\(\left(5x+1\right)^2=\dfrac{36}{49}\)
Những câu hỏi liên quan
\(5x-9=5+3x;2^3+0,5x=1,5;\left(5x+1\right)^2=\dfrac{36}{49};\left(\dfrac{-3}{81}\right)^x=-27;2^{x-1}=16\)
Tìm x, biết:a) left(5x+1right)^2dfrac{36}{49}b) left[left(-0,5right)^3right]^xdfrac{1}{64}c) 2020^{left(x-2right).left(2x+3right)}1d) left(x+1right)^{x+10}left(x+1right)^{x+4} với xin Ze) dfrac{3}{4}sqrt{x}-dfrac{1}{2}dfrac{1}{3}
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Tìm x, biết:
a) \(\left(5x+1\right)^2=\dfrac{36}{49}\)
b) \(\left[\left(-0,5\right)^3\right]^x=\dfrac{1}{64}\)
c) \(2020^{\left(x-2\right).\left(2x+3\right)}=1\)
d) \(\left(x+1\right)^{x+10}=\left(x+1\right)^{x+4}\) với \(x\in Z\)
e) \(\dfrac{3}{4}\sqrt{x}-\dfrac{1}{2}=\dfrac{1}{3}\)
\(a,\Rightarrow\left[{}\begin{matrix}5x+1=\dfrac{6}{7}\\5x+1=-\dfrac{6}{7}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}5x=\dfrac{1}{7}\\5x=-\dfrac{13}{7}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{35}\\x=-\dfrac{13}{35}\end{matrix}\right.\\ b,\Rightarrow\left(-\dfrac{1}{8}\right)^x=\dfrac{1}{64}=\left(-\dfrac{1}{8}\right)^2\Rightarrow x=2\\ c,\Rightarrow\left(x-2\right)\left(2x+3\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=2\\x=-\dfrac{3}{2}\end{matrix}\right.\\ d,\Rightarrow\left(x+1\right)^{x+10}-\left(x+1\right)^{x+4}=0\\ \Rightarrow\left(x+1\right)^{x+4}\left[\left(x+1\right)^6-1\right]=0\\ \Rightarrow\left[{}\begin{matrix}x+1=0\\\left(x+1\right)^6=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x+1=0\\x+1=1\\x+1=-1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-1\\x=0\\x=-2\end{matrix}\right.\\ e,\Rightarrow\dfrac{3}{4}\sqrt{x}=\dfrac{5}{6}\left(x\ge0\right)\\ \Rightarrow\sqrt{x}=\dfrac{10}{9}\Rightarrow x=\dfrac{100}{81}\)
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Tìm x:
a/\(\left(5x+1\right)^2=\dfrac{36}{49}\)
b/\(\left(x-\dfrac{2}{9}\right)^3=\left(\dfrac{2}{3}\right)^6\)
c/\(\left(8x+1\right)^{2n+1}=5^{2n+1}\left(n\in N\right)\)
a: =>5x+1=6/7 hoặc 5x+1=-6/7
=>5x=-1/7 hoặc 5x=-13/7
=>x=-1/35 hoặc x=-13/35
b: =>x-2/9=4/9
=>x=6/9=2/3
c: =>8x+1=5
=>8x=4
hay x=1/2
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1 Tìm x biết :
a) \(\left(5x+1\right)^2\) = \(\dfrac{36}{49}\)
b) \(\left(x-\dfrac{2}{9}\right)^3\) = \(\left(\dfrac{2}{3}\right)^6\)
c) \(\left(8x-1\right)^{2n+1}\) = \(5^{2n+1}\) \(\left(n\in N\right)\)
a) \(\left(5x+1\right)^2=\dfrac{36}{49}\)
\(\left(5x+1\right)^2=\left(\pm\dfrac{6}{9}\right)\)\(^2\)
\(5x+1=\pm\dfrac{6}{9}\)
+) \(5x+1=\dfrac{6}{9}\)
\(5x=\dfrac{6}{9}-1=\dfrac{6}{9}-\dfrac{9}{9}\)
\(5x=\dfrac{-5}{9}\)
\(x=\dfrac{-5}{9}:5=\dfrac{-1}{45}\)
+) \(5x+1=\dfrac{-6}{9}\)
\(5x=\dfrac{-6}{9}-1=\dfrac{-6}{9}-\dfrac{9}{9}\)
\(5x=\dfrac{-5}{3}\)
\(x=\dfrac{-5}{3}:5=\dfrac{-5}{15}\)
vậy \(x\in\left\{\dfrac{-5}{15};\dfrac{-1}{45}\right\}\)
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Bài 1: Tìm x , biết :
a. 172 . x2- 79: 983 = \(\left(\dfrac{1}{2}\right)^3\)
b. (5x + 1)2 = \(\dfrac{36}{49}\)
c, \(\left(x-\dfrac{2}{3}\right)^3\)= \(\left(\dfrac{2}{3}\right)^6\)
b, \(\left(5x+1\right)^2=\dfrac{36}{49}\)
\(\left(5x+1\right)^2=\left(\pm\dfrac{6}{7}\right)^2\)
\(\Rightarrow\left[{}\begin{matrix}5x+1=\dfrac{6}{7}\\5x+1=-\dfrac{6}{7}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}5x=-\dfrac{1}{7}\\5x=\dfrac{-13}{7}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{35}\\x=\dfrac{-13}{35}\end{matrix}\right.\)
Vậy .....
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Nguyễn Thanh Hằng ;Hồng Phúc Nguyễn ;Mới vô; ... các bn giúp mik vs mik đang cần gấp !
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\(\left(5x+1\right)^2=\frac{36}{49}\)
ko ghi lại đề bài lm luôn:
\(\left(5x+1\right)^2=\left(\frac{6}{7}\right)^2\)
\(5x+1=\frac{6}{7}\)
\(5x=\frac{-1}{7}\)
\(x=\frac{-1}{35}\)
tôi lm đại thôi nhưng chắc đ đấy
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1: Tính : \(\dfrac{\left(\dfrac{2}{5}\right)^7.5^7+\left(\dfrac{9}{4}\right)^3:\left(\dfrac{3}{16}\right)^3}{2^7.5^2+512}\) 2 . Tìm x biết :\(\left(5.x+1\right)^2=\dfrac{36}{49}\)
Bài 1:
\(\dfrac{\left(\dfrac{2}{5}\right)^7\cdot5^7+\left(\dfrac{9}{4}\right)^3:\left(\dfrac{3}{16}\right)^3}{2^7\cdot5^2+512}\)
\(=\dfrac{\left(\dfrac{2}{5}\cdot5\right)^7+\left(\dfrac{9}{4}:\dfrac{3}{16}\right)^3}{2^7\cdot5^2+512}\)
\(=\dfrac{2^7+12^3}{2^7\cdot5^2+512}\)
\(=\dfrac{1856}{3712}\)
\(=0,5\)
Bài 2:
\(\left(5x+1\right)^2=\dfrac{36}{49}\)
\(\Rightarrow5x+1=\dfrac{6}{7}\)
\(\Rightarrow5x=\dfrac{-1}{7}\)
\(\Rightarrow x=\dfrac{-1}{35}\)
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Tính hợp lý
\(A = \left(1-\dfrac{1}{25}\right)\left(1-\dfrac{1}{36}\right)\left(1-\dfrac{1}{49}\right)...\left(1-\dfrac{1}{10000}\right)\) B= \(\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{6}\right)\left(1-\dfrac{1}{10}\right)...\left(1-\dfrac{1}{50.101}\right)\)
Hàng dzề, hàng dzề!!! Mại dzô, mại dzô các cậu ơi...Nhanh tay lên kẻo cháy hàng, híhí........................
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Hàng dzề, hàng dzề!!! Mại dzô, mại dzô các cậu ơi...Nhanh tay lên kẻo cháy hàng, híhí........................
Giải các phương trình sau
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