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\(a,BD=\sqrt{AB^2+AD^2}=\sqrt{5^2+12^2}=13\left(cm\right)\left(pytago\right)\)
Áp dụng HTL tam giác
\(\left\{{}\begin{matrix}AB^2=BH\cdot BD\\AD^2=DH\cdot BD\\AH^2=BH\cdot HD\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}BH=\dfrac{AB^2}{BD}=\dfrac{25}{13}\left(cm\right)\\DH=\dfrac{AD^2}{BD}=\dfrac{144}{13}\left(cm\right)\\AH=\sqrt{\dfrac{25\cdot144}{13^2}}=\dfrac{60}{13}\left(cm\right)\end{matrix}\right.\)
\(b,\widehat{MAN}=\widehat{ANM}=\widehat{AMN}\left(=90^0\right)\\ \Rightarrow AMHN.là.hcn\\ \Rightarrow AH=MN=\dfrac{60}{13}\left(cm\right)\)
\(c,\) Vì \(AMHN\) là hcn nên \(\widehat{MAH}=\widehat{ANM}\)
Mà \(\widehat{MAH}=\widehat{ADB}\left(cùng.phụ.\widehat{HAD}\right)\)
\(\Rightarrow\widehat{ANM}=\widehat{ADB}\)
\(\left\{{}\begin{matrix}\widehat{ANM}=\widehat{ADB}\\\widehat{BAD}.chung\end{matrix}\right.\Rightarrow\Delta AMN\sim\Delta ADB\left(g.g\right)\\ \Rightarrow\dfrac{AM}{AD}=\dfrac{AN}{AB}\Rightarrow AM\cdot AB=AN\cdot AD\)
a/ Xét △AHB và △DAB ta được: △AHB đồng dạng △DAB (g.g) (Tự chứng minh)
\(\Rightarrow\dfrac{AH}{AD}=\dfrac{HB}{AB}=\dfrac{AB}{BD}\left(a\right)\)
Áp dụng định lí Pytago vào △ADB được: \(BD=\sqrt{12^2+5^2}=13\left(cm\right)\). Thay vào (a) được:
\(\dfrac{AH}{AD}=\dfrac{HB}{AB}=\dfrac{AB}{13}\) hay \(\dfrac{AH}{12}=\dfrac{HB}{5}=\dfrac{5}{13}\)
\(\Rightarrow\left[{}\begin{matrix}AH=\dfrac{5.12}{13}\approx4,62\left(cm\right)\\HB=\dfrac{5^2}{13}\approx1,92\left(cm\right)\\HD=13-1,92=11,08\left(cm\right)\end{matrix}\right.\)
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b/ \(\begin{matrix}\hat{A}=90\text{°}\\\hat{AMH}=90\text{°}\\\hat{ANH}=90\text{°}\end{matrix}\) ⇒ AMHN là hình chữ nhật ⇒ \(AH=MN\approx4,92\left(cm\right)\)
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c/ Ta có: △AMN = △HNA (c.g.c) (Tự chứng minh)
Ta cũng có: △HNA đồng dạng △AHB (g.g) (Tự chứng minh) ⇒ △HNA đồng dạng △DAB (cùng đồng dạng △AHB) ⇒ △AMN đồng dạng △DAB
Vậy: \(\dfrac{AM}{AN}=\dfrac{AB}{AD}\) hay \(AM.AD=AN.AB\left(đpcm\right)\)
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\(M=a^3+b^3+3ab\left(a^2+b^2\right)+6a^2b^2\left(a+b\right)\)
\(=\left(a+b\right)^3-3ab\left(a+b\right)+3ab\cdot\left(-2ab\right)+6a^2b^2\)
\(=1-3ab-6a^2b^2+6a^2b^2\)
\(=1-3ab\)
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Vì tổng các góc của hình tứ giác là 360o
Nên 3x + 5x + 2x +60o = 360o
\(\Rightarrow x=30^o\)
\(10x=300\)
nên x=30
=>\(\widehat{A}=150^0;\widehat{B}=90^0;\widehat{D}=60^0\)
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a: \(\Leftrightarrow20\left(5x-5+2\right)-15\left(9x-1\right)=24\left(3x+1\right)-60\)
\(\Leftrightarrow100x-60-135x+15=72x+24-60\)
=>72x-36=-35x-45
=>107x=-9
=>x=-9/107
b: \(\Leftrightarrow15\left(x-2\right)+2\left(5x-9.5\right)=16\left(x+2\right)+100\)
=>15x-30+10x-19=16x+32+100
=>25x-49=16x+132
=>9x=181
=>x=181/9
c: \(\Leftrightarrow3\left(9x+3+2\right)-18=4\left(5x+1\right)-3x-1\)
=>3(9x+5)-18=20x+4-3x-1=17x+3
=>27x+15-18=17x+3
=>10x=3-15+18=6
=>x=3/5
d: \(\Leftrightarrow8x+4+9\left(4x+1\right)=2\left(3x+2x+2\right)+11x+5\)
=>8x+4+36x+9=10x+4+11x+5
=>44x+13=21x+9
=>23x=-4
=>x=-4/23
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a: =>12x+5(x-1)=100-10(3x-1)
=>12x+5x-5=100-30x+30
=>17x-5=-30x+130
=>47x=135
=>x=135/47
b: \(\Leftrightarrow6\left(2x-1\right)+2\left(5-x\right)=24-9\left(x+1\right)\)
=>12x-6+10-2x=24-9x-9
=>10x+4=-9x+15
=>19x=11
=>x=11/19
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\(P=\left(\dfrac{x^2+1}{x^2-9}-\dfrac{x}{x+3}+\dfrac{5}{3-x}\right):\left(\dfrac{2x+10}{x+3}-1\right)\)
\(=\left(\dfrac{x^2+1}{\left(x-3\right)\left(x+3\right)}-\dfrac{x\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}-\dfrac{5\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}\right):\left(\dfrac{2x+10}{x+3}-\dfrac{x+3}{x+3}\right)\)
\(=\left(\dfrac{x^2+1-x^2+3x-5x-15}{\left(x-3\right)\left(x+3\right)}\right):\left(\dfrac{2x+10-x-3}{x+3}\right)\)
\(=\left(\dfrac{-2x-14}{\left(x-3\right)\left(x+3\right)}\right):\left(\dfrac{x+7}{x+3}\right)\)
\(=\dfrac{-2\left(x+7\right)}{\left(x-3\right)\left(x+3\right)}.\dfrac{x+3}{x+7}\)
\(=\dfrac{-2}{x-3}\)
đk : x khác -3 ; 3 ; -7
\(P=\left(\dfrac{x^2+1+x\left(x-3\right)+5x+15}{x^2-9}\right):\left(\dfrac{2x+10-x-3}{x+3}\right)\)
\(=\dfrac{2x^2+1+2x+15}{x^2-9}:\dfrac{x+7}{x+3}=\dfrac{2x^2+2x+16}{\left(x-3\right)\left(x+7\right)}\)
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a: Xét tứ giác HMKN có
I là trung điểm của HK
I là trung điểm của MN
Do đó: HMKN là hình bình hành
ờmm
đề bài là tìm x à bn ?
nếu tìm x thì:
ta có: (x+13):2=20
x+13=40
x=27(cm)
vậy x=17cm