\(\dfrac{3x+4}{8}=\dfrac{3-7x}{6}\) giup em vs mn
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TA
14 tháng 6 2021
`(3x+2)/3 <= (x-4)/7`
`<=>7(3x+2) <= 3(x-4)`
`<=> 21x+14<=3x-12`
`<=>18x <= -26`
`<=> x <=-13/9`
T
1 tháng 9 2023
\(a,3-x=x+1,8\)
\(\Rightarrow-x-x=1,8-3\)
\(\Rightarrow-2x=-1,2\)
\(\Rightarrow x=0,6\)
\(b,2x-5=7x+35\)
\(\Rightarrow2x-7x=35+5\)
\(\Rightarrow-5x=40\)
\(\Rightarrow x=-8\)
\(c,2\left(x+10\right)=3\left(x-6\right)\)
\(\Rightarrow2x+20=3x-18\)
\(\Rightarrow2x-3x=-18-20\)
\(\Rightarrow-x=-38\)
\(\Rightarrow x=38\)
\(d,8\left(x-\dfrac{3}{8}\right)+1=6\left(\dfrac{1}{6}+x\right)+x\)
\(\Rightarrow8x-3+1=1+6x+x\)
\(\Rightarrow8x-3=7x\)
\(\Rightarrow8x-7x=3\)
\(\Rightarrow x=3\)
\(e,\dfrac{2}{9}-3x=\dfrac{4}{3}-x\)
\(\Rightarrow-3x+x=\dfrac{4}{3}-\dfrac{2}{9}\)
\(\Rightarrow-2x=\dfrac{10}{9}\)
\(\Rightarrow x=-\dfrac{5}{9}\)
T
1 tháng 9 2023
\(g,\dfrac{1}{2}x+\dfrac{5}{6}=\dfrac{3}{4}x-\dfrac{1}{2}\)
\(\Rightarrow\dfrac{1}{2}x-\dfrac{3}{4}x=-\dfrac{1}{2}-\dfrac{5}{6}\)
\(\Rightarrow-\dfrac{1}{4}x=-\dfrac{4}{3}\)
\(\Rightarrow x=\dfrac{16}{3}\)
\(h,x-4=\dfrac{5}{6}\left(6-\dfrac{6}{5}x\right)\)
\(\Rightarrow x-4=5-x\)
\(\Rightarrow x+x=5+4\)
\(\Rightarrow2x=9\)
\(\Rightarrow x=\dfrac{9}{2}\)
\(k,7x^2-11=6x^2-2\)
\(\Rightarrow7x^2-6x^2=-2+11\)
\(\Rightarrow x^2=9\Rightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)
\(m,5\left(x+3\cdot2^3\right)=10^2\)
\(\Rightarrow5\left(x+24\right)=100\)
\(\Rightarrow x+24=20\)
\(\Rightarrow x=-4\)
\(n,\dfrac{4}{9}-\left(\dfrac{1}{6^2}\right)=\dfrac{2}{3}\left(x-\dfrac{2}{3}\right)^2+\dfrac{5}{12}\)
\(\Rightarrow\dfrac{2}{3}\left(x-\dfrac{2}{3}\right)^2+\dfrac{5}{12}=\dfrac{4}{9}-\dfrac{1}{36}\)
\(\Rightarrow\dfrac{2}{3}\left(x-\dfrac{2}{3}\right)^2+\dfrac{5}{12}=\dfrac{5}{12}\)
\(\Rightarrow\dfrac{2}{3}\left(x-\dfrac{2}{3}\right)^2=0\)
\(\Rightarrow x-\dfrac{2}{3}=0\Rightarrow x=\dfrac{2}{3}\)
#\(Urushi\text{☕}\)
em c.ơn trước
9 tháng 3 2018
a, \(\dfrac{t+3}{t-2}+\dfrac{t-2}{t+3}=\dfrac{5t+15}{t^2+t-6}\) ĐKXĐ: t\(\ne\)2,t\(\ne\)-3
\(\Leftrightarrow\dfrac{t+3}{t-2}+\dfrac{t-2}{t+3}=\dfrac{5t+15}{\left(t+3\right)\left(t-2\right)}\)
\(\Rightarrow\left(t+3\right)\left(t+3\right)+\left(t-2\right)\left(t-2\right)=5t+15\)
\(\Leftrightarrow t^2+6t+9+t^2-4t+4-5t-15=0\)
\(\Leftrightarrow-3t-2=0\)
\(\Leftrightarrow-3t=2\)
\(\Leftrightarrow t=\dfrac{-2}{3}\) (tđk)
\(\Rightarrow S=\left\{\dfrac{-2}{3}\right\}\)
b, \(\left(2x+3\right)\left(\dfrac{3x+8}{2-7x}+1\right)=\left(x-5\right)\left(\dfrac{3x+8}{2-7x}+1\right)\)ĐKXĐ: x\(\ne\)\(\dfrac{2}{7}\)
\(\Leftrightarrow\) \(\left(2x+3\right)\left(\dfrac{3x+8}{2-7x}+1\right)-\left(x-5\right)\left(\dfrac{3x+8}{2-7x}+1\right)=0\)
\(\Rightarrow\left(\dfrac{3x+8}{2-7x}+1\right)\left(2x+3-x+5\right)=0\)
\(\Leftrightarrow\) \(\Rightarrow\left(\dfrac{3x+8}{2-7x}+1\right)\left(x+8\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{3x+8}{2-7x}+1=0\\x+8=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x+8+2-7x=0\\x=-8\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}-4x+10=0\\x=-8\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-8\end{matrix}\right.\)
\(\Rightarrow S=\left\{\dfrac{5}{2};-8\right\}\)
ND
9 tháng 3 2018
ĐKXĐ: x khác 2 và x khác -3
\(\dfrac{t+3}{t-2}+\dfrac{t-2}{t+3}=\dfrac{5t+15}{t^2+t-6}\)
\(\Leftrightarrow\dfrac{\left(t+3\right)\left(t+3\right)}{\left(t+3\right)\left(t-2\right)}+\dfrac{\left(t-2\right)\left(t-2\right)}{\left(t+3\right)\left(t-2\right)}=\dfrac{5t+15}{t^2+t-6}\)
\(\Rightarrow t^2+6t+9+t^2-4=5t+15\)
\(\Leftrightarrow2t^2+t-10=0\)
\(\Leftrightarrow2t^2-4t+5t-10=0\)
\(\Leftrightarrow2t\left(t-2\right)+5\left(t-2\right)=0\)
\(\Leftrightarrow\left(2t+5\right)\left(t-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}t=2\left(loại\right)\\t=\dfrac{-5}{2}\end{matrix}\right.\)
Vậy..................
28 tháng 1 2021
1/ ĐKXĐ : \(x\ne1\)
\(\dfrac{7x-3}{x-1}=\dfrac{2}{3}\)
\(\Leftrightarrow21x-9=2x-2\)
\(\Leftrightarrow19x=7\Leftrightarrow x=\dfrac{7}{19}\left(tm\right)\)
Vậy...
b/ \(\dfrac{2\left(3-7x\right)}{1+x}=\dfrac{1}{2}\) ĐKXĐ : \(x\ne-1\)
\(\Leftrightarrow12-28x=1+x\)
\(\Leftrightarrow11=29x\Leftrightarrow x=\dfrac{11}{29}\) \(\left(tm\right)\)
Vậy....
c/ ĐKXĐ : \(x\ne0\)
\(\dfrac{x^2-6}{x}=x+\dfrac{3}{2}\)
\(\Leftrightarrow\dfrac{x^2-6}{x}=\dfrac{2x+3}{2}\)
\(\Leftrightarrow2x^2-12=2x^2+3x\)
\(\Leftrightarrow3x=-12\Leftrightarrow x=-4\) \(\left(tm\right)\)
Vậy...
4/ ĐKXĐ : \(x\ne-\dfrac{2}{3}\)
\(\dfrac{5}{3x+2}=2x-1\)
\(\Leftrightarrow\left(2x-1\right)\left(3x+2\right)=5\)
\(\Leftrightarrow6x^2+4x-3x-2=5\)
\(\Leftrightarrow6x^2+x-7=0\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{7}{6}\\x=1\end{matrix}\right.\)
Vậy....
5,6 Tương tự nhé !
28 tháng 1 2021
1)ĐKXĐ: \(x\ne1\)
Ta có: \(\dfrac{7x-3}{x-1}=\dfrac{2}{3}\)
\(\Leftrightarrow3\left(7x-3\right)=2\left(x-1\right)\)
\(\Leftrightarrow21x-9=2x-2\)
\(\Leftrightarrow21x-9-2x+2=0\)
\(\Leftrightarrow19x-7=0\)
\(\Leftrightarrow19x=7\)
\(\Leftrightarrow x=\dfrac{7}{19}\)(nhận)
Vậy: \(S=\left\{\dfrac{7}{19}\right\}\)
2) ĐKXĐ: \(x\ne-1\)
Ta có: \(\dfrac{2\left(3-7x\right)}{1+x}=\dfrac{1}{2}\)
\(\Leftrightarrow4\left(3-7x\right)=x+1\)
\(\Leftrightarrow12-28x-x-1=0\)
\(\Leftrightarrow-29x+11=0\)
\(\Leftrightarrow-29x=-11\)
\(\Leftrightarrow x=\dfrac{11}{29}\)
Vậy: \(S=\left\{\dfrac{11}{29}\right\}\)
3) ĐKXĐ: \(x\ne0\)
Ta có: \(\dfrac{x^2-6}{x}=x+\dfrac{3}{2}\)
\(\Leftrightarrow\dfrac{x^2-6}{x}=\dfrac{2x+3}{2}\)
\(\Leftrightarrow2\left(x^2-6\right)=x\left(2x+3\right)\)
\(\Leftrightarrow2x^2-12=2x^2+6x\)
\(\Leftrightarrow2x^2-12-2x^2-6x=0\)
\(\Leftrightarrow-6x-12=0\)
\(\Leftrightarrow-6x=12\)
\(\Leftrightarrow x=-2\)
Vậy: S={-2}
OP
29 tháng 1 2022
1.
<=> \(\left[{}\begin{matrix}4-3x=0\\10-5x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{4}{3}\\x=2\end{matrix}\right.\)
2.
<=>\(\left[{}\begin{matrix}7-2x=0\\4+8x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=-\dfrac{1}{2}\end{matrix}\right.\)
3.
<=>\(\left[{}\begin{matrix}9-7x=0\\11-3x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{9}{7}\\x=\dfrac{11}{3}\end{matrix}\right.\)
4.
<=>\(\left[{}\begin{matrix}7-14x=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=2\end{matrix}\right.\)
5.
<=>\(\left[{}\begin{matrix}\dfrac{7}{8}-2x=0\\3x+\dfrac{1}{3}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{16}\\x=-\dfrac{1}{9}\end{matrix}\right.\)
6,7. ko đủ điều kiện tìm
16 tháng 1 2023
1: \(=-3x^3-21x^2+x\)
2: \(=-15x^4y^7+10x^5y^6+5x^3y^5\)
3: \(=x^7y^4-2x^4y^2-3x^3y\)
5: \(=15x-6x^2\)
6: \(=4x^3-8x^2+10x\)
7: \(=-8x^5y^3+16x^7y^2-12x^3y^4\)
8: \(=x^7y^4-2x^4y^2-3x^3y\)