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a: \(\left[\left(10-x\right)\cdot2+51\right]:3-2=3\)
=>\(\left[2\left(10-x\right)+51\right]:3=5\)
=>\(\left[2\left(10-x\right)+51\right]=15\)
=>\(2\left(10-x\right)=15-51=-36\)
=>10-x=-36/2=-18
=>\(x=10-\left(-18\right)=10+18=28\)
b: \(\left(x-12\right)-15=20-\left(17+x\right)\)
=>\(x-12-15=20-17-x\)
=>\(x-27=3-x\)
=>\(2x=30\)
=>\(x=\dfrac{30}{2}=15\)
c: \(720-\left[41-\left(2x-5\right)\right]=2^3\cdot5\)
=>\(720-\left[41-2x+5\right]=8\cdot5=40\)
=>\(\left[41-2x+5\right]=720-40=680\)
=>-2x+46=680
=>-2x=680-46=634
=>\(x=\dfrac{634}{-2}=-317\)
Bài 1L
a) \(\left(x-7\right)\left(x+3\right)< 0\)
TH1:
\(\hept{\begin{cases}x-7>0\\x+3< 0\end{cases}\Leftrightarrow\hept{\begin{cases}x>7\\x< -3\end{cases}}}\)( loại )
TH2:
\(\hept{\begin{cases}x-7< 0\\x+3>0\end{cases}\Leftrightarrow\hept{\begin{cases}x< 7\\x>-3\end{cases}\Leftrightarrow}-3< x< 7}\)( chọn )
Vậy \(-3< x< 7\)
Bài 2:
a) \(\left(5x+8\right)-\left(2x-15\right)+21=2x-5\)
\(\Leftrightarrow5x+8-2x+15+21=2x-5\)
\(\Leftrightarrow5x-2x-2x=-5-21-8-15\)
\(\Leftrightarrow x=-49\)
Vậy ...
1)
a) 2x + 5 = 3⁴ : 3²
2x + 5 = 3²
2x + 5 = 9
2x = 9 - 5
2x = 4
x = 4 : 2
x = 2
b) (3x - 24).73 = 2.74
(3x - 24).73 = 148
3x - 24 = 148/73
3x = 148/73 + 24
3x = 1900/73
x = 1900/73 : 3
x = 1900/219
c) [3.(42 - x)] + 15 = 23.3
126 - 3x + 15 = 69
141 - 3x = 69
3x = 141 - 69
3x = 72
x = 72 : 3
x = 24
d) 126 + (132 - x) = 300
132 - x = 300 - 126
132 - x = 174
x = 132 - 174
x = -42
2)
a) 120 - (x + 55) = 60
x + 55 = 120 - 60
x + 155 = 60
x = 60 - 55
x = 5
b) (7x - 11).3 = 25.52 + 200
(7x - 11).3 = 1500
7x - 11 = 1500 : 3
7x - 11 = 500
7x = 500 + 11
7x = 511
x = 511 : 7
x = 73
c) 2x + 2x + 4 = 544
4x = 544 - 4
4x = 540
x = 540 : 4
x = 135
a: \(4x^3+12=120\)
=>\(4x^3=108\)
=>\(x^3=27=3^3\)
=>x=3
b: \(\left(x-4\right)^2=64\)
=>\(\left[{}\begin{matrix}x-4=8\\x-4=-8\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=12\\x=-4\end{matrix}\right.\)
c: (x+1)^3-2=5^2
=>\(\left(x+1\right)^3=25+2=27\)
=>x+1=3
=>x=2
d: 136-(x+5)^2=100
=>(x+5)^2=36
=>\(\left[{}\begin{matrix}x+5=6\\x+5=-6\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=-11\end{matrix}\right.\)
e: \(4^x=16\)
=>\(4^x=4^2\)
=>x=2
f: \(7^x\cdot3-147=0\)
=>\(3\cdot7^x=147\)
=>\(7^x=49\)
=>x=2
g: \(2^{x+3}-15=17\)
=>\(2^{x+3}=32\)
=>x+3=5
=>x=2
h: \(5^{2x-4}\cdot4=10^2\)
=>\(5^{2x-4}=\dfrac{100}{4}=25\)
=>2x-4=2
=>2x=6
=>x=3
i: (32-4x)(7-x)=0
=>(4x-32)(x-7)=0
=>4(x-8)*(x-7)=0
=>(x-8)(x-7)=0
=>\(\left[{}\begin{matrix}x-8=0\\x-7=0\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=8\\x=7\end{matrix}\right.\)
k: (8-x)(10-2x)=0
=>(x-8)(x-5)=0
=>\(\left[{}\begin{matrix}x-8=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\\x=5\end{matrix}\right.\)
m: \(3^x+3^{x+1}=108\)
=>\(3^x+3^x\cdot3=108\)
=>\(4\cdot3^x=108\)
=>\(3^x=27\)
=>x=3
n: \(5^{x+2}+5^{x+1}=750\)
=>\(5^x\cdot25+5^x\cdot5=750\)
=>\(5^x\cdot30=750\)
=>\(5^x=25\)
=>x=2
\(1.\) \(-12\left(x-5\right)+7\left(3-x\right)=5\)
\(=>-12x+60+21-7x=5\)
\(=>-12x+81-7x=5\)
\(=>-12x-7x+81=5\)
\(=>-19x+81=5\)
\(=>-19x=-76\)
\(=>x=4\)
\(2.\) \(\left(x-2\right).\left(x+15\right)=0\)
\(=>\left[\begin{matrix}x-2=0\\x+15=0\end{matrix}\right.=>\left[\begin{matrix}x=2\\x=-15\end{matrix}\right.\)
\(3.\) \(\left(7-x\right).\left(x+19\right)=0\)
\(=>\left[\begin{matrix}7-x=0\\x+19=0\end{matrix}\right.=>\left[\begin{matrix}x=7\\x=-19\end{matrix}\right.\)
\(4.\) \(\left|x\right|< 3\)
Xét: x là số dương => x < 3
Xét: x là số âm => x < -3
1: =>-12x+60+21-7x=5
=>-19x=-76
hay x=4
2: =>x-2=0 hoặc x+15=0
=>x=2 hoặc x=-15
3: =>7-x=0 hoặc x+19=0
=>x=7 hoặc x=-19
1: =>-12x+60+21-7x=5
=>-19x=-76
hay x=4
2: =>x-2=0 hoặc x+15=0
=>x=2 hoặc x=-15
3: =>7-x=0 hoặc x+19=0
=>x=7 hoặc x=-19
\(5.x+x=15:3+13\\ \Rightarrow\left(5+1\right).x=5+13=18\\ \Rightarrow6x=18\\ \Rightarrow x=\dfrac{18}{6}=3\)
\(\dfrac{2}{3}x+\dfrac{1}{3}x+\dfrac{1}{2}=\dfrac{2}{3}\\ \Rightarrow x\left(\dfrac{2}{3}+\dfrac{1}{3}\right)=\dfrac{2}{3}-\dfrac{1}{2}\\ \Rightarrow x=\dfrac{4}{6}-\dfrac{3}{6}=\dfrac{1}{6}\)
\(\dfrac{2}{5}x+x=\dfrac{3}{2}\\ \Rightarrow x\left(\dfrac{2}{5}+1\right)=\dfrac{3}{2}\\ \Rightarrow x\left(\dfrac{2}{5}+\dfrac{5}{5}\right)=\dfrac{3}{2}\\ \Rightarrow x.\dfrac{7}{5}=\dfrac{3}{2}\\ \Rightarrow x=\dfrac{3}{2}:\dfrac{7}{5}\\ \Rightarrow x=\dfrac{3.5}{7.2}=\dfrac{15}{14}\)
\(5\cdot x+x=15:3+13\)
\(\Rightarrow x\cdot\left(5+1\right)=5+13\)
\(\Rightarrow x\cdot6=18\)
\(\Rightarrow x=\dfrac{18}{6}=3\)
================
\(\dfrac{2}{3}\cdot x+\dfrac{1}{3}\cdot x+\dfrac{1}{2}=\dfrac{2}{3}\)
\(\Rightarrow x\cdot\left(\dfrac{2}{3}+\dfrac{1}{3}\right)=\dfrac{2}{3}-\dfrac{1}{2}\)
\(\Rightarrow x\cdot1=\dfrac{1}{6}\)
\(\Rightarrow x=\dfrac{1}{6}\)
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\(\dfrac{2}{5}\cdot x+x=\dfrac{3}{2}\)
\(\Rightarrow x\cdot\left(\dfrac{2}{5}+1\right)=\dfrac{3}{2}\)
\(\Rightarrow x\cdot\dfrac{7}{5}=\dfrac{3}{2}\)
\(\Rightarrow x=\dfrac{3}{2}:\dfrac{7}{5}\)
\(\Rightarrow x=\dfrac{15}{14}\)