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(x+1/5)^2 =26/25-17/25
<=> (x +1/5)^2 =(3/5)^2
<=> x+1/5=3/5
=> x= 2/5
![](https://rs.olm.vn/images/avt/0.png?1311)
1.
(x + 1/5)² = 26/25 - 17/25
(x + 1/5)² = 9/25
Rút căn hai vế :
|(x + 1/5)| = 3/5
x = -4/5
hoặc
x = 2/5
2.
(x + 2) / 327 + (x + 3) / 326 + (x + 4) / 325 + (x + 5) / 324 + (x + 349) / 5 = 0
<=> (x + 2) / 327 +1+ (x + 3) / 326 +1+ (x + 4) / 325 +1+ (x + 5) / 324 +1+ (x + 349) / 5 -4 = 0
<=> (x+ 329)/327 + (x+ 329)/326 + (x+ 329)/325 + (x+ 329)//324 + (x+ 329)/5 =0
<=> (x+ 329).(1/327 + 1/ 326 + 1/325 + 1/324 +1/5) =0
Do (1/327 + 1/ 326 + 1/325 + 1/324 +1/5) >0 nên x+ 329 =0 => x= -329
Câu 1 chưa chắc đã đúng ( quên hết kiến thức lớp 6 rùi ) hihi
aaaaaaaa . chết rồi . cho mình sủa câu thứ nhất :
(x+1/5)2 + 17/25=26/25
( x + 1/5 ) 2 = 26/25 - 17/25
( x + 1/5 ) 2 = 3/52
x + 1/5 = 3/5
x = 2/5.
![](https://rs.olm.vn/images/avt/0.png?1311)
a)56+(327-x)=26.5
56+327-x=383-x=130
x=383-130=253
b) (2 . x + 3) . 52 = 54
2.x+3=54/52=52=25
2x=25-3=22
x=11
c) 45 : ( 3x - 17) = 3 2
45/9=3x-17
5=3x-17
3x=22
x=22/3
3 + 6 + 9 + 12 + ...... + 96=(96+3)((96-3)/3+1)/2
99.32/2=99.16=1584
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(\left(327x+2^4\right)\cdot7^3=7\cdot7^5\)
\(\Leftrightarrow\left(327x+16\right)\cdot7^3=7^6\)
\(\Leftrightarrow327x+16=7^3\)
\(\Leftrightarrow327x+16=343\)
\(\Leftrightarrow327x=327\)
\(\Rightarrow x=1\)
(327.x + 24).73 = 7.75
=> (327.x + 24) = \(\frac{7\cdot7^5}{7^3}=7^3\)
=> 327.x + 24 = 73
=> 327.x = 73 - 24
=> 327.x = 327
=> x = 1
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(326-2x=78\Rightarrow2x=326-78=248\)
\(\Rightarrow x=248:2=124\)
b) \(42.5-\left(3x+6\right)=3^4\Rightarrow210-\left(3x+6\right)=81\)
\(\Rightarrow3x+6=210-81=129\Rightarrow3x=129-6=123\)
\(\Rightarrow x=123:3=41\)
a, 326 -2x =78
\(\Rightarrow\)2x=326-78=248
\(\Rightarrow\)x=248/2=124
b,\(42.5-\left(3x+6\right)=3^4\)
\(\Rightarrow\)\(210-3x-6=81\)
\(\Rightarrow\)3x=123
\(\Rightarrow\)x=123/3=41
c, \(4^x:64=4^{250}\)
\(\Rightarrow\)\(4^x:4^3=4^{250}\)
\(\Rightarrow\)x=253
d, \(30x+1+2+3+...+30=495\)
\(\Rightarrow\)\(30x+465=495\)
\(\Rightarrow\)\(x=1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
c)
\(4\left(3x-4\right)-2=18\)
<=> \(12x-16-2=18\)
<=> \(12x=36\)
<=> \(x=3\)
Vậy x=3
d)
\(\left(3x-10\right):10=50\)
<=> \(3x-10=500\)
<=> \(3x=510\)
<=> x= \(170\)
Vậy x= 170
f)
\(x-\left[42+\left(-25\right)\right]=-8\)
<=> \(x-17=-8\)
<=> x= \(9\)
Vậy x=9
h)
\(x+5=20-\left(12-7\right)\)
<=> \(x+5=15\)
<=> \(x=10\)
Vậy x= 10
k)
\(\left|x-5\right|=7-\left(-3\right)\)
<=> \(\left|x-5\right|=10\)
* Với \(x>=5\) ; ta được:
\(x-5=10\)
<=> x= 15 (thoả mãn điều kiện )
*Với \(x< 5\) ; ta được:
\(-\left(x-5\right)=10\)
<=> \(-x+5=10\)
<=> \(-x=5\)
<=> \(x=-5\) (thoả mãn điều kiện)
Vậy x=15 ; x= -5
i)
\(\left|x-5\right|=\left|7\right|\)
<=> \(\left|x-5\right|=7\)
*Với \(x>=5\) ; ta được:
\(x-5=7\)
<=> \(x=12\) (thoả mãn)
*Với \(x< 5\) ; ta được:
\(-\left(x-5\right)=7\)
<=> \(-x=2\)
<=> \(x=-2\) (thoả mãn)
Vậy x= 12; x= -2
m)
\(2^{x+1}.2^{2009}=2^{2010}\)
<=> \(2^{x+1+2009}=2^{2010}\)
<=> \(2^{x+2010}=2^{2010}\)
=> \(x+2010=2010\)
=> \(x=0\)
Vậy x=0
n)
\(10-2x=25-3x\)
<=>\(x=15\)
Vậy x=15
![](https://rs.olm.vn/images/avt/0.png?1311)
a, -326-(115-326)=-326-(-326)-115=(-326+326)-115=-115
b,-13-18-(-42)-15=-13-18+42-15=-4
c,(-558+558)+(-50+50)=0+0=0
d,35-12+16=39
e,18.17-3.6.7=18.17-18.7=18.(17-7)=18.10=180
f,21.72-11.72+90.72+49.125.10
=21.72-11.72+90.72+72.1250
=72.(21-11+90+1250)
=72.1350=49.1350
=66150
\(\Rightarrow3x-16=\dfrac{22\cdot7^{327}}{7^{326}}=22\cdot7=154\\ \Rightarrow3x=170\\ \Rightarrow x=\dfrac{170}{3}\)