Tìm x :
a) 2x+3+2x=144
b)22*x-1+95=102
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13 tháng 8 2021
a: Ta có: \(7x+25=144\)
\(\Leftrightarrow7x=119\)
hay x=17
b: Ta có: \(33-12x=9\)
\(\Leftrightarrow12x=24\)
hay x=2
c: Ta có: \(128-3\left(x+4\right)=23\)
\(\Leftrightarrow3\left(x+4\right)=105\)
\(\Leftrightarrow x+4=35\)
hay x=31
d: Ta có: \(71+\left(726-3x\right)\cdot5=2246\)
\(\Leftrightarrow5\left(726-3x\right)=2175\)
\(\Leftrightarrow726-3x=435\)
\(\Leftrightarrow3x=291\)
hay x=97
e: Ta có: \(720:\left[41-\left(2x+5\right)\right]=40\)
\(\Leftrightarrow41-\left(2x+5\right)=18\)
\(\Leftrightarrow2x+5=23\)
\(\Leftrightarrow2x=18\)
hay x=9
19 tháng 12 2017
-3-2x+17=12-3x
-2x+3x=12-17+3
x=-2
b.43-2x-2=-48+x
-2x-x = -48+2-43
-3x=-89
x=-89/-3
c -253-2x=102+19-x
-2x+x=102+19+253
-x=374
x=-374
HH
24 tháng 6 2018
a/ \(\left(2x+3\right)^2-\left(2x+1\right)\left(2x-1\right)=22\)
<=> \(\left(2x+3\right)^2-\left(4x^2-1\right)=22\)
<=> \(\left(2x+3\right)^2-4x^2+1=22\)
<=> \(\left(2x+3-2x\right)\left(2x+3+2x\right)=21\)
<=> \(3\left(4x+3\right)=21\)
<=> \(4x+3=7\)
<=> \(4x=4\)
<=> \(x=1\)
BD
24 tháng 6 2018
......................?
mik ko biết
mong bn thông cảm
nha ................
15 tháng 6 2023
`@` `\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`
`=>`\(\left[{}\begin{matrix}x-4=0\\2x+11=0\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=4\\2x=-11\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=4\\x=-\dfrac{11}{2}\end{matrix}\right.\)
BD
15 tháng 6 2023
\(\dfrac{x-3}{3}=\dfrac{2x+1}{5}\\ \left(x-3\right)\cdot5=\left(2x+1\right)\cdot3\\ x5-15=6x+3\\ x5-6x=3+15\\ -x=18\\ \Rightarrow x=-18\)
\(\dfrac{x+1}{22}=\dfrac{6}{x}\\ \left(x+1\right)\cdot x=6\cdot22\\ \left(x+1\right)\cdot x=2\cdot3\cdot2\cdot11\\ \left(x+1\right)\cdot x=12\cdot11\\ \Rightarrow x=11\)
\(\dfrac{2x-1}{21}=\dfrac{3}{2x+1}\\ \left(2x-1\right)\cdot\left(2x+1\right)=21\cdot3\\ \left(2x-1\right)\cdot\left(2x+1\right)=7\cdot3\cdot3\\ \left(2x-1\right)\cdot\left(2x+1\right)=7\cdot9\\ \Rightarrow2x+1=9\\ 2x=8\\ x=4\)
18 tháng 12 2022
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
25 tháng 8 2020
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
NC
25 tháng 8 2020
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) \(\frac{10x+5}{6}=\frac{5}{x+1}\Leftrightarrow10x^2+15x-25=0\Leftrightarrow5\left(x-1\right)\left(2x+5\right)=0\Leftrightarrow\orbr{\begin{cases}x=1\\x=-\frac{5}{2}\end{cases}}\)
10 tháng 6 2016
1)a)
x/3=y/4=>x/15=y/20
y/5=z/7=>y/20=z/28
=>x/15=y/20=z/18
Áp dụng tính chất dãy tỉ số bằng nhau ta có
x/15=y/20=z/28=2x+3y-z/30+60-28=372/62=6
=>x=90
y=120
z=168
b)
2x=3y=5z
2x=3y=>x/3=y/2=>x/15=y/10
3y=5z=>y/5=z/3=>y/10=z/6
Áp dụng tính chất dãy tỉ số bằng nhau ta có
x/15=y/10=z/6=x+y-z/15+10-6=95/19=5
=>x=75
y=50
z=30
10 tháng 6 2016
a) Ta co :x/3=y/4 suy ra x/15=y/20 (1)
y/5=z/7 suy ra y/20=z/28 (2)
Tu (1) va (2) suy ra y/20=x/15=z/28
còn lại tự làm nhé dễ rùi
b)Ta co : 2x=3y=5z suy ra x phan 1/2=y phan 1/3 = z phan 1/5
de rui tu lam nha
26 tháng 4 2019
a, \(3x+2-x+1=4x+1\)
\(3x-x-4x=1-2-1\)
\(-2x=-2\)
\(x=-2:\left(-2\right)=1\)
NJ
11 tháng 2 2018
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+1 | 1 | -1 | 2 | -2 | 4 | -4 |
| y-1 | -4 | 4 | -2 | 2 | -1 | 1 |
| x | 0 | -2/5 | 1/5 | -3/5 | 3/5 | -1 |
| y | -3 | 5 | -1 | 3 | 0 | 2 |
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+1 | 1 | -1 | 2 | -2 | 4 | -4 |
| y-1 | -4 | 4 | -2 | 2 | -1 | 1 |
| x | 0 | -2 | 1/5 | -3/5 | 3/5 | -1 |
| y | -3 | 5 | -1 | 3 | 0 | 2 |
23 tháng 5 2016
a. (x - 2)2 = 1
<=> (x - 2)2 = 12 = (-1)2
<=> \(\begin{cases}x-2=1\\x-2=-1\end{cases}\Leftrightarrow\begin{cases}x=3\\x=1\end{cases}\)
Vậy x \(\in\){1; 3}.
b. (2x - 1)3 = -8
<=> (2x - 1)3 = (-2)3
<=> 2x - 1 = -2
<=> 2x = -2 + 1
<=> 2x = -1
<=> x = -1/2
Vậy x = -1/2.
c. (x + 1/2)2 = 1/16
<=> (x + 1/2)2 = (1/4)2 = (-1/4)2
<=> \(\begin{cases}x+\frac{1}{2}=\frac{1}{4}\\x+\frac{1}{2}=-\frac{1}{4}\end{cases}\Leftrightarrow\begin{cases}x=-\frac{1}{4}\\x=-\frac{3}{4}\end{cases}\)
Vậy x \(\in\){-1/4; -3/4}.
d. (x - 2)3 = -27
<=> (x - 2)3 = (-3)3
<=> x - 2 = -3
<=> x = -3 + 2
<=> x = -1
Vậy x = -1.
ML
23 tháng 5 2016
a.\(\left(x-2\right)^2\)=1
<=> x-2=1 hoặc x-2=-1
<=> x= 3 hoặc x=1
b.\(\left(2x-1\right)^3\)=-8
\(\left(2x-1\right)^3\)=\(\left(-2\right)^3\)
2x-1=-2
2x=-1
x=-1/2
c.\(\left(x+\frac{1}{2}\right)^2\)=\(\frac{1}{16}\)
\(\left(x+\frac{1}{2}\right)^2\)=\(\left(\frac{1}{4}\right)^2\)hoặc \(\left(x+\frac{1}{2}\right)^2\)=\(\left(-\frac{1}{4}\right)^2\)
x+\(\frac{1}{2}\)=\(\frac{1}{4}\) hoặc x+\(\frac{1}{2}\)=-\(\frac{1}{4}\)
x=-\(\frac{1}{4}\)hoặc x=-\(\frac{3}{4}\)
d.\(\left(x-2\right)^3\)=-27
\(\left(x-2\right)^3\)=\(\left(-3\right)^3\)
x-2=-3
x=-1
\(2^{x+3}+2^x=144\)
\(\Rightarrow2^x.2^3+2^x=144\)
\(\Rightarrow2^x\left(2^3+1\right)=144\)
\(\Rightarrow2^x.9=144\)
\(\Rightarrow2^x=16\)
\(\Rightarrow2^x=2^4\)
\(\Rightarrow x=4\)
Vậy x = 4