x:2=3
x=? giúp em với
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TP
0
RQ
3
AH
Akai Haruma
Giáo viên
24 tháng 12 2018
Lời giải:
Vì \(3x=5y\Rightarrow y=\frac{3}{5}x=0,6x\). Thay vào điều kiện thứ 2 ta có:
\(2x+3y=-39\)
\(\Leftrightarrow 2x+3.0,6x=-39\)
\(\Leftrightarrow 3,8x=-39\Rightarrow x=\frac{-195}{19}\)
\(\Rightarrow y=0,6x=0,6.\frac{-195}{19}=\frac{-117}{19}\)
Vậy \((x,y)=(\frac{-195}{19}; \frac{-117}{19})\)
S
24 tháng 12 2018
Ta có: \(3x=5y\Leftrightarrow\dfrac{x}{5}=\dfrac{y}{3}\Leftrightarrow\dfrac{2x}{10}=\dfrac{3y}{9}\)
Áp dụng tính chất dãy tỉ số bằng nhau:
\(\dfrac{2x}{10}=\dfrac{3y}{9}=\dfrac{2x+3y}{10+9}=\dfrac{-39}{19}\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{5}=\dfrac{-39}{19}\\\dfrac{y}{3}=\dfrac{-39}{19}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{-195}{19}\\y=\dfrac{-177}{19}\end{matrix}\right.\)
Vậy, ...
NH
2
11 tháng 3 2019
\(A=3x^2+\left(x-2\right)^2+1\)
\(A=3x^2+x^2-4x+4+1\)
\(A=4x^2-4x+1+4\)
\(A=\left(2x-1\right)^2+4\ge4\forall x\)
Dấu "=" xảy ra \(\Leftrightarrow x=\dfrac{1}{2}\)
11 tháng 3 2019
\(A=3x^2+\left(x-2\right)^2+1=4x^2-4x+5=\left(2x-1\right)^2+4\)
Vì \(\left(2x-1\right)^2\ge0\Rightarrow A\ge4\)
Dấu ''='' xảy ra \(\Leftrightarrow x=\frac{1}{2}\)
Vậy \(Min_A=4\Leftrightarrow x=\frac{1}{2}\)
GH
1
23 tháng 4 2020
\(C=\frac{3x^2-x+2}{\left(x-1\right)\left(x+3\right)}-\frac{x}{x-1}-\frac{x-1}{x+3}\left(x\ne1;x\ne-3\right)\)
\(=\frac{3x^2-x+2}{\left(x-1\right)\left(x+3\right)}-\frac{x\left(x+3\right)}{\left(x-1\right)\left(x+3\right)}-\frac{\left(x-1\right)^2}{\left(x-1\right)\left(x+3\right)}\)
\(=\frac{3x^2-x+2}{\left(x-1\right)\left(x+3\right)}-\frac{x^2+3x}{\left(x-1\right)\left(x+3\right)}-\frac{x^2-2x+1}{\left(x-1\right)\left(x+3\right)}\)
\(=\frac{3x^2-x+2-x^2-3x-x^2+2x-1}{\left(x-1\right)\left(x+3\right)}\)
\(=\frac{x^2-2x+1}{\left(x-1\right)\left(x+3\right)}=\frac{\left(x-1\right)^2}{\left(x-1\right)\left(x+3\right)}=\frac{x-1}{x+3}\)
Vậy C=\(\frac{x-1}{x+3}\left(x\ne1;x\ne-3\right)\)
TK
0
PL
1 tháng 8 2018
Ta có : \(\left(5x-3\right)^2-\frac{1^2}{64}=0\)
\(\Leftrightarrow\left(5x-3\right)^2=\frac{1}{64}\)
\(\Leftrightarrow\left(5x-3\right)^2=\left(\frac{1}{8}\right)^2\)
\(\Leftrightarrow\orbr{\begin{cases}5x-3=\frac{1}{8}\\5x-3=-\frac{1}{8}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}5x=\frac{1}{8}+3\\5x=-\frac{1}{8}+3\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}5x=\frac{25}{8}\\5x=\frac{23}{8}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{25}{8}.\frac{1}{5}\\x=\frac{23}{8}.\frac{1}{5}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{5}{8}\\x=\frac{23}{40}\end{cases}}\)
1 tháng 8 2018
b) 3x - 7.(5x-1) = 6 - 2.(4-3x)
=> 3x - 35x + 7 = 6 - 8 + 6x
=> 3x - 35x - 6x = 6-8 -7
-38x = -9
x = 9/38
2 tháng 5 2017
a) \(\dfrac{2}{3}x-\dfrac{3}{2}x=\dfrac{5}{12}\)
\(x\left(\dfrac{2}{3}-\dfrac{3}{2}\right)=\dfrac{5}{12}\)
\(x\cdot\left(-\dfrac{5}{6}\right)=\dfrac{5}{12}\)
\(x=\dfrac{5}{12}:\left(-\dfrac{5}{6}\right)\)
\(x=-\dfrac{1}{2}\)
Vậy \(x=-\dfrac{1}{2}\).
b) \(\dfrac{2}{5}+\dfrac{3}{5}\cdot\left(3x-3\cdot7\right)=-\dfrac{53}{10}\)
\(\dfrac{3}{5}\left(3x-3\cdot7\right)=-\dfrac{53}{10}-\dfrac{2}{5}\)
\(\dfrac{3}{5}\left(3x-3\cdot7\right)=-\dfrac{57}{10}\)
\(3x-3\cdot7=-\dfrac{57}{10}:\dfrac{3}{5}\)
\(3x-3\cdot7=-\dfrac{19}{2}\)
\(3x-21=-\dfrac{19}{2}\)
\(3x=-\dfrac{19}{2}+21\)
\(3x=\dfrac{23}{2}\)
\(x=\)\(\dfrac{23}{2}:3\)
\(x=\dfrac{23}{6}\)
Vậy \(x=\dfrac{23}{6}\).
c) \(\dfrac{7}{9}:\left(2+\dfrac{3}{4x}\right)+\dfrac{5}{3}=\dfrac{23}{27}\)
\(\dfrac{7}{9}:\left(2+\dfrac{3}{4x}\right)=\dfrac{23}{27}-\dfrac{5}{3}\)
\(\dfrac{7}{9}:\left(2+\dfrac{3}{4x}\right)=-\dfrac{22}{27}\)
\(2+\dfrac{3}{4x}=\dfrac{7}{9}:-\dfrac{22}{27}\)
\(2+\dfrac{3}{4x}=-\dfrac{21}{22}\)
\(\dfrac{3}{4x}=-\dfrac{21}{22}-2\)
\(\dfrac{3}{4x}=-\dfrac{65}{22}\)
\(4x=\dfrac{3\cdot22}{-65}\)
\(4x=-\dfrac{66}{65}\)
\(x=-\dfrac{66}{65}:4\)
\(x=-\dfrac{33}{130}\)
Vậy \(x=-\dfrac{33}{130}\).
d) \(-\dfrac{2}{3}x+\dfrac{1}{5}=\dfrac{3}{10}\)
\(-\dfrac{2}{3}x=\dfrac{3}{10}-\dfrac{1}{5}\)
\(-\dfrac{2}{3}x=\dfrac{1}{10}\)
\(x=\dfrac{1}{10}:-\dfrac{2}{3}\)
\(x=-\dfrac{3}{20}\)
Vậy \(x=-\dfrac{3}{20}\).
e) \(\left|x\right|-\dfrac{3}{4}=\dfrac{5}{3}\)
\(\left|x\right|=\dfrac{5}{3}+\dfrac{3}{4}\)
\(\left|x\right|=\dfrac{29}{12}\)
\(x=\dfrac{29}{12}\) hoặc \(=-\dfrac{29}{12}\)
Vậy \(x\in\left\{\dfrac{29}{12};-\dfrac{29}{12}\right\}\).
HN
0
x=6
lai cho teo hem
cau tra loi la 6