giúp mình giải bt:
3x+3 - 3x-1 =2160
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5)
để \(\frac{5x-3}{x+1}\)là số nguyên
\(5x-3⋮x+1\)
\(x+1⋮x+1\)
\(\Rightarrow5\left(x+1\right)⋮x+1\)
\(5x-3-\left(5x-5\right)⋮x+1\)
\(-2⋮x+1\)
\(\Rightarrow x+1\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)
x+1 | 1 | -1 | 2 | -2 |
x | 0 | -2 | 1 | -3 |
Vậy \(x\in\left\{0;-2;1;-3\right\}\)
\(=\frac{3x^2+9x-3}{x^2+x-2}-\frac{x+1}{x+2}-\frac{x-2}{x-1}\)
\(=\frac{3x^2+9x-3}{\left(x+2\right)\left(x-1\right)}-\frac{\left(x+1\right)\left(x-1\right)}{\left(x+2\right)\left(x-1\right)}-\frac{\left(x-2\right)\left(x+2\right)}{\left(x-1\right)\left(x+2\right)}\)
\(=\frac{3x^2+9x-3-\left(x^2-1\right)-\left(x^2-4\right)}{\left(x-1\right)\left(x+2\right)}\)
\(=\frac{3x^2+9x-3-x^2+1-x^2+4}{\left(x-1\right)\left(x+2\right)}\)
\(=\frac{x^2+9x+2}{\left(x-1\right)\left(x+2\right)}\)
\(|\dfrac{4}{3}x-\dfrac{3}{4}|=\left|-\dfrac{1}{3}\right|.\left|x\right|\Leftrightarrow|\dfrac{4}{3}x-\dfrac{3}{4}|=\dfrac{1}{3}.\left|x\right|\left(1\right)\)
Tìm nghiệm \(\dfrac{4}{3}x-\dfrac{3}{4}=0\Leftrightarrow\dfrac{4}{3}x=\dfrac{3}{4}\Leftrightarrow x=\dfrac{3}{4}.\dfrac{3}{4}\Leftrightarrow x=\dfrac{9}{16}\)
\(x=0\)
Lập bảng xét dấu :
\(x\) \(0\) \(\dfrac{9}{16}\)
\(\left|\dfrac{4}{3}x-\dfrac{3}{4}\right|\) \(-\) \(0\) \(-\) \(0\) \(+\)
\(\left|x\right|\) \(-\) \(0\) \(+\) \(0\) \(+\)
TH1 : \(x< 0\)
\(\left(1\right)\Leftrightarrow-\dfrac{4}{3}x+\dfrac{3}{4}=\dfrac{1}{3}.\left(-x\right)\)
\(\Leftrightarrow-\dfrac{4}{3}x+\dfrac{3}{4}=-\dfrac{1}{3}.x\)
\(\Leftrightarrow\dfrac{4}{3}x-\dfrac{1}{3}x=\dfrac{3}{4}\)
\(\Leftrightarrow x=\dfrac{3}{4}\) (loại vì không thỏa \(x< 0\))
TH2 : \(0\le x\le\dfrac{9}{16}\)
\(\left(1\right)\Leftrightarrow-\dfrac{4}{3}x+\dfrac{3}{4}=\dfrac{1}{3}x\)
\(\Leftrightarrow\dfrac{4}{3}x+\dfrac{1}{3}x=\dfrac{3}{4}\)
\(\Leftrightarrow\dfrac{5}{3}x=\dfrac{3}{4}\Leftrightarrow x=\dfrac{3}{4}.\dfrac{3}{5}\Leftrightarrow x=\dfrac{9}{20}\) (thỏa điều kiện \(0\le x\le\dfrac{9}{16}\))
TH3 : \(x>\dfrac{9}{16}\)
\(\left(1\right)\Leftrightarrow\dfrac{4}{3}x-\dfrac{3}{4}=\dfrac{1}{3}x\)
\(\Leftrightarrow\dfrac{4}{3}x-\dfrac{1}{3}x=\dfrac{3}{4}\Leftrightarrow x=\dfrac{3}{4}\) (thỏa điều kiện \(x>\dfrac{9}{16}\))
Vậy \(x\in\left\{\dfrac{9}{20};\dfrac{3}{4}\right\}\)
\(3.\left(3x-\frac{1}{2}\right)+\frac{1}{9}=0\)
\(3.\left(3x-\frac{1}{2}\right)=0-\frac{1}{9}\)
\(3.\left(3x-\frac{1}{2}\right)=\frac{-1}{9}\)
\(3x-\frac{1}{2}=\frac{-1}{9}:3\)
\(3x-\frac{1}{2}=\frac{-1}{27}\)
\(3x=\frac{-1}{27}+\frac{1}{2}\)
\(3x=\frac{25}{54}\)
\(x=\frac{25}{54}:3\)
\(x=\frac{25}{162}\)
~ Hok tốt ~
\(3\cdot\left(3x-\frac{1}{2}\right)+\frac{1}{9}=0\)
\(3\cdot\left(3x-\frac{1}{2}\right)=0-\frac{1}{9}\)
\(3\cdot\left(3x-\frac{1}{2}\right)=\frac{-1}{9}\)
\(3x-\frac{1}{2}=\frac{-1}{9}\div3\)
\(3x-\frac{1}{2}=\frac{-1}{27}\)
\(3x=\frac{-1}{27}+\frac{1}{2}\)
\(3x=\frac{25}{54}\)
\(x=\frac{25}{54}\div3\)
\(x=\frac{25}{162}\)
#Tề _ Thiên
\(1,\\ a,=x^2+6x+9-x^2-6x=9\\ b,=3x-1+6x-9x^2+x-10=-9x^2+10x-11\\ 2,\\ a,=4xy\left(x^2-2xy+y^2\right)=4xy\left(x-y\right)^2\)
\(\frac{4}{3x7}\)+\(\frac{4}{7x11}\)+....+\(\frac{4}{\left(3x-1\right)\left(3x+3\right)}\)= \(\frac{3}{10}\)
=\(\frac{1}{3}\)-\(\frac{1}{7}\)+\(\frac{1}{7}\)-\(\frac{1}{11}\)+...+\(\frac{1}{3x-1}\)- \(\frac{1}{3x+3}\)= \(\frac{3}{10}\)
=> \(\frac{1}{3}\)- \(\frac{1}{3x+3}\)=\(\frac{3}{10}\)
=> \(\frac{1}{3}\)-\(\frac{3}{10}\)=\(\frac{1}{3x+3}\)
=> \(\frac{1}{30}\)=\(\frac{1}{3x+3}\)
nhân chéo ta có: 3x+3=30
=> 3x=30-3=27
=> x= 27:3
=>x=9
nhé ^_^ mơn nhiều
tạm thời cái đề bài là tìm x vậy
tiếp tục thui
=) (6x-10)-(6x-3)\(⋮\)2x-1
=)6x-10-6x+3\(⋮\)2x-1
=) (6x-6x)-(10-3)\(⋮\)2x-1
=)0-7\(⋮\)2x-1
=)-7\(⋮\)2x-1=)2x-1\(\in\)Ư(-7)={-7;-1;1;7}
=)2x\(\in\){-6;0;2;8}
=)x\(\in\){-3;0;1;4}
x4 - 3x3 - 6x2 + 3x + 1= 0
<=> (x4 - 4x3 - x2) + (x3 - 4x2 - x) + (-x2 + 4x + 1) = 0
<=> (x2 - 4x - 1)(x2 + x - 1) = 0
Bài giải
\(3^{x+3}-3^{x-1}=2160\)
\(3^{x-1}\left(3^4-1\right)=2160\)
\(3^{x-1}\cdot80=2160\)
\(3^{x-1}=2160\text{ : }80\)
\(3^{x-1}=27\)
\(3^{x-1}=3^3\)
\(\Rightarrow\text{ }x-1=3\)
\(x=3+1\)
\(x=4\)