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\(c,10⋮2x+1\)
\(\Rightarrow2x+1\inƯ\left(10\right)=\left\{\pm1;\pm2;\pm5;\pm10\right\}\)
Ta có bảng
2x+1 | 1 | -1 | 2 | -2 | 5 | -5 | -10 | 10 |
2x | 0 | -2 | 1 | -3 | 4 | -6 | -11 | 9 |
x | 0 | -1 | 1/2 | -3/2 | 2 | -3 | -11/2 | 9/2 |
\(d,x+13⋮x+1\)
\(x+1+12⋮x+1\)
\(\Rightarrow x+1⋮x+1\)
\(\Rightarrow12⋮x+1\)
\(\Rightarrow x+1\inƯ\left(12\right)=\left\{\pm1;\pm2;\pm3;\pm4;\pm6;\pm12\right\}\)
Ta có bảng
x+1 | 1 | -1 | 2 | -2 | 3 | -3 | 4 | -4 | 6 | -6 | 12 | -12 |
x | 0 | -2 | 1 | -3 | 2 | -4 | 3 | -5 | 5 | -7 | 11 | -13 |
Bn tự KL cả 2 phần ...
\(f,2x+108⋮2x+3\)
\(\Rightarrow\left(2x+3\right)+105⋮2x+3\)
\(\Rightarrow2x+3⋮2x+3\)
\(\Rightarrow105⋮2x+3\)
\(\Rightarrow2x+3\inƯ\left(105\right)=\left\{\pm1;\pm3;\pm7;\pm15;\pm21;\pm35;\pm105\right\}\)
Ta lập bảng xét
2x+3 | 1 | -1 | 3 | -3 | 7 | -7 | 15 | -15 | 21 | -21 | 35 | -35 | 105 | -105 |
2x | -2 | -4 | 0 | -6 | 4 | -10 | 12 | -18 | 18 | -24 | 32 | -38 | 102 | -108 |
x | -1 | -2 | 0 | -3 | 2 | -5 | 6 | -9 | 9 | -12 | 16 | -19 | 51 | -54 |
Tự KL ....
\(a,\Rightarrow x=3\\ b,\Rightarrow2x-1=2\Rightarrow x=\dfrac{3}{2}\\ c,\Rightarrow\left[{}\begin{matrix}x-2=4\\x-2=-4\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=6\\x=-2\end{matrix}\right.\\ d,\Rightarrow\left[{}\begin{matrix}2x-3=3\\2x-3=-3\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\\x=0\end{matrix}\right.\\ e,\Rightarrow2x+5=3^2=9\Rightarrow x=2\)
a) 3x – 15 = 25 – 5x
=> 3x + 5x = 25 + 15
=> 8x = 40
=> x = 5
b) 3x - 17 = 2x – 7
=> 3x - 2x = -7 + 17
=> x = 10
c) 2x – 17 = – (3x – 18)
=> 2x - 17 = -3x + 18
=> 2x + 3x = 18 + 17
=> 5x = 35
=> x = 7
d) 3x – 14 = 2(x – 9) + 1
=> 3x - 14 = 2x - 18 + 1
=> 3x - 2x = -18 + 1 + 14
=> x = -3
f) (x – 5)2 = 9
\(\Rightarrow\left[{}\begin{matrix}x-5=3\\x-5=-3\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=8\\x=2\end{matrix}\right.\)
a) Ta có: \(3x-15=25-5x\)
\(\Leftrightarrow3x-15-25+5x=0\)
\(\Leftrightarrow8x-40=0\)
\(\Leftrightarrow8x=40\)
hay x=5
Vậy: x=5
b) Ta có: \(3x-17=2x-7\)
\(\Leftrightarrow3x-17-2x+7=0\)
\(\Leftrightarrow x-10=0\)
hay x=10
Vậy: x=10
c) Ta có: \(2x-17=-\left(3x-18\right)\)
\(\Leftrightarrow2x-17=-3x+18\)
\(\Leftrightarrow2x-17+3x-18=0\)
\(\Leftrightarrow5x-35=0\)
\(\Leftrightarrow5x=35\)
hay x=7
Vậy: x=7
d) Ta có: \(3x-14=2\left(x-9\right)+1\)
\(\Leftrightarrow3x-14=2x-18+1\)
\(\Leftrightarrow3x-14-2x+18-1=0\)
\(\Leftrightarrow x+3=0\)
\(\Leftrightarrow x=-3\)
Vậy: x=-3
f) Ta có: \(\left(x-5\right)^2=9\)
\(\Leftrightarrow\left[{}\begin{matrix}x-5=3\\x-5=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\\x=2\end{matrix}\right.\)
Vậy: \(x\in\left\{2;8\right\}\)
a, 36:(x–5) = 2 2
(x–5) = 9
x = 14
b, [3.(70–x)+5]:2 = 46
[3.(70–x)+5] = 92
70–x = 29
x = 41
c, 450:[41–(2x–5)] = 3 2 .5
41–(2x–5) = 10
2x–5 = 31
2x = 36
x = 18
d, 230+[ 2 4 +(x–5)] = 315. 2018 0
16+(x–5) = 315–230
x–5 = 85–16
x = 69+5
x = 74
e, 2 x + 2 x + 1 = 48
2 x .(2+1) = 48
2 x = 16 = 2 4
x = 4
f, 3 x + 2 + 3 x = 2430
3 x . 3 2 + 1 = 2430
3 x = 2430:10 = 243 = 3 5
x = 5
a) | 5/4x -7/2| - | 5/8x + 3/5| = 0
|5/4x - 7/2| = | 5/8x + 3/5|
TH1: 5/4x - 7/2 = 5/8x + 3/5
=> 5/4x - 5/8x = 3/5 +7/2
5/8x = 41/10
x = 41/10:5/8
x = 164/25
TH2: 5/4x - 7/2 = -5/8x - 3/5
=> 5/4x + 5/8x = -3/5 +7/2
15/8x = 29/10
x = 29/10 : 15/8
x = 116/75
KL: x = 164/25 hoặc x = 116/75
các bài cn lại b lm tương tự nha! h lm dài lắm!
e) \(2^x+2^{x+3}=144\)
\(=>2^x+2^x.2^3=144\)
\(=>2^x.\left(1+2^3\right)=144\)
\(=>2^x.9=144\)
\(=>2^x=144:9\)
\(=>2^x=16=2^4\)
\(=>x=4\)
__________
f) \(3^x+3^{x+1}=108\)
\(=>3^x+3^x.3=108\)
\(=>3^x.\left(1+3\right)=108\)
\(=>3^x.4=108\)
\(=>3^x=108:4\)
\(=>3^x=27=3^3\)
\(=>x=3\)
\(#Wendy.Dang\)