)
\(\Leftrightarrow\frac{124-x}{25}+\frac{124-x}{23}+\frac{124-x}{21}+\frac{124-x}{19}=0\)
\(\Leftrightarrow\left(124-x\right)\left(\frac{1}{25}+\frac{1}{23}+\frac{1}{21}+\frac{1}{19}\right)=0\)
Vì \(\frac{1}{25}+\frac{1}{23}+\frac{1}{21}+\frac{1}{19}\ne0\)
Nên : 124 - x = 0
<=> x = 124
Vậy x = 124
16 tháng 8 2022
Bài 2:
a: \(\Leftrightarrow\left(x-5\right)\left(x+5\right)-\left(x+5\right)=0\)
=>(x+5)(x-6)=0
=>x=-5 hoặc x=6
b: \(\Leftrightarrow4x^2-4x+1-4x^2+1=0\)
=>-4x+2=0
hay x=1/2
c: \(\Leftrightarrow\left(x^2+4\right)\left(x^2-1\right)=0\)
=>x=1 hoặc x=-1
10 tháng 8 2018
\(\left(9^{30}-27^{19}\right):3^{57}+\left(125^9-25^{12}\right):5^{24}\)
\(=\left(3^{60}-3^{57}\right):3^{57}+\left(5^{27}-5^{24}\right):5^{24}\)
\(=3^{57}\left(3^3-1\right):3^{57}+5^{24}\left(5^3-1\right):5^{24}\)
\(=3^3-1+5^3-1\)
\(=27-1+125-1\)
\(=150\)
2 )
\(x^2-25-\left(x+5\right)=0\)
\(\Leftrightarrow\left(x+5\right)\left(x-5\right)-\left(x+5\right)=0\)
\(\Leftrightarrow\left(x+5\right)\left(x-5-1\right)=0\)
\(\Leftrightarrow\left(x+5\right)\left(x-6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+5=0\\x-6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=6\end{matrix}\right.\)
Vậy ...
b )
\(\left(2x-1\right)^2-\left(4x^2-1\right)=0\)
\(\Leftrightarrow4x^2-4x+1-4x^2+1=0\)
\(\Leftrightarrow2-4x=0\)
\(\Leftrightarrow4x=2\)
\(\Leftrightarrow x=\dfrac{1}{2}\)
Vậy ...
c )
\(x^2\left(x^2+4\right)-x^2-4=0\)
\(\Leftrightarrow x^2\left(x^2+4\right)-\left(4+x^2\right)=0\)
\(\Leftrightarrow\left(x^2-1\right)\left(x^2+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-1=0\\x^2+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x^2=1\\x^2=-4\left(L\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)
Vậy ...
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
10 tháng 7 2023
\(x^3\) + 125 + (\(x\) + 5)(\(x\) - 25) = 0
(\(x^3\) + 53) + (\(x\) + 5)(\(x\) - 25) = 0
(\(x\) + 5)(\(x^2\) - 5\(x\) + 25) + (\(x\) + 5)(\(x\) - 25) =0
(\(x\) + 5)(\(x^2\) - 5\(x\) + 25 + \(x\) - 25) = 0
(\(x\) + 5)(\(x^2\) - 4\(x\)) = 0
\(x\)(\(x\) + 5)(\(x\) - 4) = 0
\(\left[{}\begin{matrix}x=0\\x+5=0\\x-4=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=0\\x=-5\\x=4\end{matrix}\right.\)
20 tháng 8 2021
1, \(2x^3-50x=0\Leftrightarrow2x\left(x^2-25\right)=0\Leftrightarrow x=0;x=\pm5\)
2, \(5x^2-4\left(x^2-2x+1\right)-5=0\)
\(\Leftrightarrow5\left(x-1\right)\left(x+1\right)-4\left(x-1\right)^2=0\)
\(\Leftrightarrow\left(x-1\right)\left[5\left(x+1\right)-4\left(x-1\right)\right]=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+9\right)=0\Leftrightarrow x=-9;x=1\)
3, \(6x\left(x-2\right)=x-2\Leftrightarrow\left(6x-1\right)\left(x-2\right)=0\Leftrightarrow x=\frac{1}{6};x=2\)
4, \(7\left(x-2020\right)^2-x+2020=0\Leftrightarrow7\left(x-2020\right)^2-\left(x-2020\right)=0\)
\(\Leftrightarrow\left(x-2020\right)\left[7\left(x-2020\right)-1\right]=0\Leftrightarrow x=2020;x=\frac{14141}{7}\)
5, \(x^2-10x=-25\Leftrightarrow x^2-10x+25=0\Leftrightarrow\left(x-5\right)^2=0\Leftrightarrow x=5\)
6, \(x^2-2x-3=0\Leftrightarrow\left(x+1\right)\left(x-3\right)=0\Leftrightarrow x=-1;x=3\)
20 tháng 8 2021
\(1,\)
\(2x^3-50x=0\)
\(\Leftrightarrow2x\left(x^2-25\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x^2-25=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=\pm5\end{cases}}\)
\(2,\)
\(5x^2-4\left(x^2-2x+1\right)-5=0\)
\(\Leftrightarrow5x^2-4x^2+8x-4-5=0\)
\(\Leftrightarrow x^2+8x-9=0\)
\(\Leftrightarrow x^2-x+9x-9=0\)
\(\Leftrightarrow x\left(x-1\right)+9\left(x-1\right)=0\)
\(\Leftrightarrow\left(x+9\right)\left(x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+9=0\\x-1=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=-9\\x=1\end{cases}}\)
\(3,\)
\(6x\left(x-2\right)=x-2\)
\(\Leftrightarrow6x\left(x-2\right)-\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(6x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=2\\x=\frac{1}{6}\end{cases}}\)
\(4,\)
\(7\left(x-2020\right)^2-x+2020=0\)
\(\Leftrightarrow7\left(x-2020\right)^2-\left(x-2020\right)=0\)
\(\Leftrightarrow\left(x-2020\right)[7\left(x-2020\right)-1]=0\)
\(\Leftrightarrow\left(x-2020\right)[7x-14141]=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=2020\\7x=14141\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=2020\\x=\frac{14141}{7}\end{cases}}\)
\(5,\)
\(x^2-10x=-25\)
\(\Leftrightarrow x^2-10x+25=0\)
\(\Leftrightarrow\left(x-5\right)^2=0\)
\(\Leftrightarrow x-5=0\)
\(\Leftrightarrow x=5\)
\(6,\)
\(x^2-2x-3=0\)
\(\Leftrightarrow x^2-3x+x-3=0\)
\(\Leftrightarrow x\left(x-3\right)+\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-3=0\\x+1=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=3\\x=-1\end{cases}}\)
21 tháng 9 2021
\(\dfrac{37\cdot5^4}{25^2}=\dfrac{37\cdot5^4}{5^4}=37\\ \dfrac{2^4\cdot2^6\cdot3^8\cdot9^2}{4^4\cdot3^{11}}=\dfrac{2^{10}\cdot3^8\cdot3^4}{2^8\cdot3^{11}}=2^2\cdot3=12\\ \dfrac{3\cdot9^4\cdot9^3}{3^2\cdot9}=\dfrac{3\cdot3^8\cdot3^6}{3^2\cdot3^2}=3^{11}\\ \dfrac{125\cdot5\cdot64-25^3\cdot10\cdot4}{5^7\cdot8}=\dfrac{5^3\cdot5\cdot2^6-5^6\cdot2\cdot5\cdot2^2}{5^7\cdot2^3}=\dfrac{5^4\cdot2^3\left(2^3-5^3\right)}{5^7\cdot2^3}=\dfrac{8-125}{5^3}=\dfrac{-117}{125}\)
\(x-\frac{10}{19}\)\(+\)\(x-\frac{125}{5}\)\(+\)\(x-\frac{200}{2020}\)\(=\)\(25\) \(\Leftrightarrow\)\(3x-\frac{10}{19}\)\(-\) \(25\) \(-\frac{10}{101}\) \(=\) \(25\) \(\Leftrightarrow\)\(3x-25,62532569\)\(=\) \(25\) \(\Leftrightarrow\) \(3x=50,62532569\) \(\Leftrightarrow\)\(x=\left(50,62532569\right):3\) \(\Leftrightarrow\) \(x=16,87510856\)
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