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7 tháng 8 2018

\(\left(7x-11\right)^3=2^5.5^2+200\)

<=>\(\left(7x-11\right)^3=1000=10^3\)

<=>\(7x-11=10\)

<=>\(x=3\)

4 tháng 10 2015

(7x - 11)= 2. 5+ 200

(7x - 11)= 32 . 25 + 200

(7x - 11)= 800 + 200

(7x - 11)= 1000

(7x - 11)= 10

7x - 11 = 10

      7x  = 10 + 11

      7x  = 21

        x  = 21 : 7

        x = 3

4 tháng 10 2015

=>(7x-11)^3=1000

=>\(\left(7x-11\right)^3=10^3\)

=>7x-11=10

=>7x=10+11

=>7x=21

=>x=21:7

=>x=3

13 tháng 10 2015

( 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

25 tháng 1 2016

( 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

12 tháng 2 2017

\(a,3\frac{1}{3}x+16\frac{3}{4}=-13,25\)

       \(\frac{10}{3}x+\frac{67}{4}=-13,25\)

                    \(\frac{10}{3}x=-13,25-\frac{67}{4}\)

                    \(\frac{10}{3}x=-30\)

                            \(x=\left(-30\right):\frac{10}{3}\)

                            \(x=-9\)

\(b,\left(7x-11\right)^3=2^5.5^2+200\)

    \(\left(7x-11\right)^3=32.25+200\)

     \(\left(7x-11\right)^3=1000\)

 \(\Rightarrow\left(7x-11\right)^3=10^3\)

 \(\Rightarrow7x-11=10\)

\(\Rightarrow7x=10+11\)

\(\Rightarrow7x=21\)

\(\Rightarrow x=3\)

12 tháng 2 2017

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

13 tháng 6 2017

2n.1+4.2n=5.25

2n.(1+4)=5.25

2n=25

=>n=5

13 tháng 6 2017

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

25 tháng 8 2017

\(a,\left(7x-11\right)^3=2^5.5^2+200.\)

\(\left(7x+11\right)^3=32.25+200.\)

\(\left(7x+11\right)^3=800+200.\)

\(\left(7x-11\right)^3=1000.\)

\(\left(7x-11\right)^3=10^3.\)

\(\Rightarrow7x-11=10.\)

\(\Rightarrow x=\left(10+11\right):3=7\in Z.\)

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

10 tháng 7 2017

b) Ta có : \(\left(x-\frac{1}{3}\right)^2-\frac{1}{4}=0\)

\(\Rightarrow\left(x-\frac{1}{3}\right)^2=\frac{1}{4}\)

\(\Rightarrow\orbr{\begin{cases}\left(x-\frac{1}{3}\right)^2=\left(\frac{1}{2}\right)^2\\\left(x-\frac{1}{3}\right)^2=\left(-\frac{1}{2}\right)^2\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x-\frac{1}{3}=\frac{1}{2}\\x-\frac{1}{3}=-\frac{1}{2}\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{5}{6}\\x=-\frac{1}{6}\end{cases}}\)

10 tháng 7 2017

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) \(\frac{x+5}{2}=\frac{8}{x+5}\)

\(\Rightarrow\left(x+5\right)^2=16\)

\(\Rightarrow\orbr{\begin{cases}x+5=16\\x+5=-16\end{cases}\Rightarrow\orbr{\begin{cases}x=11\\x=-21\end{cases}}}\)