Tìm X:
a) x : 9 = 103 b) X x 4 = 848
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(a,\left(x-3\right)\left(x-1\right)=\left(x-3\right)^2\\ \Leftrightarrow\left(x-3\right)\left(x-1-x+3\right)=0\\ \Leftrightarrow2\left(x-3\right)=0\\ \Leftrightarrow x=3\)
\(b,4x^2-9=0\\ \Leftrightarrow\left(2x-3\right)\left(2x+3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}2x-3=0\\2x+3=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)
\(c,x^2+6x+9=0\\ \Leftrightarrow\left(x+3\right)^2=0\\ \Leftrightarrow x+3=0\\ \Leftrightarrow x=-3\)
a. \(\left(x-3\right)\left(x-1\right)=\left(x-3\right)^2\)
\(\Leftrightarrow\left(x-3\right)\left(x-1-x+3\right)=0\)
\(\Leftrightarrow2\left(x-3\right)=0\)
\(\Leftrightarrow x-3=0\)
\(\Leftrightarrow x=3\)
\(a,x-\dfrac{1}{4}=\dfrac{7}{2}.\dfrac{-3}{5}\\ \Rightarrow x-\dfrac{1}{4}=\dfrac{-21}{10}\\ \Rightarrow x=\dfrac{-21}{10}+\dfrac{1}{4}\\ \Rightarrow x=\dfrac{-37}{20}\\ b,\dfrac{x}{134}=\dfrac{9}{7}.\dfrac{5}{-11}\\ \Rightarrow\dfrac{x}{134}=\dfrac{-45}{77}\\ \Rightarrow x=\dfrac{-45}{77}.134\\ \Rightarrow x=\dfrac{-6030}{77}\)
\(x-\dfrac{1}{4}=\dfrac{7}{2}\cdot\dfrac{-3}{5}\)
\(x-\dfrac{1}{4}=\dfrac{-21}{10}\)
\(x=\dfrac{-21}{10}+\dfrac{1}{4}\)
\(x=\dfrac{-37}{20}\)
b ) \(\dfrac{x}{134}=\dfrac{9}{7}\cdot\dfrac{5}{-11}\)
\(\dfrac{x}{134}=\dfrac{9}{7}\cdot\dfrac{-5}{11}\)
\(\dfrac{x}{134}=\dfrac{-45}{77}\)
\(x=\dfrac{-45}{77}\cdot134\)
\(x=-\dfrac{6034}{77}\)
\(a,\sqrt{9x^2}=2x+1\\ \Leftrightarrow\left[{}\begin{matrix}3x=2x+1,\forall x\ge0\\-3x=2x+1,\forall x< 0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=1,\forall x\ge0\left(N\right)\\x=-1,\forall x< 0\left(N\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)
\(b,\sqrt{x^2+6x+9}=3x-1\\ \Leftrightarrow\sqrt{\left(x+3\right)^2}=3x-1\\ \Leftrightarrow\left[{}\begin{matrix}x+3=3x-1,\forall x+3\ge0\\x+3=1-3x,\forall x+3< 0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=2,\forall x\ge-3\left(N\right)\\x=-\dfrac{1}{2},\forall x< -3\left(L\right)\end{matrix}\right.\Leftrightarrow x=2\)
\(c,\sqrt{x^2-2x+4}=2x-3\left(x\in R\right)\\ \Leftrightarrow x^2-2x+4=\left(2x-3\right)^2\\ \Leftrightarrow x^2-2x+4=4x^2-12x+9\\ \Leftrightarrow3x^2-10x+5=0\\ \Delta=100-4\cdot3\cdot5=40\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{10-\sqrt{40}}{6}\\x=\dfrac{10+\sqrt{40}}{6}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5-\sqrt{10}}{3}\\x=\dfrac{5+\sqrt{10}}{3}\end{matrix}\right.\)
\(a.\sqrt{9x^2}=2x+1\)
<=> \(\sqrt{9}x=2x+1\)
<=> 3x = 2x + 1
<=> 3x - 2x = 1
<=> x = 1
a) \(\dfrac{-x}{4}=\dfrac{-9}{x}\)
\(\Rightarrow-x^2=-36\)
\(\Rightarrow x^2=36\)
\(\Rightarrow\left[{}\begin{matrix}x=6\\x=-6\end{matrix}\right.\)
Vậy: \(x\in\left\{6;-6\right\}\)
b) \(\dfrac{5}{9}+\dfrac{x}{-1}=-\dfrac{1}{3}\)
\(\Rightarrow\dfrac{5}{9}+\dfrac{-9x}{9}=\dfrac{-3}{9}\)
\(\Rightarrow5-9x=-3\)
\(\Rightarrow-9x=-8\)
\(\Rightarrow x=\dfrac{8}{9}\)
Vậy: \(x=\dfrac{8}{9}\)
c) \(x:3\dfrac{1}{5}=1\dfrac{1}{2}\)
\(\Rightarrow x:\dfrac{16}{5}=\dfrac{3}{2}\)
\(\Rightarrow x=\dfrac{3}{2}.\dfrac{16}{5}\)
\(\Rightarrow x=\dfrac{24}{5}\)
Vậy: \(x=\dfrac{24}{5}\)
d) \(\dfrac{3x-1}{-5}=\dfrac{-5}{3x-1}\)
\(\Rightarrow\left(3x-1\right)^2=25\)
\(\Rightarrow\left[{}\begin{matrix}3x-1=5\\3x-1=-5\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}3x=6\\3x=-4\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=2\\x=-\dfrac{4}{3}\end{matrix}\right.\)
Vậy: \(x\in\left\{2;-\dfrac{4}{3}\right\}\)
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) \(-\dfrac{15}{4}.\left(\dfrac{-7}{15}\right).\left(-2\dfrac{2}{5}\right)\)
\(=\dfrac{7}{4}.\dfrac{-12}{5}\)
\(=\dfrac{-21}{5}\)
\(#Wendy.Dang\)
\(a,\Leftrightarrow\left(x-2\right)\left(5x-2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{2}{5}\end{matrix}\right.\\ b,\Leftrightarrow2x^2+2x-x^2+4x-4-6=0\\ \Leftrightarrow x^2+6x-10=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-3+\sqrt{19}\\x=-3-\sqrt{19}\end{matrix}\right.\\ c,\Leftrightarrow2x^2-2x+9x-9=0\\ \Leftrightarrow\left(2x+9\right)\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{9}{2}\\x=1\end{matrix}\right.\)
a: Ta có: \(71-\left(x+33\right)=26\)
\(\Leftrightarrow x+33=45\)
hay x=12
b: Ta có: \(\left(x+73\right)-26=76\)
\(\Leftrightarrow x+73=102\)
hay x=29
c: Ta có: \(45-\left(x+9\right)=6\)
\(\Leftrightarrow x+9=39\)
hay x=30
a) 71 - ( 33 + x ) = 26
(33 + x ) = 71 -26
33 + x = 45
x = 45 - 33 = 12
b) ( x + 73 ) - 26 = 76
( x + 73 ) = 102
x = 102 - 73 = 29
c) 45 - ( x + 9 ) = 6
( x + 9 ) = 45 - 6 = 39
x = 39 - 9 = 30
d) 89 - ( 73 - x ) = 20
73 - x = 89 - 20
73 - x = 69
x = 73 - 69 = 4
e) 4(x+41) = 400
x + 41 = 400 : 4 = 100
x = 100 - 41 = 59
f) 11(x-9 ) = 77
x - 9 = 77 : 11 = 7
x = 7 + 9 = 16
g) x + 7 = 25 + 13 = 38
x = 38 - 7 = 31
h) x + 4 = 198 - 120 = 78
x = 78 - 4 = 74
i) x - 9 = 350 : 5 = 70
x = 70 + 9 = 79
j) 2x - 49 = 5 . 9 = 45
2x = 45 + 49 = 94
x = 94 : 2 = 47
k) 25 + 3( x - 8 ) = 106
3(x-8 ) = 106 - 25 = 81
x - 8 = 81 : 3 = 27
x = 27 +8= 35
l) 9( x + 4 ) - 25 = 20
9( x + 4 ) = 20 + 25 = 45
x + 4 =45 : 9 = 5
x = 5 - 4 = 1
m) 200 - ( 2x + 6 ) = 64
2x + 6 = 200 - 64 = 136
2x = 136 - 6 = 130
x = 130 : 2 = 65
a) x : 9 = 103 b) X x 4 = 848
x = 103 x 9 x = 848 : 4
x = 927 x = 212