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a: Xét ΔBAE và ΔBDE có
BA=BD
\(\widehat{ABE}=\widehat{DBE}\)
BE chung
Do đó: ΔBAE=ΔBDE
b: ΔBAE=ΔBDE
=>\(\widehat{BAE}=\widehat{BDE}\)
=>\(\widehat{BDE}=90^0\)
=>DE\(\perp\)BC tại D
XétΔBHF vuông tại H và ΔBHC vuông tại H có
BH chung
\(\widehat{HBF}=\widehat{HBC}\)
Do đó ΔBHF=ΔBHC
c: Xét ΔBFC có
BH,CA là các đường cao
BH cắt CA tại E
Do đó: E là trực tâm của ΔBFC
=>FE\(\perp\)BC
mà DE\(\perp\)BC
và FE,DE có điểm chung là E
nên F,E,D thẳng hàng
\(d.\dfrac{59-x}{41}+\dfrac{57-x}{43}=\dfrac{41-x}{59}+\dfrac{43-x}{57}\\ \left(\dfrac{59-x}{41}+1\right)+\left(\dfrac{57-x}{43}+1\right)=\left(\dfrac{41-x}{59}+1\right)+\left(\dfrac{43-x}{57}+1\right)\\ \dfrac{100-x}{41}+\dfrac{100-x}{43}=\dfrac{100-x}{59}+\dfrac{100-x}{57}\\ \left(100-x\right)\left(\dfrac{1}{41}+\dfrac{1}{43}-\dfrac{1}{59}-\dfrac{1}{57}\right)=0\\ 100-x=0\\ x=100\)
bài 4:
\(C=\left(1+\dfrac{1}{1\cdot3}\right)\left(1+\dfrac{1}{2\cdot4}\right)\cdot...\cdot\left(1+\dfrac{1}{98\cdot100}\right)\)
\(=\left(1+\dfrac{1}{2^2-1}\right)\left(1+\dfrac{1}{3^2-1}\right)\cdot...\cdot\left(1+\dfrac{1}{99^2-1}\right)\)
\(=\dfrac{2^2}{2^2-1}\cdot\dfrac{3^2}{3^2-1}\cdot...\cdot\dfrac{99^2}{99^2-1}\)
\(=\dfrac{2\cdot3\cdot...\cdot99}{1\cdot2\cdot3\cdot...\cdot98}\cdot\dfrac{2\cdot3\cdot...\cdot99}{3\cdot4\cdot...\cdot100}=\dfrac{99}{1}\cdot\dfrac{2}{100}=\dfrac{99}{50}\)
Bài 5:
\(B=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{205}}{\dfrac{204}{1}+\dfrac{203}{2}+...+\dfrac{1}{204}}\)
\(=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{205}}{\left(1+\dfrac{203}{2}\right)+\left(1+\dfrac{202}{3}\right)+...+\left(\dfrac{1}{204}+1\right)+1}\)
\(=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{205}}{\dfrac{205}{2}+\dfrac{205}{3}+...+\dfrac{205}{205}}=\dfrac{1}{205}\)
Bài 5
Ta có:
\(x^2-x-6=\left(x-3\right)\left(x+2\right)\) và đa thức chia bậc 2 nên dư là \(ax+b\)
Vậy \(f\left(x\right)=\left(x-3\right)\left(x+2\right)\left(x^2+4\right)+ax+b\)
Theo định lí Bezout, dư trong phép chia \(f\left(x\right)\) cho \(x-3\) là \(f\left(3\right)=21\) cho \(x+2\) là \(f\left(-2\right)=4\) nên ta có: \(\left\{{}\begin{matrix}3a+b=21\\-2a+b=-4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=5\\b=6\end{matrix}\right.\)
Đa thức cần tìm là \(\left(x+2\right)\left(x-3\right)\left(x^2+4\right)+5x+6=x^4-x^3-2x^2+x-18\)
Bài 4:
\(2n^2+6n-7⋮n-2\)
=>\(2n^2-4n+10n-20+13⋮n-2\)
=>\(13⋮n-2\)
=>\(n-2\in\left\{1;-1;13;-13\right\}\)
=>\(n\in\left\{3;1;15;-11\right\}\)
\(a.\dfrac{\dfrac{3}{4}-\dfrac{3}{5}+\dfrac{3}{7}+\dfrac{3}{13}}{\dfrac{11}{4}-\dfrac{11}{5}+\dfrac{11}{7}+\dfrac{11}{13}}:\dfrac{\dfrac{3}{5}-\dfrac{3}{8}+\dfrac{3}{11}}{\dfrac{7}{5}-\dfrac{7}{9}+\dfrac{7}{11}}\\ =\dfrac{3\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{7}+\dfrac{1}{11}\right)}{11\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{7}+\dfrac{1}{11}\right)}:\dfrac{3\left(\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{11}\right)}{7\left(\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{11}\right)}\\ =\dfrac{3}{11}:\dfrac{3}{7}\\ =\dfrac{3}{11}\cdot\dfrac{7}{3}\\ =\dfrac{7}{11}\\ b.\dfrac{1}{1\cdot3}+\dfrac{1}{3\cdot5}+\dfrac{1}{5\cdot7}+...+\dfrac{1}{19\cdot21}\\ =\dfrac{1}{2}\cdot\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{19\cdot21}\right)\\ =\dfrac{1}{2}\cdot\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{19}-\dfrac{1}{21}\right)\\ =\dfrac{1}{2}\cdot\left(1-\dfrac{1}{21}\right)\\ =\dfrac{1}{2}\cdot\dfrac{20}{21}=\dfrac{10}{21}\)
AB//CD
=>\(y=\widehat{BDC}\)(hai góc so le trong)
=>\(y=45^0\)
AB//CD
=>\(x+100^0=180^0\)
=>\(x=80^0\)
\(x-y=80^0-45^0=35^0\)
xx'//yy'
=>\(\widehat{xAB}+\widehat{yBz}=180^0\)(hai góc trong cùng phía)
=>\(\widehat{yBz}+70^0=180^0\)
=>\(\widehat{yBz}=110^0\)
xx'//yy'
=>\(\widehat{xAB}=\widehat{yBz'}\)(hai góc đồng vị)
=>\(\widehat{yBz'}=70^0\)
Ta có: \(\widehat{MAB}=\widehat{ABC}\)
mà hai góc này là hai góc ở vị trí so le trong
nên MA//BC
Ta có: \(\widehat{NAC}=\widehat{ACB}\)
mà hai góc này là hai góc ở vị trí so le trong
nên NA//BC
Ta có: MA//BC
NA//BC
MA,NA có điểm chung là A
Do đó: M,A,N thẳng hàng
a: \(\left(-\dfrac{2}{5}-\dfrac{4}{3}+\dfrac{1}{4}\right)-\left(\dfrac{3}{5}-\dfrac{1}{3}-\dfrac{3}{4}\right)\)
\(=-\dfrac{2}{5}-\dfrac{4}{3}+\dfrac{1}{4}-\dfrac{3}{5}+\dfrac{1}{3}+\dfrac{3}{4}\)
\(=\left(-\dfrac{2}{5}-\dfrac{3}{5}\right)+\left(-\dfrac{4}{3}+\dfrac{1}{3}\right)+\left(\dfrac{1}{4}+\dfrac{3}{4}\right)\)
=-1-1+1=-1
b: \(\dfrac{2}{5}-\left(\dfrac{4}{3}+\dfrac{4}{5}\right)-\left(-\dfrac{1}{9}-0,4\right)+\dfrac{11}{9}\)
\(=\dfrac{2}{5}-\dfrac{4}{3}-\dfrac{4}{5}+\dfrac{1}{9}+\dfrac{2}{5}+\dfrac{11}{9}\)
\(=\left(\dfrac{2}{5}-\dfrac{4}{5}+\dfrac{2}{5}\right)+\left(-\dfrac{4}{3}+\dfrac{1}{9}+\dfrac{11}{9}\right)\)
\(=0+\left(-\dfrac{4}{3}+\dfrac{12}{9}\right)=0\)
c: \(\dfrac{11}{8}\cdot\left[\left(-\dfrac{5}{11}:\dfrac{13}{8}-\dfrac{5}{11}:\dfrac{13}{5}\right)+\dfrac{-6}{33}\right]+\dfrac{3}{4}\)
\(=\dfrac{11}{8}\cdot\left[-\dfrac{5}{11}\cdot\dfrac{8}{13}-\dfrac{5}{11}\cdot\dfrac{5}{13}-\dfrac{2}{11}\right]+\dfrac{3}{4}\)
\(=\dfrac{11}{8}\cdot\left[-\dfrac{5}{11}\left(\dfrac{8}{13}+\dfrac{5}{13}\right)-\dfrac{2}{11}\right]+\dfrac{3}{4}\)
\(=\dfrac{11}{8}\cdot\left(-\dfrac{5}{11}-\dfrac{2}{11}\right)+\dfrac{3}{4}=\dfrac{-7}{8}+\dfrac{3}{4}=-\dfrac{1}{8}\)
d: \(A=\dfrac{4}{9}:\left(\dfrac{1}{15}-\dfrac{2}{3}\right)+\dfrac{4}{9}:\left(\dfrac{1}{11}-\dfrac{5}{22}\right)\)
\(=\dfrac{4}{9}:\left(\dfrac{1}{15}-\dfrac{10}{15}\right)+\dfrac{4}{9}:\left(\dfrac{2}{22}-\dfrac{5}{22}\right)\)
\(=\dfrac{4}{9}:\dfrac{-9}{15}+\dfrac{4}{9}:\dfrac{-3}{22}\)
\(=\dfrac{4}{9}\cdot\dfrac{-5}{3}+\dfrac{4}{9}\cdot\dfrac{-22}{3}=\dfrac{4}{9}\cdot\left(-\dfrac{5}{3}-\dfrac{22}{3}\right)=\dfrac{4}{9}\left(-9\right)=-4\)
a: Ta có: \(\widehat{xOy}=\widehat{mOn}\)(hai góc đối đỉnh)
mà \(\widehat{xOy}=50^0\)
nên \(\widehat{mOn}=50^0\)
Ta có: \(\widehat{xOy}+\widehat{mOy}=180^0\)(hai góc kề bù)
=>\(\widehat{mOy}+50^0=180^0\)
=>\(\widehat{mOy}=130^0\)
Ta có: \(\widehat{xOn}=\widehat{mOy}\)(hai góc đối đỉnh)
mà \(\widehat{mOy}=130^0\)
nên \(\widehat{xOn}=130^0\)
b: Oa là phân giác của góc xOy
=>\(\widehat{yOa}=\dfrac{\widehat{xOy}}{2}=25^0\)
Ta có: Ob là phân giác của góc yOm
=>\(\widehat{yOb}=\dfrac{\widehat{yOm}}{2}=65^0\)
Ta có: \(\widehat{aOb}=\widehat{aOy}+\widehat{bOy}=25^0+65^0=90^0\)