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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
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Gọi số số hạng của dãy trên là n (n thuộc N*)
Theo công thức, ta có:
[11 + (x - 3)]n : 2 = 0
=> [11 + (x - 3)] = 0
=> 11 + (x - 3) = 0 (Vì n khác 0)
=> x - 3 = -11
=> x = -8
Vậy...
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\(284:\left[314-\left(x+3\right)\right]=12.\)
\(\Leftrightarrow\left[314-\left(x+3\right)\right]=\frac{71}{3}\)
\(\Leftrightarrow\left(x+3\right)=\frac{871}{3}\)
\(\Leftrightarrow x=\frac{862}{3}\)
\(14.\left[251-\left(x-5\right)\right]=1470\)
\(\Leftrightarrow\left[251-\left(x-5\right)\right]=105\)
\(\Leftrightarrow x-5=146\)
\(\Leftrightarrow x=151\)
\(\left(120+x\right):x-15=16\)
\(\Leftrightarrow\left(120+x\right):x=31\)
\(\Leftrightarrow\left(120+x\right).\frac{1}{x}=31\)
\(\Leftrightarrow\frac{1}{x}.x+120x=31\)
\(\Leftrightarrow1+120x=31\)
\(\Leftrightarrow120x=30\)
\(\Leftrightarrow x=4\)
\(\left(135-x\right):x+24=28\)
\(\Leftrightarrow\left(135-x\right):x=4\)
\(\Leftrightarrow\left(135-x\right).\frac{1}{x}=4\)
\(\Leftrightarrow\frac{1}{x}.135-\frac{1}{x}x=4\)
\(\Leftrightarrow\frac{1}{x}135-1=4\)
\(\Leftrightarrow\frac{1}{x}135=5\)
\(\Leftrightarrow\frac{1}{x}=\frac{1}{27}\)
\(\Leftrightarrow x=27\)
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a. (x-12) - 15 = 13 - ( 18 + x) b. ( 3x - 7) + 2( 5 - 2x) + 5x = 19
x - 12 -15 = 13 - 18 - x 3x - 7 + 10 - 4x + 5x = 19
x - 27 = -5 - x (3x - 4x + 5x) - (7 - 10) = 19
x + x = -5 + 27 4x - (-3) = 19
2x = 22 4x = 19 + (-3)
=> x = 11 4x = 16
=> x = 4
c. 2( x+ 6) + 6(x - 10 ) = 8
2x + 12 + 6x -60 = 8
(2x + 6x ) + ( 12 - 60 ) =8
8x + (-48) = 8
8x = 8 - (-48)
8x = 56
=> x = 7
Nhớ và kết bạn nha !
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`@` `\text {Ans}`
`\downarrow`
`c)`
`( 34 - 2x ) * ( 2x - 6 ) = 0`
`=>`\(\left[{}\begin{matrix}34-2x=0\\2x-6=0\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}2x=34-0\\2x=0+6\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}2x=34\\2x=6\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=34\div2\\x=6\div2\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=17\\x=3\end{matrix}\right.\)
Vậy, `x \in {17; 3}`
`d)`
\(\left(2019-x\right)\left(3x-12\right)=0\)
`=>`\(\left[{}\begin{matrix}2019-x=0\\3x-12=0\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=2019-0\\3x=0+12\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=2019\\3x=12\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=2019\\x=12\div3\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=2019\\x=4\end{matrix}\right.\)
Vậy,` x \in {2019; 4}`
p/s: Bài này hnhu mk làm r mà ạ?
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a) x^2(3-x)=0
=> TH1 : x^2 =0 => x=0
TH2 : 3-x=0 => x= 3-0=3
Vậy x=0; x=3
b) x(x-4) <0
=> TH1 : x<0
TH2 : x-4< 0 => x<4
Vậy x< 0 thì thỏa mãn yêu cầu
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Ta có : \(\left|5x-4\right|=\left|x+2\right|\)
\(\Leftrightarrow\orbr{\begin{cases}5x-4=x+2\\5x-4=-x-2\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}5x-x=2+4\\5x+x=-2+4\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}4x=6\\6x=2\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{3}{2}\\x=\frac{1}{3}\end{cases}}\)
b) \(\left|2x-3\right|-\left|3x+2\right|=0\)
\(\Rightarrow\orbr{\begin{cases}2x-3=3x+2\\2x-3=-3x-2\end{cases}\Rightarrow\orbr{\begin{cases}2x-3x=2+3\\2x+3x=-2+3\end{cases}\Rightarrow}\orbr{\begin{cases}-x=5\\5x=1\end{cases}\Rightarrow}\orbr{\begin{cases}x=-5\\x=\frac{1}{5}\end{cases}}}\)
c)/2+3x/=/4x-3/
\(\Rightarrow\orbr{\begin{cases}2+3x=4x-3\\2+3x=-\left(4x-3\right)\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}3x-4x=-3-2\\3x+4x=3-2\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}-x=-5\\7x=1\end{cases}\Rightarrow\orbr{\begin{cases}x=5\\x=\frac{1}{7}\end{cases}}}\)
d)/7x+1/-/5x+6|=0
\(\Rightarrow\left|7x+1\right|=\left|5x+6\right|\)
\(\Rightarrow\orbr{\begin{cases}7x+1=5x+6\\7x+1=-\left(5x+6\right)\end{cases}\Rightarrow\orbr{\begin{cases}7x-5x=6-1\\7x+1=-5x-6\end{cases}\Rightarrow}\orbr{\begin{cases}2x=5\\7x+5x=-6-1\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{5}{2}\\x=-\frac{7}{12}\end{cases}}}\)
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3x2+12x=0
<=> 3x ( x + 4 ) = 0
<=> 3x = 0
và x+4 = 0
<=> x = 0
và x = -4
Vậy x = 0 và x = -4
3x2+12x=0
<=> 3x ( x + 4 ) = 0
<=> 3x = 0
và x+4 = 0
<=> x = 0
và x = -4
Vậy x = 0 và x = -4