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`#3107.101107`
a,
$2014^{15} . (2x - 60) = 2014^{16}$
$\Rightarrow 2x - 60 = 2014^{16} \div 2014^{15}$
$\Rightarrow 2x - 60 = 2014$
$\Rightarrow 2x = 2074$
$\Rightarrow x = 1037$
Vậy, `x = 1037`
b,
`x + 18 \div 3^2 = 5.4^2`
`\Rightarrow x + 18 \div 9 = 5.16`
`\Rightarrow x + 9 = 80`
`\Rightarrow x = 80 - 9`
`\Rightarrow x = 71`
Vậy, `x = 71`
c,
`7.(42 - x) = 5^3 + 134`
`\Rightarrow 7.(42 - x) = 125 + 134`
`\Rightarrow 7(42 - x) = 259`
`\Rightarrow 42 - x = 259 \div 7`
`\Rightarrow 42 - x = 37`
`\Rightarrow x = 42 - 37`
`\Rightarrow x = 5`
Vậy, `x = 5.`
![](https://rs.olm.vn/images/avt/0.png?1311)
a: 3x+21=57
=>3x=57-21
=>3x=36
=>\(x=\dfrac{36}{3}=12\)
b: \(2x-138=2^3\cdot3^2\)
=>\(2x-138=8\cdot9=72\)
=>2x=72+138=210
=>x=105
c: \(4x^3+12=120\)
=>\(4x^3=120-12=108\)
=>\(x^3=27=3^3\)
=>x=3
\(a)3x+21=57\\ \Leftrightarrow3x=57-21\\ \Leftrightarrow3x=36\\ \Leftrightarrow x=\dfrac{36}{3}\\ \Leftrightarrow x=12\\ b)2x-138=2^3\cdot3^2\\ \Leftrightarrow2x-138=72\\ \Leftrightarrow2x=72+138\\ \Leftrightarrow2x=210\\ \Leftrightarrow x=\dfrac{210}{2}\\ \Leftrightarrow x=105\\ c)4x^3+12=120\\ \Leftrightarrow4x^3=120-12\\ \Leftrightarrow4x^3=108\\ \Leftrightarrow x^3=\dfrac{108}{4}\\ \Leftrightarrow x^3=27\\ \Leftrightarrow x^3=3^3\\ \Leftrightarrow x=3.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Xét tổng : 1+3+...+149+151
Số số hạng dãy trên :
(151-1):2+1=76(số)
Tổng dãy trên :
(151+1)x76:2=5738
Lại có : (-1)+(-3)+...+(-149)+(-151)
= -(1+3+...+149+151)
= -5738
ta có, dãy số trên có số số hạng là:
(151 - 1) : 2 + 1 = 76 (số hạng)
tổng các dãy số trên là: (vì là phép cộng của các số âm nên ta bỏ dấu âm rồi tính như bình thường, kết quả nhận lại thêm vào dấu âm.)
(151 + 1) x 76 : 2 = 5776
=> (-1) + (-3) +....+ (-149) + (-151) = (-5776)
![](https://rs.olm.vn/images/avt/0.png?1311)
`@` `\text {Ans}`
`\downarrow`
`a)`
\(5^x+5^{x+2}=650\)
`\Rightarrow 5^x + 5^x . 5^2 = 650`
`\Rightarrow 5^x . (1 + 5^2) = 650`
`\Rightarrow 5^x . 26 = 650`
`\Rightarrow 5^x = 650 \div 26`
`\Rightarrow 5^x = 25`
`\Rightarrow 5^x = 5^2`
`\Rightarrow x = 2`
Vậy, `x = 2`
`b)`
`(4x + 1)^2 = 25.9`
`\Rightarrow (4x + 1)^2 = 225`
`\Rightarrow (4x + 1)^2 = (+-15^2)`
`\Rightarrow`\(\left[{}\begin{matrix}4x-1=15\\4x-1=-15\end{matrix}\right.\)
`\Rightarrow `\(\left[{}\begin{matrix}4x=16\\4x=-14\end{matrix}\right.\)
`\Rightarrow `\(\left[{}\begin{matrix}x=4\\x=-\dfrac{7}{2}\end{matrix}\right.\)
Vậy, `x \in`\(\left\{-\dfrac{7}{2};4\right\}\)
`c)`
\(2^x+2^{x+3}=144\)
`\Rightarrow 2^x + 2^x . 2^3 = 144`
`\Rightarrow 2^x . (1 + 2^3) = 144`
`\Rightarrow 2^x . 9 = 144`
`\Rightarrow 2^x = 144 \div 9`
`\Rightarrow 2^x = 16`
`\Rightarrow 2^x = 2^4`
`\Rightarrow x = 4`
Vậy, `x = 4`
`d)`
\(3^{2x+2}=9^{x+3}\)
`\Rightarrow `\(3^{2x+2}=\left(3^2\right)^{x+3}\)
`\Rightarrow `\(3^{2x+2}=3^{2x+6}\)
`\Rightarrow 2x + 2 = 2x + 6`
`\Rightarrow 2x - 2x = 6 - 2`
`\Rightarrow 0 = 4 (\text {vô lý})`
Vậy, `x` không có giá trị nào thỏa mãn.
`e)`
\(x^{15}=x^2\)
`\Rightarrow `\(x^{15}-x^2=0\)
`\Rightarrow `\(x^2\cdot\left(x^{13}-1\right)=0\)
`\Rightarrow `\(\left[{}\begin{matrix}x^2=0\\x^{13}-1=0\end{matrix}\right.\)
`\Rightarrow `\(\left[{}\begin{matrix}x=0\\x^{13}=1\end{matrix}\right.\)
`\Rightarrow `\(\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
Vậy, `x \in`\(\left\{0;1\right\}.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,5^x+5^{x+2}=650\\ \Rightarrow5^x+5^x.5^2=650\\ \Rightarrow5^x\left(1+5^2\right)=650\\ \Rightarrow5^x.26=650\\ \Rightarrow5^x=25\\ \Rightarrow5^x=5^2\\ \Rightarrow x=2\)
\(b,\left(4x+1\right)^2=25.9\\\Rightarrow\left(4x+1\right)^2=225\\ \Rightarrow\left[{}\begin{matrix}4x+1=15\\4x+1=-15\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=-4\end{matrix}\right.\)
\(c,2^x+2^{x+3}=144\\ \Rightarrow2^x+2^x.2^3=144\\ \Rightarrow2^x\left(1+2^3\right)=144\\ \Rightarrow2^x=144:\left(1+2^3\right)\\ \Rightarrow2^x=16\\ \Rightarrow2^x=2^4\\ \Rightarrow x=4\)
\(d,3^{x+2}=9^{x+3}\\ \Rightarrow3^{x+2}=\left(3^2\right)^{x+3}\\ \Rightarrow3^{x+2}=3^{2x+6}\\ \Rightarrow x+2=2x+6\\ \Rightarrow x-2x=6-2\\ \Rightarrow-x=4\\ \Rightarrow x=-4\)
\(e,x^{15}=x^2\\ \Rightarrow x^{15}-x^2=0\\ \Rightarrow x^2\left(x^{13}-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x^2=0\\x^{13}-1=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
a: =>5^x+5^x*25=650
=>5^x*26=650
=>5^x=25
=>x=2
b: =>4x+1=15 hoặc 4x+1=-15
=>4x=-16 hoặc 4x=14
=>x=7/2 hoặc x=-8
c: =>2^x*9=144
=>2^x=16
=>x=4
d: =>2x+2=2x+6
=>2=6(loại)
e: =>x^2(x^13-1)=0
=>x=0 hoặc x=1
![](https://rs.olm.vn/images/avt/0.png?1311)
\(f,\left(2x+1\right)^3=343\\\Rightarrow \left(2x+1\right)^3=7^3\\ \Rightarrow2x+1=7\\ \Rightarrow2x=6\\ \Rightarrow x=3\\ g,\left(x-1\right)^3=125\\ \Rightarrow\left(x-1\right)^3=5^3\\ \Rightarrow x-1=5\\ \Rightarrow x=6\\ h,2^{x+2}-2^x=96\\ \Rightarrow2^x.2^2-2^x=96\\ \Rightarrow2^x.\left(2^2-1\right)=96\\ \Rightarrow2^x=96:\left(2^2-1\right)\\ \Rightarrow2^x=32\\ \Rightarrow2^x=2^5\\ \Rightarrow x=5\)
\(i,\left(x-5\right)^4=\left(x-5\right)^6\\ \Rightarrow\left(x-5\right)^4\left(1-\left(x-5\right)^2\right)=0\\\Rightarrow\left[{}\begin{matrix}\left(x-5\right)^4=0\\1-\left(x-5\right)^2=0\end{matrix}\right. \\ \Rightarrow\left[{}\begin{matrix}x-5=0\\x-5=1\\x-5=-1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=5\left(t/m\right)\\x=6\left(t|m\right)\\x=4\left(loại\right)\end{matrix}\right.\)
\(j,720:\left[41-\left(2x-5\right)\right]=2^3.5\\ \Rightarrow720:\left(41-2x+5\right)=40\\ \Rightarrow\left(46-2x\right)=720:40\\ \Rightarrow46-2x=18\\ \Rightarrow2x=46-18=28\\ \Rightarrow x=28:2=14\)
@seven
f: =>2x+1=7
=>2x=6
=>x=3
g: =>x-1=5
=>x=6
h: =>2^x*3=96
=>2^x=32
=>x=5
i: =>(x-5)^4*[(x-5)^2-1]=0
=>(x-5)(x-4)(x-6)=0
=>x=5;x=4;x=6
j: =>41-(2x-5)=720:40=18
=>2x-5=23
=>2x=28
=>x=14
![](https://rs.olm.vn/images/avt/0.png?1311)
f)
`(2x+1)^3=343`
`(2x+1)^3=7^3`
`=>2x+1=7`
`2x=7-1`
`2x=6`
`x=6:2`
`x=3`
g)
`(x-1)^3 =125`
`(x-1)^3 =5^3`
`=>x-1=5`
`x=6`
h)
`2^(x+2)-2^x=96`
`2^x *2^2 -2^x =96`
`2^x (2^2 -1)=96`
`2^x *3=96`
`2^x =32`
`2^x =2^5`
`=>x=5`
i)
`(x-5)^4 =(x-5)^6` (`x>=5`)
`(x-5)^6 -(x-5)^4 =0`
`(x-5)^4 [(x-5)^2 -1]=0`
`=>x-5=0` hoặc `(x-5)^2 -1=0`
`<=>x=5` hoặc `(x-5)^2 =1`
`<=>x=5` hoặc `x-5=1` hoặc `x-5=-1`
`<=>x=5` hoặc `x=6` hoặc `x=4`
j)
`720:[41-(2x-5)]=2^3 *5`
`720:[41-(2x-5)]=8*5`
`720:[41-(2x-5)]=40`
`41-(2x-5)=720:40`
`41-(2x-5)=18`
`2x-5=41-18`
`2x-5=23`
`2x=28`
`x=14`
![](https://rs.olm.vn/images/avt/0.png?1311)
xy=-11, x<y
x,y nguyên => \(\hept{\begin{cases}x=-11\\y=1\end{cases}}\)
Vậy (x;y)=(-11;1)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{7}{1}:\dfrac{4}{1}=\dfrac{7}{1}.\dfrac{1}{4}=\dfrac{7}{4}\)
\(\dfrac{11}{7}:\dfrac{37}{9}=\dfrac{11}{7}.\dfrac{9}{37}=\dfrac{99}{259}\)
a; a - b = 6 và \(\overline{4a7}\) + \(\overline{1b5}\) ⋮ 9
Để \(\overline{4a7}\) + \(\overline{1b5}\) ⋮ 9 ⇔ 4 + a + 7 + 1 + b + 5 ⋮ 9
⇒(4 + 7 + 1 + 5) + a + b ⋮ 9
⇒ 17 + a + b ⋮ 9
⇒ 8 + a + b ⋮ 9
Vì a + b ≤ 18 ⇒ 8 + a + b ≤ 26 ⇒ 8 + a + b = 9; 18 (1)
a - b = 6 ⇒ a = 6 + b Thay a = 6 + b vào (1) ta có:
8 + 6 + b + b = 9; 18 ⇒ (8 + 6) + (b + b) = 9; 18
⇒ 14 + 2b = 9; 18
Lập bảng ta có:
Theo bẳng trên ta có (a; b) = (8; 2)
b; a - b = 6 và \(\overline{7a5}\) + \(\overline{8b4}\) ⋮ 9
\(\overline{7a5}\) + \(\overline{8b4}\) ⋮ 9 ⇔ 7 + a + 5 + 8 + b + 4 ⋮ 9 ⇒ (7 + 5 + 8 + 4) + a + b⋮ 9
⇒ (12 + 8 + 4) + a + b ⋮ 9 ⇒ (20 + 4) + a + b ⋮ 9 ⇒ 24 + a + b ⋮ 9
⇒ 6 + a + b ⋮ 9 vì 0 ≤ a + b ≤ 18 ⇒ 6 ≤ 6 + a + b ≤ 24
⇒ 6 + a + b = 9; 18 (1)
a - b = 6 ⇒ a = 6 + b thay a = b + 6 vào (1) ta có:
6 + 6 + b + b = 9; 18 ⇒ (6 + 6) + (b + b) = 9; 18 ⇒ 12 +2b = 9; 18
Lập bảng ta có:
Theo bảng trên ta có:
(a; b) = (9; 3)