Tìm x:
a) x-3=8
b) 5.x=10
c) x+6= 17
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S
1
CD
8 tháng 12 2021
a, <=> x2 -2x +1 + 5x -x2 =8
<=> 3x +1 =8
<=> 3x = 7
<=> x= 7/3
b, thiếu đề
c, <=> 2x3 -1 + 2x(4 -x2) = 7
<=> 2x3 + 8x -23 = 8
<=> 8x =8
<=> x=1
TN
a) X + 1/4 = 5/8
b) X - 3/5 = 1/10
c) X x 2/7 = 6/11
d) X : 3/2 = 1/4
đ) 7/4 - X = 5/7
e) 5/4 : x = 1/8
1
Y
7 tháng 4 2022
\(a,x+\dfrac{1}{4}=\dfrac{5}{8}\)
\(\Leftrightarrow x=\dfrac{5}{8}-\dfrac{1}{4}\)
\(\Leftrightarrow x=\dfrac{5}{8}-\dfrac{2}{8}\)
\(\Leftrightarrow x=\dfrac{3}{8}\)
\(b,x-\dfrac{3}{5}=\dfrac{1}{10}\)
\(\Leftrightarrow x=\dfrac{1}{10}+\dfrac{3}{5}\)
\(\Leftrightarrow x=\dfrac{1}{10}+\dfrac{6}{10}\)
\(\Leftrightarrow x=\dfrac{7}{10}\)
\(c,x\times\dfrac{2}{7}=\dfrac{6}{11}\)
\(\Leftrightarrow x=\dfrac{6}{11}:\dfrac{2}{7}\)
\(\Leftrightarrow x=\dfrac{6}{11}\times\dfrac{7}{2}\)
\(\Leftrightarrow x=\dfrac{42}{22}\)
\(\Leftrightarrow x=\dfrac{21}{11}\)
\(d,x:\dfrac{3}{2}=\dfrac{1}{4}\)
\(\Leftrightarrow x=\dfrac{1}{4}\times\dfrac{3}{2}\)
\(\Leftrightarrow x=\dfrac{3}{8}\)
\(đ,\dfrac{7}{4}-x=\dfrac{5}{7}\)
\(\Leftrightarrow x=\dfrac{7}{4}-\dfrac{5}{7}\)
\(\Leftrightarrow x=\dfrac{49}{28}-\dfrac{20}{28}\)
\(\Leftrightarrow x=\dfrac{29}{28}\)
\(e,\dfrac{5}{4}:x=\dfrac{1}{8}\)
\(x=\dfrac{5}{4}:\dfrac{1}{8}\)
\(\Leftrightarrow x=\dfrac{5}{4}\times8\)
\(\Leftrightarrow x=\dfrac{40}{4}\)
\(\Leftrightarrow x=10\)
15 tháng 8 2021
a) x=0
b) x=0
c) x là 1 số bất kì
d) x= 1 và x -4/9
e)x=1
CI
1
10 tháng 10 2021
\(a,\Rightarrow x^2+4x+4+x^2-2x+1+x^2-9-3x^2=-8\\ \Rightarrow2x=-4\\ \Rightarrow x=-2\\ b,\Rightarrow2021x\left(x-2020\right)-\left(x-2020\right)=0\\ \Rightarrow\left(2021x-1\right)\left(x-2020\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-2020=0\\2021x-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2020\\x=\dfrac{1}{2021}\end{matrix}\right.\)
CI
1
12 tháng 10 2021
a) \(\Rightarrow x^2+4x+4+x^2-2x+1+x^2-9-3x^2=-8\)
\(\Rightarrow2x=-4\Rightarrow x=-2\)
b) \(\Rightarrow2021x\left(x-2020\right)-\left(x-2020\right)=0\)
\(\Rightarrow\left(x-2020\right)\left(2021x-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=2020\\x=\dfrac{1}{2021}\end{matrix}\right.\)
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 12 2021
\(a.12+2x=6^3:6^2\\ 12+2x=6\\ 2x=12-6\\ 2x=6\\ x=6:2\\ x=3\)
\(b.48:x+17=33\\ 48:x=33-17\\ 48:x=16\\ x=48:16\\ x=3\)
\(c.\left(9x+2\right).5+28=83\\ \left(9x+2\right).5=83-28\\ \left(9x+2\right).5=55\\ 9x+2=55:5\\ 9x+2=11\\ 9x=11-2\\ 9x=9\\ x=9:9\\ x=1\)
\(d.187-2.\left(x+5\right)=125\\ 2.\left(x+5\right)=187-125\\ 2.\left(x+5\right)=62\\ x+5=62:2\\ x+5=31\\ x=31-5\\ x=26\)
a)x=11
b)x=2
c)x=9
a) x-3=8
x=3+8
x=11
b) 5.x=10
x=10:5
x=2
b) x+6=17
x=17-6
x=11