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a, Vì tam giác ABC cân tại A nên \(\widehat{NBM}=\widehat{ACB}\)
Mà \(\widehat{ACB}=\widehat{PCQ}\left(đối.đỉnh\right)\Rightarrow\widehat{NBM}=\widehat{PCQ}\)
Mà \(\widehat{NMB}=\widehat{CPQ}=90^0;BM=PC\)
Do đó \(\Delta BMN=\Delta CPQ\left(g.c.g\right)\)
b, Vì \(BM//PQ\left(\perp BP\right)\) nên \(\widehat{MNI}=\widehat{IQP}\)
Mà \(\widehat{NMI}=\widehat{IPQ}=90^0;MN=PQ\left(\Delta BMN=\Delta CPQ\right)\)
Do đó \(\Delta IMN=\Delta IPQ\left(g.c.g\right)\)
\(\Rightarrow IN=IQ\)
c, Vì IK là đường cao cũng là trung tuyến tam giác KNQ nên tam giác KNQ cân tại K
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
1)xét tứ giác EACD
EA//DC,ED//AC
=>EACD hình bình hành
E=C=40(hai góc đối)
ta có DAC=BAC/2=60/2=30(AD là tia pg)
mà ED//AC
=>ADE=DAC=30(so le)
xét tg EAD
E+ADE+EAD=180
EAD=180-ADE-E=180-30-40=110
2)
a)xét tgAHB và tgDHB
BAH=BDH=90,ABH=HBD(BH là tia pg),BH chung
=>tgAHB=tgDHB(cạnh huyền góc nhọn)
=>AH=HD,BA=BD
b)xét tg BDE và tgBAC
BA=BD,ABC chung,BAC=BDE=90
=>tgBDE=tgBAC(gcg)
=>BE=BC
xét tg BEC
BA/BE=BD/BC=>AD//EC(ta lét đảo)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 4:
Ta có: \(\dfrac{a}{b}=\dfrac{c}{d}\)
nên \(\dfrac{a}{c}=\dfrac{b}{d}\)
\(\Leftrightarrow\dfrac{c}{a}=\dfrac{d}{b}\)
hay \(\dfrac{a+c}{a}=\dfrac{b+d}{b}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\text{Bài 1:a)}25\dfrac{3}{19}.\left(-\dfrac{4}{5}\right)-35\dfrac{3}{19}.\left(-\dfrac{4}{5}\right)\)
\(=\dfrac{478}{19}.\left(-\dfrac{4}{5}\right)-\dfrac{668}{19}.\left(-\dfrac{4}{5}\right)\)
\(=\left(-\dfrac{4}{5}\right).\left(\dfrac{478}{19}-\dfrac{668}{19}\right)\)
\(=\left(-\dfrac{4}{5}\right).\left(\dfrac{-190}{19}\right)\)
\(=\left(-\dfrac{4}{5}\right).\left(-10\right)=8\)
\(\text{b)}5:\left(-\dfrac{5}{2}\right)^2+\dfrac{2}{15}.\sqrt{\dfrac{9}{4}}-\left(-2021\right)^0+0,25\)
\(=5:\dfrac{25}{4}+\dfrac{2}{15}.\dfrac{3}{2}-1+\dfrac{1}{4}\)
\(=\dfrac{4}{5}+\dfrac{1}{5}-1+\dfrac{1}{4}\)
\(=1-1+\dfrac{1}{4}\)
\(=0+\dfrac{1}{4}=\dfrac{1}{4}\)
\(\text{Bài 2:a)}\dfrac{8}{5}-\dfrac{3}{5}:x=0,4\)
\(\dfrac{3}{5}:x=\dfrac{8}{5}-0,4=\dfrac{6}{5}\)
\(x=\dfrac{3}{5}.\dfrac{5}{6}=\dfrac{1}{2}\)
\(\text{b)}\left(3x-\dfrac{1}{2}\right)^2+\dfrac{21}{25}=1\)
\(\left(3x-\dfrac{1}{2}\right)^2\) \(=1-\dfrac{21}{25}=\dfrac{4}{25}=\pm\left(\dfrac{2}{5}\right)^2\)
\(\text{Vậy }3x-\dfrac{1}{2}=\dfrac{2}{5}\)
\(3x\) \(=\dfrac{2}{5}+\dfrac{1}{2}=\dfrac{9}{10}\)
\(x\) \(=\dfrac{9}{10}.\dfrac{1}{3}=\dfrac{3}{10}\)
\(\text{hoặc }3x-\dfrac{1}{2}=\dfrac{-2}{5}\)
\(3x\) \(=\left(\dfrac{-2}{5}\right)+\dfrac{1}{2}=\dfrac{1}{10}\)
\(x\) \(=\dfrac{1}{10}.\dfrac{1}{3}=\dfrac{1}{30}\)
\(\Rightarrow x\in\left\{\dfrac{3}{10};\dfrac{1}{30}\right\}\)
Bài 2:
a: =>3/5:x=6/5
hay x=3/5:6/5=1/2
b: \(\Leftrightarrow\left(3x-\dfrac{1}{2}\right)^2=\dfrac{4}{5}\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-\dfrac{1}{2}=\dfrac{2}{5}\\3x-\dfrac{1}{2}=-\dfrac{2}{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{10}\\x=\dfrac{1}{30}\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 5:
a: Bậc của M là 5
b: Các hạng tử là \(x^3yz;-x^5;3\)
Bài 6:
\(N=x^2y-5x^2y-4x^3+7x^2+3xy^2-\dfrac{3}{4}=-4x^2y-4x^3+7x^2+3xy^2-\dfrac{3}{4}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
4:
a: =>2/5x+7/20-2/20=1/10
=>2/5x+5/20=1/10
=>2/5x=1/10-1/4=4/40-10/40=-6/40=-3/20
=>x=-3/20:2/5=-3/20*5/2=-15/40=-3/8
b: 3/2-1/2x=-1/3+3=8/3
=>1/2x=3/2-8/3=9/6-16/6=-7/6
=>x=-7/6*2=-7/3
c: 15/8-1/8:(1/4x-0,5)=5/4
=>1/8:(1/4x-1/2)=15/8-5/4=15/8-10/8=5/8
=>1/4x-1/2=1/8:5/8=1/5
=>1/4x=1/5+1/2=7/10
=>x=7/10*4=28/10=2,8
d: \(\Leftrightarrow\left[\left(x+\dfrac{1}{2}\right)^3-\dfrac{5}{4}\right]=\dfrac{11}{4}-\dfrac{5}{8}=\dfrac{22-5}{8}=\dfrac{17}{8}\)
=>\(\left(x+\dfrac{1}{2}\right)^3=\dfrac{17}{8}+\dfrac{5}{4}=\dfrac{27}{8}\)
=>x+1/2=3/2
=>x=1
Bài 1 :
Để \(A=\frac{x+15}{x-2}\)là số nguyên thì :
x + 15 ⋮ x - 2
=> ( x - 2 ) + 17 ⋮ x - 2
Mà x - 2 ⋮ x - 2 ∀ x ∈ Z
=> 17 ⋮ x - 2
=> x - 2 ∈ { -17 ; -1 ; 1 ; 17 }
=> x ∈ { -15 ; 1 ; 3 ; 19 }
Để \(B=\frac{3x+4}{x-3}\)nhận giá trị nguyên thì :
3x + 4 ⋮ x - 3
=> 3( x - 3 ) + 13 ⋮ x - 3
Mà 3( x - 3 ) ⋮ x - 3 ∀ x ∈ Z
=> 13 ⋮ x - 3
=> x - 3 ∈ { -13 ; -1 ; 1 ; 13 }
=> x ∈ { -10 ; 2 ; 4 ; 16 }