\(2x.\left(x-\frac{1}{7}\right)=0\)
Những câu hỏi liên quan
Bài 1:Giải phương trình
a)10x^2-5xleft(2x+3right)15
b)3x-7-left(3-4xright)left(2x+1right)4xleft(2x-7right)
c)left(4x-5right)^2-left(7-2xright)4left(2x-4right)^2+6x
Bài 2:Giải phương trình
a)frac{3left(x-1right)}{2}+4frac{2x}{3}+frac{4-5x}{6}
b)frac{4-x}{7}-frac{1}{7}left(frac{7+3x}{9}+frac{5-2x}{2}right)4-frac{4x}{3}
c)frac{2}{9}left(2x-5right)-frac{5}{3}left[left(x-2right)-frac{7}{12}right]frac{3}{4}left(x-3right)
Bài 3:Giải phương trình
a)left(x-6right)left(2x-5right)left(3x+9right)0...
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Bài 1:Giải phương trình
a)\(10x^2-5x\left(2x+3\right)=15\)
b)\(3x-7-\left(3-4x\right)\left(2x+1\right)=4x\left(2x-7\right)\)
c)\(\left(4x-5\right)^2-\left(7-2x\right)=4\left(2x-4\right)^2+6x\)
Bài 2:Giải phương trình
a)\(\frac{3\left(x-1\right)}{2}+4=\frac{2x}{3}+\frac{4-5x}{6}\)
b)\(\frac{4-x}{7}-\frac{1}{7}\left(\frac{7+3x}{9}+\frac{5-2x}{2}\right)=4-\frac{4x}{3}\)
c)\(\frac{2}{9}\left(2x-5\right)-\frac{5}{3}\left[\left(x-2\right)-\frac{7}{12}\right]=\frac{3}{4}\left(x-3\right)\)
Bài 3:Giải phương trình
a)\(\left(x-6\right)\left(2x-5\right)\left(3x+9\right)=0\)
b)\(2x\left(x-3\right)+5\left(x-3\right)=0\)
c)\(\left(x^2-4\right)-\left(x-2\right)\left(3-2x\right)=0\)
Bài 4:Tìm m để phương trình sau có nghiệm bằng 7:\(\left(2m-5\right)x-2m^2+8=43\)
Bài 5:Giải phương trình
a)\(\left(2x-1\right)^2-\left(2x+1\right)^2=0\)
b)\(\frac{1}{27}\left(x-3\right)^3-\frac{1}{125}\left(x-5\right)^3=0\)
https://i.imgur.com/u6zkAVa.jpg
Bài 3:
a) \(\left(x-6\right).\left(2x-5\right).\left(3x+9\right)=0\)
\(\Leftrightarrow\left(x-6\right).\left(2x-5\right).3.\left(x+3\right)=0\)
Vì \(3\ne0.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-6=0\\2x-5=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\2x=5\\x=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=\frac{5}{2}\\x=-3\end{matrix}\right.\)
Vậy phương trình có tập hợp nghiệm là: \(S=\left\{6;\frac{5}{2};-3\right\}.\)
b) \(2x.\left(x-3\right)+5.\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right).\left(2x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\2x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\2x=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\frac{5}{2}\end{matrix}\right.\)
Vậy phương trình có tập hợp nghiệm là: \(S=\left\{3;-\frac{5}{2}\right\}.\)
c) \(\left(x^2-4\right)-\left(x-2\right).\left(3-2x\right)=0\)
\(\Leftrightarrow\left(x^2-2^2\right)-\left(x-2\right).\left(3-2x\right)=0\)
\(\Leftrightarrow\left(x-2\right).\left(x+2\right)-\left(x-2\right).\left(3-2x\right)=0\)
\(\Leftrightarrow\left(x-2\right).\left(x+2-3+2x\right)=0\)
\(\Leftrightarrow\left(x-2\right).\left(3x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\3x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\3x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\frac{1}{3}\end{matrix}\right.\)
Vậy phương trình có tập hợp nghiệm là: \(S=\left\{2;\frac{1}{3}\right\}.\)
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Bài 4 xem lại đề nhé bác
Xem thêm câu trả lời
tim x
a)\(2x\left(x-\frac{1}{7}\right)=0\)
b)\(\left(x-9\right).\left(x+\frac{3}{5}\right)=0\)
c)\(\left(\frac{-4}{7}-2x\right)\left(x-\frac{5}{4}\right)=0\)
a) \(2x\left(x-\frac{1}{7}\right)=0\)
\(x\left(x-\frac{1}{7}\right)=0\)
\(\Rightarrow2x-2.\frac{1}{7}=0\)
\(2x-\frac{2}{7}=0\)
=> \(2x=\frac{2}{7}\)
=> x=\(\frac{1}{7}\)
b) (x-9)(\(x+\frac{3}{5}\))=0
\(\Rightarrow\orbr{\begin{cases}x-9=0\\x+\frac{3}{5}=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=\frac{-3}{5}\end{cases}}\)
Vậy x=0 hoặc x=-3/5
c) \(\left(\frac{-4}{7}-2x\right)\left(x-\frac{5}{4}\right)=0\)
\(\Rightarrow\orbr{\begin{cases}\frac{-4}{7}-2x=0\\x-\frac{5}{4}=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{-2}{7}\\x=\frac{5}{4}\end{cases}}\)
Vậy x=-2/7 hoặc x=5/4
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a, => x.(x-1/7) = 0:2 = 0
=> x=0 hoặc x-1/7=0
=> x=0 hoặc x=1/7
Vậy x thuộc {0;1/7}
b, => x-9=0 hoặc x+3/5=0
=> x=9 hoặc x=-3/5
Vậy x thuộc {-3/5;9}
c, => -4/7-2x=0 hoặc x-5/4=0
=> x=-2/7 hoặc x=5/4
Vậy x thuộc {-2/7;5/4}
Tk mk nha
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3) frac{x-2}{x-5} -frac{5}{x^2-5x}frac{1}{x}
Leftrightarrow frac{x-2}{x-5}-frac{5}{x.left(x-5right)}frac{1}{x}
Leftrightarrowfrac{left(x-2right).left(x+5right)}{x.left(x-5right)}-frac{5}{x.left(x-5right)}frac{1.left(x+5right)}{x.left(x-5right)}
Leftrightarrow x^2+5x-2x-10-51x+5
Leftrightarrow x^2+5x-2x-1x-10-5-5 0
Leftrightarrow x^2+2x-200
Leftrightarrow x^2+2x-10x-200
Leftrightarrow (x^2 + 2x) - (10x + 20) 0
Leftrightarrow x.(x + 2) - 10.(x + 2) 0
Leftrightarrow
4) frac{x-4}{x+7}-f...
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3) \(\frac{x-2}{x-5}\) \(-\frac{5}{x^2-5x}=\frac{1}{x}\)
\(\Leftrightarrow\) \(\frac{x-2}{x-5}-\frac{5}{x.\left(x-5\right)}=\frac{1}{x}\)
\(\Leftrightarrow\frac{\left(x-2\right).\left(x+5\right)}{x.\left(x-5\right)}-\frac{5}{x.\left(x-5\right)}=\frac{1.\left(x+5\right)}{x.\left(x-5\right)}\)
\(\Leftrightarrow x^2+5x-2x-10-5=1x+5\)
\(\Leftrightarrow x^2+5x-2x-1x-10-5-5\) = 0
\(\Leftrightarrow\) \(x^2+2x-20=0\)
\(\Leftrightarrow x^2+2x-10x-20=0\)
\(\Leftrightarrow\) (x\(^2\) + 2x) - (10x + 20) = 0
\(\Leftrightarrow\) x.(x + 2) - 10.(x + 2) = 0
\(\Leftrightarrow\)
4) \(\frac{x-4}{x+7}-\frac{1}{x}=\frac{-7}{x^2+7x}\)
\(\Leftrightarrow\frac{x-4}{x+7}-\frac{1}{x}=\frac{-7}{x\left(x+7\right)}\)
\(\Leftrightarrow\frac{\left(x-4\right).\left(x+7\right)}{x.\left(x+7\right)}-\frac{1.\left(x+7\right)}{x.\left(x+7\right)}=\frac{-7}{x.\left(x+7\right)}\)
\(\Leftrightarrow\) \(x^2+7x-4x-28-x-7=-7\)
\(\Leftrightarrow x^2+7x-4x-x-28-7+7=0\)
\(\Leftrightarrow\) x\(^2\) + 2x - 28 = 0
\(\Leftrightarrow\) x\(^2\) + 2x - 14x - 28 = 0
\(\Leftrightarrow\) (x\(^2\) + 2x) - (14x + 28) = 0
\(\Leftrightarrow\) x.(x + 2) - 14.(x + 2) = 0
\(\Leftrightarrow\) (x - 14) = 0 hoặc (x + 2) = 0
\(\Leftrightarrow\) x = 4 (Nhận) hoặc x = -2 (Loại)
5) \(\frac{x+2}{x-2}+\frac{x-2}{x+2}=\frac{8x}{x^2-4}\)
\(\Leftrightarrow\) \(\frac{\left(x+2\right).\left(x+2\right)}{\left(x-2\right).\left(x+2\right)}+\frac{\left(x-2\right).\left(x-2\right)}{\left(x+2\right).\left(x-2\right)}=\frac{8x}{\left(x-2\right).\left(x+2\right)}\)
\(\Leftrightarrow x^2+2x+2x+4+x^2-2x-2x+4=8x\)
\(\Leftrightarrow\) \(x^2+x^2+2x+2x-2x-2x-8x+4+4=0\)
\(\Leftrightarrow2x^2-8x+8=0\)
\(\Leftrightarrow\) 2x\(^2\) - 2x - 8x + 8 = 0
\(\Leftrightarrow\) 2x(x - 1) - 8(x - 1) = 0
\(\Leftrightarrow\) 2x - 8 = 0 hoặc x - 1 = 0
\(\Leftrightarrow\) 2x = 8 hoặc x = 1
\(\Leftrightarrow\) x = 4 (Nhận) hoặc x = 1 (Nhận)
Vậy S = {4; 1}
6) \(\frac{x+1}{x-1}-\frac{x-1}{x+1}=\frac{4}{x^2-1}\)
\(\Leftrightarrow\) \(\frac{\left(x+1\right).\left(x+1\right)}{\left(x-1\right).\left(x+1\right)}-\frac{\left(x-1\right).\left(x-1\right)}{\left(x+1\right).\left(x-1\right)}=\frac{4}{\left(x-1\right).\left(x+1\right)}\)
\(\Leftrightarrow\) x\(^2\) + x + x + 1 - x\(^2\) + x + x - 1 = 4
\(\Leftrightarrow\) 4x - 4 = 0
\(\Leftrightarrow\) 4 (x - 1) =0
\(\Leftrightarrow\) x - 1 = 0 / 4 = 0
\(\Leftrightarrow\) x = 1 (Nhận)
Vậy S = {1}
7) \(\frac{x+1}{x-1}+\frac{-4x}{x^2-1}=\frac{x-1}{x+1}\)
\(\Leftrightarrow\) \(\frac{\left(x+1\right).\left(x+1\right)}{\left(x-1\right).\left(x+1\right)}+\frac{-4x}{\left(x-1\right).\left(x+1\right)}=\frac{\left(x-1\right).\left(x-1\right)}{\left(x+1\right).\left(x+1\right)}\)
\(\Leftrightarrow x^2+x+x+1-4x=x^2-x-x+1\)
\(\Leftrightarrow\) 0
Vậy S ={\(\varnothing\)}
\(\frac{-3x.\left(5x+3\right)}{1+3x}>=0\)\(\frac{-2x^2+5x-3}{-x.\left(3x+7\right)}>0\)\(\frac{1}{x-2}-\frac{4}{x^2-4}< \frac{1}{3}\)\(x^2-20x+51>0\)\(\left(x-3\right).\left(2x+1\right)\left(1-5x\right)< 0\)\(\left(x-2\right)\left(x+3\right)=< 0\)
Bài 1 : Tìm x biết :a, left|frac{3}{2}x+frac{1}{2}right|left|4x-1right|b, left|frac{5}{4}x-frac{7}{2}right|-left|frac{5}{8}x+frac{3}{5}right|0c,left|frac{7}{5}x+frac{2}{3}right|left|frac{4}{3}x-frac{1}{4}right|Bài 2 : Tìm x biết :a, | 2x - 5 | x +1 b, | 3x - 2 | -1 x c, | 3x - 7 | 2x + 1d, | 2x-1 | +1 x
Đọc tiếp
Bài 1 : Tìm x biết :
a, \(\left|\frac{3}{2}x+\frac{1}{2}\right|=\left|4x-1\right|\)
b, \(\left|\frac{5}{4}x-\frac{7}{2}\right|-\left|\frac{5}{8}x+\frac{3}{5}\right|=0\)
c,\(\left|\frac{7}{5}x+\frac{2}{3}\right|=\left|\frac{4}{3}x-\frac{1}{4}\right|\)
Bài 2 : Tìm x biết :
a, | 2x - 5 | = x +1
b, | 3x - 2 | -1 = x
c, | 3x - 7 | = 2x + 1
d, | 2x-1 | +1 = x
1a) \(\left|\frac{3}{2}x+\frac{1}{2}\right|=\left|4x-1\right|\)
=> \(\orbr{\begin{cases}\frac{3}{2}x+\frac{1}{2}=4x-1\\\frac{3}{2}x+\frac{1}{2}=1-4x\end{cases}}\)
=> \(\orbr{\begin{cases}-\frac{5}{2}x=-\frac{3}{2}\\\frac{11}{2}x=\frac{1}{2}\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{5}{3}\\x=\frac{1}{11}\end{cases}}\)
b) \(\left|\frac{5}{4}x-\frac{7}{2}\right|-\left|\frac{5}{8}x+\frac{3}{5}\right|=0\)
=>\(\left|\frac{5}{4}x-\frac{7}{2}\right|=\left|\frac{5}{8}x+\frac{3}{5}\right|\)
=> \(\orbr{\begin{cases}\frac{5}{4}x-\frac{7}{2}=\frac{5}{8}x+\frac{3}{5}\\\frac{5}{4}x-\frac{7}{2}=-\frac{5}{8}x-\frac{3}{5}\end{cases}}\)
=> \(\orbr{\begin{cases}\frac{5}{8}x=\frac{41}{10}\\\frac{15}{8}x=\frac{29}{10}\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{164}{25}\\x=\frac{116}{75}\end{cases}}\)
c) TT
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a, \(\left|\frac{3}{2}x+\frac{1}{2}\right|=\left|4x-1\right|\)
=> \(\orbr{\begin{cases}\frac{3}{2}x+\frac{1}{2}=4x-1\\-\frac{3}{2}x-\frac{1}{2}=4x-1\end{cases}}\)
=> \(\orbr{\begin{cases}\frac{3}{2}x+\frac{1}{2}-4x=-1\\-\frac{3}{2}x-\frac{1}{2}-4x=-1\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{3}{5}\\x=\frac{1}{11}\end{cases}}\)
\(b,\left|\frac{5}{4}x-\frac{7}{2}\right|-\left|\frac{5}{8}x+\frac{3}{5}\right|=0\)
=> \(\left|\frac{5}{4}x-\frac{7}{2}\right|-0=\left|\frac{5}{8}x+\frac{3}{5}\right|\)
=> \(\frac{\left|5x-14\right|}{4}=\frac{\left|25x+24\right|}{40}\)
=> \(\frac{10(\left|5x-14\right|)}{40}=\frac{\left|25x+24\right|}{40}\)
=> \(\left|50x-140\right|=\left|25x+24\right|\)
=> \(\orbr{\begin{cases}50x-140=25x+24\\-50x+140=25x+24\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{164}{25}\\x=\frac{116}{75}\end{cases}}\)
c, \(\left|\frac{7}{5}x+\frac{2}{3}\right|=\left|\frac{4}{3}x-\frac{1}{4}\right|\)
=> \(\orbr{\begin{cases}\frac{7}{5}x+\frac{2}{3}=\frac{4}{3}x-\frac{1}{4}\\-\frac{7}{5}x-\frac{2}{3}=\frac{4}{3}x-\frac{1}{4}\end{cases}}\)
=> \(\orbr{\begin{cases}x=-\frac{55}{4}\\x=-\frac{25}{164}\end{cases}}\)
Bài 2 : a. |2x - 5| = x + 1
TH1 : 2x - 5 = x + 1
=> 2x - 5 - x = 1
=> 2x - x - 5 = 1
=> 2x - x = 6
=> x = 6
TH2 : -2x + 5 = x + 1
=> -2x + 5 - x = 1
=> -2x - x + 5 = 1
=> -3x = -4
=> x = 4/3
Ba bài còn lại tương tự
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TÌM X
a,\(\left(\frac{1}{7}x-\frac{2}{7}\right).\left(\frac{1}{5}x+\frac{3}{5}\right).\left(\frac{1}{3}x+\frac{4}{3}\right)=6\)
b,\(\left(x^2-4\right).\left(2x+\frac{4}{3}\right)=0\)
Bài 1 : Giải bất phương trình sau
1 , left(2x+3right)left(5x-7right)ge0
2 , left(3-2xright)left(4x+3right) 0
3 , left(2x+5right)left(3-xright)left(5x-1right)le0
4 , x^2-3x+2 0
5 , -x^2+12x+130
6 , x^2+6x+9le0
7 , frac{x+2}{3x+1}frac{x-2}{2x-1}
8 , frac{1}{x+2} frac{3}{x-3}
9 , frac{5x-6}{2x-5}le6
10 , left(x+1right)left(x-1right)left(3x-6right)0
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Bài 1 : Giải bất phương trình sau
1 , \(\left(2x+3\right)\left(5x-7\right)\ge0\)
2 , \(\left(3-2x\right)\left(4x+3\right)< 0\)
3 , \(\left(2x+5\right)\left(3-x\right)\left(5x-1\right)\le0\)
4 , \(x^2-3x+2< 0\)
5 , \(-x^2+12x+13>0\)
6 , \(x^2+6x+9\le0\)
7 , \(\frac{x+2}{3x+1}>\frac{x-2}{2x-1}\)
8 , \(\frac{1}{x+2}< \frac{3}{x-3}\)
9 , \(\frac{5x-6}{2x-5}\le6\)
10 , \(\left(x+1\right)\left(x-1\right)\left(3x-6\right)>0\)
16 tháng 9 2020 lúc 14:50
GPT
a) \(sin\left(2x+1\right)+cos\left(3x-1\right)=0\)
b) \(sin\left(2x-\frac{\pi}{6}\right)=-sin\left(x-\frac{\pi}{4}\right)\)
c) \(sin\left(3x+\frac{2\pi}{3}\right)+sin\left(x-\frac{7\pi}{5}\right)=0\)
d) \(cos\left(4x+\frac{\pi}{3}\right)+sin\left(x-\frac{\pi}{4}\right)=0\)
a.
\(sin\left(2x+1\right)=-cos\left(3x-1\right)\)
\(\Leftrightarrow sin\left(2x+1\right)=sin\left(3x-1-\frac{\pi}{2}\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-1-\frac{\pi}{2}=2x+1+k2\pi\\3x-1-\frac{\pi}{2}=\pi-2x-1+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{2}+2+k2\pi\\x=\frac{3\pi}{10}+\frac{k2\pi}{5}\end{matrix}\right.\)
b.
\(sin\left(2x-\frac{\pi}{6}\right)=sin\left(\frac{\pi}{4}-x\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-\frac{\pi}{6}=\frac{\pi}{4}-x+k2\pi\\2x-\frac{\pi}{6}=\frac{3\pi}{4}+x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{5\pi}{36}+\frac{k2\pi}{3}\\x=\frac{11\pi}{12}+k2\pi\end{matrix}\right.\)
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c.
\(\Leftrightarrow sin\left(3x+\frac{2\pi}{3}\right)=-sin\left(x-\frac{2\pi}{5}-\pi\right)\)
\(\Leftrightarrow sin\left(3x+\frac{2\pi}{3}\right)=sin\left(x-\frac{2\pi}{5}\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}3x+\frac{2\pi}{3}=x-\frac{2\pi}{5}+k2\pi\\3x+\frac{2\pi}{3}=\frac{7\pi}{5}-x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{8\pi}{15}+k\pi\\x=\frac{11\pi}{60}+\frac{k\pi}{2}\end{matrix}\right.\)
d.
\(\Leftrightarrow cos\left(4x+\frac{\pi}{3}\right)=sin\left(\frac{\pi}{4}-x\right)\)
\(\Leftrightarrow cos\left(4x+\frac{\pi}{3}\right)=cos\left(\frac{\pi}{4}+x\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}4x+\frac{\pi}{3}=\frac{\pi}{4}+x+k2\pi\\4x+\frac{\pi}{3}=-\frac{\pi}{4}-x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{\pi}{36}+\frac{k2\pi}{3}\\x=-\frac{7\pi}{60}+\frac{k2\pi}{5}\end{matrix}\right.\)
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Bài 1: Tìm x biết:
a, \(x.\cdot\left(\frac{1}{4}+\frac{1}{5}\right)-\left(\frac{1}{7}+\frac{1}{8}\right)=0\)
b, \(\left(5x-1\right).\left(2x-\frac{1}{3}\right)=0\)
c, \(\frac{3x+2}{5x+7}=\frac{3x-1}{5x+1}\)
d, \(\frac{x+1}{2x+1}=\frac{0,5x+2}{x+3}\)
e, \(\frac{-3}{4}-\left|\frac{4}{5}-x\right|=-1\)
b) \(\left(5x-1\right)\left(2x-\frac{1}{3}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}5x-1=0\\2x-\frac{1}{3}=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}5x=1\\2x=\frac{1}{3}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{5}\\x=\frac{1}{6}\end{matrix}\right.\)
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e, \(-\frac{3}{4}-\left|\frac{4}{5}-x\right|=-1\)
\(\Leftrightarrow\left|\frac{4}{5}-x\right|=-\frac{3}{4}-\left(-1\right)\)
\(\Leftrightarrow\left|\frac{4}{5}-x\right|=\frac{1}{4}\)
\(\Leftrightarrow\left[{}\begin{matrix}\frac{4}{5}-x=\frac{1}{4}\\\frac{4}{5}-x=-\frac{1}{4}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{7}{15}\\x=1,05\end{matrix}\right.\)
Vậy ....
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a)\(\left(\frac{5}{7}x-\frac{1}{4}\right)\left(\frac{-3}{4}x+\frac{1}{2}\right)=0\)
b)\(\left(\frac{4}{5}+x\right).\left(x-\frac{8}{13}\right)=0\)
c)\(\left(2x-\frac{1}{2}\right).\left(x-3\right)=0\)
d)\(x+3\frac{1}{2}x+x=\frac{1}{2}\)
a) \(\left(\frac{5}{7}x-\frac{1}{4}\right)\left(\frac{-3}{4}x+\frac{1}{2}\right)=0\)
\(\Rightarrow\orbr{\begin{cases}\frac{5}{7}x-\frac{1}{4}=0\\\frac{-3}{4}x+\frac{1}{2}=0\end{cases}}\Rightarrow\orbr{\begin{cases}\frac{5}{7}x=\frac{1}{4}\\\frac{-3}{4}x=\frac{-1}{2}\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{7}{20}\\x=\frac{2}{3}\end{cases}}\)
Vậy \(x=\frac{7}{20}\) hoặc x=\(\frac{2}{3}\)
b) \(\left(\frac{4}{5}+x\right)\left(x-\frac{8}{13}\right)=0\)
\(\Rightarrow\orbr{\begin{cases}\frac{4}{5}+x=0\\x-\frac{8}{13}=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{-4}{5}\\x=\frac{8}{13}\end{cases}}\)
Vậy x=-4/5 hoặc x=8/13
c) \(\left(2x-\frac{1}{2}\right)\left(x-3\right)=0\)
\(\Rightarrow\orbr{\begin{cases}2x-\frac{1}{2}=0\\x-3=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{1}{4}\\x=3\end{cases}}\)
Vậy x=1/4 hoặc x=3
\(x+\frac{7}{2}x+x=\frac{1}{2}\)
\(2x+\frac{7}{2}x=\frac{1}{2}\)
\(\left(2+\frac{7}{2}\right)x=\frac{1}{2}\)
\(\frac{11}{2}x=\frac{1}{2}\)
\(x=\frac{1}{2}:\frac{11}{2}\)
\(x=\frac{1}{11}\)
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