chết , mình lm còn sót , nhưng kia làm chắc đc rồi đấy

\(2\left(4x-3\right)-3\left(x+5\right)+4\left(x-10\right)=5\left(x+2\right)\)

\(8x-6-3x-15+4x-40=5x+10\)

Read the following passages and mark the letter A, B, C, or D on your answer sheet to indicate the correct answer to each of the following questions

The biological community changes again as one moves from the city to the suburbs. Around all cities is a biome called the "suburban forest". The trees of this forest are species that are favored by man, and most of them have been deliberately planted. Mammals such as rabbits, skunks, and opossums have moved in from the surrounding countryside. Raccoons have become experts at opening garbage cans, and in some places even deer wander suburban thoroughfares. Several species of squirrel get along nicely in suburbia, but usually only one species is predominant in any given suburb -fox squirrels in one place, red squirrels in another, gray squirrels in a third - for reasons that are little understood. The diversity of birds in the suburbs is great, and in the South, lizards thrive in gardens and even houses. Of course, insects are always present. There is an odd biological sameness in these suburban communities. True, the palms of Los Angeles are missing from the suburbs of Boston, and there are species of insects in Miami not found in Seattle. But over wide stretches of the United States, ecological conditions in suburban biomes vary much less than do those of natural biome. And unlike the natural biomes, the urban and suburban communities exist in spite of, not because of, the climate

The word "thrive" is closest in meaning to

A. remain

B. flourish

C. reproduce

D. survive

Xét \(\Delta AEB\) và \(\Delta BEK\) có

\(\widehat{E_1}=\widehat{E_2}\)(=90 độ )

\(\widehat{B_1}=\widehat{B_2}\)( BD là tia phân giác của \(\widehat{B}\))

EB chung

\(\Rightarrow\Delta ABE=\Delta KBE\)( cạnh góc vuông - góc nhọn kề cạnh đó )

\(\Rightarrow AB=BK\)( hai cạnh tương ứng )

\(\Rightarrow\) \(\Delta ABK\)cân tại B

b)Do \(\Delta ABE=\Delta KBE\)(cmt)

\(\Rightarrow\) \(AE=EK\)( hai cạnh tương ứng )

Xét \(\Delta AED\)và \(\Delta DEK\)có

\(DE\) chung

\(AE=EK\)(cmt)

\(\widehat{AED}=\widehat{DEK}\) (=90 độ )

Do đó : \(\Delta AED=\Delta KED\)( hai cạnh góc vuông )

\(\Rightarrow\widehat{ADE}=\widehat{EDK}\)( hai góc tương ứng )

Ta có : \(\widehat{DKC}=\widehat{B_2}+\widehat{BDK}\)( góc ngoài )

hay \(\widehat{DKC}=\widehat{B_1}+\widehat{ADB}\)(1)

( vì \(\widehat{B_1}=\widehat{B_2}\); \(\widehat{BDK}=\widehat{ADB}\))

Mặt khác \(\Delta ABD\) vuông tại A có \(\widehat{A}=90^o\)

\(\Rightarrow\widehat{B_1}+\widehat{ADB}=90^o\) ( định lí )(2)

Từ (1) và (2) \(\Rightarrow\widehat{DKC}=90^o\)

\(\Rightarrow DK\perp BC\)(đpcm)

Ta có : \(\Delta ABD\) đều

\(\Rightarrow\widehat{A_2}=60^o\)

\(\Delta ACE\) đều

\(\Rightarrow\widehat{A_3}=60^o\)

\(\Rightarrow\widehat{A_2}=\widehat{A_3}\)

Ta lại có : \(\widehat{A_1}+\widehat{A_2}=\widehat{DAC}\)

\(\widehat{A_1}+\widehat{A_3}=\widehat{BAE}\)

Mặt khác \(\widehat{A_1}chung\)

\(\widehat{A_2}=\widehat{A_3}\) (cmt)

Do đó : \(\widehat{BAE}=\widehat{DAC}\)

Xét \(\Delta ABE\) và \(\Delta ADC\) có:

\(AB=AD\) ( \(\Delta ABD\) đều)

\(\widehat{BAE}=\widehat{DAC}\)

\(AE=AC\)(\(\Delta ACE\) đều)

Do đó : \(\Delta ABE=\Delta ADC\)

\(\Rightarrow\widehat{AEB}=\widehat{ACD}\) ( hai góc tương ứng )

b) Gọi giao điểm của AC và BE là F

Trong \(\Delta AFE\) có :

\(\widehat{A_3}+\widehat{AFE}+\widehat{E}=180^o\) ( định lí )

Trong \(\Delta MFC\) có :

\(\widehat{MFC}+\widehat{FMC}+\widehat{FCM}=180^o\) ( định lí )

Mặt khác

\(\widehat{E}=\widehat{FCM}\)( theo câu a )

\(\widehat{MFC=}\widehat{AFE}\) ( hai góc đối đỉnh )

\(\Rightarrow\widehat{FMC}=\widehat{A_3}\)

Mà \(\widehat{A_3}=60^o\)(\(\Delta ACE\)đều )

\(\Rightarrow\)\(\widehat{FMC}=60^o\)

Ta lại có : \(\widehat{FMC}+\widehat{BMC}=180^o\)( hai góc kề bù )

hay \(60^o+\widehat{BMC}=180^o\)

\(\Rightarrow\widehat{BMC}=180^o-60^o=120^o\)(đpcm)

\(\left[{}\begin{matrix}\left\{{}\begin{matrix}x-3>0\\x+\dfrac{1}{2}>0\end{matrix}\right.\\\left\{{}\begin{matrix}x-3< 0\\x+\dfrac{1}{2}< 0\end{matrix}\right.\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>3\\x>\dfrac{-1}{2}\end{matrix}\right.\\\left\{{}\begin{matrix}x< 3\\x< -\dfrac{1}{2}\end{matrix}\right.\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x>3\\x< \dfrac{-1}{2}\end{matrix}\right.\)

Vậy x>3 hoặc x<-1/2

= \(\frac{\left(9+1\right)x9}{2}\)

= 45

\(\left(x+1\right)^2=1+3\)

\(\left(x+1\right)^2=4\)

\(\Rightarrow\left[{}\begin{matrix}x+1=4\\x+1=-4\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=4+1\\x=-4+1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=5\\x=-3\end{matrix}\right.\)

Vậy...

hai góc ở đáy bằng \(\dfrac{180-100}{2}=\dfrac{80}{2}=40\)