b.

\(\int x^3\left[\dfrac{\left(sin\dfrac{x}{2}+cos\dfrac{x}{2}\right)^2}{x^3}-2x+\dfrac{1}{x^{2024}}\right]dx\)

\(=\int\left[\left(sin\dfrac{x}{2}+cos\dfrac{x}{2}\right)^2-2x^4+\dfrac{1}{x^{2021}}\right]dx\)

\(=\int\left(sin^2\dfrac{x}{2}+cos^2\dfrac{x}{2}+2sin\dfrac{x}{2}cos\dfrac{x}{2}-2x^4+\dfrac{1}{x^{2021}}\right)dx\)

\(=\int\left(1+sinx-2x^4+\dfrac{1}{x^{2021}}\right)dx\)

\(=x-cosx-\dfrac{2}{5}x^5-\dfrac{1}{2020.x^{2020}}+C\)

a.

\(\int x\left(2024-\dfrac{1}{x^3}+\dfrac{sinx}{x}\right)dx\)

\(=\int\left(2024x-\dfrac{1}{x^2}+sinx\right)dx\)

\(=1012x^2+\dfrac{1}{x}-cosx+C\)

\(\int\dfrac{e^{2x}-1}{1-e^{-x}}dx=\int\dfrac{e^x\left(e^x-1\right)\left(e^x+1\right)}{e^x-1}dx\)

\(=\int e^x\left(e^x+1\right)dx=\int\left(e^{2x}+e^x\right)dx\)

\(=\dfrac{1}{2}e^{2x}+e^x+C\)

\(4x=3y\Rightarrow\dfrac{x}{3}=\dfrac{y}{4}\)

Áp dụng t/c dãy tỉ số bằng nhau:

\(\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{2x}{6}=\dfrac{3y}{12}=\dfrac{2x-3y}{6-12}=\dfrac{6}{-6}=-1\)

\(\Rightarrow\left\{{}\begin{matrix}x=3.\left(-1\right)=-3\\y=4.\left(-1\right)=-4\end{matrix}\right.\)

c.

Đặt \(\left\{{}\begin{matrix}a=y-xy=y\left(1-x\right)\\b=z\left(1-y\right)\\c=x\left(1-z\right)\end{matrix}\right.\) 

Ta có:

\(xyz+\left(1-x\right)\left(1-y\right)\left(1-z\right)\ge0\)

\(\Leftrightarrow1+xy+yz+zx-x-y-z\ge0\)

\(\Rightarrow x+y+z-xy-yz-zx\le1\)

Do đó:

\(ab+bc+ca=xy\left(1-x\right)\left(1-z\right)+yz\left(1-x\right)\left(1-y\right)+zx\left(1-y\right)\left(1-z\right)\)

\(\le\dfrac{1}{4}y\left(1-z\right)\left(x+1-x\right)+\dfrac{1}{4}z\left(1-x\right)\left(y+1-y\right)+\dfrac{1}{4}x\left(1-y\right)\left(z+1-z\right)\)

\(=\dfrac{1}{4}y\left(1-z\right)+\dfrac{1}{4}z\left(1-x\right)+\dfrac{1}{4}x\left(1-y\right)\)

\(=\dfrac{1}{4}\left(x+y+z-xy-yz-zx\right)\le\dfrac{1}{4}\)

\(\Rightarrow\dfrac{1}{2}\ge2\left(ab+bc+ca\right)\)

\(B=\left(1-a\right)^2+\left(1-b\right)^2+\left(1-c\right)^2\)

\(=3+a^2+b^2+c^2-2a-2b-2c\)

\(=\dfrac{5}{2}+\dfrac{1}{2}+a^2+b^2+c^2-2\left(a+b+c\right)\)

\(\ge\dfrac{5}{2}+2\left(ab+bc+ca\right)+a^2+b^2+c^2-2\left(a+b+c\right)\)

\(=\dfrac{3}{2}+\left(a+b+c-1\right)^2\ge\dfrac{3}{2}\)

b.

\(a^2b^2+b^2c^2+c^2a^2\ge abc\left(a+b+c\right)=3\)

Áp dụng BĐT chứng minh ở câu a:

\(A^2\ge\dfrac{\left(a^4+1\right)\left(b^4+1\right)\left(c^4+1\right)}{8}=\dfrac{\left(2a^4+2\right)\left(b^4c^4+b^4+c^4+1\right)}{16}\)

\(=\dfrac{\left(1+a^4+a^4+1\right)\left(b^4c^4+b^4+c^4+1\right)}{16}\)

\(\ge\dfrac{\left(b^2c^2+a^2b^2+a^2c^2+1\right)^2}{16}\ge\dfrac{\left(3+1\right)^2}{16}=1\)

\(\Rightarrow A\ge1\)

Nhiều bài nhìn ngán quá, chỗ này ngắn nhất thì làm vậy:

a.

\(2\left(x^2-x+1\right)^2=\left(x^4+1\right)+\left(x-1\right)^4\ge x^4+1\)

\(\Rightarrow x^2-x+1\ge\sqrt{\dfrac{x^4+1}{2}}\)

\(\Rightarrow\left(a^2-a+1\right)\left(b^2-b+1\right)\ge\dfrac{\sqrt{\left(a^4+1\right)\left(1+b^4\right)}}{2}\ge\dfrac{\sqrt{\left(a^2+b^2\right)^2}}{2}=\dfrac{a^2+b^2}{2}\)

\(cos\left(-x\right)=cosx\), ko phải \(cos\left(-x\right)=-cosx\)

27.

\(A=180^0-\left(B+C\right)=52^047'\)

\(\dfrac{a}{sinA}=\dfrac{c}{sinC}\Rightarrow c=\dfrac{a.sinC}{sinA}=19,9\)

28.

\(C=180^0-\left(A+B\right)=77^04'\)

\(\dfrac{AB}{sinC}=\dfrac{AC}{sinB}\Rightarrow AC=\dfrac{AB.sinB}{sinC}=68\)

25.

\(C=180^0-\left(A+B\right)=80^0\)

Áp dụng định lý sin:

\(\dfrac{BC}{sinA}=\dfrac{AB}{sinC}\Rightarrow BC=\dfrac{AB.sinA}{sinC}=\dfrac{5.sin40^0}{sin80^0}=3,3\)

26.

\(\dfrac{a}{sinA}=\dfrac{b}{sinB}=\dfrac{c}{sinC}=2R\Rightarrow\left\{{}\begin{matrix}a=2R.sinA\\b=2R.sinB\\c=2R.sinC\end{matrix}\right.\)

\(\Rightarrow2R.sinB+2R.sinC=2.2R.sinA\)

\(\Rightarrow sinB+sinC=2sinA\)