Tìm x biết:
(x-2)6=(x-2)10
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\(\text{(x+2)(y-3)=5 }\)
\(\Rightarrow\)x+2;y-3\(\in\)Ư(5)
Mà Ư(5)={1;5;-1;-5}
Có bảng:
Th1:
x+2=1;y-3=6
=>x=-3
y=9
Tương tự 3 trường hợp còn lại
1) 2(x + 5) + 3(x + 7) = 41
2x + 10 + 3x + 21 = 41
5x + 31 = 41
5x = 10
x = 2
6) 7(x - 1) + 5(3 - x) = 11x - 10
7x - 7 + 15 - 5x = 11x - 10
2x + 8 = 11x - 10
-9x = -18
x = 2
2) 5(x + 6) + 2(x - 3) = 38
5x + 30 + 2x - 6 = 38
7x + 24 = 38
7x = 14
x = 2
7) 4(2 + x) + 3(x - 2) = 12
8 + 4x + 3x - 6 = 12
7x + 2 = 12
7x = 10
x = 10/7
3) 7(5 + x) + 2(x - 10) = 15
35 + 7x + 2x - 20 = 15
9x + 15 = 15
9x = 0
x = 0
8) 5(2 + x) + 4(3 - x) = 10x - 15
10 + 5x + 12 - 4x = 10x - 15
x + 22 = 10x - 15
9x = 37
x = 37/9
4) 3(x + 4) + (8 - 2x) = 22
3x + 12 + 8 - 2x = 22
x + 20 = 22
x = 2
9) 7(x - 2) + 5(3 - x) = 11x - 6
7x - 14 + 15 - 5x = 11x - 6
2x + 1 = 11x - 6
-9x = -7
x = 7/9
5) 4(x + 5) + 3(7 - x) = 49
4x + 20 + 21 - 3x = 49
x + 41 = 49
x = 8
10) 5(3 - x) + 5(x + 4) = 6 + 4x
15 - 5x + 5x + 20 = 6 + 4x
35 = 6 + 4x
4x = 29
x = 29/4
1) 2(x + 5) + 3(x + 7) = 41
2x + 10 + 3x + 21 = 41
5x + 31 = 41
5x = 41 - 31
5x = 10
x = 10 : 5
x = 2
2) 5(x + 6) + 2(x - 3) = 38
5x + 30 + 2x - 6 = 38
7x + 24 = 38
7x = 38 - 24
7x = 14
x = 14 : 7
x = 2
3) 7(5 + x) + 2(x - 10) = 15
35 + 7x + 2x - 20 = 15
9x + 15 = 15
9x = 15 - 15
9x = 0
x = 0
4) 3(x + 4) + (8 - 2x) = 22
3x + 12 + 8 - 2x = 22
x + 20 = 22
x = 22 - 20
x = 2
5) 4(x + 5) + 3(7 - x) = 49
4x + 20 + 21 - 3x = 49
x + 41 = 49
x = 49 - 41
x = 8
6) 7(x - 1) + 5(3 - x) = 11x - 10
7x - 7 + 15 - 5x = 11x - 10
2x - 11x + 8 = -10
-9x = -10 - 8
-9x = -18
x = -18 : (-9)
x = 2
7) 4(2 + x) + 3(x - 2) = 12
8 + 4x + 3x - 6 = 12
7x + 2 = 12
7x = 12 - 2
7x = 10
x = 10/7
8) 5(2 + x) + 4(3 - x) = 10x - 15
10 + 5x + 12 - 4x = 10x - 15
10x - 15 = x + 22
10x - x = 22 + 15
9x = 37
x = 37/9
9) 7(x - 2) + 5(3 - x) = 11x - 6
7x - 14 + 15 - 5x = 11x - 6
11x - 6 = 2x + 1
11x - 2x = 1 + 6
9x = 7
x = 7/9
10) 5(3 - x) + 5(x + 4) = 6 + 4x
15 - 5x + 5x + 20 = 6 + 4x
6 + 4x = 35
4x = 35 - 6
4x = 29
x = 29/4
Lời giải:
a.
$x-\frac{1}{2}=\frac{2}{3}$
$x=\frac{2}{3}+\frac{1}{2}=\frac{7}{6}$
b.
$x\times \frac{5}{6}=\frac{1}{2}$
$x=\frac{1}{2}: \frac{5}{6}=\frac{3}{5}$
c.
$x:\frac{2}{5}=10$
$x=10\times \frac{2}{5}=4$
d.
$\frac{7}{5}-x=\frac{12}{10}$
$x=\frac{7}{5}-\frac{12}{10}=\frac{1}{5}$
\(10+2\cdot x=4^5:4^3\)
\(\Rightarrow10+2\cdot x=4^{5-3}\)
\(\Rightarrow10+2\cdot x=4^2\)
\(\Rightarrow10+2\cdot x=16\)
\(\Rightarrow2\cdot x=16-10\)
\(\Rightarrow2\cdot x=6\)
\(\Rightarrow x=\dfrac{6}{2}=3\)
__________________
\(2\cdot x-6^2:18=3\cdot2^2\)
\(\Rightarrow2\cdot x-36:18=12\)
\(\Rightarrow2\cdot x-2=12\)
\(\Rightarrow2\cdot x=12+2\)
\(\Rightarrow2\cdot x=14\)
\(\Rightarrow x=\dfrac{14}{2}=7\)
_________________
\(70-5\cdot\left(x-3\right)=3\cdot2\)
\(\Rightarrow70-5\cdot\left(x-3\right)=6\)
\(\Rightarrow70-5x+15=6\)
\(\Rightarrow-5x+15=6-70\)
\(\Rightarrow-5x+15=64\)
\(\Rightarrow-5x=64-15\)
\(\Rightarrow-5x=49\)
\(\Rightarrow x=-\dfrac{49}{5}\)
(x-2)6 = (x-2)10
<=> (x-2)6 - (x-2)10 = 0
<=> (x-2)6[1-(x-2)4] = 0
<=> \(\left[\begin{array}{nghiempt}\left(x-2\right)^6=0\\1-\left(x-2\right)^4=0\end{array}\right.\)
<=> \(\left[\begin{array}{nghiempt}x-2=0\\\left(x-2\right)^4=1\end{array}\right.\)
<=> \(\left[\begin{array}{nghiempt}x=2\\\left[\begin{array}{nghiempt}x-2=1\\x-2=-1\end{array}\right.\end{array}\right.\)
<=> \(\left[\begin{array}{nghiempt}x=2\\\left[\begin{array}{nghiempt}x=3\\x=1\end{array}\right.\end{array}\right.\)
Vậy x \(\in\){2;3;1}
x = 3