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trần thị ngọc nhi
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Huy Hoàng
24 tháng 6 2018 lúc 14:19

a/ \(\left(2x+3\right)^2-\left(2x+1\right)\left(2x-1\right)=22\)

<=> \(\left(2x+3\right)^2-\left(4x^2-1\right)=22\)

<=> \(\left(2x+3\right)^2-4x^2+1=22\)

<=> \(\left(2x+3-2x\right)\left(2x+3+2x\right)=21\)

<=> \(3\left(4x+3\right)=21\)

<=> \(4x+3=7\)

<=> \(4x=4\)

<=> \(x=1\)

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Bùi Đình Hải
24 tháng 6 2018 lúc 12:38

......................?

mik ko biết

mong bn thông cảm 

nha ................

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Khánh Vinh
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Huỳnh Quang Sang
25 tháng 8 2020 lúc 8:26

a) \(2x+\frac{3}{15}=\frac{7}{5}\) 

=> \(2x=\frac{7}{5}-\frac{3}{15}=\frac{21}{15}-\frac{3}{15}=\frac{18}{15}\)

=> \(x=\frac{18}{15}:2=\frac{18}{15}\cdot\frac{1}{2}=\frac{9}{15}\cdot\frac{1}{1}=\frac{9}{15}\)

b) \(x-\frac{2}{9}=\frac{8}{3}\)

=> \(x=\frac{8}{3}+\frac{2}{9}\)

=> \(x=\frac{24}{9}+\frac{2}{9}=\frac{26}{9}\)

c) \(\frac{-8}{x}=\frac{-x}{18}\)

=> x(-x) = (-8).18

=> -x2 = -144

=> x2 = 144(bỏ dấu âm)

=> x = \(\pm\)12

d) \(\frac{2x+3}{6}=\frac{x-2}{5}\)

=> 5(2x + 3) = 6(x - 2)

=> 10x + 15 = 6x - 12

=> 10x + 15 - 6x + 12 = 0

=> 4x + 27 = 0

=> 4x = -27

=> x = -27/4

e) \(\frac{x+1}{22}=\frac{6}{x}\)

=> x(x + 1) = 132

=> x(x + 1) = 11.12

=> x = 11

f) \(\frac{2x-1}{2}=\frac{5}{x}\)

=> x(2x - 1) = 10

=> 2x2 - x = 10

=> 2x2 - x - 10 = 0

tới đây tự làm đi nhé

g) \(\frac{2x-1}{21}=\frac{3}{2x+1}\)

=> (2x - 1)(2x + 1) = 63

=> 4x2 - 1 = 63

=> 4x2 = 64

=> x2 = 16

=> x = \(\pm\)4

h) Tương tự

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Ngô Chi Lan
25 tháng 8 2020 lúc 8:28

a) \(\frac{2x+3}{15}=\frac{7}{5}\Leftrightarrow10x+15=105\Leftrightarrow10x=90\Rightarrow x=9\)

b) \(\frac{x-2}{9}=\frac{8}{3}\Leftrightarrow3x-6=72\Leftrightarrow3x=78\Rightarrow x=26\)

c) \(\frac{-8}{x}=\frac{-x}{18}\Leftrightarrow x^2=144\Leftrightarrow\orbr{\begin{cases}x=12\\x=-12\end{cases}}\)

d) \(\frac{2x+3}{6}=\frac{x-2}{5}\Leftrightarrow10x+15=12x-12\Leftrightarrow2x=27\Rightarrow x=\frac{27}{2}\)

e) \(\frac{x+1}{22}=\frac{6}{x}\Leftrightarrow x^2+x-132=0\Leftrightarrow\left(x-11\right)\left(x+12\right)=0\Leftrightarrow\orbr{\begin{cases}x=11\\x=-12\end{cases}}\)

f) \(\frac{2x-1}{2}=\frac{5}{x}\Leftrightarrow2x^2-x-10=0\Leftrightarrow\left(x-2\right)\left(2x+5\right)=0\Leftrightarrow\orbr{\begin{cases}x=2\\x=-\frac{5}{2}\end{cases}}\)

g) \(\frac{2x-1}{21}=\frac{3}{2x+1}\Leftrightarrow4x^2=64\Leftrightarrow x^2=16\Rightarrow\orbr{\begin{cases}x=4\\x=-4\end{cases}}\)

h) \(\frac{10x+5}{6}=\frac{5}{x+1}\Leftrightarrow10x^2+15x-25=0\Leftrightarrow5\left(x-1\right)\left(2x+5\right)=0\Leftrightarrow\orbr{\begin{cases}x=1\\x=-\frac{5}{2}\end{cases}}\)

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Hà Trí Kiên
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『Kuroba ム Tsuki Ryoo...
15 tháng 6 2023 lúc 13:43

`@` `\text {Ans}`

`\downarrow`

\(\dfrac{x-3}{3}=\dfrac{2x+1}{5}\)

`=> (x-3)5 = (2x+1)3`

`=> 5x-15 = 6x+3`

`=> 5x-6x = 15+3`

`=> -x=18`

`=> x=-18`

\(\dfrac{x+1}{22}=\dfrac{6}{x}\)

`=> (x+1)x = 22*6`

`=> (x+1)x = 132`

`=> x^2 + x = 132`

`=> x^2+x-132=0`

`=> (x-11)(x+12)=0`

`=>`\(\left[{}\begin{matrix}x-11=0\\x+12=0\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=11\\x=-12\end{matrix}\right.\)

\(\dfrac{2x-1}{2}=\dfrac{5}{x}\)

`=> (2x-1)x = 2*5`

`=> 2x^2 - x =10`

`=> 2x^2 - x - 10 =0`

`=> 2x^2 + 4x - 5x - 10 =0`

`=> (2x^2 + 4x) - (5x+10)=0`

`=> 2x(x+2) - 5(x+2)=0`

`=> (2x-5)(x+2)=0`

`=>`\(\left[{}\begin{matrix}2x-5=0\\x+2=0\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}2x=5\\x=-2\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-2\end{matrix}\right.\)

\(\dfrac{2x-1}{21}=\dfrac{3}{2x+1}\)

`=> (2x-1)(2x+1)=21*3`

`=> 4x^2 + 2x - 2x - 1 = 63`

`=> 4x^2 - 1=63`

`=> 4x^2 - 1 - 63=0`

`=> 4x^2 - 64 = 0`

`=> 4(x^2 - 16)=0`

`=> 4(x^2 + 4x - 4x - 16)=0`

`=> 4[(x^2+4x)-(4x+16)]=0`

`=> 4[x(x+4)-4(x+4)]=0`

`=> 4(x-4)(x+4)=0`

`=>`\(\left[{}\begin{matrix}x-4=0\\x+4=0\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=4\\x=-4\end{matrix}\right.\)

\(\dfrac{2x+1}{9}=\dfrac{5}{x+1}\)

`=> (2x+1)(x+1) = 9*5`

`=> (2x+1)(x+1)=45`

`=> 2x^2 + 2x + x + 1 = 45`

`=> 2x^2 + 3x + 1 =45`

`=> 2x^2 + 3x + 1 - 45 =0`

`=> 2x^2+3x-44=0`

`=> 2x^2 + 11x - 8x - 44=0`

`=> (2x^2 +11x) - (8x+44)=0`

`=> x(2x+11) - 4(2x+11)=0`

`=> (x-4)(2x+11)=0`

`=>`\(\left[{}\begin{matrix}x-4=0\\2x+11=0\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=4\\2x=-11\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=4\\x=-\dfrac{11}{2}\end{matrix}\right.\)

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boi đz
15 tháng 6 2023 lúc 13:40

\(\dfrac{x-3}{3}=\dfrac{2x+1}{5}\\ \left(x-3\right)\cdot5=\left(2x+1\right)\cdot3\\ x5-15=6x+3\\ x5-6x=3+15\\ -x=18\\ \Rightarrow x=-18\)

\(\dfrac{x+1}{22}=\dfrac{6}{x}\\ \left(x+1\right)\cdot x=6\cdot22\\ \left(x+1\right)\cdot x=2\cdot3\cdot2\cdot11\\ \left(x+1\right)\cdot x=12\cdot11\\ \Rightarrow x=11\)

\(\dfrac{2x-1}{21}=\dfrac{3}{2x+1}\\ \left(2x-1\right)\cdot\left(2x+1\right)=21\cdot3\\ \left(2x-1\right)\cdot\left(2x+1\right)=7\cdot3\cdot3\\ \left(2x-1\right)\cdot\left(2x+1\right)=7\cdot9\\ \Rightarrow2x+1=9\\ 2x=8\\ x=4\)

 

 

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Vicky Lee
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KAl(SO4)2·12H2O
25 tháng 8 2019 lúc 21:50

a) \(\left(x-3\right)^2-4=0\)

\(\left(x-3\right)^2=0+4\)

\(\left(x-3\right)^2=4\)

\(\left(x-3\right)^2=\pm4\)

\(\left(x-3\right)^2=\pm2^2\)

\(\orbr{\begin{cases}x-3=2\\x-3=-2\end{cases}}\)

\(\orbr{\begin{cases}x=5\\x=1\end{cases}}\)

b) \(\left(2x+3\right)^2-\left(2x+1\right)\left(2x-1\right)=22\)

\(4x^2+12x+9-4x^2+1=22\)

\(12x+10=22\)

\(12x=22-10\)

\(12x=12\)

\(x=1\)

c) \(\left(4x+3\right)\left(4x-3\right)-\left(4x-5\right)^2=16\)

\(16x^2-9-16x^2+40x-25=16\)

\(-34+40x=16\)

\(40x=16+34\)

\(40x=50\)

\(x=\frac{50}{40}=\frac{5}{4}\)

d) \(x^3-9x^2+27x-27=-8\)

\(x^3-9x^2+27x-27+8=0\)

\(x^3-9x^2+27x-19=0\)

\(\left(x^2-8x+19\right)\left(x-1\right)=0\)

Vì \(\left(x^2-8x+19\right)>0\) nên:

\(x-1=0\)

\(x=1\)

e) \(\left(x+1\right)^3-x^2\left(x+3\right)=2\)

\(x^3+2x^2+x+x^2+2x+1-x^2-3x^2=2\)

\(3x+1=2\)

\(3x=2-1\)

\(3x=1\)

\(x=\frac{1}{3}\)

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Trinh Ngoc Tri
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Neymar Jr
11 tháng 2 2018 lúc 14:10

1.

a,  \(x-14=3x+18\)                                                                       

\(\Rightarrow x-3x=18+14\)                                                                 

\(\Rightarrow-2x=32\Rightarrow x=\frac{32}{-2}=-16\)

b, \(\left(x+7\right).\left(x-9\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x+7=0\\x-9=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-7\\x=9\end{cases}}}\)

c, \(\left|2x-5\right|-7=22\)                                                                     

\(\Rightarrow\left|2x-5\right|=22+7\)

\(\Rightarrow\left|2x-5\right|=29\)

\(\Rightarrow\orbr{\begin{cases}2x+5=29\\2x-5=29\end{cases}}\Rightarrow\orbr{\begin{cases}2x=24\\2x=34\end{cases}\Rightarrow}\orbr{\begin{cases}x=12\\x=17\end{cases}}\)

d,  \(\left(\left|2x\right|-5\right)-7=22\)

\(\Rightarrow\left(\left|2x\right|-5\right)=29\)

\(\Rightarrow\left|2x\right|=29+5\Rightarrow\left|2x\right|=34\Rightarrow x=\pm17\)

e, \(\left|x+3\right|+\left|x+9\right|+\left|x+5\right|=4x\)

Vì \(\left|x+3\right|\ge0;\left|x+9\right|\ge0;\left|x+5\right|\ge0;4x\ge0\)

Nên \(\left|x+3\right|+\left|x+9\right|+\left|x+5\right|=4x\ge0\)

\(\Rightarrow\left|x+3\right|>0\Rightarrow\left|x+3\right|=x+3\)

     \(\left|x+9\right|>0\Rightarrow\left|x+9\right|=x+9\)

      \(\left|x+5\right|>0\Rightarrow\left|x+5\right|=x+5\)

Ta có : 

\(x+3+x+9+x+5=4x\)

\(\Rightarrow3x+\left(3+9+5\right)=4x\)

\(\Rightarrow4x-3x=17\)

\(\Rightarrow x=17\)

2. a , b sai đề bn 

c, \(\left(5x+1\right).\left(y-1\right)=4\)

\(\Rightarrow\left(5x+1\right).\left(y-1\right)\inƯ\left(4\right)\)

\(\text{ }Ư\left(4\right)=\left\{1;-1;2;-2;4;-4\right\}\)

Ta có bảng sau : 

5x+11-12-24-4
y-1-44-22-11
x0-2/51/5-3/53/5-1
y-35-1302

d, \(5xy-5x+y=5\)

\(\Rightarrow\left(5xy-5x\right)+y=5\)

\(\Rightarrow5x.\left(y-1\right)+y=5\)

\(\Rightarrow\left(5x+1\right).\left(y-1\right)=4\)

\(\Rightarrow\left(5x+1\right).\left(y-1\right)\inƯ\left(4\right)\)

\(Ư\left(4\right)=\left\{1;-1;2;-2;4;-4\right\}\)

Ta có bảng sau : 

5x+11-12-24-4
y-1-44-22-11
x0-21/5-3/53/5-1
y-35-1302



 

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bin
15 tháng 4 2019 lúc 19:36

x - 14 = 3x + 18

x - 3x = 18 + 14

-2x= 32

x= 32 : (-2)

x=-16

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Phạm Thu Trang
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Trúc Giang
7 tháng 8 2020 lúc 8:32

a) \(\left(2x+1\right)\left(1-2x\right)+\left(2x-1\right)^2=22\)

\(\Rightarrow\left(1+2x\right)\left(1-2x\right)+\left[\left(2x\right)^2-2.2x+1^2\right]=22\)

\(\Rightarrow1^2-\left(2x\right)^2+\left(4x^2-4x+1\right)=22\)

\(\Rightarrow1-4x^2+4x^2-4x+1=22\)

\(\Rightarrow2-4x=22\)

\(\Rightarrow-4x=22-2=20\)

\(\Rightarrow x=20:\left(-4\right)=-5\)

b/ \(\left(x-5\right)^2+\left(x-3\right)\left(x+3\right)-2.\left(x+1\right)^2=0\)

\(\Rightarrow\left(x^2-2.x.5+5^2\right)+\left(x^2-3^2\right)+2.\left(x^2+2.x.1+1^2\right)=0\)

\(\Rightarrow x^2-10x+25+x^2-9-2\left(x^2+2x+1\right)=0\)

\(\Rightarrow x^2-10x+25+x^2-9-2x^2-4x-2=0\)

\(\Rightarrow-14x+14=0\)

\(\Rightarrow-14x=0-14=-14\)

\(\Rightarrow x=\left(-14\right):\left(-14\right)=1\)

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HUYNH NHAT TUONG VY
7 tháng 8 2020 lúc 8:45

b/\(\left(x-5\right)^2+\left(x-3\right)\left(x+3\right)-2\left(x+1\right)^2=0\)

\(\Leftrightarrow x^2-10x+25+x^2-3^2-2\left(x^2+2x+1\right)=0\)

\(\Leftrightarrow x^2-10x+25+x^2-9-2x^2-4x-2=0\)

\(\Leftrightarrow14x=14\Leftrightarrow x=1\)

c/\(\left(2x+3\right)^2+\left(2x-3\right)^2-2\left(4x^2-9\right)=0\)

\(\Leftrightarrow4x^2+12x+9+4x^2-12x+9-8x^2+18=0\)

\(\Leftrightarrow0x=-36\Leftrightarrow x=0\)

a/\(\left(2x+1\right).\left(1-2x\right)+\left(2x-1\right)^2=22\Leftrightarrow2x-4x^2+1-2x+4x^2-4x+1=22\Leftrightarrow-4x=20\Leftrightarrow x=-5\)

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thùy linh
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Nguyễn Lê Phước Thịnh
18 tháng 12 2022 lúc 11:46

3: =>x(x+1)=0

=>x=0 hoặc x=-1

4: =>(2x-3)(x+2)=0

=>x=3/2 hoặc x=-2

6: =>6x=7 hoặc 6x=-7

=>x=7/6 hoặc x==7/6

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Linh Phương
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Nguyễn Việt Hoàng
18 tháng 8 2020 lúc 16:19

+) \(2x\left(x-4\right)-x\left(2x+3\right)+22=0\)

\(\Leftrightarrow2x^2-8x-2x^2-3x+22=0\)

\(\Leftrightarrow-11x+22=0\)

\(\Leftrightarrow-11\left(x-2\right)=0\)

\(\Leftrightarrow x-2=0\Leftrightarrow x=2\)

+) \(\left(2x+3\right)\left(3x+2\right)+2\left(1-3x\right)\left(x+\frac{1}{2}\right)=1\)

\(\Leftrightarrow6x^2+4x+9x+6+\left(2-6x\right)\left(x+\frac{1}{2}\right)=1\)

\(\Leftrightarrow6x^2+13x+6+2x+1-6x^2-3x=1\)

\(\Leftrightarrow12x+7=1\)

\(\Leftrightarrow x=\frac{-1}{2}\)

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l҉o҉n҉g҉ d҉z҉
18 tháng 8 2020 lúc 19:30

2x( x - 4 ) - x( 2x + 3 ) + 22 = 0

<=> 2x2 - 8x - 2x2 - 3x + 22 = 0

<=> -11x + 22 = 0

<=> -11x = -22

<=> x = 2

( 2x + 3 )( 3x + 2 ) + 2( 1 - 3x )( x + 1/2 ) = 1

<=> 6x2 + 13x + 6 + 2( -3x2 - 1/2x + 1/2 ) = 1

<=> 6x2 + 13x + 6 - 6x2 - x + 1 = 1

<=> 12x + 7 = 1 

<=> 12x = -6

<=> x = -6/12 = -1/2

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ミ★Ƙαї★彡
18 tháng 8 2020 lúc 19:35

\(2x\left(x-4\right)-x\left(2x+3\right)+22=0\)

\(\Leftrightarrow2x^2-8x-2x^2-3x+22=0\)

\(\Leftrightarrow-11x+22=0\Leftrightarrow x=2\)

\(\left(2x+3\right)\left(3x+2\right)+2\left(1-3x\right)\left(x+\frac{1}{2}\right)=1\)

\(\Leftrightarrow6x^2+13x+6+2x+1-6x^2-3x=1\)

\(\Leftrightarrow12x+7=1\Leftrightarrow x=-\frac{1}{2}\)

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Pham Trong Bach
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Cao Minh Tâm
20 tháng 4 2018 lúc 11:33

a, 36:(x–5) =  2 2

(x–5) = 9

x = 14

b, [3.(70–x)+5]:2 = 46

[3.(70–x)+5] = 92

70–x = 29

x = 41

c, 450:[41–(2x–5)] =  3 2 .5

41–(2x–5) = 10

2x–5 = 31

2x = 36

x = 18

d, 230+[ 2 4 +(x–5)] = 315. 2018 0

16+(x–5) = 315–230

x–5 = 85–16

x = 69+5

x = 74

e,  2 x + 2 x + 1  = 48

2 x .(2+1) = 48

2 x = 16 =  2 4

x = 4

f,  3 x + 2 + 3 x  = 2430

3 x . 3 2 + 1 = 2430

3 x = 2430:10 = 243 =  3 5

x = 5

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