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vì 77+x chia hết cho x+5 và 86+x chia hết cho x+5 => (86+x)-(77+x) chia hết cho x+5
=>9 chia hết cho x+5=>x+5 thuộc Ư(9) =>x+5 thuộc{1,-1,3,-3,9,-9}
=>x thuộc{-4,-6,-2,-8,4,-14}
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Bài 1:
a)
<=> 3x - 18 - 5x + 10 = 24
<=> 3x - 5x = 24 + 18 - 10
<=> -2x = 32
<=> x = 32 : (-2)
<=> x = -16
b)
<=> -4x + 20 - 8x + 16 = 48
<=> -4x - 8x = 48 - 20 - 16
<=> -12x = 12
<=> x = 12 : (-12)
<=> x = -1
Bài 2:
\(=a^2-ab+ab-b^2\)
\(=a^2-b^2\)
a) 3(x-6)-5(x-2) = 24
<=> 3x -36 -5x + 10 =24
<=> -2x = 50
<=> x = -25
b) -4(x-5) -8(x-2) = 48
<=> -4x +20 - 8x +16 = 48
<=> -12x = 12
<=> x = -1
(a+b)(a-b) = a^2 -ab +ab -b^2 = a^2 - b^2
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|-x - 5 | + 2 = 3
<=> | -x - 5| = 1
\(\Leftrightarrow\orbr{\begin{cases}-x-5=1\\-x-5=-1\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-6\\-4\end{cases}}}\)
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Ta có: 2(x-5)-3(x-4)=-6+15(-3)
=>2x-10-3x+12=-6-45
=>-1x+2=-51
=>-1x=-53
=>x=53
Vậy x=53
Tìm x biết : 2 ( x - 5 ) - 3 ( x - 4 ) = - 6 + 15 ( - 3 )
2.(x-5)-3.(x-4)=-6+15.-3
2 (x − 5) − 3 (x − 4) = −51
(2x − 10) − (3x − 12) = −51
2x − 10 − 3x + 12 = −51
(2x − 3x) + (−10 + 12) = −51
−x + 2 = −51 −x = −53
x = 53
Vậy x = 53.
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\(150-5\left(x-2\right)^2=25\)
\(5\left(x-2\right)^2=150-25=125\)
\(\left(x-2\right)^2=125:5=25\)
\(\Rightarrow\orbr{\begin{cases}x-2=5\\x-2=-5\end{cases}\Rightarrow\orbr{\begin{cases}x=7\\x=-3\end{cases}}}\)
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a, 15/x - 1/3 = 28/57
15/x = 28/57 + 1/3
15/x = 28/57 + 19/57
15/x = 47/57
x . 47 = 15 . 57
x = 855/47
b, x/2 - 2/5 = 1/10
x/2 = 1/10 + 2/5
x/2 = 1/10 + 4/10
x/2 = 5/10 = 1/2
x/2 = 1/2
=> x=1
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A)\(\left|x+1\right|+\left|x+1\right|=2\)
\(\Rightarrow2.\left|x+1\right|=2\)
\(\Rightarrow\left|x+1\right|=2:2\)
\(\Rightarrow\left|x+1\right|=1\)
\(\Rightarrow x+1=1\) hoặc \(x+1=-1\)
1)x+1=1 2)x+1=-1
\(\Rightarrow x=1-1\) \(\Rightarrow x=-1-1\)
\(\Rightarrow x=0\) \(\Rightarrow x=-2\)
Vậy \(x\in\left\{0;-2\right\}\)
b) x-[-x+(x+3)]-[(x+3)-(x-2)]=0
\(\Rightarrow x-\left[-x+x+3\right]-\left[x+3-x+2\right]=0\)
\(\Rightarrow x-3-5=0\)
\(\Rightarrow x=0+3+5\)
\(\Rightarrow x=8\)
Vậy x=8
c)\(\left(3x+1\right)^2+\left|y-5\right|=1\)
+)Giả sử 3x+1 là số âm
\(\Rightarrow\left(3x+1\right)^2\)là số dương(1)
+)Lại giả sử 3x+1 là số dương
\(\Rightarrow\left(3x+1\right)^2\)là số dương(2)
+)Từ (1) và (2)
\(\Rightarrow\left(3x+1\right)^2\)nguyên dương với mọi x
+)Ta có:\(\left(3x+1\right)^2\ge0;\left|y-5\right|\ge0\)
\(\Rightarrow\left(3x+1\right)^2=1;\left|y-5\right|=0\)
\(\Rightarrow x=0;y=5\)
+)Ta lại có:\(\left(3x+1\right)^2\ge0;\left|y-5\right|\ge0\)
\(\Rightarrow\left(3x+1\right)^2=0;\left|y-5\right|=1\)
\(\Rightarrow x=\frac{-1}{3};y\in\left\{6;4\right\}\)
Mà \(\left(x,y\right)\in Z\)
\(\Rightarrow x=0;y=5\)
Đề bạn thiếu x,y thuộc Z đó
Chúc bn học tốt
x-2 =5 hay x-2 = -5
x= 7 hay x = -3
ok
x-2=5 hoặc x-2 =-5
x=5+2 x=-5-2
x=7 x= -7
vậy x=[7,-7]