Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
87 _(321_x ) ÷ 5 = 75
(321 _ x ) ÷ 5 = 87_75
(321 _ x ) ÷5 = 12
(321 _ x) = 12 ×5
321 _ x = 60
X = 321_60
X = 261
2)( 5x _24) × 73 = 2 × 74
(5x _ 24 ) = 2×74÷73
( 5x _ 16) = 2×7
5x _ 16 = 14
5x = 14+ 16
5x = 30
X = 30÷5
X = 6
3)93 + 3×(x_5) = 3×52
93+ 3× ( x _ 5) = 75
3×( x _ 5) = 75_93
3×(x _ 5 ) = _18
(X _ 5) = _18 ÷ 3
X _ 5 = _6
X = _6 + 5
X = _1
Tớ chưa kịp làm tối về giải sau
\(\left(b\right)3^2+3^4+3^x=3^{10}\)
\(\Rightarrow3^{2+4+x}=3^{10}\Rightarrow2+4+x=10\)
\(\Rightarrow6+x=10\Rightarrow x=10-6=4\)
\(\left(e\right)x.x^2.x^3.x^4=1024\)
\(\Rightarrow x^1.x^2.x^3.x^4=1024\Rightarrow x^{10}=1024\)
Mà \(1024=2^{10}\Rightarrow x=2\)
a,1+2+3+4...+x=45
có số hạng là (x-1)+1
suy ra:x.(x+1):2=45
x.(x+1)=90
x.(x+1)=9.10
suy ra:x=9
Vậy x=9
\(a,\left(7x-11\right)^3=2^5.5^2+200\)
\(\Leftrightarrow\left(7x-11\right)^3=32.25+200\)
\(\Leftrightarrow\left(7x-11\right)^3=800+200\)
\(\Leftrightarrow\left(7x-11\right)^3=1000\)
\(\Leftrightarrow\left(7x-11\right)^3=10^3\)
\(\Rightarrow7x-11=10\)
\(\Rightarrow7x=21\)
\(\Rightarrow x=3\)
\(b,3^2.x^2-2^2.x^2=5^5-\left(255:51\right).11\)
\(\Rightarrow x^2\left(3^2-2^2\right)=3125-5.11\)
\(\Rightarrow x^2\left(9-4\right)=3125-55\)
\(\Rightarrow5x^2=2970\)
\(\Rightarrow x^2=2970:5\)
\(\Rightarrow x^2=594\)
\(\Rightarrow x=3\sqrt{66}\)
\(a,\left(7x-11\right)^3=2^5.5^2+200.\)
\(\left(7x+11\right)^3=32.25+200.\)
\(\left(7x+11\right)^3=800+200.\)
\(\left(7x-11\right)^3=1000.\)
\(\left(7x-11\right)^3=10^3.\)
\(\Rightarrow7x-11=10.\)
\(\Rightarrow x=\left(10+11\right):3=7\in Z.\)
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.....
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) ( x-140):7=33 -23 -3
(x-140):7 = 27 - 8-3
(x-140):7 = 16
(x-140):7 = 16 .7
x-140 = 112
x-140 = 112+ 140
x = 252
\(a)2x^2-98=0\)
\(2x^2=0+98\)
\(2x^2=98\)
\(x^2=98:2\)
\(x^2=49\)
\(\rightarrow x^2=7^2\)
\(\rightarrow x=7\)
Vậy x = 7
a) \(93+3\left(x-5\right)=3.5^2=75\\ =>3\left(x-5\right)=75-93=-18\\ =>x-5=\dfrac{-18}{3}=-6\\ =>x=-6+5=-1\)
b, \(\left(5x^3+2^2.11\right):3^2=5\\ < =>\left(5x^3+44\right):9=5\\ =>5x^3+44=5.9=45\\ =>5x^3=45-44=1\\ =>x^3=\dfrac{1}{5}\\ =>x=\sqrt[3]{\dfrac{1}{5}}\)
ko hiểu