tim x biết \(\frac{x}{4}\)=\(\frac{27}{x}\)
(3x-1)\(^3\)=\(\frac{-8}{37}\)
ai nhanh tay mình tích cho :)
K
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VQ
16 tháng 12 2018
x-3/-3 = -27/x-3
=> (x-3)(x-3)=(-3)(-27)
=> (x-3)^2 = 81=9^2
=> x-3=9 hoặc x-3=-9
=> x=12 hoặc x=-6
HN
16 tháng 12 2018
Ta có :\(\frac{x-3}{-3}=\frac{-27}{x-3}\)
\(\Rightarrow\left(x-3\right)\left(x-3\right)=\left(-3\right).\left(-27\right)\) ( Tính chất tỉ lệ thức )
\(\Rightarrow\left(x-3\right)^2=81\)
\(\Rightarrow\left(x-3\right)^2=\left(\pm9\right)^2\)
\(\Rightarrow x-3=\pm9\)
\(\Rightarrow x=-6;12\)
4 tháng 8 2018
\(\left|x+\frac{1}{2}\right|+\left|x+\frac{2}{3}\right|+\left|x+\frac{3}{4}\right|=4x\)
\(\Rightarrow x=x\)hoặc \(x=-x\)
---Nếu x = x
\(\Rightarrow\left|x+\frac{1}{2}\right|+\left|x+\frac{2}{3}\right|+\left|x+\frac{3}{4}\right|=4x\)
\(x+\frac{1}{2}+x+\frac{2}{3}+x+\frac{3}{4}=4x\)
\(3x+\left(\frac{1}{2}+\frac{2}{3}+\frac{3}{4}\right)=4x\)
\(\left(\frac{6}{12}+\frac{8}{12}+\frac{9}{12}\right)=4x-3x\)
\(\Rightarrow x=\frac{23}{12}\)
---Nếu x = -x
\(\Rightarrow\left|-x+\frac{1}{2}\right|+\left|-x+\frac{2}{3}\right|+\left|-x+\frac{3}{4}\right|=4x\)
\(-x+\frac{1}{2}+\left(-x\right)+\frac{2}{3}+\left(-x\right)+\frac{3}{4}=4x\)
\(-3x+\left(\frac{1}{2}+\frac{2}{3}+\frac{3}{4}\right)=4x\)
\(\left(\frac{6}{12}+\frac{8}{12}+\frac{9}{12}\right)=4x+3x\)
\(\Rightarrow7x=\frac{23}{12}\)
\(\Rightarrow x=\frac{23}{12}:7\)
\(\Rightarrow x=\frac{23}{12}.\frac{1}{7}\)
\(\Rightarrow x=\frac{23}{84}\)
\(Vậyx=\frac{23}{12};x=\frac{23}{84}\)
17 tháng 4 2019
\(a,2,5\cdot x=\frac{11}{15}\)
\(\Rightarrow x=\frac{11}{15}:2,5\)
\(\Rightarrow x=\frac{11}{15}:\frac{25}{10}\)
\(\Rightarrow x=\frac{11}{15}\cdot\frac{10}{25}\)
\(\Rightarrow x=\frac{11}{3}\cdot\frac{2}{25}=\frac{22}{75}\)
\(b,x-15\%\cdot x=\frac{1}{3}\)
\(\Rightarrow x-\frac{15}{100}\cdot x=\frac{1}{3}\)
\(\Rightarrow x-\frac{3}{20}\cdot x=\frac{1}{3}\)
\(\Rightarrow\frac{20x}{20}-\frac{3x}{20}=\frac{1}{3}\)
\(\Rightarrow\frac{17x}{20}=\frac{1}{3}\)
\(\Rightarrow17x\cdot3=20\)
\(\Rightarrow17x=\frac{20}{3}\)
\(\Rightarrow x=\frac{20}{3}:17=\frac{20}{3}\cdot\frac{1}{17}=\frac{20}{51}\)
Câu c mk làm sau :v
17 tháng 4 2019
c,\(\left[\frac{3x}{7}+1\right]:\left[-4\frac{1}{7}\right]=\frac{-3}{28}\)
\(\Rightarrow\left[\frac{3x}{7}+1\right]:\left[-\frac{29}{7}\right]=\frac{-3}{28}\)
\(\Rightarrow\frac{3x}{7}+1=-\frac{3}{28}\cdot-\frac{29}{7}=\frac{87}{196}\)
\(\Rightarrow\frac{3x}{7}=\frac{87}{196}-1\)
Tìm nốt :v
2 tháng 11 2019
Bài 1:
\(4.\left(\frac{-1}{2}\right)^2-2.\left(\frac{-1}{2}\right)^2+3.\left(\frac{-1}{2}\right)+1\)
\(=4.\frac{1}{4}-2.\frac{1}{4}+3.\left(\frac{-1}{2}\right)+1\)
\(=1-\frac{1}{2}-\frac{3}{2}+1\)
\(=0\)
2 tháng 11 2019
Bài 2:
a) \(\frac{37-x}{x+13}=\frac{3}{7}\)
\(\Rightarrow7\left(37-x\right)=3\left(x+13\right)\)
\(\Rightarrow259-7x=3x+39\)
\(\Rightarrow259-39=3x+7x\)
\(\Rightarrow220=10x\)
\(\Rightarrow x=22\)
d) \(\frac{3^2.3^8}{27^3}=3^x\)
\(\Rightarrow\frac{3^{10}}{\left(3^3\right)^3}=3^x\)
\(\frac{\Rightarrow3^{10}}{3^9}=3^x\)
\(\Rightarrow3=3^x\)
\(\Rightarrow x=1\)
Hok tốt nha^^
KN
14 tháng 7 2019
a) \(\left(\frac{4}{9}\right)^x=\left(\frac{8}{27}\right)^6\)
\(\Leftrightarrow\left(\frac{2}{3}\right)^{2x}=\left(\frac{2}{3}\right)^{18}\)
\(\Leftrightarrow2x=18\)
\(\Leftrightarrow x=9\)
KN
14 tháng 7 2019
b) \(\left(\frac{1}{9}\right)^x=\left(\frac{1}{27}\right)^{22}\)
\(\Leftrightarrow\left(\frac{1}{9}\right)^x=\left(\frac{1}{3}\right)^{66}\)
\(\Leftrightarrow x=66\)
22 tháng 8 2017
\(\frac{2}{x+6}+\frac{1}{x+1}=\frac{3x+8}{x^2+7x+6}\left(ĐKXĐ:x\ne-6;-1\right)\)
\(\Leftrightarrow\frac{2\left(x+1\right)+x+6}{x^2+7x+6}=\frac{3x+8}{x^2+7x+6}\)
\(\Leftrightarrow2x+2+x+6=3x+8\)
\(\Leftrightarrow3x+8=3x+8\)
\(\Rightarrow\)x vô số nghiệm
17 tháng 1 2021
Đk: x \(\ne\)0; x \(\ne\)\(\pm\)3
Ta có: A = \(\left(\frac{1}{3}+\frac{3}{x^2-3x}\right):\left(\frac{x^2}{27-3x^2}+\frac{1}{x+3}\right)\)
A = \(\frac{x^2-3x+9}{3x\left(x-3\right)}:\frac{x^2+3\left(3-x\right)}{3\left(x+3\right)\left(3-x\right)}\)
A = \(\frac{x^2-3x+9}{3x\left(x-3\right)}\cdot\frac{3\left(3-x\right)\left(x+3\right)}{x^2-3x+9}\)
A = \(\frac{-\left(x+3\right)}{x}\)
Để A < -1 <=> \(-\frac{\left(x+3\right)}{x}< -1\) <=> \(\frac{-x-3}{x}+1< 0\)
<=> \(\frac{-x-3+x}{x}< 0\) <=> \(-\frac{3}{x}< 0\)
Do -3 <0 => x> 0
Vậy Để A < -1 <=> x > 0 và x khác 3
Ta có: \(\frac{x}{4}=\frac{27}{x}\Leftrightarrow x^2=108\Leftrightarrow x=\sqrt{108}\)
\(\left(3x-1\right)^3=\frac{-8}{27}\)
Mà \(\frac{-8}{27}=\left(\frac{-2}{3}\right)^3\Rightarrow\left(3x-1\right)^3=\left(\frac{-2}{3}\right)^3\)
\(\Rightarrow3x-1=\frac{-2}{3}\Rightarrow3x=\frac{1}{3}\Rightarrow x=\frac{1}{9}\)
Vậy x = 1/9
+) \(\frac{x}{4}=\frac{27}{x}\)
\(\Rightarrow x\times x=27\times4\)
\(\Rightarrow x^2=108\)
\(\Rightarrow x=\sqrt{108}\)
Vậy \(x=\sqrt{108}\)
+) \(\left(3x-1\right)^3=\frac{-8}{27}\)
\(\Rightarrow\left(3x-1\right)^3=\left(\frac{-2}{3}\right)^3\)
\(\Rightarrow3x-1=\frac{-2}{3}\)
\(\Rightarrow3x=\frac{-2}{3}+1\)
\(\Rightarrow3x=\frac{1}{3}\)
\(\Rightarrow x=\frac{1}{3}:3\)
\(\Rightarrow x=\frac{1}{9}\)
Vậy \(x=\frac{1}{9}\)
_Chúc bạn học tốt_