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\(\left(x+1\right)\left(x+7\right)< 0\)
thì \(x+1;x+7\)khác dấu
th1\(\hept{\begin{cases}x+1< 0\\x+7>0\end{cases}\Leftrightarrow\hept{\begin{cases}x< -1\\x>-7\end{cases}\Rightarrow}-7< x< -1\left(tm\right)}\)
th2\(\hept{\begin{cases}x+1>0\\x+7< 0\end{cases}\Leftrightarrow\hept{\begin{cases}x>-1\\x< -7\end{cases}\Rightarrow}-1< x< -7\left(vl\right)}\)
vậy với\(-7< x< -1\)thì \(\left(x+1\right)\left(x+7\right)< 0\)
a) (2x - 3) = 5
<=> 2x - 3 = 5
<=> 2x = 5 + 3
<=> 2x = 8
<=> x = 4
=> x = 4
b) (5x - 3) = 1/2
<=> 5x - 3 = 1/2
<=> 5x = 1/2 + 3
<=> 5x = 7/2
<=> x = 7/10
=> x = 7/10
c) (x + 1)(x + 7) < 0
<=> x = -1; -7
<=> x < -7 <=> x = -8 <=> (-8 + 1)(-8 + 7) < 0 <=> 7 < 0 (loại)
<=> -7 < x < -1 <=> x = -6 <=> (-6 + 1)(-6 + 7) < 0 <=> -5 < 0 (nhận)
<=> x > -1 <=> x = 0 <=> (x + 1)(x + 7) < 0 <=> 7 < 0 (loại)
Vậy: -7 < x < -1
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a) 2x . 4 = 128
2x = 128 : 4
2x = 32
2x = 25
=> x = 5
b) 2x . 24 = 26
=> x + 4 = 6
x = 6 - 4
x = 2
c) 5x + x = 39 - 311 : 39
6x = 39 - 32
6x = 39 - 9
6x = 30
x = 30 : 6
x = 5
d) 9x - 1 = 81
9x - 1 = 92
=> x - 1 = 2
x = 2 + 1
x = 3
e) 6x = 521 : 519 + 3 . 22 - 70
6x = 52 + 3 . 4 - 1
6x = 25 + 12 - 1
6x = 37 - 1
6x = 36
x = 36 : 6
x = 6
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a) \(x\in B\left(3\right)=\left\{0;3;6;9;12;15;18;21;24;...;63;66;...\right\}\)
Mà 21 \(\le x\le\)65 => \(x\notin\left\{0;3;6;9;12;15;18;66;...\right\}\)
Vậy \(x\in\left\{21;24;...;63\right\}\)
b) \(x⋮17\)
=> x là bội của 17 => x \(\in B\left(17\right)=\left\{0;17;34;51;68;...\right\}\)
Mà \(0\le x\le60\Rightarrow x\in\left\{0;17;34;51\right\}\)
Vậy : ...
c) \(12⋮x\)=> x \(\inƯ\left(12\right)=\left\{1;2;3;4;6;12\right\}\)
d) \(x\inƯ\left(30\right)=\left\{1;2;3;5;6;10;15;30\right\}\)
Mà x \(\ge0\)thì nguyên dàn x đã tìm ở trên :)
e) \(x⋮7\)
=> x là bội của 7 => x \(\in\)B(7) = {0;7;14;21;28;35;42;49;56;...}
Mà x \(\le\)50 thì x \(\in\){0;7;14;21;28;35;42;49}
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a) \(3x-\frac{3}{2}-5x+\frac{10}{3}=1\)
\(3x-5x=1+\frac{3}{2}-\frac{10}{3}\)
\(-2x=-\frac{5}{6}\)
\(x=\frac{5}{12}\)
b) \(\left|x-1\right|=5-x\)
Th1:
\(x-1=5-x\)
\(x+x=5+1\)
\(2x=6\)
\(x=3\)
Th2:
\(-\left(x-1\right)=5-x\)
\(x+1=5-x\)
\(x+x=5-1\)
\(2x=4\)
\(x=2\)
Vậy \(x=3\)và \(x=2\)
a) 3x - 5x = 1 + 3/2 - 10/3
-2x = -5/6
x = -5/6 : ( - 2 )
x = 5/12
b) |x-1|= 5-x
Nếu x \(\ge\)1 \(\Rightarrow\)x - 1 \(\ge\)0 \(\Rightarrow\)x - 1 = 5 - x.
2x = 6
x = 3.
Nếu x < 1 \(\Rightarrow\)x - 1 < 0 \(\Rightarrow\)Ix-1I = 1 - x
\(\Rightarrow\)1 - x = 5 - x \(\Rightarrow\)vô lý.
Vậy x = 3
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a, \(x^2-9=0\Rightarrow x^2=9\Rightarrow x\pm3\)
b, \(\left(x-3\right)^2-25=0\Rightarrow\left(x-3\right)^2=25\)
\(\Rightarrow\left\{{}\begin{matrix}x-3=5\\x-3=-5\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=8\\x=-2\end{matrix}\right.\)
c, \(\left(x-3\right)\left(2x-5\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-3=0\\2x-5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\\2x=5\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{5}{2}\end{matrix}\right.\)
d, \(\left(x-3\right)x-2\left(x-3\right)=0\)
\(\Rightarrow\left(x-3\right)\left(x-2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-3=0\\x-2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\\x=2\end{matrix}\right.\)
e, \(3x\left(x-1\right)-5\left(1-x\right)=0\)
\(\Rightarrow3x\left(x-1\right)+5\left(x-1\right)=0\)
\(\Rightarrow\left(x-1\right)\left(3x+5\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-1=0\\3x+5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{5}{3}\end{matrix}\right.\)
g, \(x^2+6x-7=0\)
\(\Rightarrow x^2-x+7x-7=0\)
\(\Rightarrow x.\left(x-1\right)+7.\left(x-1\right)=0\)
\(\Rightarrow\left(x-1\right)\left(x+7\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-1=0\\x+7=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=-7\end{matrix}\right.\)
h,\(2x^2+5x-7=0\)
\(\Rightarrow2x^2-2x+7x-7=0\)
\(\Rightarrow2x.\left(x-1\right)+7.\left(x-1\right)=0\)
\(\Rightarrow\left(x-1\right)\left(2x+7\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-1=0\\2x+7=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{7}{2}\end{matrix}\right.\)
Chúc bạn học tốt!!!
a) \(x^2-9=0\Leftrightarrow x^2=9\Leftrightarrow\left\{{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\) vậy \(x=3;x=-3\)
b) \(\left(x-3\right)^2-25=0\Leftrightarrow\left(x-3\right)^2=25\Leftrightarrow\left\{{}\begin{matrix}x-3=5\\x-3=-5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=8\\x=-2\end{matrix}\right.\)
vậy \(x=8;x=-2\)
c) \(\left(x-3\right)\left(2x-5\right)=0\Leftrightarrow\left\{{}\begin{matrix}x-3=0\\2x-5=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=3\\x=\dfrac{5}{2}\end{matrix}\right.\)
vậy \(x=3;x=\dfrac{5}{2}\)
d)\(\left(x-3\right).x-2\left(x-3\right)=0\Leftrightarrow\left(x-2\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-2=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\x=3\end{matrix}\right.\) vậy \(x=2;x=3\)
e) \(3x\left(x-1\right)-5\left(1-x\right)=0\Leftrightarrow\left(3x+5\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}3x+5=0\\x-1=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{-5}{3}\\x=1\end{matrix}\right.\) vậy \(x=\dfrac{-5}{3};x=1\)
câu e t thấy sai sai nhưng vẫn làm ; bn coi lại đề nha
g) \(x^2+6x-7=0\Leftrightarrow x^2-x+7x-7=0\)
\(\Leftrightarrow x\left(x-1\right)+7\left(x-1\right)=0\Leftrightarrow\left(x+7\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+7=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-7\\x=1\end{matrix}\right.\) vậy \(x=-7;x=1\)
h) \(2x^2+5x-7=0\Leftrightarrow2x^2-2x+7x-7=0\)
\(\Leftrightarrow2x\left(x-1\right)+7\left(x-1\right)=0\Leftrightarrow\left(2x+7\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x+7=0\\x-1=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{-7}{2}\\x=1\end{matrix}\right.\) vậy \(x=\dfrac{-7}{2};x=1\)
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x +555 = 666
x = 666 - 555
x = 111
k minh nha cac bn minh k lai cho
\(hua\)
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b) \(x^2-7x=0\)
\(\Rightarrow x\left(x-7\right)=0\)
\(\Rightarrow\left\{\begin{matrix}x=0\\x-7=0\end{matrix}\right.\Rightarrow\left\{\begin{matrix}x=0\\x=7\end{matrix}\right.\)
Vậy \(x\in\left\{0;7\right\}\)
c) \(x^2=-5x\)
\(\Rightarrow x^2+5x=0\)
\(\Rightarrow x\left(x+5\right)=0\)
\(\Rightarrow\left\{\begin{matrix}x=0\\x+5=0\end{matrix}\right.\Rightarrow\left\{\begin{matrix}x=0\\x=-5\end{matrix}\right.\)
Vậy \(x\in\left\{0;-5\right\}\)
a) \(\left(x-5\right)\left(x+4\right)=0\)
\(\Leftrightarrow\left[\begin{matrix}x-5=0\\x+4=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{\begin{matrix}x=5\\x=-4\end{matrix}\right.\)
Vậy...
b) \(x^2-7x=0\)
\(\Leftrightarrow x\left(x-7\right)=0\)
\(\Leftrightarrow\left[\begin{matrix}x=0\\x-7=0\Leftrightarrow x=7\end{matrix}\right.\)
Vậy...
c) \(x^2=-5x\)
\(\Leftrightarrow x=-5\)
d) \(x^3=x\)
\(\Leftrightarrow\left[\begin{matrix}x=1\\x=-1\\x=0\end{matrix}\right.\)
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a)
(x+2)2+(y-3)2+(z-2)2=0
\(\Rightarrow\hept{\begin{cases}\left(x+2\right)^2=0\\\left(y-3\right)^2=0\\\left(z-2\right)^2=0\end{cases}\Rightarrow\hept{\begin{cases}x=-2\\y=3\\z=2\end{cases}}}\)
Vậy...
b)
(x-3).y-x=5
xy - 3x - x = 5
xy - 4x = 5
x(y - 4) = 5 = 1.5 = (-1).(-5)
TH1:
\(\Rightarrow\hept{\begin{cases}x=1\\y-4=5\end{cases}\Rightarrow\hept{\begin{cases}x=1\\y=9\end{cases}}}\)
TH2:
\(\Rightarrow\hept{\begin{cases}x=5\\y-4=1\end{cases}\Rightarrow\hept{\begin{cases}x=5\\y=5\end{cases}}}\)
TH3:
\(\Rightarrow\hept{\begin{cases}x=-1\\y-4=-5\end{cases}\Rightarrow\hept{\begin{cases}x=-1\\y=-1\end{cases}}}\)
TH4:
\(\Rightarrow\hept{\begin{cases}x=-5\\y-4=-1\end{cases}\Rightarrow\hept{\begin{cases}x=-5\\y=3\end{cases}}}\)
Vậy...
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a/\(5x\cdot\left(x-\frac{1}{3}\right)=0\)
Chia làm 2 TH :
TH 1: \(5x=0\Rightarrow x=0\)
TH 2:\(x-\frac{1}{3}=0\Rightarrow x=\frac{1}{3}\)
\(\Rightarrow x\in\left\{0;\frac{1}{3}\right\}\)
b/\(\left(x+\frac{1}{4}\right)\cdot\left(x-\frac{3}{7}\right)=0\)
Chia làm 2 Th
Th1 : \(x+\frac{1}{4}=0\Rightarrow x=-\frac{1}{4}\)
Th2 :\(x-\frac{3}{7}=0\Rightarrow x=\frac{3}{7}\)
\(\Rightarrow x\in\left\{-\frac{1}{4};\frac{3}{7}\right\}\)
1) \(5x\left(x-\frac{1}{3}\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}5x=0\\x-\frac{1}{3}=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{1}{3}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x=\frac{1}{3}\end{cases}}\)
2) \(\left(x+\frac{1}{4}\right)\left(x-\frac{3}{7}\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+\frac{1}{4}=0\\x-\frac{3}{7}=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-\frac{1}{4}\\x=\frac{3}{7}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=-\frac{1}{4}\\x=\frac{3}{7}\end{cases}}\)
x2+5x=0
=>x.(x+5)=0
=>x=0 hoặc x=-5
x = -5
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