tìm x biết 1/6x +1/10x-4/15x+1=0
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b, \(\frac{1}{6}x+\frac{1}{10}x-\frac{4}{15}x+1=0\)
\(x\left(\frac{1}{6}+\frac{1}{10}-\frac{4}{15}\right)+1=0\)
\(x.0=-1\)
\(\Rightarrow x\in rỗng\)
a)\(\dfrac{1}{6}x+\dfrac{1}{10}x-\dfrac{4}{15}x+1=0\)
\(\left(\dfrac{1}{6}+\dfrac{1}{10}-\dfrac{4}{15}\right).x+1=0\)
\(\left(\dfrac{5}{30}+\dfrac{3}{30}-\dfrac{8}{30}\right).x+1=0\)
\(0.x+1=0\)
\(0.x=-1\)
=> Không có giá trị nào của x.
Vậy...
b)\(\left(\dfrac{1}{7}x-\dfrac{2}{7}\right).\left(-\dfrac{1}{5}x+\dfrac{3}{5}\right).\left(\dfrac{1}{3}x+\dfrac{4}{3}\right)=0\)
=> \(\dfrac{1}{7}x-\dfrac{2}{7}=0hoặc-\dfrac{1}{5}x+\dfrac{3}{5}=hoăc\dfrac{1}{3}x+\dfrac{4}{3}=0\)
+)\(~\dfrac{1}{7}x-\dfrac{2}{7}=0\) +) \(-\dfrac{1}{5}x+\dfrac{3}{5}=0\) +) \(\dfrac{1}{3}x+\dfrac{4}{3}=0\)
\(\dfrac{1}{7}x=-\dfrac{2}{7}\) \(-\dfrac{1}{5}x=-\dfrac{3}{5}\) \(\dfrac{1}{3}x=-\dfrac{4}{3}\)
\(x=2\) \(x=3\) \(x=-4\)
Vậy...
a 1/6x+1/10x-4/15x+1=0
(1/6+1/10-4/15)x+1=0
0x+1=0
0x=-1
x=-1/0
Vậy không có x (vì không có số nào chia cho 0)
1, \(3x\left(x-7\right)+2x-14=0\)
\(\Rightarrow3x\left(x-7\right)+2\left(x-7\right)=0\)
\(\Rightarrow\left(x-7\right)\left(3x+2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=7\\x=\frac{-2}{3}\end{cases}}\)
2, \(x^3+3x^2-\left(x+3\right)=0\)
\(\Rightarrow x^2\left(x+3\right)-\left(x+3\right)=0\)
\(\Rightarrow\left(x+3\right)\left(x^2-1\right)=0\)
\(\Rightarrow\left(x+3\right)\left(x-1\right)\left(x+1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=-3\\x=\pm1\end{cases}}\)
3, \(15x-5+6x^2-2x=0\)
\(\Rightarrow\left(15x-5\right)+\left(6x^2-2x\right)=0\)
\(\Rightarrow5\left(3x-1\right)+2x\left(3x-1\right)=0\)
\(\Rightarrow\left(3x-1\right)\left(5+2x\right)=0\)
\(\Rightarrow\orbr{\begin{cases}3x-1=0\\5+2x=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{1}{3}\\x=\frac{-5}{2}\end{cases}}\)
4, \(5x-2-25x^2+10x=0\)
\(\Rightarrow\left(5x-25x^2\right)-\left(2-10x\right)=0\)
\(\Rightarrow5x\left(1-5x\right)-2\left(1-5x\right)=0\)
\(\Rightarrow\left(1-5x\right)\left(5x-2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}1-5x=0\\5x-2=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{1}{5}\\x=\frac{2}{5}\end{cases}}\)
a, \(\left(x+2\right)^3-x\left(x^2+6x-3\right)=0\Leftrightarrow x^3+4x^2+4x+2x^2+8x+8-x^3-6x^2+3x=0\)
\(\Leftrightarrow15x+8=0\Leftrightarrow x=-\frac{8}{15}\)
b, \(\left(x+4\right)^3-x\left(x+6\right)^2=7\Leftrightarrow12x+64=0\Leftrightarrow x=-\frac{19}{4}\)làm tắt:P
Tự làm nốt nhé
a, \(x^3+3x^2-\left(x+3\right)=0\Leftrightarrow x^2\left(x+3\right)-\left(x+3\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(x+3\right)=0\Leftrightarrow x=1;x=-1;x=-3\)
b, \(15x-5+6x^2-2x=0\Leftrightarrow5\left(3x-1\right)+2x\left(3x-1\right)=0\)
\(\Leftrightarrow\left(2x+5\right)\left(3x-1\right)=0\Leftrightarrow x=-\frac{5}{2};x=\frac{1}{3}\)
c, \(5x-2-25x^2+10x=0\)
\(\Leftrightarrow\left(5x-2\right)-5x\left(5x-2\right)=0\Leftrightarrow\left(1-5x\right)\left(5x-2\right)=0\Leftrightarrow x=\frac{2}{5};x=\frac{1}{5}\)
Ta có:
\(15x^4y^4-M=10x^2y^4+6x^2y^4\)
\(\Leftrightarrow M=15x^4y^4-\left(10x^2y^4+6x^2y^2\right)\)
\(\Leftrightarrow M=15x^4y^4-16x^2y^4\)
Thay \(x=-\dfrac{1}{2};x=2\) vào M ta có:
\(M=15\cdot\left(-\dfrac{1}{2}\right)^4\cdot2^4-16\cdot\left(-\dfrac{1}{2}\right)^2\cdot2^4=-49\)
nha nha\
\(\frac{1}{6x}+\frac{1}{10x}-\frac{4}{15x}+1=0\)
\(\Rightarrow\frac{1}{x}\left(\frac{1}{6}+\frac{1}{10}-\frac{4}{15}\right)+1=0\)
\(\Rightarrow\frac{1}{x}.0+1=0\)
\(\Rightarrow1=0\)( vô lý )
=> không có giá trị x thỏa mãn đề bài
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