Tìm x biết:
0,5x=3/5
Đáp số: x=
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a) \(\Leftrightarrow36+3-11x=0\)
\(\Leftrightarrow-11x=-39\)
\(\Leftrightarrow x=\frac{39}{11}\)
b) \(x^2-2x\frac{1}{4}+\frac{1}{16}-\frac{81}{16}=0\)
\(\left(x-\frac{1}{4}\right)^2=\frac{81}{16}\)
\(x-\frac{1}{4}=\frac{9}{4}\)
\(x=\frac{10}{4}=\frac{5}{2}\)
c) \(x^2\left(x-3\right)-4\left(x-3\right)=0\)
\(\left(x^2-4\right)\left(x-3\right)=0\)
\(\left(x-2\right)\left(x+2\right)\left(x-3\right)=0\)
x = 2 hoặc x = - 2 hoặc x = 3
a) \(\frac{8}{2}\)
b) \(\frac{5}{2}\)
c) x=2 hoạc x=-2 hoặc x=3
=>1/2:(0,5x-1,5)=0,2-0,35=-0,15=-3/20
=>0,5x-1,5=1/2:(-3/20)=-1/2*20/3=-10/3
=>0,5x=-10/3+3/2=-11/6
=>x=-11/3
c: Ta có: \(\left|\dfrac{1}{2}x-2\right|-\left|x+3\right|=0\)
\(\Leftrightarrow\left|\dfrac{1}{2}x-2\right|=\left|x+3\right|\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{2}x-2=x+3\\\dfrac{1}{2}x-2=-x-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x\cdot\dfrac{-1}{2}=5\\x\cdot\dfrac{3}{2}=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-10\\x=-\dfrac{2}{3}\end{matrix}\right.\)
tìm nghiệm của đa thức một biến:
a, x(x-5)-(x-5)*(3x+2)
b,2*(0,3x-0,5x)-0,5(3-x)
c,(x-1)-(x-2)=(-x-40-8
a. \(x\left(x-5\right)-\left(x-5\right)\left(3x+2\right)=0\)
\(\Rightarrow\left(x-5\right)\left[x-\left(3x+2\right)\right]=0\)
\(\Rightarrow\left(x-5\right)\left(x-3x-2\right)=0\)
\(\Rightarrow\left(x-5\right)\left(-2x-2\right)=0\)
\(\Rightarrow-2\left(x-5\right)\left(x+1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}-2\ne0\\x-5=0\\x+1=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}-2\ne0\\x=5\\x=-1\end{matrix}\right.\)
b. \(2\left(0,3x-0,5x\right)-0,5\left(3-x\right)=0\)
\(\Rightarrow-0,4x-1,5+0,5x=0\)
\(\Rightarrow0,1x-1,5=0\)
\(\Rightarrow0,1x=1,5\)
\(\Rightarrow x=\dfrac{1,5}{0,1}=15\)
c. \(\left(x-1\right)-\left(x-2\right)=\left(-x-40\right)-8\)
\(\Rightarrow x-1-x+2=-x-40-8\)
\(\Rightarrow1=-x-48\)
\(\Rightarrow1+48=-x\)
\(\Rightarrow x=-49\)
\(a)\) \(\frac{x+1}{2x+1}=\frac{0,5x+2}{x+3}\)
\(\Leftrightarrow\)\(\left(2x+1\right)\left(0,5x+2\right)=\left(x+1\right)\left(x+3\right)\)
\(\Leftrightarrow\)\(2x\left(0,5x+2\right)+0,5x+2=x\left(x+1\right)+3\left(x+1\right)\)
\(\Leftrightarrow\)\(x^2+4x+0,5+2=x^2+x+3x+3\)
\(\Leftrightarrow\)\(x^2+4x-x^2-4x=3-0,5-2\)
\(\Leftrightarrow\)\(0=0,5\) ( vô lí :'< )
Vậy không có x thoả mãn đề bài
\(a,0,5x-\frac{2}{3}x=\frac{7}{12}\Rightarrow\frac{1}{2}x-\frac{2}{3}x=\frac{7}{12}\)
\(\Rightarrow x\left(\frac{1}{2}-\frac{2}{3}\right)=\frac{7}{12}\Rightarrow x\cdot\left(\frac{3}{6}-\frac{4}{6}\right)=\frac{7}{12}\)
\(\Rightarrow x\cdot\left(-1\right)=\frac{7}{12}\Rightarrow x=\frac{7}{12}:\left(-1\right)=\frac{7}{-12}\)
\(c,\frac{\left(x-5\right)}{12}\cdot\frac{9}{29}=\frac{-6}{29}\Rightarrow\frac{\left(x-5\right)}{12}=\frac{-6}{29}:\frac{9}{26}\)
\(\frac{\Rightarrow\left(x-5\right)}{12}=\frac{-6}{9}=\frac{-2}{3}\Rightarrow x-5=-\frac{2}{3}\cdot12\)
\(\Rightarrow x-5=\frac{-24}{3}=-8\Rightarrow x=-8+5=-3\)
\(a,0,5x-\frac{2}{3}x=\frac{7}{12}\)
\(\Rightarrow\frac{1}{2}x-\frac{2}{3}x=\frac{7}{12}\)
\(\Rightarrow-\frac{1}{6}x=\frac{7}{12}\)
\(\Rightarrow x=-\frac{7}{2}\)
\(c,\frac{x-5}{12}\cdot\frac{9}{29}=-\frac{6}{29}\)
\(\Rightarrow\frac{x-5}{12}=-\frac{2}{3}\)
\(\Rightarrow x-5=12.\left(-\frac{2}{3}\right)\)
\(\Rightarrow x-5=-8\)
\(\Rightarrow x=-3\)
\(0,5x-75\%x=\left(\frac{1}{3}-1\right)^2\)
\(\Leftrightarrow\frac{1}{2}x-\frac{3}{4}x=\left(-\frac{2}{3}\right)^2\)
\(\Leftrightarrow-\frac{1}{4}x=\frac{4}{9}\)
\(\Leftrightarrow x=-\frac{16}{9}\)
\(0,5x-75\%x=\left(\frac{1}{3}-1\right)^2\)
\(< =>\frac{1}{2}x-\frac{3}{4}x=\left(-\frac{2}{3}\right)^2\)
\(< =>-\frac{1}{4}x=\frac{4}{9}\)
\(=>x=\frac{-16}{9}\)
áp dụng tính chất của dãy tỉ số bằng nhau ta có:
\(\frac{x+1}{2x+1}=\frac{0,5+2}{x+3}=\frac{\left(x+1\right)-2.\left(0,5x+2\right)}{\left(2x+1\right)-2.\left(x+3\right)}=\frac{x+1-x-4}{2x+1-2x-6}=\frac{-3}{-5}=\frac{3}{5}\)
suy ra:
\(\frac{x+1}{2x+1}=\frac{3}{5}\Rightarrow5.\left(x+1\right)=3.\left(2x+1\right)\)
=>5x+5=6x+3
5x-6x=3-5
-x=-2
x=2
0,5x = 3/5
1/2x = 3/5
x = 3/5 : 1/2
x = 3/5 x 2
x = 6/5
X=6/5