\(x^2+5x=0\)

Minh tich lai nha Hung

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24 tháng 1 2016

x2+5x=0

=>x.(x+5)=0

=>x=0 hoặc x=-5

24 tháng 1 2016

x = -5

tick mình nha bạn thân

\(\left(x+1\right)\left(x+7\right)< 0\)

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 \(\left(x+1\right)\left(x+7\right)< 0\)

16 tháng 7 2019

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

14 tháng 10 2017

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

20 tháng 8 2020

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}

26 tháng 6 2016

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 \(x=2\)

26 tháng 6 2016

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 \(\ge\)\(\Rightarrow\)x - 1 \(\ge\)\(\Rightarrow\)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

16 tháng 7 2017

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!!!

16 tháng 7 2017

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 \(x=\dfrac{-7}{2};x=1\)

5 tháng 12 2016

111 nha bạn 

tk nha 

Xin đó

5 tháng 12 2016

x +555 = 666 

x = 666 - 555

x = 111

k minh nha cac bn minh k lai cho 

\(hua\)

17 tháng 1 2017

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 \(x\in\left\{0;-5\right\}\)

17 tháng 1 2017

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) \(x^3=x\)

\(\Leftrightarrow\left[\begin{matrix}x=1\\x=-1\\x=0\end{matrix}\right.\)

9 tháng 1 2019

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...

9 tháng 1 2019

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...

3 tháng 8 2019

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 :\(x-\frac{3}{7}=0\Rightarrow x=\frac{3}{7}\)

\(\Rightarrow x\in\left\{-\frac{1}{4};\frac{3}{7}\right\}\)

3 tháng 8 2019

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) \(\left(x+\frac{1}{4}\right)\left(x-\frac{3}{7}\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+\frac{1}{4}=0\\x-\frac{3}{7}=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-\frac{1}{4}\\x=\frac{3}{7}\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=-\frac{1}{4}\\x=\frac{3}{7}\end{cases}}\)