8/25=2x/5x-1
tìm x
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=>10x+15y=5m và -10x+2y=-2
=>17y=5m-2 và -5x+y=-1
=>y=5/17m-2/17 và 5x-y=1
=>y=5/17m-2/17 và 5x=1+y=5/17m+15/17
=>y=5/17m-2/17 và x=1/17m+5/17
x>0; y>0
=>5m-2>0 và m+5>0
=>m>2/5
a, 117 - \(x\) = 28 - (-7)
117 - \(x\) = 28 + 7
117 - \(x\) = 35
\(x\) = 117 - 35
\(x\) = 82
b, \(x\) - (-38 - 2\(x\)) = (-3) - 8 + 2\(x\)
\(x\) + 38 + 2\(x\) = - 11 + 2\(x\)
3\(x\) + 38 = - 11 + 2\(x\)
3\(x\) - 2\(x\) = - 11 - 38
\(x\) = - 49
a) \(\text{5x(x-2)+(2-x)=0}\)
\(\Rightarrow5x\left(x-2\right)-\left(x-2\right)=0\\ \Rightarrow\left(x-2\right)\left(5x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-2=0\\5x-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{1}{5}\end{matrix}\right.\)
b) \(\text{x(2x-5)-10x+25=0}\)
\(\Rightarrow x\left(2x-5\right)-5\left(2x-5\right)=0\\ \Rightarrow\left(x-5\right)\left(2x-5\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-5=0\\2x-5=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=5\\x=2,5\end{matrix}\right.\)
c) \(\dfrac{25}{16}-4x^2+4x-1=0\)
\(\Rightarrow\dfrac{9}{16}-4x^2+4x=0\)
\(\Rightarrow-4x^2+4x+\dfrac{9}{16}=0\)
\(\Rightarrow-4x^2-\dfrac{1}{2}x+\dfrac{9}{2}x+\dfrac{9}{16}=0\)
\(\Rightarrow\left(-4x^2-\dfrac{1}{2}x\right)+\left(\dfrac{9}{2}x+\dfrac{9}{16}\right)=0\)
\(\Rightarrow-\dfrac{1}{2}x\left(8x+1\right)+\dfrac{9}{16}\left(8x+1\right)=0\)
\(\Rightarrow\left(-\dfrac{1}{2}x+\dfrac{9}{16}\right)\left(8x+1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}-\dfrac{1}{2}x+\dfrac{9}{16}=0\\8x+1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{9}{8}\\x=\dfrac{-1}{8}\end{matrix}\right.\)
\(\Delta=\left(-5\right)^2-4\cdot2\cdot m=25-8m\)
\(f\left(x\right)\ge0\)
\(\Leftrightarrow\Delta\le0\)
\(\Leftrightarrow25-8m\le0\)
\(\Leftrightarrow m\ge\dfrac{25}{8}\)
a) \(3\frac{1}{3}\left(3\frac{1}{4}+2x\right)=6\frac{2}{3}\)
\(3\frac{1}{3}\times3\frac{1}{4}+2x=6\frac{2}{3}\)
\(10\frac{5}{6}+2x=6\frac{2}{3}\)
\(2\times x=6\frac{2}{3}+10\frac{5}{6}=17,5\)
\(x=17,5\div2=8,75\)
Vậy x = 8,75
b) \(x-25\%x=\frac{6}{11}\left(\frac{1}{2}+\frac{3}{4}-\frac{1}{3}\right)\)
\(x-\frac{25}{100}x=\frac{6}{11}\left(\frac{1}{2}+\frac{3}{4}-\frac{1}{3}\right)\)
\(x-\frac{1}{4}\times x=\frac{6}{11}\times1\frac{7}{12}=\frac{19}{22}\)
\(x\times x=\frac{19}{22}+\frac{1}{4}=\frac{49}{44}\)
\(\Rightarrow2x\left(x\times x\right)=\frac{49}{44}\)
\(x=\frac{49}{44}\div2=\frac{49}{88}\)
Vậy x = \(\frac{49}{88}\)
c) \(\left(4,5-2x\right)\times1\frac{4}{7}=\frac{11}{14}\)
\(4,5-2x\times1\frac{4}{7}=\frac{11}{14}\)
\(-2x\times1\frac{4}{7}=\frac{11}{14}-4,5=-3\frac{5}{7}\)
\(-2\times x=-3\frac{5}{7}\div1\frac{4}{7}=-2\frac{4}{11}\)
\(x=-2\frac{4}{11}\div\left(-2\right)=1\frac{2}{11}\)
Vậy x = \(1\frac{2}{11}\)
d) \(-3^2-|2x+3|=4\)
\(9-|2x+3|=4\)
\(-|2x+3|=4-9=-5\)
\(-|2x|=-5-|3|=-8\)
\(-|x|=-8\div2=-4\)
\(-x=4\Rightarrow x=-4\)
Vậy x = -4 (-x được xem là số đối của x)
\(8x-75=5x+21\)
\(8x-5x=75+21\)
\(3x=96\)
\(x=32\)
Vậy \(x=32.\)
\(9x+25=-\left(2x-58\right)\)
\(9x+25=-2x+58\)
\(9x+2x=-25+58\)
\(11x=33\)
\(x=3\)
Vậy \(x=3.\)
\(15-\left|2x-1\right|=-8\)
\(\left|2x-1\right|=23\)
\(\Rightarrow\orbr{\begin{cases}2x-1=23\\2x-1=-23\end{cases}\Rightarrow}\orbr{\begin{cases}2x=24\\2x=-22\end{cases}}\Rightarrow\orbr{\begin{cases}x=12\\x=-11\end{cases}}\)
Vậy \(x\in\left\{12;-11\right\}\)
8x-5x=75+21
3x=96
x=32
9x+25=-2x+58
9x+2x=58-25
11x=33
x=3
+) (5x-1). (2x+3)-3. (3x-1)=0
10x^2+15x-2x-3 - 9x+3=0
10x^2 +8x=0
2x(5x+4)=0
=> x=0 hoặc x= -4/5
+) x^3 (2x-3)-x^2 (4x^2-6x+2)=0
2x^4 -3x^3 -4x^4 + 6x^3 - 2x^2=0
-2x^4 + 3x^3-2x^2=0
x^2(-2x^2+x-2)=0
-2x^2(x-1)^2=0
=> x=0 hoặc x=1
+) x (x-1)-x^2+2x=5
x^2 -x -x^2+2x=5
x=5
+) 8 (x-2)-2 (3x-4)=25
8x - 16-6x+8=25
2x=33
x=33/2
a, \(5^0=1\)
b, \(12.35+35.182-35.66\)
\(=35.\left(12+182-66\right)\)
\(=35.128\)
\(=4480\)
c, \(x+2x=6\)
\(x.1+2.x=6\)
\(x\left(1+2\right)=6\)
\(x.3=6\)
\(x=2\)
\(15-\left(x-4\right)=8\)
\(x-4=15-8\)
\(x-4=7\)
\(x=7+4\)
\(x=11\)
Ta có \(\frac{8}{25}=\frac{2^3}{5^2}\)
Do đó x = 3 và x - 1 = 2
Kết luận : x = 3
\(\Leftrightarrow\frac{2^x}{5^x}.5=\frac{8}{25}\Leftrightarrow\left(\frac{2}{5}\right)^x=\frac{8}{125}=\left(\frac{2}{5}\right)^3\Leftrightarrow x=3\)