Giải phương trình: 2 x - 50 = 0
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Cao Minh Tâm
22 tháng 11 2019
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Những câu hỏi liên quan
N
2
16 tháng 2 2021
Ta có: \(\dfrac{x+1}{99}+\dfrac{x+2}{98}+...+\dfrac{x+50}{50}+50=0\)
\(\Leftrightarrow\dfrac{x+1}{99}+1+\dfrac{x+2}{98}+1+...+\dfrac{x+50}{50}+1=0\)
\(\Leftrightarrow\dfrac{x+100}{99}+\dfrac{x+100}{98}+...+\dfrac{x+100}{50}=0\)
\(\Leftrightarrow\left(x+100\right)\left(\dfrac{1}{99}+\dfrac{1}{98}+...+\dfrac{1}{50}\right)=0\)
mà \(\dfrac{1}{99}+\dfrac{1}{98}+...+\dfrac{1}{50}>0\)
nên x+100=0
hay x=-100
Vậy: S={-100}
16 tháng 2 2021
\(\dfrac{x+1}{99}+\dfrac{x+2}{98}+...+\dfrac{x+50}{50}+50=0\)
\(\Leftrightarrow\left(\dfrac{x+1}{99}+1\right)+\left(\dfrac{x+2}{98}+1\right)+\left(\dfrac{x+3}{97}+1\right)+...+\left(\dfrac{x+50}{50}+1\right)=0\)
\(\Leftrightarrow\dfrac{x+100}{99}+\dfrac{x+100}{98}+...+\dfrac{x+100}{50}=0\)
\(\Leftrightarrow\left(x+100\right).\left(\dfrac{1}{99}+\dfrac{1}{98}+\dfrac{1}{97}+...+\dfrac{1}{50}\right)=0\)
\(\Leftrightarrow x+100=0\) (vì \(\dfrac{1}{99}+\dfrac{1}{98}+\dfrac{1}{97}+...+\dfrac{1}{50}>0\) )
\(\Leftrightarrow x=-100\)
L
1
17 tháng 10 2021
Đk: \(x\ne5;x\ne-10\)
Pt: \(\Rightarrow\dfrac{\left(x-2\right)\left(x+5\right)}{x^2}-\dfrac{40}{\left(x-5\right)\left(x+10\right)}=0\)
\(\Rightarrow\left(x-2\right)\left(x+5\right)\left(x-5\right)\left(x+10\right)-40x^2=0\)
\(\Rightarrow\left(x^2-12x+20\right)\left(x^2-25\right)-40x^2=0\)
\(\Rightarrow x^4-12x^3-45x^2+300x=500\)
\(\Rightarrow\left\{{}\begin{matrix}x=5\left(loại\right)\\x=-5\left(tm\right)\end{matrix}\right.\)
ND
1
MT
19 tháng 12 2015
2) 2x4-21x3+74x2-105x+50=0
<=>(2x4-2x3)+(-19x3+19x2)+(55x2-55x)+(-50x+50)=0
<=>2x3.(x-1)-19x2.(x-1)+55x.(x-1)-50.(x-1)=0
<=>(x-1)(2x3-19x2+55x-50)=0
<=>(x-1)[(2x3-20x2+50x)+(x2+5x-50)]=0
<=>(x-1)[2x.(x-5)2+(x2-5x+10x-50)]=0
<=>(x-1){2x.(x-5)2+[x.(x-5)+10.(x-5)]}=0
<=>(x-1)[2x.(x-5)2+(x-5)(x+10)]=0
<=>(x-1)(x-5)(2x2-10x+x+10)=0
<=>(x-1)(x-5)(2x2-5x-4x+10)=0
<=>(x-1)(x-5)[x.(2x-5)-2.(2x-5)]=0
<=>(x-1)(x-5)(x-2)(2x-5)=0
<=>x=1 hoặc x=5 hoặc x=2 hoặc x=5/2
9 tháng 2 2021
Ta có : \(\dfrac{x-50}{50}+\dfrac{x-51}{49}+\dfrac{x-52}{49}+\dfrac{x-53}{47}+\dfrac{x-200}{25}=0\)
\(\Leftrightarrow\dfrac{x-50}{50}-1+\dfrac{x-51}{49}-1+\dfrac{x-52}{49}-1+\dfrac{x-53}{47}-1+\dfrac{x-200}{25}+4=0\)
\(\Leftrightarrow\dfrac{x-100}{50}+\dfrac{x-100}{49}+\dfrac{x-100}{49}+\dfrac{x-100}{47}+\dfrac{x-100}{25}=0\)
\(\Leftrightarrow\left(x-100\right)\left(\dfrac{1}{50}+\dfrac{1}{49}+\dfrac{1}{48}+\dfrac{1}{47}+\dfrac{1}{25}\right)=0\)
<=> x - 100 = 0
<=> x = 100
Vậy ..
9 tháng 2 2021
Ta có: \(\dfrac{x-50}{50}+\dfrac{x-51}{49}+\dfrac{x-52}{48}+\dfrac{x-53}{47}+\dfrac{x-200}{25}=0\)
\(\Leftrightarrow\dfrac{x-50}{50}-1+\dfrac{x-51}{49}-1+\dfrac{x-52}{48}-1+\dfrac{x-53}{47}-1+\dfrac{x-200}{25}+4=0\)
\(\Leftrightarrow\dfrac{x-100}{50}+\dfrac{x-100}{49}+\dfrac{x-100}{48}+\dfrac{x-100}{47}+\dfrac{x-100}{25}=0\)
\(\Leftrightarrow\left(x-100\right)\left(\dfrac{1}{50}+\dfrac{1}{49}+\dfrac{1}{48}+\dfrac{1}{47}+\dfrac{1}{25}\right)=0\)
mà \(\dfrac{1}{50}+\dfrac{1}{49}+\dfrac{1}{48}+\dfrac{1}{47}+\dfrac{1}{25}>0\)
nên x-100=0
hay x=100
Vậy: S={100}
16 tháng 8 2023
\(\dfrac{1}{5}\sqrt[]{25x+50}-5\sqrt[]{x+2}+\sqrt[]{9x+18}+9=0\)
\(\Leftrightarrow\dfrac{1}{5}\sqrt[]{25\left(x+2\right)}-5\sqrt[]{x+2}+\sqrt[]{9\left(x+2\right)}+9=0\)
\(\Leftrightarrow\dfrac{1}{5}.5\sqrt[]{x+2}-5\sqrt[]{x+2}+3\sqrt[]{x+2}+9=0\)
\(\Leftrightarrow\sqrt[]{x+2}-5\sqrt[]{x+2}+3\sqrt[]{x+2}+9=0\)
\(\Leftrightarrow\sqrt[]{x+2}\left(1-5+3\right)+9=0\)
\(\Leftrightarrow-\sqrt[]{x+2}+9=0\)
\(\Leftrightarrow\sqrt[]{x+2}=9\)
\(\Leftrightarrow x+2=81\)
\(\Leftrightarrow x=79\)
11 tháng 3 2021
\(PT\Leftrightarrow\left(\dfrac{x-70}{130}-1\right)+\left(\dfrac{x-25}{175}-1\right)+\left(\dfrac{x-50}{150}-1\right)+\left(\dfrac{x-275}{25}+3\right)=0\)
\(\Leftrightarrow\left(x-200\right)\left(\dfrac{1}{130}+\dfrac{1}{175}+\dfrac{1}{150}+\dfrac{1}{25}\right)=0\Leftrightarrow x=200\).
Vậy...
VL
2
4 tháng 3 2021
a) \(4x-9=0\) \(\Leftrightarrow4x=9\) \(\Leftrightarrow x=\dfrac{9}{4}\)
Vậy \(x=\dfrac{9}{4}\)
b) \(-2x+50=0\) \(\Leftrightarrow2x=50\) \(\Leftrightarrow x=25\)
Vậy \(x=25\)
c) \(3x+11=0\) \(\Leftrightarrow3x=-11\) \(\Leftrightarrow x=-\dfrac{11}{3}\)
Vậy \(x=-\dfrac{11}{3}\)
HP
3 tháng 1 2021
1.
\(\sqrt{50}-3\sqrt{8}+\sqrt{32}=5\sqrt{2}-6\sqrt{2}+4\sqrt{2}=3\sqrt{2}\)
2.
a, ĐK: \(x\in R\)
\(pt\Leftrightarrow\sqrt{\left(x-2\right)^2}=1\)
\(\Leftrightarrow\left|x-2\right|=1\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=1\\x-2=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=1\end{matrix}\right.\)
b, ĐK: \(x\ge3\)
\(pt\Leftrightarrow\sqrt{x-3}\left(\sqrt{x}-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-3}=0\\\sqrt{x}-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\left(tm\right)\\x=1\left(l\right)\end{matrix}\right.\)
VL
1
