A>5/3>5/4=>A>5/4 chứ mị

mk nhìn nhầm

\(1-\frac{5}{3}=\frac{1}{5}-x\)

\(\frac{1}{5}-x=1-\frac{5}{3}\)

\(\frac{1}{5}-x=\frac{-2}{3}\)

\(x=\frac{1}{5}-\frac{-2}{3}\)

\(x=\frac{3}{15}+\frac{10}{15}\)

\(x=\frac{13}{15}\)

\(1-\frac{5}{3}=\frac{1}{5}-x\)

\(\Rightarrow\frac{2}{3}=\frac{1}{5}-x\)

\(\Rightarrow x=\frac{1}{5}-\frac{2}{3}\)

\(\Rightarrow x=\frac{-7}{15}\)

-Ngan chan viec pha rung

-Han che viec khai thac cac loai thuc vat quy hiem

-Xay dung vuon thuc vat,vuon quoc gia,cac khu bao ton

Cam buon ban ,xuat khau cac loai thuc vat quy hiem

Tuyen truyen giao duc rong de m.n cung tham gia bao ve rung

-Tham gia trong rung ,bao ve rung va moi truong song cua thuc vat

a ) Do Ox và Oy là 2 tia đối nhau 

=> \(\widehat{xOy}=180^o\)

Có : \(\widehat{xOt}\)và \(\widehat{yOt}\)là hai góc kề bù

=> \(\widehat{xOt}+\widehat{yOt}=180^o\)

     \(\widehat{xOt}+130^o=180^o\)

=> \(\widehat{xOt}=180^o-130^o=50^o\)

Có : \(\widehat{xOt}\)và \(\widehat{tOz}\)là 2 góc kề nhau

=> \(\widehat{xOt}+\widehat{tOz}=\widehat{xOz}\)

      \(50^o+\widehat{tOz}=100^o\)

=> \(\widehat{tOz}=100^o-50^o=50^o\)

Lại có : \(\widehat{yOz}\)và \(\widehat{tOz}\)là 2 góc kề nhau

=> \(\widehat{yOz}+\widehat{tOz}=\widehat{yOt}\)

     \(\widehat{yOz}+50^o=130^o\)

=> \(\widehat{yOz}=130^o-50^o=80^o\)

Vậy \(\widehat{xOt}=50^o\)và \(\widehat{yOz}=80^o\)

b ) Có  : Ot nằm giữa Ox và Oz 

       và   \(\widehat{xOt}=\widehat{tOz}\left(=50^o\right)\)( đã chứng minh phần a )

Từ 2 điều này => Ot là tia p/g của \(\widehat{xOz}\)

c ) Có \(\widehat{yOm}\)và \(\widehat{mOz}\)là 2 góc kề nhau

=> \(\widehat{yOm}+\widehat{mOz}=\widehat{yOz}\)

      \(35^o+\widehat{mOz}=80^o\)   ( đã tính góc yOz ở phần a rồi nhé ! )

=> \(\widehat{mOz}=80^o-35^o=45^o\)

Vậy \(\widehat{mOz}=45^o\)

\(S=\frac{1}{31}+\frac{1}{32}+\frac{1}{33}+...+\frac{1}{60}\)

\(S=\left[\frac{1}{31}+\frac{1}{32}+...+\frac{1}{40}\right]+\left[\frac{1}{41}+\frac{1}{42}+...+\frac{1}{50}\right]+\left[\frac{1}{51}+\frac{1}{52}+...+\frac{1}{60}\right]\)

\(S< \left[\frac{1}{30}+\frac{1}{30}+...+\frac{1}{30}\right]+\left[\frac{1}{40}+\frac{1}{40}+...+\frac{1}{40}\right]+\left[\frac{1}{50}+\frac{1}{50}+...+\frac{1}{50}\right]\)

\(S< \frac{10}{30}+\frac{10}{40}+\frac{10}{50}\)

\(S< \frac{37}{60}< \frac{48}{60}=\frac{4}{5}\)

\(S=\frac{1}{31}+\frac{1}{32}+\frac{1}{33}+\frac{1}{34}+...+\frac{1}{60}\)\(CMR:S< \frac{4}{5}\)

Số số hạng của S là: (60 - 31 ) + 1 = 30 ( số ), chia thành 6 nhóm, mỗi nhóm 5 số hạng.

Ta có: 

\(\frac{1}{31}+\frac{1}{32}+\frac{1}{33}+\frac{1}{34}+\frac{1}{35}< \frac{1}{31}+\frac{1}{31}+\frac{1}{31}+\frac{1}{31}+\frac{1}{31}\)

\(\frac{1}{36}+\frac{1}{37}+\frac{1}{38}+\frac{1}{39}+\frac{1}{40}< \frac{1}{36}+\frac{1}{36}+\frac{1}{36}+\frac{1}{36}+\frac{1}{36}\)

\(\frac{1}{41}+\frac{1}{42}+\frac{1}{43}+\frac{1}{44}+\frac{1}{45}< \frac{1}{41}+\frac{1}{41}+\frac{1}{41}+\frac{1}{41}+\frac{1}{41}\)

\(\frac{1}{46}+\frac{1}{47}+\frac{1}{48}+\frac{1}{49}+\frac{1}{50}< \frac{1}{46}+\frac{1}{46}+\frac{1}{46}+\frac{1}{46}+\frac{1}{46}\)

\(\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+\frac{1}{54}+\frac{1}{55}< \frac{1}{51}+\frac{1}{51}+\frac{1}{51}+\frac{1}{51}+\frac{1}{51}\)

\(\frac{1}{56}+\frac{1}{57}+\frac{1}{58}+\frac{1}{59}+\frac{1}{60}< \frac{1}{56}+\frac{1}{56}+\frac{1}{56}+\frac{1}{56}+\frac{1}{56}\)

\(=>S=\frac{5}{31}+\frac{5}{36}+\frac{5}{41}+\frac{5}{46}+\frac{5}{51}+\frac{5}{56}\)

\(=>S< 0,78...\)\(=>S< \frac{7}{10}\)( mình ước lượng thôi nha )

Vậy \(S< \frac{4}{5}\)vì \(\frac{4}{5}=\frac{8}{10}< \frac{7}{10}\)

~UMK..., mình ko chắc đúng ko nữa~

minh nhe

len lever 39 rui

1+3x0x0...x0=0

mình nè 

lever 29

\(\frac{7}{9}\cdot\frac{8}{11}+\frac{7}{9}\cdot\frac{3}{11}-\frac{45}{19}\)

\(=\frac{7}{9}\left[\frac{8}{11}+\frac{3}{11}\right]-\frac{45}{19}\)

\(=\frac{7}{9}\cdot1-\frac{45}{19}=-\frac{272}{171}\)

\(1,4\cdot\frac{15}{49}-\left[20\%+\frac{2}{3}\right]:\frac{11}{5}\)

\(=\frac{14}{10}\cdot\frac{15}{49}-\left[\frac{20}{100}+\frac{2}{3}\right]\cdot\frac{5}{11}\)

\(=\frac{7}{5}\cdot\frac{15}{49}-\left[\frac{1}{5}+\frac{2}{3}\right]\cdot\frac{5}{11}\)

\(=\frac{1}{1}\cdot\frac{3}{7}-\frac{13}{15}\cdot\frac{5}{11}\)

\(=\frac{3}{7}-\frac{13}{3}\cdot\frac{1}{11}\)

\(=\frac{3}{7}-\frac{13}{33}=\frac{8}{231}\)

\(-\frac{7}{6}-x=\frac{1}{6}\)

\(\Rightarrow x=-\frac{7}{6}-\frac{1}{6}\)

\(\Rightarrow x=-\frac{4}{3}\)

 \(30\%x=-\frac{6}{5}\)

\(\Rightarrow\frac{30}{100}x=-\frac{6}{5}\)

\(\Rightarrow\frac{3}{10}x=-\frac{6}{5}\Leftrightarrow x=-4\)

Làm nốt hai bài còn lại

1943*a^2-1943*a-54*a+54=0

1943*a(a-1)-54(a-1)=0

(1943*a-54)(a-1)=0

1943a=54 hoặc a=1

a=1943/54 hoặc a=1

A = 1 + 1.2 + 2 + 2.3 + 3 + ....+ 98.99 + 99 + 99.100 + 100

A = ( 1.2 + 2.3 + 3.4+ ...+99.100) + ( 1+2+3+...+100)

A = 99 . 100 . 101 : 3 + 100.101:2

A = 100 . 101 . 33 + 101.50

A = 101.50.(66+1)

A= 101.50.67

Trả lời :

B.\(\frac{2}{7},\frac{2}{5},\frac{1}{2}\)

~ Hok tốt ~

B=\(\frac{2}{7},\frac{2}{5},\frac{1}{2}\)