Tìm n :
5 mũ 2 x n - 3 - 2 x 5 mũ 2 = 5 mũ 2 x 3
Khó hiểu thì cứ hỏi mk nhé!
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TL:
a.\(2^6.2^n=2^{11}\)
\(2^{6+n}=2^{11}\)
\(\Rightarrow n=5\)
b. \(3^7:3^n=3^4\)
\(3^{7-n}=3^4\)
\(\Rightarrow n=3\)
c.\(2^n.32=2^{10}\)
\(2^{n+5}=2^{10}\)
\(\Rightarrow n=5\)
a)2^x-15=17
2^x=17+15=32
2^x=32=>x=5(vì 2^5=32)
b)mk ko bt cách giải
c)7^2-(15+x)=5.2^2
49-(15+x)=5.4=20
15+x=49-20
15+x=29
d)mk cx ko bt cách giải nhưng mk bt x=2(bn thử lại là bt)
b) (7x-11)3= 25.52+200
(7x-11)3 = 25.52 + 23.52
(7x-11)3 = 23.52( 22+1)
(7x-11)3 = 8 . 25 . 5
(7x-11)3 = 1000
(7x-11)3 = 103
=> 7x-11 = 10
7x = 10+11
7x = 21
x= 3
d) (2x+1)3 = 125
(2x+1)3 = 53
=> 2x + 1 = 5
=> 2x = 5-1
2x = 4
x=2
Bài 6 :
a) \(\dfrac{625}{5^n}=5\Rightarrow\dfrac{5^4}{5^n}=5\Rightarrow5^{4-n}=5^1\Rightarrow4-n=1\Rightarrow n=3\)
b) \(\dfrac{\left(-3\right)^n}{27}=-9\Rightarrow\dfrac{\left(-3\right)^n}{\left(-3\right)^3}=\left(-3\right)^2\Rightarrow\left(-3\right)^{n-3}=\left(-3\right)^2\Rightarrow n-3=2\Rightarrow n=5\)
c) \(3^n.2^n=36\Rightarrow\left(2.3\right)^n=6^2\Rightarrow\left(6\right)^n=6^2\Rightarrow n=6\)
d) \(25^{2n}:5^n=125^2\Rightarrow\left(5^2\right)^{2n}:5^n=\left(5^3\right)^2\Rightarrow5^{4n}:5^n=5^6\Rightarrow\Rightarrow5^{3n}=5^6\Rightarrow3n=6\Rightarrow n=3\)
Bài 7 :
a) \(3^x+3^{x+2}=9^{17}+27^{12}\)
\(\Rightarrow3^x\left(1+3^2\right)=\left(3^2\right)^{17}+\left(3^3\right)^{12}\)
\(\Rightarrow10.3^x=3^{34}+3^{36}\)
\(\Rightarrow10.3^x=3^{34}\left(1+3^2\right)=10.3^{34}\)
\(\Rightarrow3^x=3^{34}\Rightarrow x=34\)
b) \(5^{x+1}-5^x=100.25^{29}\Rightarrow5^x\left(5-1\right)=4.5^2.\left(5^2\right)^{29}\)
\(\Rightarrow4.5^x=4.25^{2.29+2}=4.5^{60}\)
\(\Rightarrow5^x=5^{60}\Rightarrow x=60\)
c) Bài C bạn xem lại đề
d) \(\dfrac{3}{2.4^x}+\dfrac{5}{3.4^{x+2}}=\dfrac{3}{2.4^8}+\dfrac{5}{3.4^{10}}\)
\(\Rightarrow\dfrac{3}{2.4^x}-\dfrac{3}{2.4^8}+\dfrac{5}{3.4^{x+2}}-\dfrac{5}{3.4^{10}}=0\)
\(\Rightarrow\dfrac{3}{2}\left(\dfrac{1}{4^x}-\dfrac{1}{4^8}\right)+\dfrac{5}{3.4^2}\left(\dfrac{1}{4^x}-\dfrac{1}{4^8}\right)=0\)
\(\Rightarrow\left(\dfrac{1}{4^x}-\dfrac{1}{4^8}\right)\left(\dfrac{3}{2}+\dfrac{5}{3.4^2}\right)=0\)
\(\Rightarrow\dfrac{1}{4^x}-\dfrac{1}{4^8}=0\)
\(\Rightarrow\dfrac{4^8-4^x}{4^{x+8}}=0\Rightarrow4^8-4^x=0\left(4^{x+8}>0\right)\Rightarrow4^x=4^8\Rightarrow x=8\)
1.
a) \(3^2\cdot2^5-\left(3\cdot6^2-x\right)=120\\ 9\cdot32-\left(3\cdot36-x\right)=120\\ 288-\left(108-x\right)=120\\ 288-108+x=120\\ 180+x=120\\ \Rightarrow x=-60\)
Vậy x = -60
b) \(\left(x+3\right)\cdot2^3-2^2\cdot5=2^2\cdot3^5\\ \left(x+3\right)\cdot8-4\cdot5=4\cdot243\\ \left(x+3\right)\cdot8-20=972\\ \Rightarrow8\left(x+3\right)=992\\ \Rightarrow x+3=124\\ \Rightarrow x=121\)
Vậy x = 121
2.
a) \(5^{x-1}-13=612\\ \Rightarrow5^{x-1}=625=5^3\\ \Rightarrow x-1=3\\ \Rightarrow x=4\)
Vậy x = 4
b) \(5^x\cdot5^3=125\\ 5^{x+3}=5^3\\ \Rightarrow x+3=3\\ \Rightarrow x=0\)
Vậy x = 0
a) x4+x3+2x2+x+1=(x4+x3+x2)+(x2+x+1)=x2(x2+x+1)+(x2+x+1)=(x2+x+1)(x2+1)
b)a3+b3+c3-3abc=a3+3ab(a+b)+b3+c3 -(3ab(a+b)+3abc)=(a+b)3+c3-3ab(a+b+c)
=(a+b+c)((a+b)2-(a+b)c+c2)-3ab(a+b+c)=(a+b+c)(a2+2ab+b2-ac-ab+c2-3ab)=(a+b+c)(a2+b2+c2-ab-ac-bc)
c)Đặt x-y=a;y-z=b;z-x=c
a+b+c=x-y-z+z-x=o
đưa về như bài b
d)nhóm 2 hạng tử đầu lại và 2hangj tử sau lại để 2 hạng tử sau ở trong ngoặc sau đó áp dụng hằng đẳng thức dề tính sau đó dặt nhân tử chung
e)x2(y-z)+y2(z-x)+z2(x-y)=x2(y-z)-y2((y-z)+(x-y))+z2(x-y)
=x2(y-z)-y2(y-z)-y2(x-y)+z2(x-y)=(y-z)(x2-y2)-(x-y)(y2-z2)=(y-z)(x2-2y2+xy+xz+yz)
\(5^2.n-3-2.5^2=5^2.3\)
\(\Rightarrow25n-3-2.25=25.3\)
\(\Rightarrow25.n-3-50=75\)
\(\Rightarrow25.n-53=75\)
\(\Rightarrow25.n=128\)
\(\Rightarrow n=\frac{128}{25}\)
\(5^2.n-3-2.5^2=5^2.3\)
\(\Rightarrow5^2.\left(n-2\right)-3=5^2.3\)
\(\Rightarrow\left(n-2\right)-3=3\)
\(\Rightarrow n-2=3+3\)
\(\Rightarrow n-2=6\)
\(\Rightarrow n=6+2\)
\(\Rightarrow n=8\)