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Câu 4:
1. Hiển nhiên $AD\parallel BC$. Áp dụng định lý Talet:
$\frac{BM}{AN}=\frac{PM}{PN}$
$\frac{CM}{NE}=\frac{PM}{PN}$
$\Rightarrow \frac{BM}{AN}=\frac{CM}{NE}$. Mà $BM=CM$ do $M$ là trung điểm $BC$ nên $AN=NE$. $N$ thì nằm giữa $A,E$ (dễ cm)
Do đó $N$ là trung điểm $AE$
2.
Xét tam giác $ABC$ và $DCA$ có:
$\widehat{ABC}=\widehat{DCA}=90^0$
$\widehat{BCA}=\widehat{CAD}$ (so le trong)
$\Rightarrow \triangle ABC\sim \triangle DCA$ (g.g)
3. Theo định lý Pitago:
Từ tam giác đồng dạng phần 2 suy ra:
$\frac{AC}{DA}=\frac{BC}{CA}$
$\Rightarrow AD=\frac{AC^2}{BC}=\frac{6^2}{4}=9$ (cm)
4,Theo phần 1 thì:
$\frac{PM}{PN}=\frac{BM}{AN}=\frac{CM}{AN}$
Mà cũng theo định lý Talet: $\frac{CM}{AN}=\frac{QM}{QN}$
$\Rightarrow \frac{PM}{PN}=\frac{QM}{QN}$
(đpcm)
Bài 3:
Xét ΔIAB có
\(\widehat{AIB}+\widehat{IAB}+\widehat{IBA}=180^0\)
\(\Leftrightarrow\widehat{IAB}+\widehat{IBA}=115^0\)
hay \(\widehat{DAB}+\widehat{ABC}=230^0\)
Xét tứ giác ABCD có
\(\widehat{D}+\widehat{C}+\widehat{DAB}+\widehat{CBA}=360^0\)
\(\Leftrightarrow\widehat{D}+\widehat{C}=150^0\)
mà \(\widehat{C}-\widehat{D}=10^0\)
nên \(2\cdot\widehat{C}=160^0\)
\(\Leftrightarrow\widehat{C}=80^0\)
\(\Leftrightarrow\widehat{D}=70^0\)
Ta có 2003.2005=2003.(2004+1)=2003.2004+2003
2004^2=2004.2004=2004.(2003+1)=2003.2004+2004
Vì 2003<2004 nên 2003.2004+2003<2003.2004+2004
Vậy 2003.2005<2004^2
Ta có A=2003.2005=2003.(2004+1)=2003.2004+2003A=2003.2005=2003.(2004+1)=2003.2004+2003
B=20042=2004.2004=2004.(2003+1)=2003.2004+2004B=20042=2004.2004=2004.(2003+1)=2003.2004+2004
Vì 2003<2004 nên 2003.2004+2003<2003.2004+2004
Vậy A<B
tick nha để mk làm câu b
1: Ta có: \(a^2+2ab+b^2-12a-12b+50\)
\(=\left(a+b\right)^2-12\left(a+b\right)+50\)
\(=2^2-12\cdot2+50\)
=54-24
=30
b: Xét ΔBID có \(\widehat{DBI}=\widehat{DIB}\left(=\widehat{IBC}\right)\)
nên ΔBID cân tại D
Xét ΔEIC có \(\widehat{EIC}=\widehat{ECI}\left(=\widehat{ICB}\right)\)
nên ΔEIC cân tại E
c: Ta có: DE=DI+IE
mà DI=DB
và EC=IE
nên DE=DB+EC
2:
a: ĐKXĐ: \(x\notin\left\{0;-4\right\}\)
\(\dfrac{6}{x^2+4x}+\dfrac{3}{2x+8}\)
\(=\dfrac{6}{x\left(x+4\right)}+\dfrac{3}{2\left(x+4\right)}\)
\(=\dfrac{12+3x}{2x\left(x+4\right)}=\dfrac{3\left(x+4\right)}{2x\left(x+4\right)}=\dfrac{3}{2x}\)
b: ĐKXĐ: \(x\notin\left\{2;-2\right\}\)
\(\dfrac{x+1}{x-2}+\dfrac{x-2}{x+2}+\dfrac{x-14}{x^2-4}\)
\(=\dfrac{\left(x+1\right)\cdot\left(x+2\right)+\left(x-2\right)^2+x-14}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{x^2+3x+2+x^2-4x+4+x-14}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{2x^2-8}{x^2-4}=2\)
c: ĐKXĐ: \(x\notin\left\{1;-1\right\}\)
\(\dfrac{2}{x+1}+\dfrac{-4}{1-x}+\dfrac{5x+1}{1-x^2}\)
\(=\dfrac{2}{x+1}+\dfrac{4}{x-1}-\dfrac{5x+1}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{2x-2+4x+4-5x-1}{\left(x+1\right)\left(x-1\right)}=\dfrac{x+1}{\left(x+1\right)\left(x-1\right)}=\dfrac{1}{x-1}\)
d: ĐKXĐ: \(x\ne\pm y\)
\(\dfrac{x}{x^2+xy}+\dfrac{x-3y}{y^2-x^2}+\dfrac{x}{xy-x^2}\)
\(=\dfrac{x}{x\left(x+y\right)}-\dfrac{x-3y}{\left(x-y\right)\left(x+y\right)}-\dfrac{x}{x\left(x-y\right)}\)
\(=\dfrac{1}{x+y}-\dfrac{x-3y}{\left(x-y\right)\left(x+y\right)}-\dfrac{1}{x-y}\)
\(=\dfrac{x-y-x+3y-x-y}{\left(x-y\right)\left(x+y\right)}=\dfrac{-x+y}{\left(x-y\right)\left(x+y\right)}=\dfrac{-1}{x+y}\)
e: ĐKXĐ: \(\left\{{}\begin{matrix}x< >0\\y< >0;x\ne y\end{matrix}\right.\)
\(\dfrac{y}{x^2-xy}+\dfrac{x}{y^2-xy}\)
\(=\dfrac{y}{x\left(x-y\right)}-\dfrac{x}{y\left(x-y\right)}\)
\(=\dfrac{y^2-x^2}{xy\left(x-y\right)}=\dfrac{-\left(x-y\right)\left(x+y\right)}{xy\left(x-y\right)}=\dfrac{-x-y}{xy}\)
f: ĐKXĐ: x<>1
\(\dfrac{11x-4}{x-1}+\dfrac{10x+4}{2-2x}\)
\(=\dfrac{11x-4}{x-1}-\dfrac{5x+2}{x-1}\)
\(=\dfrac{11x-4-5x-2}{x-1}=\dfrac{6x-6}{x-1}=6\)
lm hết aèk bẹn
Bài 1:
1: \(\left(2a+b\right)^2=4a^2+4ab+b^2\)
2: \(\left(a-3b\right)^2=a^2-6ab+9b^2\)
4: \(\left(3x-5y\right)^2=9x^2-30xy+25y^2\)
7: \(\left(3x-1\right)^2=9x^2-6x+1\)
9: \(\left(x+\dfrac{1}{2}\right)^2=x^2+x+\dfrac{1}{4}\)
10: \(\left(3x-\dfrac{1}{2}\right)^2=9x^2-3x+\dfrac{1}{4}\)
11: \(\left(4-\dfrac{1}{2}x\right)^2=16-4x+\dfrac{1}{4}x^2\)
12: \(\left(3x-0.5\right)^2=9x^2-3x+\dfrac{1}{4}\)
13: \(\left(4x-0.25\right)^2=16x^2-2x+\dfrac{1}{16}\)