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Giải:
\(a,2^x-15=17.\)
\(2^x=17+15.\)
\(2^x=32.\)
\(2^x=2^5.\)
\(\Rightarrow x=5\in N.\)
Vậy \(x=5.\)
\(b,\left(7x-11\right)^3=2^5.5^2+200.\)
\(\left(7x-11\right)^3=32.25+200.\)
\(\left(7x-11\right)^3=800+200.\)
\(\left(7x-11\right)^3=1000.\)
\(\left(7x-11\right)^3=10^3.\)
\(\Rightarrow7x-11=10.\)
\(\Rightarrow7x=10+11=21.\)
\(\Rightarrow x=21:7=3\in N.\)
Vậy \(x=3.\)
\(c,3^x+25=26.2^2+2.3^0.\)
\(3^x+25=26.2^2+2.1.\)
\(3^x+25=2\left(26.2+1\right).\)
\(3^x+25=2.53.\)
\(3^x+25=106.\)
\(3^x=106-25.\)
\(3^x=81.\)
\(3^x=3^4\Rightarrow x=4\in N.\)
Vậy \(x=4.\)
\(d,49.7^x=2041.\)(đề này sai).
Sửa đề:
\(49.7^x=2401.\)
\(7^x=2401:49.\)
\(7^x=49.\)
\(7^x=7^2\Rightarrow x=2\in N.\)
Vậy \(x=2.\)
\(e,3^x=243.\)
\(3^x=3^5\Rightarrow x=5\in N.\)
Vậy \(x=5.\)
~ Học tốt!!! ~
a, \(2^x-15=17\)
\(\Rightarrow2^x=32\Rightarrow2^x=2^5\)
Vì \(2\ne-1;2\ne0;2\ne1\) nên \(x=5\)
Vậy....
b, \(\left(7x-11\right)^3=2^5.5^2+200\)
\(\Rightarrow\left(7x-11\right)^3=32.25+200\)
\(\Rightarrow\left(7x-11\right)^3=1000=10^3\)
\(\Rightarrow7x-11=10\Rightarrow7x=21\Rightarrow x=3\)
Vậy.....
c, \(3^x+25=26.2^2+2.3^0\)
\(\Rightarrow3^x+25=26.4+2\)
\(\Rightarrow3^x=104+2-25\)
\(\Rightarrow3^x=81=3^4\)
Vì \(3\ne-1;3\ne0;3\ne1\) nên x=4
Vậy.....
d,Sửa đề:\(49.7^x=2401\)
\(\Rightarrow7^2.7^x=7^4\Rightarrow7^{2+x}=7^4\)
Vì \(7\ne-1;7\ne0;7\ne1\) nên \(2+x=4\Rightarrow x=2\)
Vậy.....
Câu e làm tương tự! Chúc bạn học tốt!!!
ne ban minh biet cau tra loi nhung lam the nao bam ngoac vuong
k) \(2x-49=5.3^2\)
\(2x-49=45\)
\(2x=49+45\)
\(2x=94\)
\(x=47\)
l) \(3^2.\left(x+14\right)-5^2=5.2^2\)
\(9.\left(x+14\right)-25=20\)
\(9.\left(x+14\right)=45\)
\(x+14=5\)
\(x=-9\)
m) \(6x+x=5^{11}:5^9+3^1\)
\(7x=5^{11-9}+3\)
\(7x=5^2+3\)
\(7x=28\)
\(x=4\)
n) \(7x-x=5^{21}:5^{19}+3.2^2-\left(-7^{-0}\right)\)
\(6x=5^{21-19}+12-1\)
\(6x=5^2+11\)
\(6x=36\)
\(x=6\)
o) \(7x-2x=6^{17}:6^{15}+44:11\)
\(5x=6^{17-15}+4\)
\(5x=6^2+4\)
\(5x=40\)
\(x=8\)
o)7x-x=521:519+3.22-70
=> 6x = 5^2 + 12 -1
=> 6x = 36
=> x = 36/6 = 6
Kết quả 6
Học tốt
Bài 1 :
a, Ta có : \(\left(-123\right)+\left|-13\right|+\left(-7\right)\)
= \(\left(-123\right)+13+\left(-7\right)=\left(-117\right)\)
b, Ta có : \(\left|-10\right|+\left|45\right|+\left(-\left|-455\right|\right)+\left|-750\right|\)
= \(10+45-455+750=350\)
c, Ta có : \(-\left|-33\right|+\left(-15\right)+20-\left|45-40\right|-57\)
= \(\left(-33\right)+\left(-15\right)+20-5-57=-90\)
(7x - 11)3 = 25.52 + 200
=> (7x - 11)3 = 800 + 200
=> (7x - 11)3 = 1000
=> (7x - 11)3 = 103
=> 7x - 11 = 10
=> 7x = 10 + 11
=> 7x = 21
=> x = 21 : 7 = 3
3x + 25 = 26.22 + 2.30
=> 3x + 25 = 104 + 2
=> 3x + 25 = 106
=> 3x = 106 - 25
=> 3x = 81
=> 3x = 34
=> x = 4
a)\(\left(7x-11\right)^3=2^5.5^2+200\) b) \(3^x+25=26.2^2+2.3^0\)
\(\left(7x-11\right)^3=32.25+200\) \(3^x+25=26.4+2.1\)
\(\left(7x-11\right)^3=800+200\) \(3^x+25=104+2\)
\(\left(7x-11\right)^3=1000\) \(3^x+25=106\)
\(\Rightarrow\left(7x-11\right)^3=10^3\) \(3^x=106-25\)
\(\Rightarrow7x-11=10\) \(3^x=81\)
\(7x=10+11\) \(3^x=3^4\)
\(x=\frac{21}{7}\) \(\Rightarrow x=4\)
\(x=3\) Vậy x = 4
Vậy x = 3
A.(x+2)x-1=150
=>A.(x+2)x-1=1
=> x + 2 = 1 hoặc x + 2 = -1 hoặc x - 1 = 0
=> x = -1 hoặc x = -3 hoặc x = 1.
B. (5-x)x=1(x<5)
=> 5 - x = 1 hoặc 5 - x = -1 hoặc x = 0
=> x = 4 hoặc x = 6 hoặc x = 0.
C.15x-2=225
=> 15x-2=152
=> x - 2 = 2 => x = 4.
D.(x+2)2.(x+1)=64
=>(x+2).(x+2).(x+1)=64 = 1.2.32 = 2.2.16 = ...
Mà x + 2 và x + 2 và x + 1 chỉ hơn kém nhau 1 đơn vị nên không có x nào thỏa mãn.
E.(x-5)3.(x-5)=16
=>(x-5)4=16=24
=>x-5=2=>x=7.
Trả lời:
H.(7x-11)3 =25.52 +200
=(7x-11)3 =32.25 +200 =(7x-11)3 =800 +200
=(7x-11)3 =1000 =(7x-11)3 = 103
= 7x-11 = 10 = 7x = 10 + 11
= 7x = 21 = x = 21:7
= x = 3
I.3x +25 = 26.22+2.30
=3x +25 = 26.4 +2.1 =3x +25 = 106
=3x = 106-25 =3x = 81
=3x = 34 => x =4
K.27.3x= 243
= 3x =243:27
= 3x = 9
= 3x = 32
=> x = 2
Mấy câu khác cứ thế làm nha
a: \(\dfrac{4^5+4^5+4^5+4^5}{3^5+3^5+3^5+3^5}\cdot\dfrac{6^5+6^5+6^5+6^5+6^5+6^5}{2^5+2^5+2^5+2^5+2^5+2^5}=2^x\)
\(\Leftrightarrow2^x=\dfrac{4^5}{3^5}\cdot\dfrac{6^5}{2^5}=4^5=2^{10}\)
=>x=10
b: \(\left(x-1\right)^{x+4}=\left(x-1\right)^{x+2}\)
\(\Leftrightarrow\left(x-1\right)^{x+2}\left[\left(x-1\right)^2-1\right]=0\)
\(\Leftrightarrow x\left(x-1\right)^{x+2}\cdot\left(x-2\right)=0\)
hay \(x\in\left\{0;1;2\right\}\)
c: \(6\left(6-x\right)^{2003}=\left(6-x\right)^{2003}\)
\(\Leftrightarrow5\cdot\left(6-x\right)^{2003}=0\)
\(\Leftrightarrow6-x=0\)
hay x=6
\(a,\left(7x-11\right)^3=2^5.5^2+200.\)
\(\left(7x+11\right)^3=32.25+200.\)
\(\left(7x+11\right)^3=800+200.\)
\(\left(7x-11\right)^3=1000.\)
\(\left(7x-11\right)^3=10^3.\)
\(\Rightarrow7x-11=10.\)
\(\Rightarrow x=\left(10+11\right):3=7\in Z.\)
Vậy.....
\(b,3^x+25=26.2^2+2.3^0.\)
\(3^x+25=26.4+2.\)
\(3^x+25=104+2.\)
\(3^x+25=106.\)
\(3^x=106-25.\)
\(3^x=81.\)
\(3^x=3^4\Rightarrow x=4\in Z.\)
Vậy.....
\(c,2^x+3.2=64.\)(có vấn đề).
\(d,5^{x+1}+5^x=750.\)
\(5^x.5^1+5^x+1=750.\)
\(5^x\left(5^1+1\right)=750.\)
\(5^x\left(5+1\right)=750.\)
\(5^x.6=750.\)
\(5^x=750:6.\)
\(5^x=125.\)
\(5^x=5^3\Rightarrow x=3\in Z.\)
Vậy.....
\(e,x^{15}=x.\)
\(\Rightarrow x\left(x^{14}-1\right)=0\Rightarrow\left\{{}\begin{matrix}x=0\\x=1\end{matrix}\right..\)
\(f,\left(x-5\right)^4=\left(x-5\right)^6.\)
\(\Leftrightarrow\left(x-5\right)^4-\left(x-5^6\right)=0.\)
\(\Leftrightarrow\left(x-5\right)^4\left[1-\left(x-5\right)^2\right]=0.\)
\(\Leftrightarrow\left(x-5\right)^4\left(1-x+5\right)\left(1+x-5\right)=0.\)
\(\Leftrightarrow\left(x-5\right)^4\left(6-x\right)\left(x-4\right)=0.\)
\(\Leftrightarrow\left(x-5\right)^4=0\Rightarrow x-5=0\Rightarrow x=5\in Z.\)
\(6-x=0\Rightarrow x=6\in Z.\)
\(x-4=0\Rightarrow x=4\in Z.\)
Vậy.....