\(\int\limits^2_0\left[f\left(x\right)-2g\left(x\right)\right]dx=\int\limits^2_0f\left(x\right)dx-2\int\limits^2_0g\left(x\right)dx=3+2=5\)

\(_{u_k=u_{k-1}+4\left(k-1\right)+3=u_{k-2}+4\left(k-2\right)+4\left(k-1\right)+2.3=.....=u_1+4\left(1+2+3+...+k-1\right)+3\left(k-1\right)=\left(k-1\right)\left(2k+3\right)}\)

\(\Rightarrow lim\dfrac{\sqrt{u_{kn}}}{n}=\dfrac{\sqrt{\left(2kn+3\right)\left(kn-1\right)}}{n}=\sqrt{\left(2k+\dfrac{3}{n}\right)\left(k-\dfrac{1}{n}\right)}=k\sqrt{2}\)

\(\Rightarrow lim_P=\dfrac{1\sqrt{2}+4\sqrt{2}+4^2\sqrt{2}+...+4^{2022}\sqrt{2}}{1\sqrt{2}+2\sqrt{2}+2^2\sqrt{2}+....+2^{2022}\sqrt{2}}=\dfrac{4^{2023}-1}{3.\left(2^{2023}-1\right)}=\dfrac{2^{2023}+1}{3}\)

\(\Rightarrow a+b-c=0\)

Tìm số nguyên n>=17 thỏa mãn\(C^0_{17}C^{17}_n+C^1_{17}C^{16}_n+.....+C^{17}_{17}C^0_n=\dfrac{1}{2}C^{18}_{2n}\)

Tính \(\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2}+\sqrt{2+....}}}}\)

giải pt : sinx^4 + cosx^4-cosx^2+ 1/(4.sin2x^2) -1 =0

cho pt : m +sinx +cosx +cosx.sin x =0. Tìm m để pt có 2 nghiệm x thuộc [ 0,2pi] sao cho 2 nghiệm hơn kém nhau pi/2

giải pt \(\sqrt{x-2}+\sqrt{4-x}+\sqrt{2x-5}=2x^2-5x\)

2) \(x^2+x+2=\sqrt{5x+5}+\sqrt{3x+2}\)

giải pt \(\dfrac{1}{\sqrt{2x-1}}+\dfrac{1}{\sqrt{2x+1}}=\dfrac{1}{\sqrt{x+1}}+\dfrac{1}{\sqrt{3x-1}}\)

à bạn học góc bên trong đường tròn chưa nhỉ ?