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b) Để g(x) có nghiệm
\(\Leftrightarrow\left(x-1\right)\left(2-3x\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\2-3x=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=1\\x=\frac{2}{3}\end{cases}}\)
Vậy \(x\in\left\{1;\frac{2}{3}\right\}\)là nghiệm của đa thức g(x)
c) Để k(x) có nghiệm
\(\Leftrightarrow x^2-3x-4=0\)
\(\Leftrightarrow x^2+x-4x-4=0\)
\(\Leftrightarrow x\left(x+1\right)-4\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x-4\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+1=0\\x-4=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-1\\x=4\end{cases}}}\)
Vậy \(x\in\left\{-1;4\right\}\)là nghiệm của đa thức
\(\frac{x-1}{4}=\frac{2x+1}{5}\)
\(\Rightarrow5\left(x-1\right)=4\left(2x+1\right)\)
\(\Rightarrow5x-5=8x+4\)
\(\Rightarrow5x-8x=4+5\)
\(\Rightarrow-3x=9\)
\(\Rightarrow x=-3\)
vậy_
\(\frac{x+2}{x-1}=\frac{x-3}{x+1}\)
\(\Rightarrow\left(x+2\right)\left(x+1\right)=\left(x-1\right)\left(x-3\right)\)
\(\Rightarrow x^2+x+2x+2=x^2-3x-x+3\)
\(\Rightarrow x^2+x+2x-x^2+3x+x=3-2\)
\(\Rightarrow7x=1\)
\(\Rightarrow x=\frac{1}{7}\)
vậy_
\(\frac{3}{5}x-\frac{13}{9}:\left(\frac{131313}{151515}+\frac{131313}{353535}+\frac{131313}{636363}\right)=-10\)
<=> \(\frac{3}{5}x-\frac{13}{9}:\left(\frac{13}{15}+\frac{13}{35}+\frac{13}{63}\right)=-10\)
<=> \(\frac{3}{5}x-\frac{13}{9}:13:\left(\frac{1}{15}+\frac{1}{35}+\frac{1}{63}\right)=-10\)
<=> \(\frac{3}{5}x-\frac{1}{9}:\left(\frac{21}{315}+\frac{9}{315}+\frac{5}{315}\right)=-10\)
<=> \(\frac{3}{5}x-\frac{1}{9}:\frac{35}{315}=-10\)
<=> \(\frac{3}{5}x-\frac{1}{9}:\frac{1}{9}=-10\)
<=> \(\frac{3}{5}x-1=-10\)
<=> \(\frac{3}{5}x=-9\)
<=> \(x=-15\)
Vậy x = -15.
\(\frac{3}{5}x-1\frac{4}{9}:\left(\frac{131313}{151515}+\frac{131313}{353535}+\frac{131313}{636363}\right)=-10\)
\(\Leftrightarrow\frac{3}{5}x-\frac{13}{9}:\left(\frac{13}{15}+\frac{13}{35}+\frac{13}{63}=-10\right)\)
\(\Leftrightarrow\frac{3}{5}x-\frac{13}{9}:\left[\frac{13}{2}\left(\frac{2}{15}+\frac{2}{35}+\frac{2}{63}\right)\right]=-10\)
\(\Leftrightarrow\frac{3}{5}x-\frac{13}{9}:\left[\frac{13}{2}\left(\frac{2}{3.5}+\frac{2}{.57}+\frac{2}{7.9}\right)\right]=-10\)
\(\Leftrightarrow\frac{3}{5}x-\frac{13}{9}:\left[\frac{13}{2}\left(\frac{1}{3}-\frac{1}{9}\right)\right]=-10\)
\(\Leftrightarrow\frac{3}{5}x-\frac{13}{9}:\left(\frac{13}{2}.\frac{2}{9}\right)=-10\)
\(\Leftrightarrow\frac{3}{5}x-\frac{13}{9}:\frac{26}{18}=-10\)
\(\Leftrightarrow\frac{3}{5}x-1=-10\)
\(\Leftrightarrow\frac{3}{5}x=-10+1\)
\(\Leftrightarrow\frac{3}{5}x=-9\)
\(\Rightarrow x=-9:\frac{3}{5}\)
\(\Rightarrow x=-15\)
Vậy \(x=-15\)
\(xy+3x-y=6\)
=> \(xy+3x-y-3=3\)
=> \(\left(xy+3x\right)-\left(y+3\right)=3\)
=> \(x\left(y+3\right)-\left(y+3\right)=3\)
=> \(\left(y+3\right)\left(x-1\right)=3\)
Mà x, y nguyên
=> \(x-1\)và \(y+3\)là số nguyên
=> \(\hept{\begin{cases}x-1=1\\y+3=3\end{cases}}\); \(\hept{\begin{cases}x-1=3\\y+3=1\end{cases}}\)và \(\hept{\begin{cases}x-1=-1\\y+3=-3\end{cases}}\)
=> \(\hept{\begin{cases}x=2\\y=0\end{cases}}\); \(\hept{\begin{cases}x=4\\y=-2\end{cases}}\)và \(\hept{\begin{cases}x=0\\y=-6\end{cases}}\)
Vậy cặp số nguyên (x;y) thỏa mãn là (2;0), (4;-2) và (0;-6)
\(6-2\left|1+3x\right|\le6\)'
Max \(A=6\Leftrightarrow1+3x=0\)
\(\Rightarrow3x=-1\)
\(\Rightarrow x=\frac{-1}{3}\)
\(\left|x-2\right|+\left|x-5\right|\ge0\)
Max \(B=0\Leftrightarrow\hept{\begin{cases}x-2=0\\x-5=0\end{cases}\Rightarrow\hept{\begin{cases}x=2\\x=5\end{cases}}}\)