\(2^{x+1}+5.2^{x-2}=\dfrac{7}{32}\) ( đề sai )

Ta có:\(2^{x+1}+5.2^{x-1}=\dfrac{7}{32}\)

\(2^{x+1}.\left(1+5\right)=\dfrac{7}{32}\)

\(2^{x-1}=\dfrac{7}{32}:6\)

\(2^{x-1}=\dfrac{7}{192}\)

mà \(2^{x-1}\ne\dfrac{7}{192}\)

\(\Rightarrow x\in\varnothing\)

Bài 1:

\(a,\dfrac{2^{19}.27^3+15.4^9.9^4}{6^9.2^{10}+12^{10}}\)

\(=\dfrac{2^{19}.\left(3^3\right)^3+3.5.\left(2^2\right)^9.\left(3^2\right)^4}{2^9.3^9.2^{10}+3^{10}.\left(2^2\right)^{10}}\)

\(=\dfrac{2^{19}.3^9+3.5.2^{18}.3^8}{2^9.3^9.2^{10}+3^{10}.2^{20}}\)

\(=\dfrac{2^{19}.3^9+5.2^{18}.\left(3^8.3\right)}{2^9.2^{10}.3^9+3^{10}.2^{20}}\)

\(=\dfrac{2^{19}.3^9+5.2^{18}.3^9}{2^{19}.3^9+3^{10}.2^{20}}\)

\(=\dfrac{1+5.1.1}{1+3.2^2}=\dfrac{6}{13}\)

\(8^2.5^3=8.8.125=8.1000=8000\)

ngại làm wá bạn thông cảm

1, So sánh

\(a,-\dfrac{1}{5}\) và \(\dfrac{1}{1000}\)

Cách 1: \(-\dfrac{1}{5}=-\dfrac{1.200}{5.200}=-\dfrac{200}{1000}\)

Ta có: \(-\dfrac{200}{1000}< \dfrac{1}{1000}\Rightarrow-\dfrac{1}{5}< \dfrac{1}{1000}\)

Cách 2: Ta có:

\(-\dfrac{1}{5}\) là số âm, \(\dfrac{1}{1000}\) là số dương

âm < dương \(\Rightarrow-\dfrac{1}{5}< \dfrac{1}{1000}\)

\(b,\dfrac{267}{-268}\) và \(\dfrac{-1347}{1343}\)

Ta có: \(\dfrac{267}{-268}< -1\left(1\right)\), \(\dfrac{-1347}{1343}>-1\left(2\right)\)

Từ \(\left(1\right)\) và \(\left(2\right)\)\(\Rightarrow\dfrac{267}{-268}>\dfrac{-1347}{1343}\)

\(c,\dfrac{-13}{38}\) và \(\dfrac{29}{-88}\)

Ta có: \(\dfrac{-13}{38}=-\dfrac{13.44}{38.44}=-\dfrac{572}{1672}\)

\(\dfrac{29}{-88}=-\dfrac{29.19}{88.19}=-\dfrac{551}{1672}\)

\(\Rightarrow-\dfrac{572}{1672}< -\dfrac{551}{1672}\) hay \(-\dfrac{13}{38}< \dfrac{29}{-88}\)

Bài 2: Tìm x:

\(a,5-\left|x+\dfrac{1}{2}\right|=1\)

\(\Rightarrow\left|x+\dfrac{1}{2}\right|=4\)

Ta có \(\left\{{}\begin{matrix}x+\dfrac{1}{2}=4\\x+\dfrac{1}{2}=-4\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{7}{2}\\x=-\dfrac{9}{2}\end{matrix}\right.\)

Vậy \(x\in\left\{-\dfrac{9}{2};\dfrac{7}{2}\right\}\)

\(b,\dfrac{4}{3}+\left|2-\dfrac{1}{2}x\right|=7\)

\(\Rightarrow\left|2-\dfrac{1}{2}x\right|=7-\dfrac{4}{3}\)

\(\left|2-\dfrac{1}{2}x\right|=\dfrac{17}{3}\)

Ta có: \(\left\{{}\begin{matrix}2-\dfrac{1}{2}x=\dfrac{17}{3}\\2-\dfrac{1}{2}x=-\dfrac{17}{3}\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=\left(2-\dfrac{17}{3}\right):\dfrac{1}{2}=-\dfrac{22}{3}\\x=\left[2-\left(-\dfrac{17}{3}\right)\right]:\dfrac{1}{2}=\dfrac{46}{3}\end{matrix}\right.\)

Vậy \(x\in\left\{-\dfrac{22}{3};\dfrac{46}{3}\right\}\)

So be nhat la 58 ko nho hon 56 

So lon nhat  la 3000 ko lon hon 3000

So pahn tu cua tap hop A la : 

(3000-58):2+1=1472 (phan tu )

Vay so phan tu cua tap hop A la 1472 phan tu 

v hỏi

Mình không phải mấy bạn bạn vừa đọc mình giải được khôngGiải

a) Ta có: \(\widehat{A}+\widehat{B}+\widehat{C}=180^0\)

\(\Rightarrow\widehat{A}=180^0-\widehat{B}-\widehat{C}\)

Thay số:\(\widehat{A}=180^0-60^0-55^0=65^0\)

b) Ta lại có: \(\widehat{B}=\widehat{mEf}=60^0\) ( đồng vị )

c) Gọi đoạn thẳng AB là d

đoạn thẳng EF là g

đoạn thẳng BC là h

Ta sẽ có:\(d\cap g=\left\{E\right\}\) Và \(d\cap h=\left\{B\right\}\)

Và \(\widehat{B}=\widehat{E}=60^0\)( đồng vị, theo c/m a)

\(\Rightarrow g//h\) hay \(EF//BC\)

d) Ta lại có: \(\widehat{C}=\widehat{F}=\widehat{N}=55^0\)( đồng vị )

\(\Rightarrow\widehat{aNm}=55^0\)

Chúc bạn học tốt

\(1,\)\(a,\dfrac{2+3}{9+3}=\dfrac{5}{12}\)

\(\Leftrightarrow\dfrac{5}{12}=\dfrac{5.3}{12.3}=\dfrac{15}{36}\) và \(\dfrac{6}{9}=\dfrac{24}{36}\)

\(\dfrac{15}{36}< \dfrac{24}{36}\Rightarrow\dfrac{5}{12}< \dfrac{6}{9}\) hay \(\dfrac{2+3}{9+3}< \dfrac{6}{9}\)

\(\dfrac{2}{3}=\dfrac{8}{12}\Rightarrow\dfrac{5}{12}< \dfrac{2}{3}\) hay \(\dfrac{2+3}{9+3}< \dfrac{2}{3}\)

\(b,\dfrac{6-2}{9-3}=\dfrac{4}{6}=\dfrac{2}{3}\Rightarrow\dfrac{6-2}{9-3}=\dfrac{2}{3}\)

\(\dfrac{2}{3}=\dfrac{6}{9}\Rightarrow\dfrac{6-2}{9-3}=\dfrac{6}{9}\)

\(2,a,\dfrac{3+12}{5+20}=\dfrac{15}{25}=\dfrac{3}{5}\Rightarrow\dfrac{3+12}{5+20}=\dfrac{3}{5}\)

\(\dfrac{3}{5}=\dfrac{3.4}{5.4}=\dfrac{12}{20}\Rightarrow\dfrac{3+12}{5+20}=\dfrac{12}{20}\)

\(b,\dfrac{3-12}{5-20}=\dfrac{-9}{-15}=\dfrac{9}{15}=\dfrac{3}{5}\Rightarrow\dfrac{3-12}{5-20}=\dfrac{3}{5}\)

\(\dfrac{3}{5}=\dfrac{3.4}{5.4}=\dfrac{12}{20}\Rightarrow\dfrac{3-12}{5-20}=\dfrac{12}{20}\)