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5.3x = 8.310 - 3.310
<=> 5.3x = ( 8-3 ) . 310
<=> 5.3x = 5 .310
<=> x = 10
5.3x = 8.310 - 3.310
<=> 5.3x = ( 8-3 ) . 310
<=> 5.3x = 5 .310
<=> x = 10
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\(5.3^x=8.3^{10}-3.3^{10}\)
\(\Rightarrow5.3^x=\left(8-3\right).3^{10}\)
\(\Rightarrow5.3^x=5.3^{10}\)
\(\Rightarrow x=10\)
a. 37-7.(x+1)=40-8.3
37 - 7(x+1) = 16
7(x+1) = 21
x+1 = 3
x=2
b. (x+5).3=11=10+(3+4).4
(sao có 2 dấu bằng, dấu nào là dấu cộng do viết nhầm ???)
c. 25-5.3+4.10=100-5.x
50 = 100 - 5x
5x = 50
x=10
d. 36+2.(x-7)=12+8.(3+5)-36
36 + 2(x-7) = 40
2(x-7) = 4
x-7 =2
x=9
e. 3.(x+7)-9=11.5-5-8
3(x+7) - 9 = 42
3(x+7) = 51
x+7 = 17
x=10
1/ ta co
vi x \(\in Z\Rightarrow x\in\left\{-9;-8;..;9;10\right\}\)
Tong cac so x thoa man la
-9+(-8)+(-7)+....+9+10
=(-9+9)+(-8+8)+...+(-1+1)+0
=0+0+0+..+0+0
=0
vay tong cac so ma x thoa man la 0
2/ ta co
vi x \(\in Z\Rightarrow x\in\left\{-8;-7;..;5;6;7\right\}\)
Tong cac so ma x thoa man la
-8+(-7)+(-6)+...+6+7
=-8+0+(-7+7)+(-6+6)+(-5+5)+...+(-1+1)
=-8+0+0+0+...+0
=-8
vay tong cac gia tri ma x thoa man la -8
3/ ta co
vi x \(\in Z\Rightarrow x\in\left\{-22;-21;...;22;23\right\}\)
Tong cac gia tri ma x thoa man la
(-22)+(-21)+....+22+23
=23+0+(-21+21)+(-22+22)+...+(-1+1)
=23+0+0+0+...+0
=23
vay tong cac gia tri ma x thoa man la 23
4/ ta co :
vi |x|\(\le2\Rightarrow\left|x\right|\in\left\{1;2\right\}hay.x\in\left\{2;1;-1;-2\right\}\)
Tong cac gia tri ma x thoa man la :
2+1+(-1)+(-2)
=3+(-3)
=0
vay tong cac gia tri ma x thoa man la 0
5/ ta co
│-x│< 13 nen |x| \(\in\left\{12;11;10;..;2;1;0;-1;-2;...\right\}\)
hay x \(\in\left\{12;11;10;9;...;1;0;-12;-13;...;-1\right\}\)
Tong cac so ma x thoa man la
12+13+14+15+....+1+0+(-1)+(-2)+....+(-12)
=(-12+12)+(-13+13)+...+(-1+1)+0
=0+0+0+0+...+0+0
=0
Vay tong cac gia tri ma x thoa man la 0
7) \(\dfrac{-5}{17}+\dfrac{3}{17}\le\dfrac{x}{17}\le\dfrac{13}{17}+\dfrac{-11}{17}\)
\(\Rightarrow\dfrac{-2}{17}\le\dfrac{x}{17}\le\dfrac{2}{17}\)
\(\Rightarrow-2\le x\le2\)
\(\Rightarrow x\in\left\{-2;-1;0;1;2\right\}\)
8) \(\dfrac{2}{3}\left(\dfrac{1}{2}+\dfrac{3}{4}-\dfrac{1}{3}\right)\le\dfrac{x}{18}\le\dfrac{7}{3}\left(\dfrac{1}{2}-\dfrac{1}{6}\right)\)
\(\Rightarrow\dfrac{2}{3}\left(\dfrac{6}{12}+\dfrac{9}{12}-\dfrac{4}{12}\right)\le\dfrac{x}{18}\le\dfrac{7}{3}\left(\dfrac{6}{12}-\dfrac{2}{12}\right)\)
\(\Rightarrow\dfrac{2}{3}\cdot\dfrac{11}{12}\le\dfrac{x}{18}\le\dfrac{7}{3}\cdot\dfrac{4}{12}\)
\(\Rightarrow\dfrac{22}{36}\le\dfrac{x}{18}\le\dfrac{28}{36}\)
\(\Rightarrow\dfrac{11}{18}\le\dfrac{x}{18}\le\dfrac{14}{18}\)
\(\Rightarrow x\in\left\{11;12;13;14\right\}\)
8) \(\dfrac{2}{3}\left(\dfrac{1}{2}+\dfrac{3}{4}-\dfrac{1}{3}\right)\le\dfrac{x}{18}\le\dfrac{7}{3}\left(\dfrac{1}{2}-\dfrac{1}{6}\right)\\ \dfrac{2}{3}\left(\dfrac{6}{12}+\dfrac{9}{12}-\dfrac{4}{12}\right)\le\dfrac{x}{18}\le\dfrac{7}{3}\left(\dfrac{3}{6}-\dfrac{1}{6}\right)\\ \dfrac{2}{3}.\dfrac{11}{12}\le\dfrac{x}{18}\le\dfrac{7}{3}.\dfrac{2}{6}\\ \dfrac{11}{18}\le\dfrac{x}{18}\le\dfrac{14}{18}\\ \Rightarrow11\le x\le14\\ \Rightarrow x\in\left\{11;12;13;14\right\}\)
a. 8 + (x - 9) = 125 - 64
8 + (x - 9) = 61
x - 9 = 53
x = 62
b. 5 x (X + 7) - 10 = 8 x 5
5 x (X + 7) - 10 = 40
5 x (X + 7) = 50
X + 7 = 10
X = 3
\(\dfrac{-8}{13}+\dfrac{-7}{17}+\dfrac{21}{13}\le x\le\dfrac{-9}{14}+3+\dfrac{5}{-14}\)
=> \(\dfrac{10}{17}\le x\le2\)
=> \(\dfrac{10}{17}\le\dfrac{17x}{17}\le\dfrac{34}{17}\)
=> 10 \(\le17x\le34\)
=> x = 1; 2 (thỏa mãn)
@Khánh Linh
a) \(2^x=32\)
Ta có: \(2^5=32\)
\(\Rightarrow2^x=2^5\)
\(\Rightarrow x=5\)
b) Sửa đề tí: \(9< 3^x< 81\)
\(\Rightarrow3^2< 3^x< 3^4\)
\(\Rightarrow2< x< 4\)
\(\Rightarrow x=\left\{3\right\}\)
Vậy x = 3
c) Ta có: \(25\le5^x\le125\)
\(\Rightarrow5^2\le5^x\le5^3\)
\(\Rightarrow2\le x\le3\)
\(\Rightarrow x=\left\{2;3\right\}\)
Vậy x = 2 hoặc x = 3
d) \(\left(x-2\right)^3\times5=40\)
\(\Rightarrow\left(x-2\right)^3=8\)
Mà \(8=2^3\Rightarrow\left(x-2\right)^3=2^3\)
Suy ra: x - 2 = 2
Vậy x = 4
13, ta có 25\(\le5^x\le125\)
Hay 52 \(\le\) 5x \(\le\) 53
suy ra 2 \(\le\) x \(\le\) 3
Cái này chắc là làm đến đây thôi vì có thể x là phân số hoặc số thập phân thì chịu !!!!
13) \(25\le5^x\le125\\ 5^2\le5^x\le5^3\\ \Rightarrow2\le x\le3\\ \Rightarrow x\in\left\{\varnothing\right\}\)