mn giúp mik câu này vs
10x(x-150)+150-x=0
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bạn ơi giúp mình vs
tìm x,y, z nguyên thỏa mãn
x^3 + xyz = 957
y^3 + xyz = 759
z^3 + xyz = 579
\(\Leftrightarrow x-3\sqrt{x}-\sqrt{x-8}+1=0\)
\(\Leftrightarrow x=9\left(tm\right)\)
a. (x-3)(x\(^2\)+6x+9)(x-1)(x\(^2\)+2x+1)(-x\(^2\)+2x+3)=0
\(\Leftrightarrow\)(x-3)(x\(^2\)+6x+9)(x-1)(x\(^2\)+2x+1)(x-3)(x+1)=0
a, \(\frac{3}{35}-\left(\frac{3}{5}+x\right)=\frac{2}{7}\)
\(\frac{3}{5}+x=\frac{3}{35}-\frac{2}{7}=-\frac{1}{5}\)
\(x=-\frac{1}{5}-\frac{3}{5}\)
\(x=-\frac{4}{5}\)
b,\(\left(5x-1\right)\left(2x-\frac{1}{3}\right)=0\)
=> \(\left(5x-1\right)=0\) hoặc \(\left(2x-\frac{1}{3}\right)=0\)
=> \(5x=1\) hoặc \(2x=\frac{1}{3}\)
=> \(x=\frac{1}{5}\) hoặc \(x=\frac{1}{6}\)
\(\dfrac{1}{2}\left(x-2\right)+\dfrac{1}{3}\left(2-x\right)=x\\ \Leftrightarrow\dfrac{1}{2}\left(x-2\right)-\dfrac{1}{3}\left(x-2\right)=x\\ \Leftrightarrow\left(x-2\right).\left(\dfrac{1}{2}-\dfrac{1}{3}\right)=x\\ \Leftrightarrow\left(x-2\right).\left(\dfrac{3-2}{6}\right)=x\\ \Leftrightarrow\left(x-2\right).\dfrac{1}{6}=x\\ \Leftrightarrow\dfrac{1}{6}x-\dfrac{1}{3}-x=0\\ \Leftrightarrow\left(\dfrac{1}{6}-1\right)x=\dfrac{1}{3}\\ \Leftrightarrow\left(\dfrac{1-6}{6}\right)x=\dfrac{1}{3}\\ \Leftrightarrow\dfrac{-5}{6}x=\dfrac{1}{3}\\ \Leftrightarrow x=\dfrac{1}{3}:\left(-\dfrac{5}{6}\right)\\ \Leftrightarrow x=-\dfrac{2}{5}\)
Vậy \(x=-\dfrac{2}{5}\)
\(\frac{y+z+1}{x}=\frac{x+z+2}{y}=\frac{x+y-3}{z}=\frac{\left(y+z+1\right)+\left(x+z+2\right)+\left(x+y-3\right)}{x+y+z}=\frac{2\left(x+y+z\right)}{x+y+z}=2=\frac{1}{x+y+z}\)
=> x+y+z =1/2
+y+z+1=2x => x+y+z +1 =3x => 3x =1/2 +1 =3/2 => x =1/2
+x+y+2 =2y => x+y+z+2 =3y => 3y = 1/2 +2 = 5/2 => y =5/6
+z =1/2 -x-y =1/2 -1/2 -5/6 =-5/6
3.(x - 2) + 150 = 240
3.(x - 2) = 240 - 150
3.(x - 2) = 90
x - 2 = 90 : 3
x - 2 = 30
x = 30 + 2
x = 32
Vậy x = 32
Ủng hộ mk nha ^_-
ta lấy 240-150=90 ra 3.(x-2)=90 rồi ta lấy 90:3=30 xong ta lấy 30 +2=32 vậy x = 32.sorry mình ko thể trình bày được.
10x(x-150)+150-x=0
<=> 10x(x-150)-(x-150)=0
<=> (x-150)(10x-1)=0
<=> \(\orbr{\begin{cases}x=150\\x=\frac{1}{10}\end{cases}}\)
vậy.
\(10x\left(x-150\right)+150-x=0\)
\(\Leftrightarrow10x^2-1500x+150-x=0\)
\(\Leftrightarrow10x^2-1501x+150=0\)
\(\Leftrightarrow10x^2-1501x=-150\Leftrightarrow x\left(10x-1501\right)=-150\)
\(\Leftrightarrow\hept{\begin{cases}x=-150\\10x-1501=-150\end{cases}\Leftrightarrow\hept{\begin{cases}x=-150\\x=\frac{-1651}{10}\end{cases}}}\)