Tìm x
a)\(\dfrac{6}{x-3}=\dfrac{9}{2x-7}\)
b)\(\dfrac{-7}{x+1}=\dfrac{6}{x+27}\)
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VN
Võ Ngọc Phương
VIP
11 tháng 9 2023
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) \(-\dfrac{15}{4}.\left(\dfrac{-7}{15}\right).\left(-2\dfrac{2}{5}\right)\)
\(=\dfrac{7}{4}.\dfrac{-12}{5}\)
\(=\dfrac{-21}{5}\)
\(#Wendy.Dang\)
26 tháng 4 2022
a. 5 - 3(x + 4) = -1
⇔ 5 - 3x - 12 = -1
⇔ 3x = -1 - 5 + 12
⇔ 3x = 6
⇔ x = 2
TT
26 tháng 4 2022
\(d,2x^2-3=5\)
\(\Leftrightarrow2x^2=8\)
\(\Leftrightarrow x^2=4\)
\(\Leftrightarrow x=\pm2\)
\(e,x\left(2x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\x=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=1\\x=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=0\end{matrix}\right.\)
17 tháng 7 2021
a) Ta có: \(\dfrac{1}{4}-\left|x+\dfrac{1}{2}\right|=\dfrac{1}{8}\)
\(\Leftrightarrow\left|x+\dfrac{1}{2}\right|=\dfrac{1}{8}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=\dfrac{1}{8}\\x+\dfrac{1}{2}=-\dfrac{1}{8}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-3}{8}\\x=\dfrac{-5}{8}\end{matrix}\right.\)
NH
15 tháng 12 2022
a)
\(\left|x-2\right|-\dfrac{3}{5}=\dfrac{1}{2}\\ \left|x-2\right|=\dfrac{1}{2}+\dfrac{3}{5}\\ \left|x-2\right|=\dfrac{11}{10}\\ =>\left[{}\begin{matrix}x-2=\dfrac{11}{10}\\x-2=-\dfrac{11}{10}\end{matrix}\right.\left[{}\begin{matrix}x=\dfrac{31}{10}\\x=\dfrac{9}{10}\end{matrix}\right.\)
b)
\(\left(x-\dfrac{7}{3}\right):\dfrac{-1}{3}=0,4\\ x-\dfrac{7}{3}=0,4\cdot\dfrac{-1}{3}\\ x-\dfrac{7}{3}=-\dfrac{2}{15}\\ x=-\dfrac{2}{15}+\dfrac{7}{3}\\ x=\dfrac{11}{5}\)
c)
\(\left|x-3\right|=5\\ =>\left[{}\begin{matrix}x-3=5\\x-3=-5\end{matrix}\right.\left[{}\begin{matrix}x=5+3\\x=-5+3\end{matrix}\right.\left[{}\begin{matrix}x=8\\x=-2\end{matrix}\right.\)
d)
\(\left(2x+3\right)^2=25\\ =>\left[{}\begin{matrix}2x+3=5\\2x+3=-5\end{matrix}\right.\left[{}\begin{matrix}2x=2\\2x=-8\end{matrix}\right.\left[{}\begin{matrix}x=1\\x=-4\end{matrix}\right.\)
e)
\(\dfrac{3}{4}+\dfrac{1}{4}:x=\dfrac{2}{5}\)
\(\dfrac{1}{4}:x=\dfrac{2}{5}-\dfrac{3}{4}\)
\(\dfrac{1}{4}:x=-\dfrac{7}{20}\)
\(x=\dfrac{1}{4}:\dfrac{-7}{20}\\ x=-\dfrac{5}{7}\)
f)
\(\left(x-\dfrac{1}{2}\right)^3=\dfrac{1}{27}\\ =>x-\dfrac{1}{2}=\dfrac{1}{3}\\ x=\dfrac{1}{3}+\dfrac{1}{2}\\ x=\dfrac{5}{6}\)
DT
Dang Tung
CTVHS
31 tháng 1 2023
\(< =>\dfrac{13\left(x+3\right)}{\left(2x+7\right)\left(x-3\right)\left(x+3\right)}+\dfrac{x^2-9}{\left(2x+7\right)\left(x-3\right)\left(x+3\right)}=\dfrac{6\left(2x+7\right)}{\left(2x+7\right)\left(x-3\right)\left(x+3\right)}\left(ĐK:x\ne\left\{-\dfrac{7}{2};3;-3\right\}\right)\\ =>13x+39+x^2-9=12x+42\\ < =>x^2+x-12=0\\ < =>\left(x+4\right)\left(x-3\right)=0\\ =>\left[{}\begin{matrix}x=-4\left(TM\right)\\x=3\left(KTM\right)\end{matrix}\right.\\ =>S=\left\{-4\right\}\)
NT
Nguyễn thành Đạt
CTVHS
31 tháng 1 2023
\(ĐKXĐ:x\ne\dfrac{7}{2}\) và \(x\ne\pm3\)
mẫu chung : \(\left(2x+7\right)\left(x+3\right)\left(x-3\right)\)
Khử mẫu ta được :
\(13\left(x+3\right)+\left(x+3\right)\left(x-3\right)=6\left(2x+7\right)\)
\(\Leftrightarrow x^2+x-12=0\)
\(\Leftrightarrow\left(x+4\right)\left(x-3\right)=0\)
\(x=\left\{{}\begin{matrix}-4\\3\end{matrix}\right.\)
do \(x=3\) không thỏa mãn điều kiện thích hợp nên pt có nghiệm duy nhất là : \(-4\)
\(Vậy...\)
E
14 tháng 9 2023
\(a,\dfrac{5}{8}-x=\dfrac{4}{7}\times\dfrac{1}{4}\\ \dfrac{5}{8}-x=\dfrac{4\times1}{7\times4}\\ \dfrac{5}{8}-x=\dfrac{4}{28}\\ \dfrac{5}{8}-x=\dfrac{1}{7}\\ x=\dfrac{5}{8}-\dfrac{1}{7}\\ x=\dfrac{5\times7}{8\times7}-\dfrac{1\times8}{7\times8}\\ x=\dfrac{35}{56}-\dfrac{8}{56}\\ x=\dfrac{35-8}{56}\\ x=\dfrac{27}{56}\\ b,x:\dfrac{4}{5}=\dfrac{3}{4}+\dfrac{1}{6}\\ x:\dfrac{4}{5}=\dfrac{3\times3}{4\times3}+\dfrac{1\times2}{6\times2}\\ x:\dfrac{4}{5}=\dfrac{9}{12}+\dfrac{2}{12}\\ x:\dfrac{4}{5}=\dfrac{9+2}{12}\\ x:\dfrac{4}{5}=\dfrac{11}{12} \\ x=\dfrac{11}{12}\times\dfrac{4}{5}\\ x=\dfrac{11\times4}{12\times5}\\ x=\dfrac{44}{60}=\dfrac{11}{15}\)
14 tháng 9 2023
a: =>5/8-x=1/7
=>x=5/8-1/7=35/56-8/56=27/56
b: =>x:4/5=9/12+2/12=11/12
=>x=11/12*4/5=44/48=11/12
12 tháng 11 2021
a) \(\Leftrightarrow\dfrac{3}{2}:x=\dfrac{1}{2}\\ \Leftrightarrow x=\dfrac{3}{2}:\dfrac{1}{2}\\ \Leftrightarrow x=3\)
b) \(\Leftrightarrow x=\dfrac{7}{9}-\dfrac{2}{3}\\ \Leftrightarrow x=\dfrac{1}{9}\)
c) \(\Leftrightarrow x=\dfrac{8}{7}:\dfrac{6}{7}\\ \Leftrightarrow x=\dfrac{4}{3}\)
d) \(\Leftrightarrow x=\dfrac{9}{5}-\dfrac{3}{7}\\ \Leftrightarrow x=\dfrac{48}{35}\)
15 tháng 11 2021
a) x = 3
b) x = \(\dfrac{1}{9}\)
c) x = \(\dfrac{4}{3}\)
d)\(\dfrac{48}{35}\)
L
19 tháng 5 2022
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) \(\dfrac{x+3}{15}=\dfrac{1}{3}=>\left(x+3\right).3=15=>x+3=5=>x=2\)
a, \(\dfrac{6}{x-3}=\dfrac{9}{2x-7}\)
=> 6(2x-7) = 9(x-3)
=> 12x - 42 = 9x - 27
=> 12x - 9x = -27 + 42
=> 3x = 15
=> x = 5
Vậy x = 5
b, \(\dfrac{-7}{x+1}=\dfrac{6}{x+27}\)
=> -7(x + 27) = 6(x + 1)
=> -7x - 189 = 6x + 6
=> -7x - 6x = 6 + 189
=> -13x = 195
=> x = -15
Vậy x = -15
a) Ta có: \(\dfrac{6}{x-3}=\dfrac{9}{2x-7}\)
\(\Leftrightarrow6\left(2x-7\right)=9\left(x-3\right)\)
\(\Leftrightarrow12x-42=9x-27\)
\(\Leftrightarrow12x-9x=-27+42\)
\(\Leftrightarrow3x=15\)
hay x=5
Vậy: x=5
b) Ta có: \(\dfrac{-7}{x+1}=\dfrac{6}{x+27}\)
\(\Leftrightarrow6\left(x+1\right)=-7\left(x+27\right)\)
\(\Leftrightarrow6x+6=-7x+189\)
\(\Leftrightarrow6x+7x=189-6\)
\(\Leftrightarrow13x=183\)
hay \(x=\dfrac{183}{13}\)
Vậy: \(x=\dfrac{183}{13}\)