tìm x:
a) \(\left(2x+3\right)^2=\frac{9}{121}\)
b)\(\left(3x-1\right)^3=-\frac{8}{27}\)
Những câu hỏi liên quan
\(\left(2x+3\right)^2=\frac{9}{121}\\ \left(3x-1\right)^3=-\frac{8}{27}\)
a)\(\left(2x+3\right)^2=\frac{9}{121}\\ \Leftrightarrow\left(2x+3\right)^2=\left(\pm\frac{3}{11}\right)^2\\ \Rightarrow\left\{{}\begin{matrix}2x+3=\frac{3}{11}\\2x+3=\frac{-3}{11}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\frac{-15}{11}\\x=\frac{-18}{11}\end{matrix}\right.\)
Vậy...
b)\(\left(3x-1\right)^3=\frac{-8}{27}\\ \Leftrightarrow\left(3x-1\right)^3=\left(\frac{-2}{3}\right)^3\\ 3x-1=\frac{-2}{3}\\ \Rightarrow x=\frac{1}{9}\)
Vậy...
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a) \(\left(2x+3\right)^2=\frac{9}{121}\)
\(\Rightarrow2x+3=\pm\frac{3}{11}\)
\(\Rightarrow\left[{}\begin{matrix}2x+3=\frac{3}{11}\\2x+3=-\frac{3}{11}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=\frac{3}{11}-3=-\frac{30}{11}\\2x=\left(-\frac{3}{11}\right)-3=-\frac{36}{11}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\left(-\frac{30}{11}\right):2\\x=\left(-\frac{36}{11}\right):2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-\frac{15}{11}\\x=-\frac{18}{11}\end{matrix}\right.\)
Vậy \(x\in\left\{-\frac{15}{11};-\frac{18}{11}\right\}.\)
b) \(\left(3x-1\right)^3=-\frac{8}{27}\)
\(\Rightarrow\left(3x-1\right)^3=\left(-\frac{2}{3}\right)^3\)
\(\Rightarrow3x-1=-\frac{2}{3}\)
\(\Rightarrow3x=\left(-\frac{2}{3}\right)+1\)
\(\Rightarrow3x=\frac{1}{3}\)
\(\Rightarrow x=\frac{1}{3}:3\)
\(\Rightarrow x=\frac{1}{9}\)
Vậy \(x=\frac{1}{9}.\)
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Tính:a) frac{frac{3}{5}+frac{3}{27}-frac{3}{9}-frac{3}{11}}{frac{4}{5}+frac{4}{27}-frac{4}{9}-frac{4}{11}}b) frac{5-frac{5}{3}-frac{5}{27}}{8-frac{8}{3}-frac{8}{27}}:frac{15+frac{15}{121}-frac{15}{11}}{16+frac{16}{121}-frac{16}{11}}c) frac{1}{2}:left(frac{-3}{2}right):frac{4}{3}:left(frac{-5}{4}right):frac{6}{5}:left(frac{-7}{6}right):...:left(frac{-101}{100}right)Mọi người nhớ đánh nhanh câu trả lời giúp mình nhé!!
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Tính:
a) \(\frac{\frac{3}{5}+\frac{3}{27}-\frac{3}{9}-\frac{3}{11}}{\frac{4}{5}+\frac{4}{27}-\frac{4}{9}-\frac{4}{11}}\)
b) \(\frac{5-\frac{5}{3}-\frac{5}{27}}{8-\frac{8}{3}-\frac{8}{27}}:\frac{15+\frac{15}{121}-\frac{15}{11}}{16+\frac{16}{121}-\frac{16}{11}}\)
c) \(\frac{1}{2}:\left(\frac{-3}{2}\right):\frac{4}{3}:\left(\frac{-5}{4}\right):\frac{6}{5}:\left(\frac{-7}{6}\right):...:\left(\frac{-101}{100}\right)\)
Mọi người nhớ đánh nhanh câu trả lời giúp mình nhé!!
Mọi người đánh giúp mình nhé! Hạn là tối nay!!
Tìm x:a) 3xleft(3x-8right)-9x^2+80b)6x-15-xleft(5-2xright)0c) x^3-16x0d) 2x^2+3x-50e) 3x^2-xleft(3x-6right)36f) left(x+2right)^2-left(x-5right)left(x+1right)17g) left(x-4right)^2-xleft(x+6right)9h) 4xleft(x-1000right)-x+10000i) x^2-360j) x^2y-2+x+x^2-2y+xy0k) xleft(x+1right)-left(x-1right).left(2x-3right)0l) 3x^3-27x0
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Tìm x:
a) \(3x\left(3x-8\right)-9x^2+8=0\)
b)\(6x-15-x\left(5-2x\right)=0\)
c) \(x^3-16x=0\)
d) \(2x^2+3x-5=0\)
e) \(3x^2-x\left(3x-6\right)=36\)
f) \(\left(x+2\right)^2-\left(x-5\right)\left(x+1\right)=17\)
g) \(\left(x-4\right)^2-x\left(x+6\right)=9\)
h) \(4x\left(x-1000\right)-x+1000=0\)
i) \(x^2-36=0\)
j) \(x^2y-2+x+x^2-2y+xy=0\)
k) \(x\left(x+1\right)-\left(x-1\right).\left(2x-3\right)=0\)
l) \(3x^3-27x=0\)
tìm x biết :
a) \(\left(x+\frac{1}{5}\right)^2+\frac{17}{25}=\frac{26}{25}\)
b) \(-1\frac{5}{27}-\left(3x-\frac{7}{9}\right)^3=-\frac{24}{27}\)
c) \(\left(x+\frac{1}{2}\right).\left(\frac{2}{3}-2x\right)=0\)
\(\left(x+\frac{1}{5}\right)^2+\frac{17}{25}=\frac{26}{25}\\ \left(x+\frac{1}{5}\right)^2=\frac{26}{25}-\frac{17}{25}\\ \left(x+\frac{1}{5}\right)^2=\frac{9}{25}\\ \left|\left(x+\frac{1}{5}\right)\right|=\frac{3}{5}\)
TH1: \(x=\frac{3}{5}-\frac{1}{5}\\ x=\frac{2}{5}\)
TH2: \(\left|\left(x+\frac{1}{5}\right)\right|=-\frac{3}{5}\\ x=-\frac{3}{5}-\frac{1}{5}\\ x=-\frac{4}{5}\)
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\(a,\left(x+\frac{1}{5}\right)^2+\frac{17}{25}=\frac{26}{25}\)
\(\Rightarrow\left(x+\frac{1}{5}\right)^2=\frac{9}{25}\)
\(\Rightarrow\left(x+\frac{1}{5}\right)^2=\left(\frac{3}{5}\right)^2\)
\(\Rightarrow x+\frac{1}{5}=\frac{3}{5}\)
\(\Rightarrow x=\frac{2}{5}\)
\(b,-1\frac{5}{27}-\left(3x-\frac{7}{9}\right)^3=-\frac{24}{27}\)
\(\Rightarrow-\frac{32}{27}-\left(3x-\frac{7}{9}\right)^3=-\frac{24}{27}\)
\(\Rightarrow\left(3x-\frac{7}{9}\right)^3=-\frac{32}{27}+\frac{24}{27}\)
\(\Rightarrow\left(3x-\frac{7}{9}\right)^3=-\frac{8}{27}\)
\(\Rightarrow\left(3x-\frac{7}{9}\right)^3=\left(-\frac{2}{3}\right)^3\)
\(\Rightarrow3x-\frac{7}{9}=-\frac{2}{3}\)
\(\Rightarrow3x=-\frac{2}{3}+\frac{7}{9}\)
\(\Rightarrow3x=\frac{1}{9}\)
\(\Rightarrow x=\frac{1}{27}\)
\(c,\left(x+\frac{1}{2}\right)\left(\frac{2}{3}-2x\right)=0\)
\(\Rightarrow\) \(\left[\begin{array}{nghiempt}x+\frac{1}{2}=0\\\frac{2}{3}-2x=0\end{array}\right.\) \(\Rightarrow\) \(\left[\begin{array}{nghiempt}x=-\frac{1}{2}\\2x=\frac{2}{3}\end{array}\right.\) \(\Rightarrow\) \(\left[\begin{array}{nghiempt}x=-\frac{1}{2}\\x=\frac{1}{3}\end{array}\right.\)
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Bổ sung câu a: \(\Rightarrow\) \(\left[\begin{array}{nghiempt}\left(x+\frac{1}{5}\right)^2=\left(\frac{3}{5}\right)^2\\\left(x+\frac{1}{5}\right)^2=\left(-\frac{3}{5}\right)^2\end{array}\right.\)\(\Rightarrow\) \(\left[\begin{array}{nghiempt}x+\frac{1}{5}=\frac{3}{5}\\x+\frac{1}{5}=-\frac{3}{5}\end{array}\right.\) \(\Rightarrow\) \(\left[\begin{array}{nghiempt}x=\frac{2}{5}\\x=-\frac{4}{5}\end{array}\right.\)
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Tìm x:
a)\(2.\left(3x-\frac{1}{2}\right)-2x=\frac{1}{2}\left(2x-3\right)\)
b)\(\left(2x-\frac{3}{5}\right)^2=\frac{4}{25}\)
c)\(\left(3x-1\right)^3=27\)
d)\(5-\left|x\right|=2\)
e)|2x+1|-3=3
f)|3-2x|=5
\(\left(5-x\right)\left(3x-\frac{1}{4}\right)=0\)
Câu 1: Tìm x
1/\(\frac{1+3x}{2}=\frac{7}{3}-\frac{x+1}{6}\)
2,\(\left|\frac{2x+1}{3}\right|=\left|\frac{x-1}{5}\right|\)
3,\(\left(\frac{4}{3}+2x\right)^3=\frac{8}{-27}\)
Bài 1:Giải phương trình
a)10x^2-5xleft(2x+3right)15
b)3x-7-left(3-4xright)left(2x+1right)4xleft(2x-7right)
c)left(4x-5right)^2-left(7-2xright)4left(2x-4right)^2+6x
Bài 2:Giải phương trình
a)frac{3left(x-1right)}{2}+4frac{2x}{3}+frac{4-5x}{6}
b)frac{4-x}{7}-frac{1}{7}left(frac{7+3x}{9}+frac{5-2x}{2}right)4-frac{4x}{3}
c)frac{2}{9}left(2x-5right)-frac{5}{3}left[left(x-2right)-frac{7}{12}right]frac{3}{4}left(x-3right)
Bài 3:Giải phương trình
a)left(x-6right)left(2x-5right)left(3x+9right)0...
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Bài 1:Giải phương trình
a)\(10x^2-5x\left(2x+3\right)=15\)
b)\(3x-7-\left(3-4x\right)\left(2x+1\right)=4x\left(2x-7\right)\)
c)\(\left(4x-5\right)^2-\left(7-2x\right)=4\left(2x-4\right)^2+6x\)
Bài 2:Giải phương trình
a)\(\frac{3\left(x-1\right)}{2}+4=\frac{2x}{3}+\frac{4-5x}{6}\)
b)\(\frac{4-x}{7}-\frac{1}{7}\left(\frac{7+3x}{9}+\frac{5-2x}{2}\right)=4-\frac{4x}{3}\)
c)\(\frac{2}{9}\left(2x-5\right)-\frac{5}{3}\left[\left(x-2\right)-\frac{7}{12}\right]=\frac{3}{4}\left(x-3\right)\)
Bài 3:Giải phương trình
a)\(\left(x-6\right)\left(2x-5\right)\left(3x+9\right)=0\)
b)\(2x\left(x-3\right)+5\left(x-3\right)=0\)
c)\(\left(x^2-4\right)-\left(x-2\right)\left(3-2x\right)=0\)
Bài 4:Tìm m để phương trình sau có nghiệm bằng 7:\(\left(2m-5\right)x-2m^2+8=43\)
Bài 5:Giải phương trình
a)\(\left(2x-1\right)^2-\left(2x+1\right)^2=0\)
b)\(\frac{1}{27}\left(x-3\right)^3-\frac{1}{125}\left(x-5\right)^3=0\)
https://i.imgur.com/u6zkAVa.jpg
Bài 3:
a) \(\left(x-6\right).\left(2x-5\right).\left(3x+9\right)=0\)
\(\Leftrightarrow\left(x-6\right).\left(2x-5\right).3.\left(x+3\right)=0\)
Vì \(3\ne0.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-6=0\\2x-5=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\2x=5\\x=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=\frac{5}{2}\\x=-3\end{matrix}\right.\)
Vậy phương trình có tập hợp nghiệm là: \(S=\left\{6;\frac{5}{2};-3\right\}.\)
b) \(2x.\left(x-3\right)+5.\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right).\left(2x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\2x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\2x=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\frac{5}{2}\end{matrix}\right.\)
Vậy phương trình có tập hợp nghiệm là: \(S=\left\{3;-\frac{5}{2}\right\}.\)
c) \(\left(x^2-4\right)-\left(x-2\right).\left(3-2x\right)=0\)
\(\Leftrightarrow\left(x^2-2^2\right)-\left(x-2\right).\left(3-2x\right)=0\)
\(\Leftrightarrow\left(x-2\right).\left(x+2\right)-\left(x-2\right).\left(3-2x\right)=0\)
\(\Leftrightarrow\left(x-2\right).\left(x+2-3+2x\right)=0\)
\(\Leftrightarrow\left(x-2\right).\left(3x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\3x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\3x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\frac{1}{3}\end{matrix}\right.\)
Vậy phương trình có tập hợp nghiệm là: \(S=\left\{2;\frac{1}{3}\right\}.\)
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Bài 4 xem lại đề nhé bác
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tìm x:
a) \(\left(2x+5\right)\left(4x-10\right)+4x.\left(3-2x\right)^2=0\)
b) \(\left(2x-3\right)^2-\left(3x+1\right)^2=0\)
a: =>8x^2-20x+20x-50+4x(4x^2-12x+9)=0
=>8x^2-50+8x^3-48x^2+36x=0
=>8x^3-40x^2+36x-50=0
=>\(x\simeq4,29\)
b: =>(2x-3-3x-1)(2x-3+3x+1)=0
=>(-x-4)(5x-2)=0
=>x=2/5 hoặc x=-4
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Tìm số nguyên N biết:
\(a,\left(\frac{1}{3}\right)^n=\frac{1}{81}\)
\(b,\frac{-512}{343}=\left(\frac{-8}{7}\right)^n\)
\(c,\left(\frac{-3}{4}\right)^n=\frac{81}{256}\)
\(d,\left(2x+3\right)^2=\frac{9}{121}^n\)
a) \(\left(\frac{1}{3}\right)^n=\frac{1}{81}\)
\(\Rightarrow\left(\frac{1}{3}\right)^n=\frac{1^4}{3^4}\)
\(\Rightarrow\left(\frac{1}{3}\right)^n=\left(\frac{1}{3}\right)^4\)
\(\Rightarrow n=4\)
Vậy n = 4
b) \(\frac{-512}{343}=\left(\frac{-8}{7}\right)^n\)
\(\Rightarrow\frac{-8^3}{7^3}=\left(\frac{-8}{7}\right)^n\)
\(\Rightarrow\left(\frac{-8}{7}\right)^3=\left(\frac{-8}{7}\right)^n\)
\(\Rightarrow n=3\)
Vậy n = 3
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