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a)(4-x)2+15=40
(4-x)2=40-15
(4-x)2=25
(4-x)2=52
=>4-x=5
x=5-4
x=1
a) ( 4 - x )2 + 15 = 40
( 4 - x )2 = 40 - 15
( 4 - x )2 = 25
( 4 - x )2 = 52
4 - x = 5
x = 4 - 5
x = ( - 1 )
b, ( 3x - 2 )10 = ( 3x - 2 )4
=> x \(\in\left\{0;1\right\}\)
c, X = 1 + 4 + 42 + ......... + 42017
4X = 4 + 41 + ......... + 42018
4X - X = ( 4 + 41 + ........ + 42018 ) - ( 1 + 4 + 42 + ......... + 42017 )
4X - X = 4 + 41 + ......... + 42018 - 1 - 4 - 42 - ......... - 42017
=> 3X = 42018 - 1
=> X = \(\frac{4^{2018}-1}{3}\)
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Giải:
a) \(\left(3x^2-5\right)+3^4+6^0=5^3\)
\(\Leftrightarrow3x^2-5+3^4+6^0=5^3\)
\(\Leftrightarrow3x^2-5+81+1=125\)
\(\Leftrightarrow3x^2=125+5-81-1\)
\(\Leftrightarrow3x^2=48\)
\(\Leftrightarrow x^2=\dfrac{48}{3}=16\)
\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-4\end{matrix}\right.\)
Vậy ...
b) \(3x+2x\left(2^3.5-3^2.4\right)+5^2=4^4\)
\(\Leftrightarrow3x+2x\left(8.5-9.4\right)+25=256\)
\(\Leftrightarrow3x+2x\left(40-36\right)+25=256\)
\(\Leftrightarrow3x+2x.4+25=256\)
\(\Leftrightarrow3x+8x+25=256\)
\(\Leftrightarrow11x+25=256\)
\(\Leftrightarrow11x=256-25=231\)
\(\Leftrightarrow x=\dfrac{231}{11}\)
\(\Leftrightarrow x=21\)
Vậy ...
Chúc bạn học tốt!
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a, \(\left(3x^2-5\right)+3^4+6^0=5^3\)
\(\left(3x^2-5\right)+81+1=125\)
\(3x^2-5=43\)
\(3x^2=48\)
\(x^2=16\)
\(\Rightarrow\orbr{\begin{cases}x=-4\\x=4\end{cases}}\)
Vậy ...
b,\(3x+2x\left(2^3.5-3^2.4\right)+5^2=4^4\)
\(3x+2x\left(8.5-9.4\right)+25=256\)
\(3x+2x.9=231\)
\(21x=231\)
x=11
![](https://rs.olm.vn/images/avt/0.png?1311)
\(16^x< 128^4\)
=> \(\left[2^4\right]^x< \left[2^7\right]^4\)
=> \(2^{4x}< 2^{28}\)
=> 4x < 28
=> x < 7
Đến đây tìm x được rồi
\(\left[3x^2-5\right]+3^4+6^0=5^3\)
=> \(\left[3x^2-5\right]=5^3-6^0-3^4=43\)
=> \(3x^2-5=43\)
=> \(3x^2=48\)
=> \(x^2=16\)
=> \(x=\pm4\)
\(3x+2x\left[2^3\cdot5-3^2\cdot4\right]+5^2=4^4\)
=> \(3x+2x\left[8\cdot5-9\cdot4\right]+25=256\)
=> \(3x+2x\cdot4+25=256\)
=> \(3x+2x\cdot4=231\)
Đến đây tìm x
![](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
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Mình làm được mỗi câu đầu à!!!!
16x < 1284
= ( 24 )x < ( 27)4
= 24x < 228
= 4x < 28
=> x < 7
Vậy x ∈ { 0; 1; 2; 3; 4; 5; 6 }.
\(\left(4+3x\right)^2=\left(4+3x\right)^4\)
\(\Leftrightarrow1=\left(4x+3\right)^2\)
\(\Leftrightarrow\left(4+3x\right)^2=1\)
\(\Leftrightarrow\orbr{\begin{cases}4+3x=1\\4+3x=-1\end{cases}\Leftrightarrow}\orbr{\begin{cases}3x=-3\\3x=-5\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=-1\\x=-\frac{5}{3}\end{cases}}\)
Vậy \(x=\left\{-1,\frac{-5}{3}\right\}\)