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VE HINH
a) Ta có : tia Oy nằm giữa hai tia Ox và Oz
=> góc xOz = góc xOy + góc yOz
=> yOz = xOz - xOy=75-35=40do
b) Ta có : góc yOt = góc xOt + góc xOy ( Vì xOt và xOy là hai góc kề bù )
=> góc xOt = góc yOt - góc xOy = 180 - 35 =145 độ ok nha bạn
hình bạn tự vẽ nha
b) Vì Ot là tia đối của tia Oy => tOy là góc bẹ mà góc bẹt có tổng số đo = 180độ
=> tOy - xOy = xOt
=> 180độ - 25độ = xOt
=> xOt = 155độ
nhớ thêm dấu góc vào nha
\(a,\frac{1}{2}+\frac{2}{3}x=\frac{4}{5}\)
=> \(\frac{2}{3}x=\frac{4}{5}-\frac{1}{2}=\frac{3}{10}\)
=> \(x=\frac{3}{10}:\frac{2}{3}=\frac{9}{20}\)
Vậy \(x\in\left\{\frac{9}{20}\right\}\)
\(b,x+\frac{1}{4}=\frac{4}{3}\)
=> \(x=\frac{4}{3}-\frac{1}{4}=\frac{13}{12}\)
Vậy \(x\in\left\{\frac{13}{12}\right\}\)
\(c,\frac{3}{5}x-\frac{1}{2}=-\frac{1}{7}\)
=> \(\frac{3}{5}x=-\frac{1}{7}+\frac{1}{2}=\frac{5}{14}\)
=> \(x=\frac{5}{14}:\frac{3}{5}=\frac{25}{42}\)
Vậy \(x\in\left\{\frac{25}{42}\right\}\)
\(d,\left|x+5\right|-6=9\)
=> \(\left|x+5\right|=9+6=15\)
=> \(\left[{}\begin{matrix}x+5=15\\x+5=-15\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=15-5=10\\x=-15-5=-20\end{matrix}\right.\)
Vậy \(x\in\left\{10;-20\right\}\)
\(e,\left|x-\frac{4}{5}\right|=\frac{3}{4}\)
=> \(\left[{}\begin{matrix}x-\frac{4}{5}=\frac{3}{4}\\x-\frac{4}{5}=-\frac{3}{4}\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=\frac{3}{4}+\frac{4}{5}=\frac{31}{20}\\x=-\frac{3}{4}+\frac{4}{5}=\frac{1}{20}\end{matrix}\right.\)
Vậy \(x\in\left\{\frac{31}{20};\frac{1}{20}\right\}\)
\(f,\frac{1}{2}-\left|x\right|=\frac{1}{3}\)
=> \(\left|x\right|=\frac{1}{2}-\frac{1}{3}\)
=> \(\left|x\right|=\frac{1}{6}\)
=> \(\left[{}\begin{matrix}x=\frac{1}{6}\\x=-\frac{1}{6}\end{matrix}\right.\)
Vậy \(x\in\left\{\frac{1}{6};-\frac{1}{6}\right\}\)
\(g,x^2=16\)
=> \(\left|x\right|=\sqrt{16}=4\)
=> \(\left[{}\begin{matrix}x=4\\x=-4\end{matrix}\right.\)
vậy \(x\in\left\{4;-4\right\}\)
\(h,\left(x-\frac{1}{2}\right)^3=\frac{1}{27}\)
=> \(x-\frac{1}{2}=\sqrt[3]{\frac{1}{27}}=\frac{1}{3}\)
=> \(x=\frac{1}{3}+\frac{1}{2}=\frac{5}{6}\)
Vậy \(x\in\left\{\frac{5}{6}\right\}\)
\(i,3^3.x=3^6\)
\(x=3^6:3^3=3^3=27\)
Vậy \(x\in\left\{27\right\}\)
\(J,\frac{1,35}{0,2}=\frac{1,25}{x}\)
=> \(x=\frac{1,25.0,2}{1,35}=\frac{5}{27}\)
Vậy \(x\in\left\{\frac{5}{27}\right\}\)
\(k,1\frac{2}{3}:x=6:0,3\)
=> \(\frac{5}{3}:x=20\)
=> \(x=\frac{5}{3}:20=\frac{1}{12}\)
Vậy \(x\in\left\{\frac{1}{12}\right\}\)
Câu 1 :
1. \(\frac{-17}{30}-\frac{11}{-15}+\frac{-7}{12}\)
\(=\frac{-17}{30}+\frac{22}{30}+\frac{-7}{12}\)
\(=\frac{2}{12}+\frac{-7}{12}\)
\(=-\frac{5}{12}\)
Câu 2 :
\(x+\frac{-7}{15}=-1\frac{1}{20}\)
\(x=-\frac{21}{20}-\frac{-7}{15}\)
\(\Rightarrow x=-\frac{7}{12}\)
1)
a) x - 4/3 = 2 1/3 b) 4/7 - 1/7x =13/14
x - 4/3 = 7/3 1/7x = 4/7 - 13/14
x = 7/3 + 4/3 1/7x = 8/14 - 13/14
x = 11/3 1/7x = -5/14
Vậy x = 11/3 Vậy x = -5/14
\(\frac{4}{7}-\frac{1}{7}x=\frac{13}{14}\)
\(\frac{4}{7}-\frac{1}{7}x=\frac{13}{14}+\frac{4}{7}=\frac{13}{14}+\frac{8}{14}\)
\(\frac{1}{7}x=\frac{21}{14}\)
\(x=\frac{21}{14}:\frac{1}{7}=\frac{21}{14}\times\frac{7}{1}\)
\(x=\frac{147}{14}\)
\(\text{Câu 1 :}\)
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{12.13}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{12}-\frac{1}{13}\)
\(=\frac{1}{1}-\frac{1}{13}\)
\(=\frac{12}{13}\)
\(\text{Câu 2 :}\)
\(\frac{5}{1.3}+\frac{5}{3.5}+\frac{5}{5.7}+...+\frac{5}{99.101}\)
\(=\frac{5}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\right)\)
\(=\frac{5}{2}.\left(\frac{1}{1}-\frac{1}{101}\right)\)
\(=\frac{5}{2}.\frac{100}{101}\)
\(=\frac{250}{101}\)
Câu 2 :
a) \(\frac{1}{4}\) + \(\frac{1}{3}\): (3x) = -5
⇒ \(\frac{1}{3}\): (3x) = -5 - \(\frac{1}{4}\)
⇒ \(\frac{1}{3}\): (3x) = \(\frac{-21}{4}\)
⇒ 3x = \(\frac{1}{3}\): \(\frac{-21}{4}\)
⇒ 3x = \(\frac{-4}{63}\)
⇒ x = \(\frac{-4}{63}\):3
⇒ x = \(\frac{-4}{189}\)
Vậy x = \(\frac{-4}{189}\)
b) (3x-4) . (5x+15)=0
⇒ xảy ra 2 trường hợp 3x-4=0 ; 5x+15=0
* 3x-4 =0
⇒ 3x =0+4
⇒ 3x =4
⇒ x =4:3
⇒ Vô lý không tính được bạn nhé
* 5x+15 = 0
⇒ 5x = 0-15
⇒ 5x = -15
⇒ x = -15:5
⇒ x = -3
Vậy x ∈ ∅ và x ∈ 3
c) |2x-1| = \(\frac{11}{2}\)
⇒ xảy ra 2 trường hợp 2x-1 = \(\frac{11}{2}\); 2x-1 = \(\frac{-11}{2}\)
* 2x-1=\(\frac{11}{2}\)
⇒ 2x = \(\frac{11}{2}\)+1
⇒ 2x = \(\frac{9}{2}\)
⇒ x = \(\frac{9}{2}\):2
⇒ x = \(\frac{9}{4}\)
* 2x-1 = \(\frac{-11}{2}\)
⇒ 2x = 1 + \(\frac{-11}{2}\)
⇒ 2x = \(\frac{-9}{2}\)
⇒ x = \(\frac{-9}{2}\):2
⇒ x = \(\frac{-9}{4}\)
Vậy x ∈ { \(\frac{9}{4}\); \(\frac{-9}{4}\)}
Câu 2:
a) \(\frac{1}{4}+\frac{1}{3}:\left(3x\right)=-5\)
\(\Leftrightarrow\frac{1}{3}\cdot\frac{1}{3x}=-5-\frac{1}{4}=-\frac{21}{4}\)
\(\Leftrightarrow\frac{1}{9x}=\frac{-21}{4}\)
\(\Leftrightarrow9x=\frac{4\cdot1}{-21}=-\frac{4}{21}\)
hay \(x=-\frac{4}{21}:9=-\frac{4}{189}\)
Vậy: \(x=-\frac{4}{189}\)
b) Ta có: \(\left(3x-4\right)\left(5x+15\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-4=0\\5x+15=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=4\\5x=-15\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{4}{3}\\x=-3\end{matrix}\right.\)
Vậy: \(x\in\left\{\frac{4}{3};-3\right\}\)
c) Ta có: \(\left|2x-1\right|=\frac{11}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=\frac{11}{2}\\2x-1=-\frac{11}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=\frac{11}{2}+1=\frac{13}{2}\\2x=-\frac{11}{2}+1=-\frac{9}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{13}{2}:2=\frac{13}{4}\\x=-\frac{9}{2}:2=-\frac{9}{4}\end{matrix}\right.\)
Vậy: \(x\in\left\{\frac{13}{4};\frac{-9}{4}\right\}\)
Câu 3:
a) Ta có: \(1-3\cdot\left[4-30:\left(-18+3\right)\right]\)
\(=1-3\cdot\left[4-30:\left(-15\right)\right]\)
\(=1-3\cdot\left[4-\left(-2\right)\right]\)
\(=1-3\cdot6=1-18=-17\)
b) Ta có: \(\frac{5\cdot7+5\cdot\left(-4\right)}{21\cdot5}\)
\(=\frac{5\cdot\left(7-4\right)}{5\cdot21}=\frac{3}{21}=\frac{1}{7}\)
c) Ta có: \(\frac{-2}{9}+\frac{5}{4}+\left(-\frac{1}{6}\right):\frac{3}{5}+\frac{1}{18}\)
\(=-\frac{2}{9}+\frac{5}{4}-\frac{5}{18}+\frac{1}{18}\)
\(=-\frac{2}{9}+\frac{5}{4}-\frac{4}{18}\)
\(=\frac{-8}{36}+\frac{45}{36}-\frac{8}{36}=\frac{29}{36}\)