284,91

417,12

a.284,91

b.417,12

Nha^^

bạn tự vẽ hình giúp mik nha

a. xét \(\Delta ADN\) và \(\Delta BAM\) có

AB=AD(gt)

\(\widehat{ADN}=\widehat{BAM}=90^o\)

DN=MA(N,M là trung điểm của cạnh DC,AD)

\(\Rightarrow\Delta ADN\sim\Delta BAM\left(c.g.c\right)\)

\(\Rightarrow\widehat{DNA}=\widehat{AMB}\)

mà:\(\widehat{DNA}+\widehat{DAN}=90^o\Rightarrow\widehat{BMA}+\widehat{DAN}=90^o\)

\(\Rightarrow\Delta MAI\) vuông tại I

\(\Rightarrow AI\perp MI\) hay \(MB\perp AN\)

b.ta có M là trung điểm của AD\(\Rightarrow AM=\dfrac{1}{2}AD=\sqrt{5}\)

trong \(\Delta MAB\) vuông tại A có

\(MB=\sqrt{AM^2+AB^2}=\sqrt{\sqrt{5^2}+\left(2\sqrt{5}\right)^2}=5\)

\(AM^2=MB.MI\Rightarrow MI=\dfrac{AM^2}{MB}=\dfrac{\sqrt{5^2}}{5^5}=0,2\)

\(AI.MB=AM.AB\Rightarrow AI=\dfrac{AM.AB}{MB}=\dfrac{\sqrt{5}.2\sqrt{5}}{5}\)=2

c.IB=MB-MI=5-0,2=4,8

\(S_{\Delta AIB}=\dfrac{AI.IB}{2}=\)\(\dfrac{2.4,8}{2}=4,8\)

\(S_{\Delta ADN}=\dfrac{AD.DN}{2}=\dfrac{2\sqrt{5}.\sqrt{5}}{2}=5\)

\(S_{\Delta ABCD}=\left(2\sqrt{5}\right)^2=20\)

\(S_{BINC}=S_{ABCD}-S_{\Delta AIB}-S_{\Delta DAN}\)=20-4,8-5=10,2

Ok mình cảm ơn bạn nha

Bài 1:

a: \(-\frac52-\frac49\cdot\frac{21}{16}\)

\(=-\frac52-\frac{4}{16}\cdot\frac{21}{9}\)

\(=-\frac52-\frac14\cdot\frac73=-\frac52-\frac{7}{12}=\frac{-30-7}{12}=-\frac{37}{12}\)

b: \(\frac{-5}{4}\cdot\frac{-3}{7}+\frac{-5}{4}\cdot\frac{17}{7}+12\frac12\)

\(=-\frac54\left(-\frac37+\frac{17}{7}\right)+12,5\)

\(=-\frac54\cdot\frac{14}{7}+12,5=-\frac54\cdot2+12,5=-\frac52+12,5=10\)

c: \(\left(-\frac13\right)^2+60\%+2\frac89-\frac85\)

\(=\frac19+\frac35+2+\frac89-\frac85\)

\(=2+\left(\frac19+\frac89\right)+\left(\frac35-\frac85\right)=2\)

d: \(15\frac37-\left(\frac89+7\frac37\right)\)

\(=15+\frac37-\frac89-7-\frac37\)

\(=8-\frac89=\frac{72-8}{9}=\frac{64}{9}\)

Bài 2:

a: \(1,75-20\%\cdot x=\frac15\)

=>\(0,2x=1,75-0,2=1,55\)

=>\(x=1,55:0,2=1,55\cdot5=7,75\)

b: \(\left(\frac23-x\right):\frac{-3}{10}=75\%\)

=>\(\frac23-x=\frac34\cdot\frac{-3}{10}=\frac{-9}{40}\)

=>\(x=\frac23+\frac{9}{40}=\frac{80+27}{120}=\frac{107}{120}\)

c: \(\left|\frac34x-\frac15\right|=0,25\)

=>\(\left[\begin{array}{l}\frac34x-\frac15=\frac14\\ \frac34x-\frac15=-\frac14\end{array}\right.\Rightarrow\left[\begin{array}{l}\frac34x=\frac15+\frac14=\frac{9}{20}\\ \frac34x=-\frac14+\frac15=\frac{-1}{20}\end{array}\right.\)

=>\(\left[\begin{array}{l}x=\frac{9}{20}:\frac34=\frac{9}{20}\cdot\frac43=\frac{36}{60}=\frac35\\ x=-\frac{1}{20}:\frac34=-\frac{1}{20}\cdot\frac43=\frac{-4}{60}=-\frac{1}{15}\end{array}\right.\)

d: \(x-\frac29x=\frac59\)

=>\(x\left(1-\frac29\right)=\frac59\)

=>\(x\cdot\frac79=\frac59\)

=>\(x=\frac59:\frac79=\frac57\)

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