cho đa thức \(f\left(x\right)=4\cdot x^2+3x+1\); \(g\left(x\right)=3x^2-2x+1\); \(k\left(x\right)=7\cdot x^2-35x+42\)
a) tính f(x)-g(x)=h(x)
b) tính nghiệm của h(x) và k(x)
c) tìm gia trị của đa thức h(x) biết:
\(\left(x^2-9\right)^{2021}=\left(\frac{3}{4}-81\right)\cdot\left(\frac{3^2}{5}-81\right)^2\cdot\left(\frac{3^2}{6}-81\right)^3\cdot\cdot\cdot\left(\frac{3^{2020}}{2023}-81\right)^{2020}\)
\(2\cdot f\left(x\right)+5\cdot f\left(\frac{1}{x}\right)=x^2\)
Tính \(f\left(\frac{1}{2}\right)\)
TÌM x
\(\left(\left(\frac{3}{4}\cdot x+5\right)-\left(\frac{2}{3}\cdot x-4\right)-\left(\frac{1}{6}\cdot x+1\right)\right)=\left(\frac{1}{3}\cdot x+4\right)-\left(\frac{1}{3}-3\right)\)
\(2\cdot f\left(x\right)+5\cdot\left(\frac{1}{x}\right)=x^2\)
Tìm x biết:
\(\frac{3}{\left(x+2\right)\cdot\left(x+5\right)}+\frac{5}{\left(x+5\right)\cdot\left(x+10\right)}+\frac{7}{\left(x+10\right)\cdot\left(x+17\right)}=\frac{x}{\left(x+2\right)\cdot\left(x+17\right)}\)
\(\left(\frac{1}{7}\cdot x-\frac{2}{7}\right)\cdot\left(-\frac{1}{5}\cdot x+\frac{3}{5}\right)\cdot\left(\frac{1}{3}\cdot x+\frac{4}{3}\right)=0\)
Tìm x biết :
a, ( 4x - 9 ) . ( 2,5 + \(\frac{-7}{3}\). x ) = 0
b, \(\frac{1}{x\cdot\left(x+1\right)}\cdot\frac{1}{\left(x+1\right)\cdot\left(x+2\right)}\cdot\frac{1}{\left(x+2\right)\cdot\left(x+3\right)}-\frac{1}{x}=\frac{1}{2015}\)
Tìm x:
a) \(\frac{3}{\left(x+2\right)\cdot\left(x+5\right)}\)+\(\frac{5}{\left(x+5\right)\cdot\left(x+10\right)}\)+\(\frac{7}{\left(x+10\right)\cdot\left(x+17\right)}\)= \(\frac{x}{\left(x+2\right)\cdot\left(x+17\right)}\)
Với x không thuộc (-2;-5;-10;-17)
b) \(\frac{2}{\left(x-1\right)\cdot\left(x-3\right)}\)+\(\frac{5}{\left(x-3\right)\cdot\left(x-8\right)}\)+\(\frac{12}{\left(x-8\right)\cdot\left(x-20\right)}\)-\(\frac{1}{20}\)= \(\frac{-3}{4}\)
Với x không thuộc (1;3;8;20)
c)\(\frac{x+1}{2019}\)+\(\frac{x+2}{2018}\)= \(\frac{x-3}{2017}\)\(\frac{x-4}{2016}\)
a)\(\frac{3}{2}-\frac{1}{3}\cdot\left(x-\frac{3}{2}\right)-\frac{1}{2}\cdot\left(2\cdot x+1\right)=5\)
b)\(\left(x+\frac{1}{2}\right)\cdot\left(x-\frac{3}{4}\right)=0\)
c)\(2x-3=x+\frac{1}{2}\)