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\(x=\frac{10^2.5^3.15^6}{3^6.5^{10}}=\frac{\left(2.5\right)^2.\left(3.5\right)^6}{3^6.5^7}=\frac{2^2.5^2.3^6.5^6}{3^6.5^7}=2^2.5=20\)
Vậy x = 20
\(5^{x+4}-3.5^{x+3}=2.5^{11}\)
\(5^{x+3}\left(5-3\right)=2.5^{11}\)
\(5^{x+3}.2=2.5^{11}\)
\(5^{x+3}=5^{11}\)
\(x+3=11\)
\(x=8\)
\(4^{x+3}-3.4^{x+1}=13.4^{11}\)
\(4^{x+1}\left(4^2-3\right)=13.4^{11}\)
\(4^{x+1}.13=13.4^{11}\)
\(4^{x+1}=4^{11}\)
\(x+1=11\)
\(x=10\)
a) \(\left|x-\frac{1}{2}\right|+3=2^2\)
\(\Leftrightarrow\left|x-\frac{1}{2}\right|+3=4\)\(\Leftrightarrow\left|x-\frac{1}{2}\right|=1\)
\(\Leftrightarrow\orbr{\begin{cases}x-\frac{1}{2}=-1\\x-\frac{1}{2}=1\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{-1}{2}\\x=\frac{3}{2}\end{cases}}\)
Vậy \(x=-\frac{1}{2}\)hoặc \(x=\frac{3}{2}\)
b) \(3^x+3^{x+2}=10^2-2.5\)
\(\Leftrightarrow3^x+3^x.3^2=100-10\)
\(\Leftrightarrow3^x+3^x.9=90\)
\(\Leftrightarrow3^x.\left(1+9\right)=90\)
\(\Leftrightarrow3^x.10=90\)
\(\Leftrightarrow3^x=9=3^2\)\(\Leftrightarrow x=2\)
Vậy \(x=2\)
c) \(3-2x^2=\frac{5}{2}\)
\(\Leftrightarrow2x^2=3-\frac{5}{2}\)\(\Leftrightarrow2x^2=\frac{1}{2}\)
\(\Leftrightarrow x^2=\frac{1}{4}\)\(\Leftrightarrow\orbr{\begin{cases}x=-\frac{1}{2}\\x=\frac{1}{2}\end{cases}}\)
Vậy \(x=-\frac{1}{2}\)hoặc \(x=\frac{1}{2}\)
\(a,\left(\frac{6^3-10.5^3}{6^2.3^3-15^2.5^2}.|x-2|\right):10=\left(1-\frac{1}{2}\right)....\left(1-\frac{1}{10}\right)\)
\(=\frac{1.2.3.4...9}{1.2.....10}=\frac{1}{10}\Leftrightarrow\frac{6^3-10.5^3}{6^2.3^3-15^2.5^2}.|x-2|=1\)
\(\Leftrightarrow\frac{6^2.6-2.5^4}{6^2.3^2-3^2.5^4}.|x-2|=1\Leftrightarrow|x-2|.\frac{2}{3}=1\Leftrightarrow|x-2|=\frac{3}{2}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=\frac{7}{2}\end{cases}}\)
\(\left(\frac{6^3-10,5^3}{6^2.3^3-15^2.5^2}.\left|x-2\right|\right):10=\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).....\left(1-\frac{1}{9}\right).\left(1-\frac{1}{10}\right)\)
\(=\frac{1.2.3.4...9}{1.2.....10}=\frac{1}{10}\)
\(\Leftrightarrow\frac{6^3-10,5^3}{6^2.3^3-15^2.5^2}.\left|x-2\right|=1\)
\(\Leftrightarrow\frac{6^2.6-2.5^4}{6^2.3^2-3^2.5^4}.\left|x-2\right|=1\)
\(\Leftrightarrow\left|x-2\right|.\frac{2}{3}=1\Leftrightarrow\left|x-2\right|=\frac{3}{2}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=\frac{7}{2}\end{cases}}\)
/X/ = \(\frac{10}{3}\)
suy ra: X = \(\frac{10}{3}\)hay X = \(\frac{-10}{3}\)
/X - \(\frac{5}{6}\)/ = \(\frac{1}{2}\)
suy ra : X - \(\frac{5}{6}\)= \(\frac{1}{2}\)hay X - \(\frac{5}{6}\)= \(\frac{-1}{2}\)
X = \(\frac{1}{2}+\frac{5}{6}\) hay X = \(\frac{-1}{2}+\frac{5}{6}\)
X = \(\frac{4}{3}\) hay X = \(\frac{1}{3}\)
/ X + \(\frac{4}{9}\)/ - \(\frac{1}{2}\)= \(\frac{3}{2}\)
/ X + \(\frac{4}{9}\)/ = \(\frac{3}{2}+\frac{1}{2}\)
/ X + \(\frac{4}{9}\)/ = \(2\)
suy ra : X + \(\frac{4}{9}\)= \(2\)hay X + \(\frac{4}{9}\)= \(-2\)
X = \(2-\frac{4}{9}\)hay X = \(\left(-2\right)-\frac{4}{9}\)
X = \(\frac{14}{9}\) hay X = \(\frac{-22}{9}\)
a. |x| = \(10/3\)
=> x = 10/3 hoặc x = -10/3
b/ |x - 5/6| = 1/2
*TH1: x - 5/6 = 1/2
=> x = 1/2 + 5/6
=> x = 4/3
*TH2: -x + 5/6 = 1/2
=> -x = 1/2 - 5/6
=> -x = -1/3
=> x = 1/3
Vậy...
Phần c mk không chắc nên để lại ạ
\(\frac{x-6}{7}+\frac{x-7}{8}+\frac{x-8}{9}=\frac{x-9}{10}+\frac{x-10}{11}+\frac{x-11}{12}\)
\(\Leftrightarrow\left(\frac{x-6}{7}+1\right)+\left(\frac{x-7}{8}+1\right)+\left(\frac{x-8}{9}+1\right)=\left(\frac{x-9}{10}+1\right)+\left(\frac{x-10}{11}+1\right)+\left(\frac{x-11}{12}+1\right)\)
\(\Leftrightarrow\frac{x+1}{7}+\frac{x+1}{8}+\frac{x+1}{9}=\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}\)
\(\Leftrightarrow\frac{x+1}{7}+\frac{x+1}{8}+\frac{x+1}{9}-\frac{x+1}{10}-\frac{x+1}{11}-\frac{x+1}{12}=0\)
\(\Leftrightarrow\left(x+1\right)\left(\frac{1}{7}+\frac{1}{8}+\frac{1}{9}-\frac{1}{10}-\frac{1}{11}-\frac{1}{12}\right)=0\)
\(\Leftrightarrow x+1=0\)( \(\frac{1}{7}+\frac{1}{8}+\frac{1}{9}-\frac{1}{10}-\frac{1}{11}-\frac{1}{12}\ne0\))
\(\Leftrightarrow x=-1\)
Vậy x=-1
mỗi phân số + 1 thì sẽ có tử chung là x + 1
chuyển vế có \(\left(x+1\right)\left(\frac{1}{7}+\frac{1}{8}+\frac{1}{9}-\frac{1}{10}-\frac{1}{11}-\frac{1}{12}\right)\)) =0
mà tổng các phân số kia khác 0 nên x+1 bằng 0
=> x=-1
câu 1 (có sai đề ko ?) vì có z nên khó tìm được x
câu 2 thì cứ biến z/5=2z/10 rồi áp dụng tính chất dãy tỉ số bằng nhau nên ta có được:
x+y+2z/2+3+10=10/15=2/3
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2015}{2017}\)
\(\Rightarrow\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left(x+1\right)}=\frac{2015}{2017}\)
\(\Rightarrow2\times\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2015}{2017}\)
\(\Rightarrow2\times\left(\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2015}{2017}\)
\(\Rightarrow2\times\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2015}{2017}\)
\(\Rightarrow2\times\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2015}{2017}\)
\(\Rightarrow1-\frac{2}{x+1}=\frac{2015}{2017}\)
\(\Rightarrow\frac{2}{x+1}=\frac{2}{2017}\Rightarrow x+1=2017\Rightarrow x=2016\)
Vậy x = 2016
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2015}{2017}\)
\(\Rightarrow2\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2015}{2017}\)
\(\Rightarrow2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2015}{2017}\)
\(\Rightarrow2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2015}{2017}\)
\(\Rightarrow2\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2015}{2017}\)
\(\Rightarrow1-\frac{2}{x+1}=\frac{2015}{2017}\)
\(\Rightarrow\frac{2}{x+1}=\frac{2}{2017}\)
\(\Rightarrow x+1=2017\)
\(\Rightarrow x=2016\)
Vậy \(x=2016\)
\(x=\frac{10^2.5^3.15^6}{3^6.5^{10}}=\frac{\left(2.5\right)^2.5^3.\left(3.5\right)^6}{3^6.5^{10}}=\frac{2^2.5^2.5^3.3^6.5^6}{3^6.5^{10}}=\frac{2^2.5^{11}.3^6}{3^6.5^{10}}=2^2.5=4.5=20\)
Vậy x=20.