tìm x thuộc z :
\(\left(5x+1\right)^2=\frac{36}{49}\)
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PU
3
20 tháng 7 2016
a) (5x + 1)2 = 36/49
(5x + 1)2 = (6/7)2
suy ra: 5x + 1= 6/7 hoặc 5x +1 = -6/7
5x = 6/7 - 1 5x = -6/7 -1
5x = -1/7 5x = -13/7
x = -1/7 : 5 x = -13/7 : 5
x = -1/35 x = -13/35
20 tháng 7 2016
a, \(\left(5x+1\right)^2=\frac{36}{49}\)
\(\left(5x+1\right)^2=\left(\frac{6}{7}\right)^2\)
=> 5x + 1 = \(\frac{6}{7}\)
\(\Rightarrow5x=\frac{6}{7}-1\)
\(\Rightarrow5x=\frac{-1}{7}\)
\(\Rightarrow x=\frac{-1}{7}:5=\frac{-1}{35}\)
b,, \(\left(2x-\frac{1}{2}\right)^2=\frac{1}{4}\)
\(\left(2x-\frac{1}{2}\right)^2=\left(\frac{1}{2}\right)^2\)
\(\Rightarrow2x-\frac{1}{2}=\frac{1}{2}\)
\(\Rightarrow2x=\frac{1}{2}+\frac{1}{2}=1\)
\(\Rightarrow x=1:2=\frac{1}{2}\)
23 tháng 11 2021
\(a,\Rightarrow\left[{}\begin{matrix}5x+1=\dfrac{6}{7}\\5x+1=-\dfrac{6}{7}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}5x=\dfrac{1}{7}\\5x=-\dfrac{13}{7}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{35}\\x=-\dfrac{13}{35}\end{matrix}\right.\\ b,\Rightarrow\left(-\dfrac{1}{8}\right)^x=\dfrac{1}{64}=\left(-\dfrac{1}{8}\right)^2\Rightarrow x=2\\ c,\Rightarrow\left(x-2\right)\left(2x+3\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=2\\x=-\dfrac{3}{2}\end{matrix}\right.\\ d,\Rightarrow\left(x+1\right)^{x+10}-\left(x+1\right)^{x+4}=0\\ \Rightarrow\left(x+1\right)^{x+4}\left[\left(x+1\right)^6-1\right]=0\\ \Rightarrow\left[{}\begin{matrix}x+1=0\\\left(x+1\right)^6=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x+1=0\\x+1=1\\x+1=-1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-1\\x=0\\x=-2\end{matrix}\right.\\ e,\Rightarrow\dfrac{3}{4}\sqrt{x}=\dfrac{5}{6}\left(x\ge0\right)\\ \Rightarrow\sqrt{x}=\dfrac{10}{9}\Rightarrow x=\dfrac{100}{81}\)
T
21 tháng 7 2019
#)Giải :
a) \(\left(5x+1\right)^2=\frac{36}{49}\Leftrightarrow\left(5x+1\right)^2=\left(\frac{6}{7}\right)^2\Leftrightarrow5x+1=\frac{6}{7}\Leftrightarrow5x=-\frac{1}{7}\Leftrightarrow x=-\frac{1}{35}\)
b) \(\left(x-\frac{2}{9}\right)^3=\left(\frac{2}{3}\right)^6\Leftrightarrow\left(x-\frac{2}{9}\right)^3=\left[\left(\frac{2}{3}\right)^2\right]^3\Leftrightarrow x-\frac{2}{9}=\left(\frac{2}{3}\right)^2=\frac{4}{9}\Leftrightarrow x=\frac{2}{3}\)
c) \(\left(8x-1\right)^{2x+1}=5^{2x+1}\Leftrightarrow8x-1=5\Leftrightarrow8x=6\Leftrightarrow x=\frac{6}{8}\)
C
21 tháng 7 2019
a) \(\left(5x+1\right)^2=\frac{36}{49}\)
\(\left(5x+1\right)^2=\frac{6^2}{7^2}\)
\(\left(5x+1\right)^2=\left(\frac{6}{7}\right)^2\)
\(\Leftrightarrow5x+1=\frac{6}{7}\)
\(5x=\frac{6}{7}-1\)
\(5x=\frac{6}{7}-\frac{7}{7}\)
\(5x=-\frac{1}{7}\)
\(x=-\frac{1}{7}\div5\)
\(x=-\frac{1}{7}\times\frac{1}{5}\)
\(x=-\frac{1}{35}\)
Vậy \(x=-\frac{1}{35}\)
NV
2
HN
5 tháng 7 2018
a) \(\left(5x+1\right)^2=\frac{36}{49}\)
\(\Rightarrow\left(5x+1\right)^2=\left(\frac{6}{7}\right)^2=\left(\frac{-6}{7}\right)^2\)
\(\Rightarrow\hept{\begin{cases}5x+1=\frac{6}{7}\\5x+1=\frac{-6}{7}\end{cases}}\Rightarrow\hept{\begin{cases}5x=\frac{-1}{7}\\5x=\frac{-13}{7}\end{cases}\Rightarrow\hept{\begin{cases}x=\frac{-1}{35}\\x=\frac{-13}{35}\end{cases}}}\)
b) \(\left(x-\frac{2}{9}\right)^3=\left(\frac{2}{3}\right)^6\)
\(\left(x-\frac{2}{9}\right)^3=\left(\frac{2}{3}^2\right)^3\)
\(\left(x-\frac{2}{9}\right)^3=\left(\frac{4}{9}\right)^3\)
\(x-\frac{2}{9}=\frac{4}{9}\)
\(x=\frac{4}{9}+\frac{2}{9}\)
\(x=\frac{6}{9}=\frac{2}{3}\)
5 tháng 7 2018
\(a.\left(5x+1\right)^2=\frac{36}{49}\)
\(\left(5x+1\right)^2=\frac{6^2}{7^2}\)
\(\left(5x+1\right)^2=\left(\frac{6}{7}\right)^2\)
\(\Rightarrow5x+1=\frac{6}{7}\)
\(\Rightarrow5x=\frac{6}{7}-1\)
\(\Rightarrow5x=-\frac{1}{7}\)
\(\Rightarrow x=-\frac{1}{7}:5\)
\(\Rightarrow x=-\frac{1}{35}\)
Vậy \(x=-\frac{1}{35}\)
\(b\left(x-\frac{2}{9}\right)^3=\left(\frac{2}{3}\right).^6\)
\(\left(x-\frac{2}{9}\right)^3=\frac{\left(2^2\right)^3}{\left(3^2\right)^3}\)
\(\left(x-\frac{2}{9}\right)^3=\frac{4^3}{9^3}\)
\(\left(x-\frac{2}{9}\right)^3=\left(\frac{4}{9}\right)^3\)
\(\Rightarrow x-\frac{2}{9}=\frac{4}{9}\)
\(\Rightarrow x=\frac{4}{9}+\frac{2}{9}\)
\(\Rightarrow x=\frac{6}{9}=\frac{2}{3}\)
Vậy \(x=\frac{2}{3}\)
26 tháng 9 2016
a) (5x +1)^2= 6^2/7^2
=> 5x+1= 6/7 hoặc -6/7 ( vì cả hai đều có mũ hai nên có thể bỏ đi - cái này mình giải thích cho bạn hỉu thui, đừng chép vào vở nhé)
Đến đây thì bạn cứ tính theo cách tìm x thông thường, cuối cùng thì ra số âm nên không có kết quả x thuộc N
11 tháng 10 2020
Bài 1:
\(\left(x-y+z\right)^2+\left(z-y\right)^2+\left(x-y+z\right)\left(2y-2z\right)\)
\(=\left(x-y+z\right)^2+2\left(x-y+z\right)\left(y-z\right)+\left(y-z\right)^2\)
\(=\left(x-y+z+y-z\right)^2\)
\(=x^2\)
Bài 2:
đk: \(x\ne\left\{0;-1;-2;-3;-4;-5\right\}\)
Xét BT trái ta có:
\(\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+...+\frac{1}{\left(x+4\right)\left(x+5\right)}\)
\(=\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+...+\frac{1}{x+4}-\frac{1}{x+5}\)
\(=\frac{1}{x}-\frac{1}{x+5}\)
\(=\frac{5}{x\left(x+5\right)}=\frac{5}{x^2+5x}\)
GT của biểu thức lớn sẽ là: \(\frac{5}{x^2+5x}\cdot\frac{x^2+5x}{5}=1\) không phụ thuộc vào biến
=> đpcm
11 tháng 10 2020
Bài 1.
( x - y + z ) + ( z - y )2 + ( x - y + z )( 2y - 2z )
= ( x - y + z ) - 2( x - y + z )( z - y ) + ( z - y )2
= [ ( x - y + z ) - ( z - y ) ]2
= ( x - y + z - z + y )2
= x2
Bài 2. ĐKXĐ tự ghi nhé :))
\(\left(\frac{1}{x^2+x}+\frac{1}{x^2+3x+2}+\frac{1}{x^2+5x+6}+\frac{1}{x^2+7x+12}+\frac{1}{x^2+9x+20}\right)\times\left(\frac{x^2+5x}{5}\right)\)
\(=\left(\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}+\frac{1}{\left(x+3\right)\left(x+4\right)}+\frac{1}{\left(x+4\right)\left(x+5\right)}\right)\times\left(\frac{x\left(x+5\right)}{5}\right)\)
\(=\left(\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+...+\frac{1}{x+4}-\frac{1}{x+5}\right)\times\left(\frac{x\left(x+5\right)}{5}\right)\)
\(=\left(\frac{1}{x}-\frac{1}{x+5}\right)\times\frac{x\left(x+5\right)}{5}\)
\(=\left(\frac{x+5}{x\left(x+5\right)}-\frac{x}{\left(x+5\right)}\right)\times\frac{x\left(x+5\right)}{5}\)
\(=\frac{x+5-x}{x\left(x+5\right)}\times\frac{x\left(x+5\right)}{5}\)
\(=\frac{5}{x\left(x+5\right)}\times\frac{x\left(x+5\right)}{5}=1\)
=> đpcm