giúp mik 3 bài này vs ak
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a) \(2^4+8\left[\left(-2\right)^2:\dfrac{1}{2}\right]^0-2^{-2}.4+\left(-2\right)^2\)
\(=2^4+8.1-\dfrac{1}{4}.4+4\)
\(=16+8-1+4\)
\(=24-1+4\)
\(=23+4\)
\(=27\)
Bài 3:
Ta có: a//b
nên \(x+y=180\)
mà \(2x-3y=0\)
nên \(\left\{{}\begin{matrix}x+y=180\\2x-3y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x+2y=180\\2x-3y=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}5y=180\\x+y=180\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=36\\x=144\end{matrix}\right.\)
1.
a//b mà b⊥CD nên a⊥CD
Do đó \(\widehat{D}=90^0\)
Góc A là góc nào??
2.
a, Vì a và b cùng vuông góc với MN nên a//b
b, a//b \(\Rightarrow\widehat{P}+\widehat{Q}=180^0\left(trong.cùng.phía\right)\Rightarrow\widehat{P}=70^0\)
10 + 20 + 10 + 90
= ( 10 + 20 ) + ( 10 + 90 )
= 30 + 100
= 130
HỌC TỐT
10 + 20 + 10 + 90
= ( 10 + 20 ) + ( 10 + 90 )
= 30 + 100
= 130
2 tá \(=24\)
Muốn mua 8 cái bút chì cần trả \(30000:24\times8=10000\left(VNĐ\right)\)
Số tiền phải trả khi mua 8 cái bút là:
\(30000:24\cdot8=10000\left(đồng\right)\)
\(\left|2x-3\right|=3-2x\)
\(ĐK:x\le\dfrac{3}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-3=3-2x\\3-2x=3-2x\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\0=0\left(đúng\right)\end{matrix}\right.\)
Vậy \(S=\left\{x\in R;x=\dfrac{3}{2}\right\}\)
a. Ta có: \(\dfrac{x}{-2}=\dfrac{y}{-3}=\dfrac{4x}{-8}=\dfrac{3y}{-9}\)
Áp dụng tính chất dãy tỉ số bằng nhau, ta được:
\(\dfrac{4x-3y}{-8-\left(-9\right)}=\dfrac{9}{1}=9\)
=> \(\dfrac{x}{-2}=\dfrac{y}{-3}=9\)
=> \(\left\{{}\begin{matrix}x=-18\\y=-27\end{matrix}\right.\)
b. Ta có: \(\dfrac{x}{2}=\dfrac{y}{-3}=\dfrac{z}{5}=\dfrac{2x}{4}=\dfrac{3y}{-9}=\dfrac{5z}{25}\)
Áp dụng tính chất dãy tỉ số bằng nhau, ta được:
\(\dfrac{2x+3y+5z}{4+\left(-9\right)+25}=\dfrac{6}{20}=\dfrac{3}{10}\)
=> \(\left\{{}\begin{matrix}x=0,6\\y=-0,9\\z=0,15\end{matrix}\right.\)
5 + 9/14 = 70/14 + 9/14 = 79/14
6/54 : 8 = 6/54 x 1/8 = 1/72
5 - 13/17 = 85/17 - 13/17 = 72/17
3 : 9/10 = 3/1 x 10/9 = 10/3
24/72 - 202/909 = 1/9
9/11 : 5/13 = 9/11 x 13/5 = 117/55
\(5 +\) \(\dfrac{9}{14}\) = \(\dfrac{70}{14}\) + \(\dfrac{9}{14}\) = \(\dfrac{79}{14}\)
\(\dfrac{6}{54}\) \(: 8 =\) \(\dfrac{1}{9}\) \(:\) \(\dfrac{8}{1}\) = \(\dfrac{1}{9}\) x \(\dfrac{1}{8}\) = \(\dfrac{1}{72}\)
\(5 -\) \(\dfrac{13}{17}\) = \(\dfrac{85}{17}\) \(-\) \(\dfrac{13}{17}\) = \(\dfrac{72}{17}\)
\(3\) \(:\) \(\dfrac{9}{10}\) = \(\dfrac{3}{1}\) \(:
\) \(\dfrac{9}{10}\) = \(\dfrac{3}{1}\) x \(\)\(\dfrac{10}{9}\) = \(\dfrac{30}{9}\) = \(\dfrac{10}{3}\)
\(\dfrac{24}{72}\) - \(\dfrac{202}{909}\) = \(\dfrac{3}{9}\) - \(\dfrac{2}{9}\) = \(\dfrac{1}{9}\)
\(\dfrac{9}{11}\) : \(\dfrac{5}{13}\) = \(\dfrac{9}{11}\) x \(\dfrac{13}{5}\) = \(\dfrac{117}{55}\)
0,15km=150m
Tổng số phần bằng nhau là 2+3=5(phần)
Chiều rộng là:
\(150:5\times2=60\left(m\right)\)
Chiều dài là 150-60=90(m)
Diện tích sân trường là:
\(60\times90=5400\left(m^2\right)=0,54\left(ha\right)\)





giúp mik bài 1 vs ak
BÀi 2:
\(\left|x-\frac{2019}{2020}\right|\ge0\forall x\)
=>\(-\frac{2020}{2019}\left|x-\frac{2019}{2020}\right|\le0\forall x\)
=>\(-\frac{2020}{2019}\left|x-\frac{2019}{2020}\right|+\frac{2019}{2020}\le\frac{2019}{2020}\forall x\)
Dấu '=' xảy ra khi \(x-\frac{2019}{2020}=0\)
=>\(x=\frac{2019}{2020}\)
Bài 1:
a: \(A=\frac{155-\frac{10}{7}-\frac{5}{11}+\frac{5}{23}}{402-\frac{26}{7}-\frac{13}{11}+\frac{13}{23}}+\frac{\frac35+\frac{3}{13}-0,9}{\frac{7}{91}+0,2-\frac{3}{10}}\)
\(=\frac{5\left(31-\frac27-\frac{1}{11}+\frac{1}{23}\right)}{13\left(31-\frac27-\frac{1}{11}+\frac{1}{23}\right)}+\frac{3\left(\frac15+\frac{1}{13}-\frac{3}{10}\right)}{\frac15+\frac{1}{13}-\frac{3}{10}}=\frac{5}{13}+3=\frac{44}{13}\)
b: \(B=-\frac54+\frac35-1\frac{3}{14}:\left|-\frac{34}{21}\right|+\frac{-5}{17}:\left|-\frac{1}{34}\right|+\frac53\cdot\left(\frac12-\frac25\right)\)
\(=-\frac54+\frac35-\frac{17}{14}\cdot\frac{21}{34}+\frac{-5}{17}\cdot34+\frac53\cdot\frac{1}{10}\)
\(=-\frac54+\frac35-\frac{3}{2\cdot2}+\left(-10\right)+\frac{5}{30}=-\frac54+\frac35-\frac34+\left(-10\right)+\frac16\)
\(=-\frac{75}{60}+\frac{36}{60}-\frac{45}{60}+\frac{\left(-600\right)}{60}+\frac{10}{60}=\frac{-674}{60}=-\frac{337}{30}\)
c: \(C=1-\frac{1}{5\cdot10}-\frac{1}{10\cdot15}-\cdots-\frac{1}{95\cdot100}\)
\(=1-\frac15\left(\frac{5}{5\cdot10}+\frac{5}{10\cdot15}+\cdots+\frac{5}{95\cdot100}\right)\)
\(=1-\frac15\left(\frac15-\frac{1}{10}+\frac{1}{10}-\frac{1}{15}+\cdots+\frac{1}{95}-\frac{1}{100}\right)\)
\(=1-\frac15\left(\frac15-\frac{1}{100}\right)=1-\frac15\cdot\frac{19}{100}=1-\frac{19}{500}=\frac{481}{500}\)
d: \(D=\frac17+\frac{1}{91}+\frac{1}{247}+\frac{1}{475}+\frac{1}{775}+\frac{1}{1147}\)
\(=\frac{1}{1\cdot7}+\frac{1}{7\cdot13}+\cdots+\frac{1}{31\cdot37}\)
\(=\frac16\left(\frac{6}{1\cdot7}+\frac{6}{7\cdot13}+\cdots+\frac{6}{31\cdot37}\right)=\frac16\left(1-\frac17+\frac17-\frac{1}{13}+\cdots+\frac{1}{31}-\frac{1}{37}\right)\)
\(=\frac16\left(1-\frac{1}{37}\right)=\frac16\cdot\frac{36}{37}=\frac{6}{37}\)