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Bài 4:
a) \(\dfrac{4}{3}+\left(1,25-x\right)=2,25\)
\(1,25-x=2,25-\dfrac{4}{3}=\dfrac{9}{4}-\dfrac{4}{3}\)
\(1,25-x=\dfrac{11}{12}\)
\(x=1,25-\dfrac{11}{12}=\dfrac{5}{4}-\dfrac{11}{12}\)
\(x=\dfrac{1}{3}\)
b) \(\dfrac{17}{6}-\left(x-\dfrac{7}{6}\right)=\dfrac{7}{4}\)
\(x-\dfrac{7}{6}=\dfrac{17}{6}-\dfrac{7}{4}=\dfrac{34}{12}-\dfrac{21}{12}\)
\(x-\dfrac{7}{6}=\dfrac{13}{12}\)
\(x=\dfrac{13}{12}+\dfrac{7}{6}=\dfrac{13}{12}+\dfrac{14}{12}\)
\(x=\dfrac{27}{12}=\dfrac{9}{4}\)
c) \(4-\left(2x+1\right)=3-\dfrac{1}{3}=\dfrac{9}{3}-\dfrac{1}{3}\)
\(4-\left(2x+1\right)=\dfrac{8}{3}\)
\(2x+1=\dfrac{8}{3}+4=\dfrac{8}{3}+\dfrac{12}{3}\)
\(2x+1=\dfrac{20}{3}\)
\(2x=\dfrac{20}{3}-1=\dfrac{20}{3}-\dfrac{3}{3}\)
\(2x=\dfrac{17}{3}\)
\(x=\dfrac{17}{3}.\dfrac{1}{2}=\dfrac{17}{6}\)
Bài 15:
a) \(\left(\dfrac{-2}{3}\right)^9:x=\dfrac{-2}{3}\)
\(x=\left(\dfrac{-2}{3}\right)^9:\dfrac{-2}{3}=\left(\dfrac{-2}{3}\right)^{9-1}\)
\(=>x=\left(\dfrac{-2}{3}\right)^8\)
b) \(x:\left(\dfrac{4}{9}\right)^5=\left(\dfrac{4}{9}\right)^4\)
\(x=\left(\dfrac{4}{9}\right)^4.\left(\dfrac{4}{9}\right)^5=\left(\dfrac{4}{9}\right)^{4+5}\)
\(=>x=\left(\dfrac{4}{9}\right)^9\)
c) \(\left(x+4\right)^3=-125\)
\(\left(x+4\right)^3=\left(-5\right)^3\)
\(=>x+4=-5\)
\(x=-5-4\)
\(=>x=-9\)
d) \(\left(10-5x\right)^3=64\)
\(\left(10-5x\right)^3=4^3\)
\(=>10-5x=4\)
\(5x=10-4\)
\(5x=6\)
\(=>x=\dfrac{6}{5}\)
e) \(\left(4x+5\right)^2=81\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(4x+5\right)^2=\left(-9\right)^2\\\left(4x+5\right)^2=9^2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}4x+5=-9\\4x+5=9\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}4x=-14\\4x=4\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-14}{4}\\x=1\end{matrix}\right.\)
Bài 16:
a) \(4-1\dfrac{2}{5}-\dfrac{8}{3}\)
\(=4-\dfrac{7}{5}-\dfrac{8}{3}\)
\(=\dfrac{60-21-40}{15}=\dfrac{-1}{15}\)
b) \(-0,6-\dfrac{-4}{9}-\dfrac{16}{15}\)
\(=\dfrac{-3}{5}+\dfrac{4}{9}-\dfrac{16}{15}\)
\(=\dfrac{\left(-27\right)+20-48}{45}=\dfrac{-55}{45}=\dfrac{-11}{9}\)
c) \(-\dfrac{15}{4}.\left(\dfrac{-7}{15}\right).\left(-2\dfrac{2}{5}\right)\)
\(=\dfrac{7}{4}.\dfrac{-12}{5}\)
\(=\dfrac{-21}{5}\)
\(#Wendy.Dang\)
\(a,2\dfrac{1}{2}-x+\dfrac{4}{5}=\dfrac{2}{3}-\left(-\dfrac{4}{7}\right)\\ \Rightarrow\dfrac{5}{2}-x+\dfrac{4}{5}=\dfrac{26}{21}\\ \Rightarrow\dfrac{5}{2}-x=\dfrac{46}{105}\\ \Rightarrow x=\dfrac{433}{210}\\ b,-\dfrac{4}{7}-x=\dfrac{3}{5}-2x\\ \Rightarrow2x-\dfrac{4}{7}-x=\dfrac{3}{5}\\ \Rightarrow2x-x=\dfrac{41}{35}\\ \Rightarrow x=\dfrac{41}{35}\\ c,\left(\dfrac{3}{8}-\dfrac{1}{5}\right)+\left(\dfrac{5}{8}-x\right)=\dfrac{1}{5}\\ \Rightarrow\dfrac{7}{40}+\dfrac{5}{8}-x=\dfrac{1}{5}\\ \Rightarrow\dfrac{4}{5}-x=\dfrac{1}{5}\\ \Rightarrow x=\dfrac{3}{5}.\)
1. Ta có: \(\dfrac{x}{-7}=\dfrac{y}{4}\Rightarrow\dfrac{2x}{-14}=\dfrac{3y}{12}\)
Áp dụng tính chất dãy tỉ số bằng nhau, ta được:
\(\dfrac{2x-3y}{-14-12}=\dfrac{-78}{-26}=3\)
=> \(\left\{{}\begin{matrix}x=-21\\y=12\end{matrix}\right.\)
2. Ta có:
- \(\dfrac{x}{y}=\dfrac{9}{7}\Rightarrow\dfrac{x}{9}=\dfrac{y}{7}\)
- \(\dfrac{y}{z}=\dfrac{7}{3}\Rightarrow\dfrac{y}{7}=\dfrac{z}{3}\)
=> \(\dfrac{x}{9}=\dfrac{y}{7}=\dfrac{z}{3}\)
Áp dụng tính chất dãy tỉ số bằng nhau, ta được:
\(\dfrac{x-y+z}{9-7+3}=\dfrac{-15}{5}=-3\)
=> \(\left\{{}\begin{matrix}x=-27\\y=-21\\z=-9\end{matrix}\right.\)
a) Ta có: \(\dfrac{4}{5}-3\left|x\right|=\dfrac{1}{5}\)
\(\Leftrightarrow3\left|x\right|=\dfrac{4}{5}-\dfrac{1}{5}=\dfrac{3}{5}\)
\(\Leftrightarrow\left|x\right|=\dfrac{1}{5}\)
hay \(x\in\left\{\dfrac{1}{5};-\dfrac{1}{5}\right\}\)
b) Ta có: \(4x-\dfrac{1}{2}x+\dfrac{3}{5}x=\dfrac{4}{5}\)
nên \(\dfrac{41}{10}x=\dfrac{4}{5}\)
hay \(x=\dfrac{8}{41}\)
c) Ta có: \(\left(2x-8\right)\left(10-5x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-8=0\\10-5x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=8\\5x=10\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=2\end{matrix}\right.\)
d) Ta có: \(\dfrac{3}{4}+\dfrac{1}{4}\left|2x-1\right|=\dfrac{7}{2}\)
\(\Leftrightarrow\dfrac{1}{4}\left|2x-1\right|=\dfrac{7}{2}-\dfrac{3}{4}=\dfrac{14}{4}-\dfrac{3}{4}=\dfrac{11}{4}\)
\(\Leftrightarrow\left|2x-1\right|=11\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=11\\2x-1=-11\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=12\\2x=-10\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-5\end{matrix}\right.\)
\(a.ĐK:x\ne3;1\)
\(\Rightarrow\dfrac{1}{2\left(x-3\right)}+\dfrac{3x-10}{\left(x-1\right)\left(x-3\right)}=\dfrac{7}{2}\)
\(\Leftrightarrow\dfrac{\left(x-1\right)+2\left(3x-10\right)}{2\left(x-1\right)\left(x-3\right)}=\dfrac{7\left(x-1\right)\left(x-3\right)}{2\left(x-1\right)\left(x-3\right)}\)
\(\Leftrightarrow x-1+2\left(3x-10\right)=7\left(x-1\right)\left(x-3\right)\)
\(\Leftrightarrow x-1+6x-20=7\left(x^2-4x+3\right)\)
\(\Leftrightarrow7x-21=7x^2-28x+21\)
\(\Leftrightarrow7x^2-35x+42=0\)
\(\Leftrightarrow7\left(x^2-5x+6\right)=0\)
\(\Leftrightarrow x^2-5x+6=0\)
\(\Leftrightarrow x^2-2x-3x+6=0\)
\(\Leftrightarrow x\left(x-2\right)-3\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\left(tm\right)\\x=3\left(ktm\right)\end{matrix}\right.\)
b.\(ĐK:x\ne2;4\)
\(\Rightarrow\dfrac{x-1}{x-2}-\dfrac{x+3}{4-x}=\dfrac{2}{\left(x-2\right)\left(4-x\right)}\)
\(\Leftrightarrow\dfrac{\left(x-1\right)\left(4-x\right)-\left(x+3\right)\left(x-2\right)}{\left(x-2\right)\left(4-x\right)}=\dfrac{2}{\left(x-2\right)\left(4-x\right)}\)
\(\Leftrightarrow\left(x-1\right)\left(4-x\right)-\left(x+3\right)\left(x-2\right)=2\)
\(\Leftrightarrow4x-x^2-4+x-x^2+2x-3x+6-2=0\)
\(\Leftrightarrow-2x^2+4x=0\)
\(\Leftrightarrow-2x\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(tm\right)\\x=2\left(ktm\right)\end{matrix}\right.\)
a: \(\Leftrightarrow\dfrac{1}{2\left(x-3\right)}+\dfrac{3x-10}{\left(x-1\right)\left(x-3\right)}=\dfrac{7}{2}\)
\(\Leftrightarrow x-1+2\left(3x-10\right)=7\left(x-1\right)\left(x-3\right)\)
\(\Leftrightarrow7\left(x^2-4x+3\right)=x-1+6x-20=7x-21\)
\(\Leftrightarrow\left(x-3\right)\left(7x-7\right)-7\left(x-3\right)=0\)
=>(x-3)(7x-14)=0
=>x=3(loại) hoặc x=2(nhận)
b: \(\Leftrightarrow\left(x-1\right)\left(x-4\right)+\left(x+3\right)\left(x-2\right)=-2\)
\(\Leftrightarrow x^2-5x+4+x^2+x-6=-2\)
\(\Leftrightarrow2x^2-4x=0\)
=>2x(x-2)=0
=>x=0(nhận) hoặc x=2(loại)
1) \(\dfrac{11}{12}-\left(\dfrac{2}{5}+x\right)=\dfrac{2}{3}\)
\(\Leftrightarrow\dfrac{11}{12}-\dfrac{2}{5}-x=\dfrac{2}{3}\)
\(\Leftrightarrow x=\dfrac{11}{12}-\dfrac{2}{5}-\dfrac{2}{3}\)
\(\Leftrightarrow x=-\dfrac{3}{20}\)
2) \(2x\left(x-\dfrac{1}{7}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=0\\x-\dfrac{1}{7}=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{7}\end{matrix}\right.\)
3) \(\dfrac{3}{4}+\dfrac{1}{4}:x=\dfrac{2}{5}\)
\(\Leftrightarrow\dfrac{1}{4x}=\dfrac{2}{5}-\dfrac{3}{4}\)
\(\Leftrightarrow\dfrac{1}{4x}=-\dfrac{7}{20}\)
\(\Leftrightarrow4x=-\dfrac{20}{7}\)
\(\Leftrightarrow x=-\dfrac{5}{7}\)
a) \(\left(8x-3\right)\left(3x+2\right)-\left(4x+7\right)\left(x+4\right)=\left(2x+1\right)\left(5x-1\right)-33\)
\(20x^2-16x-34=10x^2+3x-34\)
\(10x^2-19x=0\)
\(x\left(10x-19\right)=0\)
\(\Leftrightarrow x=0\)
hoặc \(10x-19=0\)
\(\Leftrightarrow x=\dfrac{19}{10}\)
Vạy ..............
b) \(\dfrac{x-1}{x+5}=\dfrac{6}{7}\)
\(\Leftrightarrow1-\dfrac{x-1}{x+5}=1-\dfrac{6}{7}\)
\(\Leftrightarrow\dfrac{x+5}{x+5}-\dfrac{x-1}{x+5}=\dfrac{7}{7}-\dfrac{6}{7}\)
\(\Leftrightarrow\dfrac{\left(x+5\right)-\left(x-1\right)}{x+5}=\dfrac{1}{7}\)
\(\Leftrightarrow\dfrac{x+5-x+1}{x+5}=\dfrac{1}{7}\)
\(\Leftrightarrow\dfrac{\left(x-x\right)+\left(5+1\right)}{x+5}=\dfrac{1}{7}\)
\(\Leftrightarrow\dfrac{6}{x+5}=\dfrac{1}{7}\)
\(\Leftrightarrow x+5=42\)
\(\Leftrightarrow x=37\)
a) \(\dfrac{x-4}{15}=\dfrac{5}{3}\)
\(\Leftrightarrow x-4=15.\dfrac{5}{3}\)
\(\Leftrightarrow x-4=25\)
\(\Leftrightarrow x=29\) thỏa \(x\inℤ\)
b) \(\dfrac{x}{4}=\dfrac{18}{x+1}\left(x\ne-1\right)\)
\(\Leftrightarrow x\left(x+1\right)=18.4\)
\(\Leftrightarrow x\left(x+1\right)=72\)
vì \(72=8.9=\left(-8\right).\left(-9\right)\)
\(\Leftrightarrow x\in\left\{8;-9\right\}\left(x\inℤ\right)\)
c) \(2x+3⋮x+4\) \(\left(x\ne-4;x\inℤ\right)\)
\(\Leftrightarrow2x+3-2\left(x+4\right)⋮x+4\)
\(\Leftrightarrow2x+3-2x-8⋮x+4\)
\(\Leftrightarrow-5⋮x+4\)
\(\Leftrightarrow x+4\in\left\{-1;1;-5;5\right\}\)
\(\Leftrightarrow x\in\left\{-5;-3;-9;1\right\}\)
`#3107`
a)
\(\dfrac{11}{12}-\left(\dfrac{2}{5}+\dfrac{3}{4}x\right)=\dfrac{2}{3}?\\ \Rightarrow\dfrac{2}{5}+\dfrac{3}{4}x=\dfrac{11}{12}-\dfrac{2}{3}\\ \Rightarrow\dfrac{2}{5}+\dfrac{3}{4}x=\dfrac{1}{4}\\ \Rightarrow\dfrac{3}{4}x=\dfrac{1}{4}-\dfrac{2}{5}\\ \Rightarrow\dfrac{3}{4}x=-\dfrac{3}{20}\\ \Rightarrow x=-\dfrac{3}{20}\div\dfrac{3}{4}\\ \Rightarrow x=-\dfrac{1}{5}\)
Vậy, \(x=-\dfrac{1}{5}\)
b)
\(\dfrac{-2}{5}+\dfrac{5}{3}\cdot\left(\dfrac{3}{2}-\dfrac{4}{15}x\right)=\dfrac{-7}{6}\\ \Rightarrow\dfrac{5}{3}\cdot\left(\dfrac{3}{2}-\dfrac{4}{15}x\right)=\dfrac{-7}{6}-\dfrac{-2}{5}\\ \Rightarrow\dfrac{5}{3}\cdot\left(\dfrac{3}{2}-\dfrac{4}{15}x\right)=-\dfrac{23}{30}\\ \Rightarrow\dfrac{3}{2}-\dfrac{4}{15}x=-\dfrac{23}{30}\div\dfrac{5}{3}\\ \Rightarrow\dfrac{3}{2}-\dfrac{4}{15}x=-\dfrac{23}{50}\\ \Rightarrow\dfrac{4}{15}x=\dfrac{3}{2}-\left(-\dfrac{23}{50}\right)\\ \Rightarrow\dfrac{4}{15}x=\dfrac{49}{25}\\ \Rightarrow x=\dfrac{147}{20}\)
Vậy, \(x=\dfrac{147}{20}\)
c)
\(\dfrac{1}{2}+\dfrac{3}{4}x=\dfrac{1}{4}\\ \Rightarrow\dfrac{3}{4}x=\dfrac{1}{4}-\dfrac{1}{2}\\ \Rightarrow\dfrac{3}{4}x=-\dfrac{1}{4}\\ \Rightarrow x=-\dfrac{1}{4}\div\dfrac{3}{4}\\ \Rightarrow x=-\dfrac{1}{3}\)
Vậy, \(x=-\dfrac{1}{3}.\)
\(#Emyeu1aithatroi...\)
(2/5 + 3/4 . x)= 11/12 -2/3
(2/5 +3/4 . x)= 1/4
3/4 . x = 1/4 - 2/5
3/4 . x = -3/20
x = -3/20 : 3/4
x = -1/5
Vậy .....
a . \(\dfrac{x+1}{x-2}=\dfrac{3}{4}\)
=> \(\left(x+1\right).4=3.\left(x-2\right)\)
=> \(4x+4=3x-6\)
=> \(4x-3x=-6-4\)
=> x = -10
b. \(\dfrac{2x-3}{x+1}=\dfrac{4}{7}\)
=> \(\left(2x-3\right).7=4.\left(x+1\right)\)
=> \(14x-21=4x+4\)
=> \(14x-4x=4+21\)
=> \(10x=25\)
=> \(x=\dfrac{5}{2}\)
c. \(\dfrac{2x+4}{7}=\dfrac{4x-2}{15}\)
=> \(\left(2x+4\right).15=\left(4x-2\right).7\)
=> \(30x+60=28x-14\)
=> \(30x-28x=-14-60\)
=> \(2x=-74\)
=> \(x=-37\)
#Yiin
a, \(\dfrac{x+1}{x-2}=\dfrac{3}{4}\Rightarrow4\left(x+1\right)=3\left(x-2\right)\)
\(\Rightarrow4x+4=3x-6\)
\(\Rightarrow4x-3x=-6-4\Rightarrow x=-10\)
b, \(\dfrac{2x-3}{x+1}=\dfrac{4}{7}\Rightarrow7\left(2x-3\right)=4\left(x+1\right)\)
\(\Rightarrow14x-21=4x+4\)
\(\Rightarrow14x-4x=4+21\Rightarrow10x=25\Rightarrow x=\dfrac{5}{2}\)
c, \(\dfrac{2x+4}{7}=\dfrac{4x-2}{15}\Rightarrow15\left(2x+4\right)=7\left(4x-2\right)\)
\(\Rightarrow30x+60=28x-14\)
\(\Rightarrow30x-28x=-14-60\)
\(\Rightarrow2x=-74\Rightarrow x=-37\)