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(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
=>(7x-11)^3=1000
=>\(\left(7x-11\right)^3=10^3\)
=>7x-11=10
=>7x=10+11
=>7x=21
=>x=21:7
=>x=3
( 7x-1)3=25.52+200
( 7x-1)3=32.25+200
( 7x-1)3=800+200
( 7x-1)3=1000
( 7x-1)3=103
=>7x-1=10
7x=10+1
7x=11
x=11:7
x=11/7
( 7x-1)3=25.52+200
( 7x-1)3=32.25+200
( 7x-1)3=800+200
( 7x-1)3=1000
( 7x-1)3=103
=>7x-1=10
7x=10+1
7x=11
x=11:7
x=11/7
\(a,3\frac{1}{3}x+16\frac{3}{4}=-13,25\)
\(\frac{10}{3}x+\frac{67}{4}=-13,25\)
\(\frac{10}{3}x=-13,25-\frac{67}{4}\)
\(\frac{10}{3}x=-30\)
\(x=\left(-30\right):\frac{10}{3}\)
\(x=-9\)
\(b,\left(7x-11\right)^3=2^5.5^2+200\)
\(\left(7x-11\right)^3=32.25+200\)
\(\left(7x-11\right)^3=1000\)
\(\Rightarrow\left(7x-11\right)^3=10^3\)
\(\Rightarrow7x-11=10\)
\(\Rightarrow7x=10+11\)
\(\Rightarrow7x=21\)
\(\Rightarrow x=3\)
A) (10/3)x+67/4=-53/4<=>(10/3)x=-53/4-67/4=-30<=>x=-30:(10/3)=-9 b) (7x-11)^3=1000=10^3<=>7x-11=10=>7x=21=>x=3
Ta có : 2n + 4.2n = 5.25
<=> 2n(1 + 4) = 5.25
<=> 2n.5 = 5.25
Bỏ 5 ở ca 2 về đi
=> 2n = 25
=> n = 5
\(a,\left(7x-11\right)^3=2^5.5^2+200.\)
\(\left(7x+11\right)^3=32.25+200.\)
\(\left(7x+11\right)^3=800+200.\)
\(\left(7x-11\right)^3=1000.\)
\(\left(7x-11\right)^3=10^3.\)
\(\Rightarrow7x-11=10.\)
\(\Rightarrow x=\left(10+11\right):3=7\in Z.\)
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
b) Ta có : \(\left(x-\frac{1}{3}\right)^2-\frac{1}{4}=0\)
\(\Rightarrow\left(x-\frac{1}{3}\right)^2=\frac{1}{4}\)
\(\Rightarrow\orbr{\begin{cases}\left(x-\frac{1}{3}\right)^2=\left(\frac{1}{2}\right)^2\\\left(x-\frac{1}{3}\right)^2=\left(-\frac{1}{2}\right)^2\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x-\frac{1}{3}=\frac{1}{2}\\x-\frac{1}{3}=-\frac{1}{2}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{5}{6}\\x=-\frac{1}{6}\end{cases}}\)
b) \(\left(x-\frac{1}{3}\right)^2-\frac{1}{4}=0\)
\(\Leftrightarrow\left(x-\frac{1}{3}\right)^2=\frac{1}{4}\)
\(\Rightarrow\orbr{\begin{cases}x-\frac{1}{3}=\frac{1}{4}\\x-\frac{1}{3}=-\frac{1}{4}\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{7}{12}\\x=\frac{1}{12}\end{cases}}\)
d) \(\frac{x+5}{2}=\frac{8}{x+5}\)
\(\Rightarrow\left(x+5\right)^2=16\)
\(\Rightarrow\orbr{\begin{cases}x+5=16\\x+5=-16\end{cases}\Rightarrow\orbr{\begin{cases}x=11\\x=-21\end{cases}}}\)
\(\left(7x-11\right)^3=2^5.5^2+200\)
<=>\(\left(7x-11\right)^3=1000=10^3\)
<=>\(7x-11=10\)
<=>\(x=3\)