doesn't she
isn't it
has he
wasn't he
aren't they
is it
must he
isn't she
don't they
hasn't he
do they
didn't he
did she
isn't it
didn't he
was it
does she
doesn't he
didn't he
do they
didn't they
aren't there

a) \(k=-5\)

b) 

Tìm x xong rồi tìm y

3 thì làm kiểu gì cũng được

Bài 3:

Hình 1:

Xét ΔABC có \(\widehat{A}+\widehat{B}+\widehat{C}=180^0\)

=>\(x+60^0+40^0=180^0\)

=>\(x=180^0-100^0=80^0\)

Hình 2:

Xét ΔABD có \(\widehat{ADC}\) là góc ngoài tại đỉnh D

nên \(\widehat{ADC}=\widehat{ABD}+\widehat{BAD}=110^0\)

=>\(y=110^0\)

ΔDAC cân tại D

=>\(\widehat{DAC}=\widehat{DCA}=\dfrac{180^0-\widehat{ADC}}{2}\)

=>\(x=\dfrac{180^0-110^0}{2}=35^0\)

Hình 3:

Ta có: \(\widehat{CAB}+60^0=180^0\)

=>\(\widehat{CAB}=180^0-60^0=120^0\)

Xét ΔCAB có \(\widehat{CAB}+\widehat{ABC}+\widehat{ACB}=180^0\)

=>\(x+2x+120^0=180^0\)

=>\(3x=60^0\)

=>\(x=20^0\)

Hình 4:

Ta có: \(\widehat{ADC}+\widehat{ADB}=180^0\)

=>\(\widehat{ADB}=180^0-80^0=100^0\)

Xét ΔADB có

\(\widehat{ADB}+\widehat{ABD}+\widehat{BAD}=180^0\)

=>\(x+60^0+100^0=180^0\)

=>\(x=20^0\)

ta có: \(\widehat{ACD}+135^0=180^0\)

=>\(\widehat{ACD}=180^0-135^0=45^0\)

Xét ΔACD có \(\widehat{ACD}+\widehat{ADC}+\widehat{DAC}=180^0\)

=>\(y+80^0+45^0=180^0\)

=>\(y=55^0\)

\(110-3\times\left(8+x\right)=1\)

\(3\times\left(8+x\right)=110-1\)

\(3\times\left(8+x\right)=109\)

\(8+x=\dfrac{109}{3}\)

\(x=\dfrac{109}{3}-8\)

\(x=\dfrac{85}{3}\)

Chúc bạn học tốt

$110-3\times \left(8+x\right)=1$

$3\times \left(8+x\right)=110-1$

$3\times \left(8+x\right)=109$

8+�=10938+x=3109​

�=1093−8x=3109​−8

�=853x=385​
 

trong \(\Delta ABH\) vuông tại H có

AH=AB.cosA=5.cos60=2,5

BH=\(\sqrt{AB^2-AH^2}\)(pytago)=\(\sqrt{5^2-2,5^2}\)=4,3

trong \(\Delta BHC\) vuông tại H có

\(HB^2=BC.BF\)(dl1)\(\Rightarrow BF=\dfrac{HB^2}{BC}\)=\(\dfrac{4,3^2}{5\sqrt{3}}\)=2,1

HF=\(\sqrt{HB^2-BF^2}\)=\(\sqrt{4,3^2-2,1^2}\)=3,8

a,tam giác bmh vuông tại h có đường cao mh nên ah^2=am.ab

tương tự ah^2=ac.an

suy ra đpcm

b,chứng minh dc \(\Delta AMN\sim\Delta ACB\left(g.g\right)\\\)

suy ra \(\dfrac{S_{AMN}}{S_{ABC}}=\dfrac{MN^2}{BC^2}=\dfrac{AH^2}{BC^2}=\dfrac{AH^4}{\left(AH.BC\right)^2}=\dfrac{AH^4}{\left(2S\right)^2}\\ =\dfrac{AH^4}{\left(AB.AC\right)^2}=\dfrac{AH^2}{AB^2}.\dfrac{AH^2}{AC^2}=sin^2B.sin^2C\left(đpcm\right)\)

tick mik nha