Bài 1:
Ta có:
\(A=\dfrac{x}{\left(\sqrt{x}+\sqrt{y}\right)\left(1-\sqrt{y}\right)}-\dfrac{y}{\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}+1\right)}-\dfrac{xy}{\left(\sqrt{x}+1\right)\left(1-\sqrt{y}\right)}\)
\(A=\dfrac{x\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+\sqrt{y}\right)\left(1-\sqrt{y}\right)\left(\sqrt{x}+1\right)}-\dfrac{y\left(1-\sqrt{y}\right)}{\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}+1\right)\left(1-\sqrt{y}\right)}-\dfrac{xy\left(\sqrt{x}+\sqrt{y}\right)}{\left(\sqrt{x}+1\right)\left(1-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}\)
\(A=\dfrac{x\left(\sqrt{x}+1\right)-y\left(1-\sqrt{y}\right)-xy\left(\sqrt{x}+\sqrt{y}\right)}{\left(\sqrt{x}+\sqrt{y}\right)\left(1-\sqrt{y}\right)\left(\sqrt{x}+1\right)}\)
\(A=\dfrac{x\sqrt{x}+x-y+y\sqrt{y}-xy\sqrt{x}-xy\sqrt{y}}{\left(\sqrt{x}+\sqrt{y}\right)\left(1-\sqrt{y}\right)\left(\sqrt{x}+1\right)}\)
\(A=\dfrac{\left(\sqrt{x}\right)^3+\left(\sqrt{x}\right)^2-\left(\sqrt{y}\right)^2+(\sqrt{y})^3-(xy\sqrt{x}+xy\sqrt{y})}{\left(\sqrt{x}+\sqrt{y}\right)\left(1-\sqrt{y}\right)\left(\sqrt{x}+1\right)}\)
\(A=\dfrac{[\left(\sqrt{x}\right)^3+\left(\sqrt{y}\right)^3]+[\left(\sqrt{x}\right)^2-\left(\sqrt{y}\right)^2]-(xy\sqrt{x}+xy\sqrt{y})}{\left(\sqrt{x}+\sqrt{y}\right)\left(1-\sqrt{y}\right)\left(\sqrt{x}+1\right)}\)
\(A=\dfrac{\left(\sqrt{x}+\sqrt{y}\right)[\left(\sqrt{x}\right)^2-\sqrt{xy}+\left(\sqrt{y}\right)^2]+\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)-xy(\sqrt{x}+\sqrt{y})}{\left(\sqrt{x}+\sqrt{y}\right)\left(1-\sqrt{y}\right)\left(\sqrt{x}+1\right)}\)
\(A=\dfrac{\left(\sqrt{x}+\sqrt{y}\right)[\left(\sqrt{x}\right)^2-\sqrt{xy}+\left(\sqrt{y}\right)^2+\sqrt{x}-\sqrt{y}-xy]}{\left(\sqrt{x}+\sqrt{y}\right)\left(1-\sqrt{y}\right)\left(\sqrt{x}+1\right)}\)
\(A=\dfrac{\left(\sqrt{x}\right)^2-\sqrt{xy}+\left(\sqrt{y}\right)^2+\sqrt{x}-\sqrt{y}-xy}{\left(1-\sqrt{y}\right)\left(\sqrt{x}+1\right)}\)
\(A=\dfrac{x-\sqrt{xy}+y+\sqrt{x}-\sqrt{y}-xy}{\left(1-\sqrt{y}\right)\left(\sqrt{x}+1\right)}\)
\(A=\dfrac{(x-xy)+\left(\sqrt{x}-\sqrt{xy}\right)-(\sqrt{y}-y)}{\left(1-\sqrt{y}\right)\left(\sqrt{x}+1\right)}\)
\(A=\dfrac{x(1-y)+\sqrt{x}\left(1-\sqrt{y}\right)-\sqrt{y}(1-\sqrt{y})}{\left(1-\sqrt{y}\right)\left(\sqrt{x}+1\right)}\)
\(A=\dfrac{x(1-\sqrt{y})\left(1+\sqrt{y}\right)+\sqrt{x}\left(1-\sqrt{y}\right)-\sqrt{y}(1-\sqrt{y})}{\left(1-\sqrt{y}\right)\left(\sqrt{x}+1\right)}\)
\(A=\dfrac{(1-\sqrt{y})\left(x+x\sqrt{y}\right)+\sqrt{x}\left(1-\sqrt{y}\right)-\sqrt{y}(1-\sqrt{y})}{\left(1-\sqrt{y}\right)\left(\sqrt{x}+1\right)}\)
\(A=\dfrac{(1-\sqrt{y})\left(x+x\sqrt{y}+\sqrt{x}-\sqrt{y}\right)}{\left(1-\sqrt{y}\right)\left(\sqrt{x}+1\right)}\)
\(A=\dfrac{\left(x+x\sqrt{y}+\sqrt{x}-\sqrt{y}\right)}{\left(\sqrt{x}+1\right)}\)
\(A=\dfrac{\left(x+\sqrt{x}\right)+\left(x\sqrt{y}-\sqrt{y}\right)}{\left(\sqrt{x}+1\right)}\)
\(A=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)+\sqrt{y}\left(x-1\right)}{\left(\sqrt{x}+1\right)}\)
\(A=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)+\sqrt{y}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)}\)
\(A=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)+\left(\sqrt{xy}-\sqrt{y}\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)}\)
\(A=\dfrac{(\sqrt{x}+\sqrt{xy}-\sqrt{y})\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)}\)
\(A=(\sqrt{x}+\sqrt{xy}-\sqrt{y})\)
Vậy \(A=(\sqrt{x}+\sqrt{xy}-\sqrt{y})\)
Bài 2:
a, \(\dfrac{x}{\left(a-b\right)\left(a-c\right)}+\dfrac{x}{\left(b-a\right)\left(b-c\right)}+\dfrac{x}{\left(c-a\right)\left(c-b\right)}=2\)
\(\Leftrightarrow\)\(x(\dfrac{1}{\left(a-b\right)\left(a-c\right)}+\dfrac{1}{\left(b-a\right)\left(b-c\right)}+\dfrac{1}{\left(c-a\right)\left(c-b\right)})=2\)
\(\Leftrightarrow\)\(x(\dfrac{1}{\left(a-b\right)\left(a-c\right)}-\dfrac{1}{\left(a-b\right)\left(b-c\right)}+\dfrac{1}{\left(a-c\right)\left(b-c\right)})=2\)
\(\Leftrightarrow\)\(x(\dfrac{b-c}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}-\dfrac{a-c}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}+\dfrac{a-b}{\left(a-c\right)\left(b-c\right)\left(a-b\right)})=2\)
\(\Leftrightarrow\)\(x.0=2\)
Vậy phương trình vô nghiệm
b,\(\dfrac{x}{\left(a-b\right)\left(a-c\right)}-\dfrac{2x}{\left(a-b\right)\left(a-d\right)}+\dfrac{3x}{\left(a-c\right)\left(a-d\right)}=\dfrac{4a}{\left(a-c\right)\left(a-d\right)}\)
\(\Leftrightarrow\dfrac{x\left(a-d\right)}{\left(a-b\right)\left(a-c\right)\left(a-d\right)}-\dfrac{2x\left(a-c\right)}{\left(a-b\right)\left(a-d\right)\left(a-c\right)}+\dfrac{3x\left(a-b\right)}{\left(a-c\right)\left(a-d\right)\left(a-b\right)}=\dfrac{4a\left(a-b\right)}{\left(a-c\right)\left(a-d\right)\left(a-b\right)}\)
\(\Leftrightarrow x\left(a-d\right)-2x\left(a-c\right)+3x\left(a-b\right)=4a\left(a-b\right)\)
\(\Leftrightarrow x\left[a-d-2a+2c+3a-3b\right]=4a\left(a-b\right)\)
\(\Leftrightarrow x\left[2a-3b+2c-d\right]=4a\left(a-b\right)\)
\(\Leftrightarrow x\left[2a-3b+b+d-d\right]=4a\left(a-b\right)\) (vì b+d=2c)
\(\Leftrightarrow x\left[2a-2b\right]=4a\left(a-b\right)\)
\(\Leftrightarrow2x\left(a-b\right)=4a\left(a-b\right)\)
\(\Leftrightarrow2x=4a\) (vì \(a\ne b\) nên \(a-b\ne0\) )
\(\Leftrightarrow x=2a\)
Vậy \(x=2a\) là nghiệm của phương trình
Bài 3:
Gọi vận tốc của xuồng máy là x km/h , vận tốc của dòng nước là y km/h ( x > y >0 )
Vì bè trôi trôi tự do nên vận tốc của bè trôi = vận tốc của nước.
Vận tốc xuồng máy khi xuôi dòng là: \(x+y\) (km/h)
Vận tốc xuồng máy khi ngược dòng là: \(x-y\) (km/h)
Do thời gian cả đi và về của xuồng máy là 7h nên ta có phương trình:
\(\dfrac{48}{x+y}\)+ \(\dfrac{48}{x-y}\) = 7 (1)
Thời gian xuồng máy đi xuôi dòng 48 km và ngược dòng 36kn là:
\(\dfrac{48}{x+y}\)+\(\dfrac{36}{x-y}\)
Thời gian bè trôi trôi 12km là: \(\dfrac{12}{y}\left(h\right)\)
Hai thời gian bàng nhau nên ta có phương trình:
\(\dfrac{48}{x+y}\)+ \(\dfrac{36}{x-y}\) = \(\dfrac{12}{y}\)
\(\Leftrightarrow\dfrac{4}{x+y}+\dfrac{3}{x-y}=\dfrac{1}{y}\)
\(\Leftrightarrow\dfrac{4x-4y+3x+3y}{x^2-y^2}=\dfrac{1}{y}\)
\(\Leftrightarrow\dfrac{7x-y}{x^2-y^2}=\dfrac{1}{y}\)
\(\Leftrightarrow7xy-y^2=x^2-y^2\)
\(\Leftrightarrow7xy=x^2\)
\(\Leftrightarrow7y=x\) ( Vì x > 0) (2)
Thay (2) vào (1) ta được:
\(\dfrac{48}{7y+y}+\dfrac{48}{7y-y}=7\)
\(\Leftrightarrow\) \(\dfrac{48}{8y}+\dfrac{48}{6y}=7\)
\(\Leftrightarrow\) \(\dfrac{6}{y}+\dfrac{8}{y}=7\)
\(\Leftrightarrow\)\(\dfrac{14}{y}=7\)
\(\Leftrightarrow\) y = 2
Khi đó:
\(x=7y=14\)
Vậy vận tốc riêng của xuồng máy là 14 km/h, của dòng nước là 2 km/h.
Bài 4:
:
Gọi giao điểm của AN và BC là E
Xét \(\Delta\)ABC có AE, BM, CD đồng quy nên theo định lý Xê - va ta có:
\(\dfrac{AM}{MC}.\dfrac{CE}{EB}.\dfrac{BD}{AD}=1\)
\(\Rightarrow\dfrac{CE}{EB}.\dfrac{BD}{AD}=1\) ( Vì \(\dfrac{AM}{MC}=1\) )
\(\Rightarrow\dfrac{CE}{EB}=\dfrac{AD}{BD}\) (1)
Xét \(\Delta\)ABC có phân giác CD
\(\Rightarrow\dfrac{AD}{BD}=\dfrac{AC}{BC}\) (2)
Từ (1), (2)
=> \(\dfrac{CE}{EB}=\dfrac{AC}{BC}\)
=> \(\dfrac{BE}{EC}=\dfrac{BC}{AC}\)
Vì \(\dfrac{AD}{BD}=\dfrac{AC}{BC}\)
\(\Rightarrow\dfrac{AD}{AD+BD}=\dfrac{AC}{BC+AC}\)
\(\Rightarrow\dfrac{AD}{AB}=\dfrac{AC}{BC+AC}\)
Xét \(\Delta\) BDC có A\(\in\) đường thẳng BD, N\(\in\) đường thẳng DC, E \(\in\) đường thẳng BC, A,N,E thẳng hàng nên theo định lý Mê - nê - la - uýt ta có:
\(\dfrac{CN}{ND}.\dfrac{AD}{AB}.\dfrac{BE}{CE}=1\)
hay \(\dfrac{CN}{ND}.\dfrac{AC}{BC+AC}.\dfrac{BC}{AC}=1\)
\(\Rightarrow\dfrac{CN}{ND}.\dfrac{BC}{BC+AC}=1\)
\(\Rightarrow\dfrac{CN}{ND}=\dfrac{BC+AC}{BC}\)
\(\Rightarrow\dfrac{CN}{ND}=1+\dfrac{AC}{BC}\)
\(\Rightarrow\dfrac{CN}{ND}-\dfrac{AC}{BC}=1\) (đpcm)