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a: x+2/5=1/2
=>x=1/2-2/5=5/10-4/10=1/10
b; x-2/5=2/7
=>x=2/7+2/5=10/35+14/35=24/35
c: 3/5-x=1/10
=>x=3/5-1/10=6/10-1/10=5/10=1/2
d: x*3/4=9/20
=>x=9/20:3/4=9/20*4/3=36/60=3/5
e: x:1/7=14
=>x=14*1/7=2
f: =>x+1/4=2/5:1/2=4/5
=>x=4/5-1/4=16/20-5/20=11/20
g: =>x*2/3=9/12+2/3=3/4+2/3=9/12+8/12=17/12
=>x=17/12:2/3=17/12*3/2=51/24=17/8
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a)<=>\(\dfrac{\left(2x-3\right).2}{6}-\dfrac{3.3}{6}=\dfrac{5-2x}{6}-\dfrac{1.3}{6}\)
<=>\(\dfrac{4x-6}{6}-\dfrac{9}{6}=\dfrac{5-2x}{6}-\dfrac{3}{6}\)
<=>\(\dfrac{4x-6}{6}-\dfrac{9}{6}-\dfrac{5-2x}{6}+\dfrac{3}{6}=0\)
<=>\(\dfrac{4x-6-9-5+2x+3}{6}=\dfrac{4x-17}{6}=0\)
<=>\(4x-17=0\)
<=>\(4x=17\)<=>\(x=\dfrac{17}{4}\)
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a) (2x - 3)(6 - 2x) = 0
=> \(\left[{}\begin{matrix}2x-3=0\\6-2x=0\end{matrix}\right.=>\left[{}\begin{matrix}2x=3\\2x=6\end{matrix}\right.=>\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=3\end{matrix}\right.\)
b) \(5\dfrac{4}{7}:x=13=>\dfrac{39}{7}:x=13=>x=\dfrac{39}{7}:13=>x=\dfrac{3}{7}\)
c) \(2x-\dfrac{3}{7}=6\dfrac{2}{7}=>2x-\dfrac{3}{7}=\dfrac{44}{7}=>2x=\dfrac{47}{7}=>x=\dfrac{47}{14}\)
d) \(\dfrac{x}{5}+\dfrac{1}{2}=\dfrac{6}{10}=>\dfrac{x}{5}=\dfrac{6}{10}-\dfrac{1}{2}=>\dfrac{x}{5}=\dfrac{1}{10}=>x.10=5=>x=\dfrac{1}{2}\)
e) \(\dfrac{x+3}{15}=\dfrac{1}{3}=>\left(x+3\right).3=15=>x+3=5=>x=2\)
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a) \(\left(\dfrac{3}{20}-\dfrac{1}{5}x\right)\cdot1\dfrac{2}{3}=1\dfrac{1}{4}\)
\(\left(\dfrac{3}{20}-\dfrac{1}{5}x\right)\cdot\dfrac{5}{3}=\dfrac{5}{4}\)
\(\dfrac{3}{20}-\dfrac{1}{5}x=\dfrac{5}{4}:\dfrac{5}{3}\\ \dfrac{3}{20}-\dfrac{1}{5}x=\dfrac{3}{4}\\ \dfrac{1}{5}x=\dfrac{3}{20}-\dfrac{3}{4}\\ \dfrac{1}{5}x=-\dfrac{3}{5}\\ x=-\dfrac{3}{5}:\dfrac{1}{5}\\ x=-3\)
b) \(\dfrac{-2}{3}x+\dfrac{1}{5}=\dfrac{3}{10}\)
\(-\dfrac{2}{3}x=\dfrac{3}{10}-\dfrac{1}{5}\\ \dfrac{-2}{3}x=\dfrac{1}{10}\\ x=\dfrac{1}{10}:\dfrac{-2}{3}\\ x=-\dfrac{3}{20}\)
c) \(-\dfrac{2}{3}-\dfrac{1}{3}\left(2x-7\right)=\dfrac{3}{2}\)
\(\dfrac{1}{3}\left(2x-7\right)=-\dfrac{2}{3}-\dfrac{3}{2}\\ \dfrac{1}{3}\left(2x-7\right)=-\dfrac{13}{6}\\ 2x-7=-\dfrac{13}{6}:\dfrac{1}{3}\\ 2x-7=-\dfrac{13}{2}\\ 2x=-\dfrac{13}{2}+7\\ 2x=\dfrac{1}{2}\\ x=\dfrac{1}{4}\)
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a:=>0,75x-x+1,25x=0,2
=>x=0,2
b: =>1/3-x=-3/6+4/6=1/6
=>x=1/3-1/6=1/6
c: =>(x-1)/45=-6/30=-1/5
=>x-1=-9
=>x=-8
d: =>(2/5x-1)=-5/7
=>2/5x=2/7
=>x=5/7
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Câu 1:
\(\Rightarrow \left[\begin{array}{} x+\frac{1}{2}=0\\ \frac{2}{3}-2x=0 \end{array} \right.\)
\(\Leftrightarrow \left[\begin{array}{} x=\frac{-1}{2}\\ x=\frac{1}{3} \end{array} \right.\)
Vậy phương trình có tập nghiệm S={\(\frac{-1}{2};\frac{1}{3}\)}
Câu 2:
\(\Rightarrow \left[\begin{array}{} 3x-10=0\\ 5-\frac{1}{2}x=0 \end{array} \right.\)
\(\Leftrightarrow \left[\begin{array}{} x-=\frac{10}{3}\\ x=10 \end{array} \right.\)
Vậy phương trình có tập nghiệm S={\(10;\frac{10}{3}\)}
Câu 3:
\(\Leftrightarrow \frac{1}{3}x=\frac{65}{4}-\frac{53}{4}\)
\( \Leftrightarrow \frac{1}{3}x=\frac{12}{4}\)
\(\Leftrightarrow x=9\)
Vậy phương trình có tập nghiệm S={9}
Câu 4:
\(\Leftrightarrow \frac{2}{3}x=\frac{2}{3}\)
\(\Leftrightarrow x=1\)
Vậy phương trình có tập nghiệm S={1}
Câu 5:
\(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{x(x+1)}=\frac{2010}{2011}\)
\(\Leftrightarrow 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2010}{2011}\)
\(\Leftrightarrow 1-\frac{1}{x+1}=\frac{2010}{2011}\)
\(\Leftrightarrow \frac{x}{x+1}=\frac{2010}{2011}\)
\(\Rightarrow 2010x+2010=2011x\)
\(\Leftrightarrow x=2010\)
Vậy phương trình có tập nghiệm S={2010}
cảm ơn bạn Hoàng Bình Bảo nha nhưng mà đây là toán lớp 6 mà bạn
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Bài 2: a) \(\dfrac{x-3}{x+5}=\dfrac{5}{7}\)
\(\Leftrightarrow\left(x-3\right).7=\left(x+5\right).5\)
\(\Leftrightarrow7x-21=5x+25\)
\(\Leftrightarrow7x-5x=21+25\)
\(\Leftrightarrow2x=46\)
\(\Rightarrow x=46:2=23\)
b) \(\dfrac{7}{x-1}=\dfrac{x+1}{9}\)
\(\Leftrightarrow\left(x+1\right)\left(x-1\right)=63\)
\(\Leftrightarrow x^2-1=63\)
\(\Leftrightarrow x^2=64\)
\(\Rightarrow x^2=\left(\pm8\right)^2\)
\(\Rightarrow x=8\) hoặc \(x=-8\)
2)a) \(\dfrac{x-3}{x+5}=\dfrac{5}{7}\)
\(\Leftrightarrow7\left(x-3\right)=5\left(x+5\right)\)
\(7x-21=5x+25\)
\(7x-5x+25=21\)
\(2x+25=21\)
\(2x=-4\Rightarrow x=-2\)
b) \(\dfrac{7}{x-1}=\dfrac{x+1}{9}\)
\(7.9=\left(x+1\right)\left(x-1\right)\)
\(63=x\left(x-1\right)+1\left(x-1\right)\)
\(63=x^2-x+x-1\)
\(x^2=63+1=64\)
\(x=\left\{\pm8\right\}\)
c) \(\dfrac{x+4}{20}=\dfrac{2}{x+4}\)
\(\Leftrightarrow\left(x+4\right)\left(x+4\right)=2.20=40\)
\(x\left(x+4\right)+4\left(x+4\right)=40\)
\(x^2+4x+4x+16=40\)
\(x^2+8x=40-16=24\)
\(x\left(x+8\right)=24\)
\(x\in\left\{\varnothing\right\}\)
d) \(\dfrac{x-1}{x+2}=\dfrac{x-2}{x+3}\)
\(\Leftrightarrow\left(x+2\right)\left(x-2\right)=\left(x-1\right)\left(x+3\right)\)
\(x\left(x-2\right)+2\left(x-2\right)=x\left(x+3\right)-1\left(x+3\right)\)
\(x^2-2x+2x-4=x^2+3x-x-3\)
\(\)\(x^2-4=x^2+2x-3\)
\(\Leftrightarrow x^2-x^2-2x+3=4\)
\(-2x+3=4\)
\(-2x=1\)
\(x=-\dfrac{1}{2}\)
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câu c) mang tính mua vui hay gì hả bn
mếu thật thì x=0,x=số nào cx đc(câu trả lời này mang tính mua vui thôi nhé)
\(\dfrac{x-5}{x+2}-\dfrac{10}{x}-\dfrac{20}{x^2+2x}\)
\(=\dfrac{x\left(x-5\right)-10\left(x+2\right)-20}{x\left(x+2\right)}\)
\(=\dfrac{x^2-5x-10x-20-20}{x\left(x+2\right)}\)
\(=\dfrac{x^2-15x-40}{x\left(x+2\right)}\)