Chọn B

Đáp án C

Đặt  u = ln x 2 + 9 d v = x d x ⇒ d u = 2 x x 2 + 9 d x v = x 2 + 9 2 ⇒ ∫ 0 4 x ln x 2 + 9 d x = x 2 + 9 2 ln x 2 + 9 − 0 4 ∫ 0 4 x d x

= 25 ln 5 − 9 ln 3 − 8 ⇒ a = 25 ; b = − 9 ; c = − 8 ⇒ a + b + c = 8.

Đáp án B

Đáp án B

1/\(=4a^2+4b^2+c^2+8ab-4bc-4ca+4b^2+4c^2+a^2+8bc-4ca-4ab+4a^2+4c^2+b^2+8ca-4bc-4ab=\)

\(=9a^2+9b^2+9c^2=9\left(a^2+b^2+c^2\right)\)

2/

Ta có

\(\left(a+b+c\right)^2=a^2+b^2+c^2+2\left(ab+bc+ca\right)\ge0\)

\(\Leftrightarrow a^2+b^2+c^2\ge-2\left(ab+bc+ca\right)=2\)

\(\Rightarrow P=9\left(a^2+b^2+c^2\right)\ge18\)

\(\Rightarrow P_{min}=18\)

Ta có : \(b=\dfrac{c+a}{2}\Rightarrow2b=c+a\Rightarrow a-b=b-c\)

Dó đó : \(P=\left(\dfrac{1}{\sqrt{a}+\sqrt{b}}+\dfrac{1}{\sqrt{b}+\sqrt{c}}\right)\left(\sqrt{a}+\sqrt{c}\right)\)

\(P=\left[\dfrac{\sqrt{a}-\sqrt{b}}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}+\dfrac{\sqrt{b}-\sqrt{c}}{\left(\sqrt{b}+\sqrt{c}\right)\left(\sqrt{b}-\sqrt{c}\right)}\right]\left(\sqrt{a}+\sqrt{c}\right)\)

\(P=\left[\dfrac{\sqrt{a}-\sqrt{b}}{a-b}+\dfrac{\sqrt{b}-\sqrt{c}}{b-c}\right]\left(\sqrt{a}+\sqrt{c}\right)\)

\(P=\left[\dfrac{\sqrt{a}-\sqrt{b}}{b-c}+\dfrac{\sqrt{b}-\sqrt{c}}{b-c}\right]\left(\sqrt{a}+\sqrt{c}\right)\) Vì  \(\left(a-b=b-c\right)\)

\(P=\left[\dfrac{\sqrt{a}-\sqrt{b}+\sqrt{b}-\sqrt{c}}{b-c}\right]\left(\sqrt{a}+\sqrt{c}\right)\)

\(P=\dfrac{\sqrt{a}-\sqrt{c}}{b-c}\left(\sqrt{a}+\sqrt{c}\right)\)

\(P=\dfrac{a-c}{a-b}=\dfrac{a-c}{a-\dfrac{a+c}{2}}=\dfrac{a-c}{\dfrac{2a-a-c}{2}}=\dfrac{a-c}{\dfrac{a-c}{2}}=2\)

a) phương trình \(x^3-3x^2+1\) có 3 nghiệm thực phân biệt là a,b,c(đề bài). Áp dụng Định lí Vi-ét cho đa thức bậc 3 ta có:\(\left\{{}\begin{matrix}a+b+c=3\\ab+bc+ac=0\\a.b.c=-1\end{matrix}\right.\)

ta có

      a+b+c=3

<=>\(\left(a+b+c\right)^2=9\)

<=>\(a^2+b^2+c^2+2ab+2bc+2ac=9\)

<=>\(a^2+b^2+c^2=9\)

<=>\(\left(a^2+b^2+c^2\right)^2=81\)

<=>\(a^4+b^4+c^4+2\left(a^2b^2+b^2c^2+a^2c^2\right)=81\)(1)

ta có ab+bc+ac=0

   <=>\(\left(ab+bc+ac\right)^2=0\)

   <=>\(a^2b^2+b^2c^2+a^2c^2+2abc\left(a+b+c\right)=0\)

   <=>\(a^2b^2+b^2c^2+a^2c^2-2.1.3=0\)

   <=>\(a^2b^2+b^2c^2+a^2c^2=6\)(2)

Thay (2) vào (1) ta có \(a^4+b^4+c^4+2.6=81\)

                                <=>\(a^4+b^4+c^4=69\)

b) \(\dfrac{a+1}{\left(b+c\right)\left(1-a\right)+1}=\dfrac{a+1}{\left(3-a\right)\left(1-a\right)+1}=\dfrac{a+1}{3+a^2-4a+1}=\dfrac{a+1}{a^2-4a+4}=\dfrac{a+1}{\left(a-2\right)^2}\)

cmtt =>\(B=\dfrac{a+1}{\left(a-2\right)^2}+\dfrac{b+1}{\left(b-2\right)^2}+\dfrac{c+1}{\left(c-2\right)^2}\)=\(\dfrac{1}{a-2}+\dfrac{1}{b-2}+\dfrac{1}{c-2}+3\left[\dfrac{1}{\left(a-2\right)^2}+\dfrac{1}{\left(b-2\right)^2}+\dfrac{1}{\left(c-2\right)^2}\right]\)=\(\dfrac{3\left[\left(a-2\right)\left(b-2\right)\right]^2+3\left[\left(b-2\right)\left(c-a\right)\right]^2+3\left[\left(c-2\right)\left(a-2\right)\right]^2}{\left[\left(a-2\right)\left(b-2\right)\left(c-2\right)\right]^2}\)

đặt t=(a-2)(b-2);u=(b-2)(c-2);v=(c-2)(a-2)     =>t+u+v=0

B thành \(\dfrac{3\left(t^2+u^2+v^2\right)}{t.u.v}\) bạn biến đổi để xuất hiện t+u+v

=>B=\(\dfrac{3\left(t+u+v\right)^2-6\left(t.u+u.v+t.v\right)}{t.u.v}=\dfrac{-6.\left(a-2\right)\left(b-2\right)\left(c-2\right)\left(a-2+b-2+c-2\right)}{t.u.v}=\dfrac{18}{\left(a-2\right)\left(b-2\right)\left(c-2\right)}\)

(a-2)(b-2)(c-2)= abc-2(ab+bc+ac)+4(a+b+c)-8=12-9=3

Vậy B=3

c) ta có \(\dfrac{a^3}{a^2+2bc}=\dfrac{a^3}{a^2-2ac-2ab}=\dfrac{a^2}{a-2c-2b}=\dfrac{a^2}{3a-2\left(a+b+c\right)}=\dfrac{a^2}{3\left(a-2\right)}\)

cmtt =>C=\(\dfrac{a^2}{3\left(a-2\right)}+\dfrac{b^2}{3\left(b-2\right)}+\dfrac{c^2}{3\left(c-2\right)}=\dfrac{a^2\left(b-2\right)\left(c-2\right)+b^2\left(a-2\right)\left(c-2\right)+c^2\left(a-2\right)\left(b-2\right)}{3\left(a-2\right)\left(b-2\right)\left(c-2\right)}\)

bạn nhân vô thì ra C=\(\dfrac{4a^2-2a\left(ab+ac\right)-a+4b^2-2b\left(bc+ab\right)-b+4c^2-2c\left(ac+bc\right)-c}{3\left(a-2\right)\left(b-2\right)\left(c-2\right)}=\dfrac{ }{ }4\dfrac{ }{ }=\dfrac{4\left(a^2+b^2+c^2\right)-\left(a+b+c\right)+6abc}{3\left(a-2\right)\left(b-2\right)\left(c-2\right)}=\dfrac{4.9-3-6}{3.3}=\dfrac{27}{9}=3\)

\(I_1=\int\limits^2_0f\left(2x\right)dx\)

Đặt \(2x=t\Rightarrow dx=\frac{dt}{2}\) ; \(\left\{{}\begin{matrix}x=0\Rightarrow t=0\\x=2\Rightarrow t=4\end{matrix}\right.\)

\(\Rightarrow I_1=\int\limits^4_0f\left(t\right).\frac{dt}{2}=\frac{1}{2}\int\limits^4_0f\left(t\right)dt=\frac{1}{2}\int\limits^4_0f\left(x\right)dx=\frac{1}{2}.2018=1009\)

\(I_2=\int\limits^2_{-2}f\left(2-x\right)dx\)

Đặt \(2-x=t\Rightarrow dx=-dt\); \(\left\{{}\begin{matrix}x=-2\Rightarrow t=4\\x=2\Rightarrow t=0\end{matrix}\right.\)

\(\Rightarrow I_2=\int\limits^0_4f\left(t\right).\left(-dt\right)=\int\limits^4_0f\left(t\right)dt=\int\limits^4_0f\left(x\right)dx=2018\)

\(\Rightarrow I=I_1+I_2=1009+2018=3027\)

Đáp án B

\(a^2+b^2+c^2=ab+bc+ca\Rightarrow2a^2+2b^2+2c^2=2ab+2bc+2ca\)

\(\Rightarrow\left(2a^2+2b^2+2c^2\right)-\left(2ab+2bc+2ca\right)=0\)

\(\Rightarrow\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ca+a^2\right)=0\)

\(\Rightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)

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

\(\Rightarrow P=\left(a-b\right)^{2015}+\left(b-c\right)^{2016}+\left(c-a\right)^{2017}=0\)

cảm ơn bạn nha