Vòng 2

Bài 1:

ĐKXĐ \(\left\{{}\begin{matrix}x,y>0\\x,y\ne1\end{matrix}\right.\)

Đặt \(a=\sqrt{x};b=\sqrt{y}\). Ta có:

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

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

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

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

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

\(=\dfrac{\left(a+b\right)\left[a^2+b^2-ab+\left(a-b\right)-\left(ab\right)^2\right]}{\left(a+b\right)\left(a+1\right)\left(1-b\right)}\)

\(=\dfrac{\left[a^2-\left(ab\right)^2+b^2-1\right]+\left(a-ab-b+1\right)}{\left(a+1\right)\left(1-b\right)}\)

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

\(=\dfrac{\left(a+1\right)\left(a-1\right)\left(1+b\right)\left(1-b\right)+\left(a+1\right)\left(1-b\right)}{\left(a+1\right)\left(1-b\right)}\)

\(=\dfrac{\left(a+1\right)\left(1-b\right)\left[\left(a-1\right)\left(b+1\right)+1\right]}{\left(a+1\right)\left(1-b\right)}\)

\(=\left(a+1\right)\left(b-1\right)+1=\left(\sqrt{x}+1\right)\left(\sqrt{y}-1\right)+1\)

Bài 2:

a) \(\dfrac{x}{\left(a-b\right)\left(a-c\right)}+\dfrac{x}{\left(b-a\right)\left(b-c\right)}+\dfrac{x}{\left(c-a\right)\left(c-b\right)}=2\)

\(\Rightarrow\dfrac{x}{\left(a-b\right)\left(a-c\right)}-\dfrac{x}{\left(a-b\right)\left(b-c\right)}+\dfrac{x}{\left(a-c\right)\left(b-c\right)}=2\)

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

\(\Rightarrow\dfrac{x}{a-b}\cdot\dfrac{b-a}{\left(a-c\right)\left(b-c\right)}+\dfrac{x}{\left(a-c\right)\left(b-c\right)}=2\)

\(\Rightarrow-\dfrac{x}{a-b}\cdot\dfrac{a-b}{\left(a-c\right)\left(b-c\right)}+\dfrac{x}{\left(a-c\right)\left(b-c\right)}=2\)

\(\Rightarrow-\dfrac{x}{\left(a-c\right)\left(b-c\right)}+\dfrac{x}{\left(a-c\right)\left(b-c\right)}=2\)

\(\Rightarrow0=2\) (vô lí)   

Vậy \(x\in\varnothing\)

b, 

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

\(\Rightarrow\dfrac{x}{a-b}\left(\dfrac{1}{a-c}-\dfrac{2}{a-d}\right)=\dfrac{4a-3x}{\left(a-c\right)\left(a-d\right)}\)

\(\Rightarrow\dfrac{x}{a-b}\cdot\dfrac{a-d-2\left(a-c\right)}{\left(a-c\right)\left(a-d\right)}=\dfrac{4a-3x}{\left(a-c\right)\left(a-d\right)}\)

\(\Rightarrow\dfrac{x}{a-b}\cdot\dfrac{b-a}{\left(a-c\right)\left(a-d\right)}=\dfrac{4a-3x}{\left(a-c\right)\left(a-d\right)}\)

\(\Rightarrow-\dfrac{x}{a-b}\cdot\dfrac{a-b}{\left(a-c\right)\left(a-d\right)}=\dfrac{4a-3x}{\left(a-c\right)\left(a-d\right)}\)

\(\Rightarrow\dfrac{-x}{\left(a-c\right)\left(a-d\right)}=\dfrac{4a-3x}{\left(a-c\right)\left(a-d\right)}\)

\(\Rightarrow-x=4a-3x\Rightarrow2x=4a\)

\(\Rightarrow x=2a\)

Bài 3:

Gọi vận tốc bè trôi = vận tốc dòng nước = a (km/h); vận tốc xuồng là x (km/h) 

\(\Rightarrow\left\{{}\begin{matrix}V_{xuôi}=x+a\\V_{nguoc}=x-a\end{matrix}\right.\)

Theo đề bài ta có:

\(\left\{{}\begin{matrix}\dfrac{48}{x+a}+\dfrac{48}{x-a}=7\left(\text{*}\right)\\\dfrac{48}{x+a}+\dfrac{48-12}{x-a}=\dfrac{12}{a}\left(1\right)\end{matrix}\right.\)

Từ \(\left(1\right)\Rightarrow\dfrac{48}{x+a}+\dfrac{36}{x-a}=\dfrac{12}{a}\)

\(\Rightarrow\dfrac{48x-48a+36x+36a}{x^2-a^2}=\dfrac{12}{a}\)

\(\Rightarrow\dfrac{84x-12a}{x^2-a^2}=\dfrac{12}{a}\)

\(\Rightarrow84ax-12a^2=12x^2-12a^2\)

\(\Rightarrow84ax=12x^2\Leftrightarrow7a=x\) 

Thay x = 7a vao (*). Ta có:

\(\dfrac{48}{7a+a}+\dfrac{48}{7a-a}=7\)

\(\Rightarrow\dfrac{48}{8a}+\dfrac{48}{6a}=7\)

\(\Rightarrow\dfrac{6+8}{a}=7\Rightarrow a=\dfrac{14}{7}=2\)

\(\Rightarrow x=2\cdot7=14\)

Vậy vận tốc dòng nước là 2km/h, vận tốc riêng của xuồng là 14km/h

Bài 4:

A B C D M N G

Vẽ DG // AC

Ta có: \(\dfrac{AC}{BC}=\dfrac{AD}{BD}\) (tính chất tia phân giác) 

\(\Rightarrow1+\dfrac{AC}{BC}=1+\dfrac{AD}{BD}=\dfrac{AB}{BD}\)

Dựa vào định lí Talet, ta có:

\(\dfrac{AB}{BD}=\dfrac{AM}{DG}=\dfrac{MC}{DG}\)

Ta lại có: \(\left\{{}\begin{matrix}\widehat{MCN}=\widehat{NDG}\left(so-le-trong\right)\\\widehat{AMN}=\widehat{DGB}\left(dong-vi\right)\end{matrix}\right.\Rightarrow\widehat{DGN}=\widehat{NMC}\)

\(\Rightarrow\Delta NDG\) ~ \(\Delta NCM\left(g-g-g\right)\)

\(\Rightarrow\dfrac{NC}{CM}=\dfrac{ND}{DG}\Rightarrow\dfrac{NC}{ND}=\dfrac{CM}{DG}\)

\(\Rightarrow\dfrac{NC}{ND}=1+\dfrac{AC}{BC}\) \(\Leftrightarrow\dfrac{NC}{ND}-\dfrac{AC}{BC}=1\) (đpcm) 

Bài 5:

\(\left\{{}\begin{matrix}b\ge5\\c\ge5\end{matrix}\right.\Rightarrow b^2+c^2\ge50\Rightarrow a^2\le19\Rightarrow a\le4\)

\(\Rightarrow2a-8\le0\)

\(\left\{{}\begin{matrix}a\ge2\\b\ge5\end{matrix}\right.\Rightarrow a^2+b^2\ge29\Rightarrow c^2\le40\Rightarrow c\le6\)

\(\Rightarrow c-6\le0\)

Tương tự ta có \(b\le6< 8\)

\(\Rightarrow c-8< 0\)

\(\Rightarrow\left\{{}\begin{matrix}\left(a-2\right)\left(2a-8\right)\le0\\\left(b-5\right)\left(b-8\right)\le0\\\left(c-5\right)\left(c-6\right)\le0\end{matrix}\right.\)

\(\Rightarrow\left(a-2\right)\left(2a-8\right)+\left(b-5\right)\left(b-8\right)+\left(c-5\right)\left(c-6\right)\le0\)

\(\Rightarrow2a^2-12a+16+b^2-13b+40+c^2-11c+30\le0\)

\(\Rightarrow\left(2a^2+b^2+c^2\right)+\left(16+40+30\right)-\left(12a+13b+11c\right)\le0\)

\(\Rightarrow69+86-\left(12a+13b+11c\right)\le0\)

\(\Rightarrow155-\left(12a+13b+11c\right)\le0\)

\(\Rightarrow12a+13b+11c\ge155\)

=> GTNN của P = 155 tại \(\left\{{}\begin{matrix}2\left(a-2\right)\left(a-4\right)=0\\\left(b-5\right)\left(b-8\right)=0\\\left(c-5\right)\left(c-6\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}a-2=0\\a-4=0\end{matrix}\right.\\\left[{}\begin{matrix}b-5=0\\b-8=0\end{matrix}\right.\\\left[{}\begin{matrix}c-5=0\\c-6=0\end{matrix}\right.\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}a=2\\a=4\end{matrix}\right.\\\left[{}\begin{matrix}b=5\\b=8\end{matrix}\right.\\\left[{}\begin{matrix}c=5\\c=6\end{matrix}\right.\end{matrix}\right.\)

Nếu a = 4 => b2 + c2 = 53 \(\Rightarrow b^2\le53\Rightarrow b\le7\) => b = 5 => c2 = 69 - 2.42 - 52 = 12 => c khác 5 và 6 (loại)

Vậy a = 2

a = 2 => b2 + c2 = 61 => \(b^2\le61\Rightarrow b\le7\) => b = 5 => c2 = 69 - 2.22 - 52 = 36 => c = 6 (chọn) 

Ta được (a,b,c) = (2,5,6) 

Kết luận: GTNN của P = 155 khi và chỉ khi a = 2, b = 5, c = 6

Điểm  18

Nhận xét: Bài 1 lộn dấu dẫn đến kết quả sai hoàn toàn (-2đ ) . Bài 3 thiếu đk (-0,5đ ) . +0,5đ ở vòng 1