Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) \(x-\dfrac{3}{4}=-\dfrac{5}{8}\Rightarrow x=-\dfrac{5}{8}+\dfrac{3}{4}\Rightarrow x=\dfrac{1}{8}\)
b) \(x+\dfrac{5}{8}=-\dfrac{1}{4}\Rightarrow x=-\dfrac{1}{4}-\dfrac{5}{8}\Rightarrow x=-\dfrac{7}{8}\)
c) \(\dfrac{5}{6}+\dfrac{3}{4}x=\dfrac{5}{24}\Rightarrow x=\left(\dfrac{5}{24}-\dfrac{5}{6}\right):\dfrac{3}{4}\Rightarrow x=-\dfrac{5}{6}\)
d) \(\dfrac{3}{8}-\dfrac{2}{3}:x=-\dfrac{5}{12}\Rightarrow\dfrac{2}{3}:x=\dfrac{3}{8}+\dfrac{5}{12}\Rightarrow\dfrac{2}{3}:x=\dfrac{19}{24}\Rightarrow x=\dfrac{2}{3}:\dfrac{19}{24}=\dfrac{16}{19}\)
a) \(x-\dfrac{3}{4}=-\dfrac{5}{8}\\ \Rightarrow x=\dfrac{1}{8}\)
b) \(x+\dfrac{5}{8}=-\dfrac{1}{4}\\ \Rightarrow x=-\dfrac{7}{8}\)
c) \(\dfrac{5}{6}+\dfrac{3}{4}x=\dfrac{5}{24}\\ \Rightarrow\dfrac{3}{4}x=-\dfrac{5}{8}\\ \Rightarrow x=-\dfrac{5}{6}\)
d) \(\dfrac{3}{8}-\dfrac{2}{3}:x=-\dfrac{5}{12}\\ \Rightarrow\dfrac{2}{3}:x=\dfrac{19}{24}\\ \Rightarrow x=\dfrac{16}{19}\)
e) \(\left(6,5-2x\right):\dfrac{5}{13}=\dfrac{13}{10}\\ \Rightarrow6,5-2x=\dfrac{1}{2}\\ \Rightarrow2x=6\\ \Rightarrow x=3\)
f) \(\left|\dfrac{1}{3}x+\dfrac{1}{2}\right|-\dfrac{3}{4}=-\dfrac{1}{6}\\ \Rightarrow\left|\dfrac{1}{3}x+\dfrac{1}{2}\right|=\dfrac{7}{12}\\ \Rightarrow\left[{}\begin{matrix}\dfrac{1}{3}x+\dfrac{1}{2}=\dfrac{7}{12}\\\dfrac{1}{3}x+\dfrac{1}{2}=-\dfrac{7}{12}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{4}\\x=-\dfrac{13}{4}\end{matrix}\right.\)
g) \(\dfrac{x-3}{3}=\dfrac{2x+3}{5}\\ \Rightarrow5x-15=6x+9\\ \Rightarrow-x=24\\ \Rightarrow x=-24\)
h) \(\dfrac{x-5}{6}=\dfrac{6}{x-5}\\ \Rightarrow\left(x-5\right)^2=6^2\\ \Rightarrow\left[{}\begin{matrix}x-5=-6\\x-5=6\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-1\\x=11\end{matrix}\right.\)
Bài 4:
a) Ta có: \(\widehat{yOz}+\widehat{xOy}=180^0\)(2 góc kề bù)
\(\Rightarrow\widehat{yOz}=180^0-\widehat{xOy}=180^0-50^0=130^0\)
b) Ta có: \(\widehat{zOt}=\widehat{yOt}=\dfrac{1}{2}\widehat{yOz}=\dfrac{1}{2}.130^0=65^0\)(do Ot là tia phân giác \(\widehat{yOz}\))
c) Ta có: \(\widehat{xOt}=\widehat{yOt}+\widehat{xOy}=65^0+50^0=115^0\)
Bài 5:
a) Ta có: \(\widehat{xOz}+\widehat{xOy}=180^0\)(2 góc kề bù)
\(\Rightarrow\widehat{xOz}=180^0-\widehat{xOy}=180^0-110^0=70^0\)
b) Ta có: \(\widehat{zOt}=\dfrac{1}{2}\widehat{xOz}=\dfrac{1}{2}.70^0=35^0\)( Ot là tia phân giác \(\widehat{xOz}\))
c) Ta có: \(\widehat{xOt}=\widehat{zOt}=35^0\)( Ot là tia phân giác \(\widehat{xOz}\))
Bài 4:
a: Ta có: \(\widehat{xOy}+\widehat{yOz}=180^0\)
\(\Leftrightarrow\widehat{yOz}=180^0-50^0\)
\(\Leftrightarrow\widehat{yOz}=130^0\)
b: \(\widehat{zOt}=\dfrac{\widehat{yOz}}{2}=65^0\)
=23 x (58-30) + 28 x 77
=23 x 28 +28 x 77
=28 x (23+77)
=28 x 100
=2800
học tốt bạn nhé
bài này là dạng nâng cao về toán tính nhanh, mik nghĩ là ẽ ít bạn trả lời đc
Bài 3:
a: =>1<x<=4
hay \(x\in\left\{2;3;4\right\}\)
b: =>-104<x<-99
hay \(x\in\left\{-103;-102;-101;-100\right\}\)
c: ⇔n+2∈{1;−1;5;−5}⇔n+2∈{1;−1;5;−5}
hay n∈{−1;−3;3;−7}n∈{−1;−3;3;−7}
d: ⇔n+2∈{1;−1;2;−2;4;−4}⇔n+2∈{1;−1;2;−2;4;−4}
hay n∈{−1;−3;0;−4;2;−6}n∈{−1;−3;0;−4;2;−6}
a: ⇔n−1∈{1;−1;5;−5}⇔n−1∈{1;−1;5;−5}
hay n∈{2;0;6;−4}
b: \(\Leftrightarrow n+1\in\left\{1;-1;2;-2;4;-4\right\}\)
hay \(n\in\left\{0;-2;1;-3;3;-5\right\}\)
c: \(\Leftrightarrow n+2\in\left\{1;-1;5;-5\right\}\)
hay \(n\in\left\{-1;-3;3;-7\right\}\)
d: \(\Leftrightarrow n+2\in\left\{1;-1;2;-2;4;-4\right\}\)
hay \(n\in\left\{-1;-3;0;-4;2;-6\right\}\)
a: \(\Leftrightarrow n-1\in\left\{1;-1;5;-5\right\}\)
hay \(n\in\left\{2;0;6;-4\right\}\)
Trả lời:
1, \(27^{20}-3^{56}=\left(3^3\right)^{20}-3^{56}\)
\(=3^{60}-3^{56}\)
\(=3^{55}.\left(3^5-3\right)\)
\(=3^{55}.\left(243-3\right)\)
\(=3^{55}\times240\)\(⋮240\)
Vậy \(27^{20}-3^{56}\)chia hết cho 240
2, Ta có: \(3a+7b⋮19\)
\(\Leftrightarrow2.\left(3a+7b\right)⋮19\)
\(\Leftrightarrow6a+14b⋮19\)
\(\Leftrightarrow6a+33b-19b⋮19\)
\(\Leftrightarrow3.\left(2a+11b\right)-19b⋮19\)
Do \(19b\)chia hết cho 19. Theo t/c chia hết của 1 hiệu thì \(3.\left(2a+11b\right)⋮19\Leftrightarrow2a+11b⋮19\)
Vậy \(2a+11b\)chia hết cho 19
\(\frac{1}{4}+\frac{1}{3}:\left(2x-1\right)=-5\)
\(\frac{1}{3}:\left(2x-1\right)=-5-\frac{1}{4}\)
\(\frac{1}{3}:\left(2x-1\right)=-\frac{20}{4}-\frac{1}{4}\)
\(\frac{1}{3}:\left(2x-1\right)=-\frac{21}{4}\)
\(\left(2x-1\right)=\frac{1}{3}:-\frac{21}{4}\)
\(\left(2x-1\right)=\frac{1}{3}.-\frac{4}{21}\)
\(\left(2x-1\right)=-\frac{4}{63}\)
2x= -4/63 + 1
2x = 59/63
x = 59/63 : 2
x = 59/126
1/3:(2.x-1)=-5-1/4
1/3:(2.x-1)=-21/4
2.x-1=1/3:-21/4
2.x-1=-4/63
2.x=-4/63+1
2.x=\(3\frac{59}{63}\)
x=\(3\frac{59}{63}\):2
x=\(1\frac{61}{63}\)
h: Ta có: \(\dfrac{5}{x+3}=\dfrac{x+3}{5}\)
\(\Leftrightarrow\left(x+3\right)^2=25\)
\(\Leftrightarrow\left[{}\begin{matrix}x+3=-5\\x+3=5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-8\\x=2\end{matrix}\right.\)