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a, 117 - \(x\) = 28 - (-7)
117 - \(x\) = 28 + 7
117 - \(x\) = 35
\(x\) = 117 - 35
\(x\) = 82
b, \(x\) - (-38 - 2\(x\)) = (-3) - 8 + 2\(x\)
\(x\) + 38 + 2\(x\) = - 11 + 2\(x\)
3\(x\) + 38 = - 11 + 2\(x\)
3\(x\) - 2\(x\) = - 11 - 38
\(x\) = - 49
a) \(\dfrac{2x+5}{2x+1}=\dfrac{2x+1+4}{2x+1}=\dfrac{2x+1}{2x+1}+\dfrac{4}{2x+1}=1+\dfrac{4}{2x+1}\)
Để \(\dfrac{2x+5}{2x+1}\in Z\) thì \(\dfrac{4}{2x+1}\in Z\)
\(\Rightarrow4\) ⋮ \(2x+1\)
\(\Rightarrow2x+1\inƯ\left(4\right)=\left\{1;-1;2;-2;4;-4\right\}\)
\(\Rightarrow2x\in\left\{0;-2;1;-3;3;-5\right\}\)
\(\Rightarrow x\in\left\{0;-1;\dfrac{1}{2};-\dfrac{3}{2};\dfrac{3}{2};-\dfrac{5}{2}\right\}\)
Mà x nguyên \(\Rightarrow\text{x}\in\left\{0;-1\right\}\)
b) \(\dfrac{3x+5}{x+1}=\dfrac{3x+3+2}{x+1}=\dfrac{3\left(x+1\right)+2}{x+1}=\dfrac{3\left(x+1\right)}{x+1}+\dfrac{2}{x+1}=3+\dfrac{2}{x+1}\)
Để \(\dfrac{3x+5}{x+1}\in Z\) thì \(\dfrac{2}{x+1}\in Z\)
\(\Rightarrow2\) ⋮ \(x+1\)
\(\Rightarrow x+1\inƯ\left(2\right)=\left\{1;-1;2;-2\right\}\)
\(\Rightarrow x\in\left\{0;-2;1;-3\right\}\)
c) \(\dfrac{3x+8}{x-1}=\dfrac{3x-3+11}{x-1}=\dfrac{3\left(x-1\right)+11}{x-1}=\dfrac{3\left(x-1\right)}{x-1}+\dfrac{11}{x-1}=3+\dfrac{11}{x-1}\)
Để: \(\dfrac{3x+8}{x-1}\in Z\) thì \(\dfrac{11}{x-1}\in Z\)
\(\Rightarrow11\) ⋮ \(x-1\)
\(\Rightarrow x-1\inƯ\left(11\right)=\left\{1;-1;11;-11\right\}\)
\(\Rightarrow x\in\left\{2;0;12;-10\right\}\)
d) \(\dfrac{5x+12}{x-2}=\dfrac{5x-10+22}{x-2}=\dfrac{5\left(x-2\right)+22}{x-2}=\dfrac{5\left(x-2\right)}{x-2}+\dfrac{22}{x-2}=5+\dfrac{22}{x-2}\)
Để: \(\dfrac{5x+12}{x-2}\in Z\) thì \(\dfrac{22}{x-2}\in Z\)
\(\Rightarrow22\) ⋮ \(x-2\)
\(\Rightarrow x-2\inƯ\left(22\right)=\left\{1;-1;2;-2;11;-11;22;-22\right\}\)
\(\Rightarrow x\in\left\{3;1;4;0;13;-9;24;-20\right\}\)
e) \(\dfrac{7x-12}{x+16}=\dfrac{7x+112-124}{x+16}=\dfrac{7\left(x+16\right)-124}{x+16}=\dfrac{7\left(x+16\right)}{x+16}-\dfrac{124}{x+16}=7-\dfrac{124}{x+16}\)
Để \(\dfrac{7x-12}{x+16}\in Z\) thì \(\dfrac{124}{x+16}\in Z\)
\(\Rightarrow124\) ⋮ \(x+16\)
\(\Rightarrow x+16\inƯ\left(124\right)=\left\{1;-1;2;-2;4;-4;31;-31;62;-62;124;-124\right\}\)
\(\Rightarrow x\in\left\{-15;-17;-14;-18;-12;-20;15;-47;46;-78;108;-140\right\}\)
Noob ơi, bạn phải đưa vào máy tính ý solve cái là ra x luôn, chỉ tội là đợi hơi lâu
a, 4.(18 - 5x) - 12(3x - 7) = 15(2x - 16) - 6(x + 14)
=> 72 - 20x - 36x + 84 = 30x - 240 - 6x - 84
=> (72 + 84) + (-20x - 36x) = (30x - 6x) + (-240 - 84)
=> 156 - 56x = 24x - 324
=> 24x + 56x = 324 + 156
=> 80x = 480
=> x = 480 : 80 = 6
Vậy x = 6
a, \(4\left(18-5x\right)-12\left(3x-7\right)=15\left(2x-16\right)-6\left(x+14\right)\)
\(\Leftrightarrow72-20x-36x+84=30x-240-6x-84\)
\(\Leftrightarrow156-56x=24x-324\)
\(\Leftrightarrow-80x+480=0\Leftrightarrow x=-6\)
b, \(5\left(3x+5\right)-4\left(2x-3\right)=5x+3\left(2x-12\right)+1\)
\(\Leftrightarrow15x+25-8x+12=5x+6x-36+1\)
\(\Leftrightarrow7x+37=11x-35\)
\(\Leftrightarrow-4x+72=0\Leftrightarrow x=18\)
c, \(2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\)
\(\Leftrightarrow10x-16-12x+15=12x-16+11\)
\(\Leftrightarrow-2x-1=12x-5\)
\(\Leftrightarrow-14x+4=0\Leftrightarrow x=\frac{2}{7}\)
d, \(5x-3\left\{4x-2\left[4x-3\left(5x-2\right)\right]\right\}=182\)
\(\Leftrightarrow5x-3\left[4x-15x+6\right]=182\)
\(\Leftrightarrow5x-3\left(-11x+6\right)=182\)
\(\Leftrightarrow5x+33x-18-182=0\)
\(\Leftrightarrow38x-200=0\Leftrightarrow x=\frac{100}{19}\)
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
a)4.(x+3)-(2x-12)=x-(-11+4)
4x+12-2x+12=x+11-4
2x+24=x+7
2x-x=-24+7
x=-17
b)-(4x-13)+(5x-4)=-3-(-15+7)
-4x+13+5x-4=-3+15-7
(-4x+5x)+13-4=12-7
x+9=5
x=-4
c)(5x-3)-(-2x+4)=6x-12
5x-3+2x-4=6x-12
5x+2x-6x=3+4-12
x=-5
d)(15x+20)-(9x-3)=5x-(-12)
15x+20-9x+3=5x+12
15x-9x-5x=-20-3+12
x=-11
e,(7x+14)+(3x-8)=-(-9x+3)
7x+14+3x-8=9x-3
7x+3x-9x=-14+8-3
x=-9
Tìm x
a, 5x - 10 . (-7) = 15 . (-3)
5x + 70 = -45
5x = -45 - 70
5x = -115
x = -115 : 5
x = -23
Vậy x = -23.
b, 2x + 12 = -8
2x = -8 - 12
2x = -20
x = -20 : 2
x = -10
Vậy x = -10.
c, 3x - 5 = -20
3x = -20 + 5
3x = -15
x = -15 : 3
x = -5
Vậy x = -5.
d, 2x - (-12) = -8 . 23
2x + 12 = -184
2x = -184 - 12
2x = -196
x = -196 : 2
x = -98
Vậy x = -98.
e, x2 = 49
x2 = 72 = (-7)2
=> x = 7 hoặc x = -7.
Vậy x thuộc {-7 ; 7}
i, (x - 2)2 + 1 = 65
(x - 2)2 = 65 - 1
(x - 2)2 = 64
(x - 2)2 = 82 = (-8)2
=> x - 2 = 8 hoặc x - 2 = -8
=> x = 10 hoăc x = -6
Vậy x thuộc {-6 ; 10}.
g, 2 (x - 3)2 - 5 = 5 . (-3)2
2 (x - 3)2 - 5 = 5 . 9
2 (x - 3)2 - 5 = 45
2 (x - 3)2 = 45 + 5
2 (x - 3)2 = 50
(x - 3)2 = 50 : 2
(x - 3)2 = 25
(x - 3)2 = 52 = (-5)2
=> x - 3 = 5 hoặc x - 3 = -5
=> x = 8 hoặc x = -2
Vậy x thuộc {-2 ; 8}.
a) 5x - 10.(-7)=15.(-3) b) 2x+12=-8 c) 3x-5=-20 d)2x-(-12)=-8.23 e)\(^{x^2}\)=49 i)\(\left(x-2\right)^2+1\)=65
5x+70=-45 2x=-8-12 3x=-20+5 2x+12=-184 x=7 \(\left(x-2\right)^2\) =64
5x=-45-70 2x=-20 3x=-15 2x=-184-12 X-2 =8
5x=-115 x=-20:2 x=-15:3 2x=-186 x =10
x=-23 x=-10 x=-5 x=-93
g)\(2.\left(x-3\right)^2-5=5.\left(-3\right)^2\)
\(2\left(x-3\right)^2-5=30\)
\(2.\left(x-3\right)^2=35\)
\(2.\left(x^2-2x.3+3^2\right)=35\)
\(2x^2-24x+18=35\)
\(2x^2-24x=17\)
\(2x\left(x-12\right)=17\)
\(\orbr{\begin{cases}2x=17\\x-12=17\end{cases}}\orbr{\begin{cases}x=\frac{17}{2}\\x=29\end{cases}}\)
X
a: =>x+38+2x=-3-8+2x
=>3x+38=2x-11
=>x=-49
b: \(\Leftrightarrow65+x-15-5x=12-5x\)
=>-4x+50=-5x+12
=>x=-38
c: \(\Leftrightarrow3x+12-7x+21=-3-5x-2=-5x-5\)
=>-4x+33=-5x-5
=>x=-38
d: \(\Leftrightarrow-123+2x+23=x-120\)
=>2x-100=x-120
=>x=-20
e: =>-45+25+5x=16-x
=>5x-20=-x+16
=>6x=36
=>x=6