\(\left[6\cdot\left(\dfrac{1}{3}-3\cdot\dfrac{-1}{3}\right)+1\right]:\dfrac{-1}{3}-1\\=\left[6\cdot\left(\dfrac{1}{3}+1\right)+1\right]:\dfrac{-1}{3}-1\\ =\left(6\cdot\dfrac{4}{3}+1\right)\cdot\left(-3\right)-1\\ =\left(8+1\right)\cdot\left(-3\right)-1\\ =9\cdot\left(-3\right)-1\\ =-27-1\\ =-28\)

\(A=x+\dfrac{0,2-0,375+\dfrac{5}{11}}{-0,3+\dfrac{9}{16}-\dfrac{15}{22}}\\ =x+\dfrac{\dfrac{2}{10}-\dfrac{3}{8}+\dfrac{5}{11}}{-\dfrac{3}{10}+\dfrac{9}{16}-\dfrac{15}{22}}\\ =x+\dfrac{\dfrac{2}{10}-\dfrac{6}{16}+\dfrac{10}{22}}{\dfrac{-3}{10}+\dfrac{9}{16}-\dfrac{15}{22}}\\ =x+\dfrac{2\left(\dfrac{1}{10}-\dfrac{3}{16}+\dfrac{5}{22}\right)}{-3\left(\dfrac{1}{10}-\dfrac{3}{16}+\dfrac{5}{22}\right)}\\ =x-\dfrac{2}{3}\)

Thay x = -1/3 vào A ta có:

A = `-1/3-2/3=-3/3=-1` 

\(\dfrac{50-\dfrac{4}{13}+\dfrac{2}{15}-\dfrac{2}{17}}{100-\dfrac{8}{13}+\dfrac{4}{15}-\dfrac{4}{17}}\\ =\dfrac{50-\dfrac{4}{13}+\dfrac{2}{15}-\dfrac{2}{17}}{2\left(50-\dfrac{4}{13}+\dfrac{2}{15}-\dfrac{2}{17}\right)}\\ =\dfrac{1}{2}\) 

\(\dfrac{3+\dfrac{3}{17}-\dfrac{3}{11}+\dfrac{3}{1001}-\dfrac{3}{13}}{\dfrac{9}{1001}-\dfrac{9}{13}+\dfrac{9}{17}-\dfrac{9}{11}+9}\\ =\dfrac{3+\dfrac{3}{17}-\dfrac{3}{11}+\dfrac{3}{1001}-\dfrac{3}{13}}{9+\dfrac{9}{17}-\dfrac{9}{11}+\dfrac{9}{1001}-\dfrac{9}{13}}\\ =\dfrac{3+\dfrac{3}{17}-\dfrac{3}{11}+\dfrac{3}{1001}-\dfrac{3}{13}}{3\left(3+\dfrac{3}{17}-\dfrac{3}{11}+\dfrac{3}{1001}-\dfrac{3}{13}\right)}\\ =\dfrac{1}{3}\)

\(\dfrac{x}{2}=\dfrac{y}{3}=>\dfrac{x}{8}=\dfrac{y}{12};\dfrac{y}{12}=\dfrac{z}{21}\\ =>\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{z}{21}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có:

\(\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{z}{21}=\dfrac{2x-y+z}{2\cdot8-12+21}=\dfrac{50}{25}=2\\ =>\dfrac{x}{8}=2=>x=2\cdot8=16\\ =>\dfrac{z}{12}=2=>z=2\cdot12=24\\ =>\dfrac{z}{21}=2=>z=2\cdot21=42\)

 tôi thấy nó raats jnjdnjnnv

\(x\left(2x+\dfrac{-4}{10}\right)\) = 0

\(\left[{}\begin{matrix}x=0\\2x-\dfrac{4}{10}=10\end{matrix}\right.\)

\(\left[{}\begin{matrix}x=0\\2x=\dfrac{4}{10}\end{matrix}\right.\)

\(\left[{}\begin{matrix}x=0\\x=\dfrac{4}{10}:2\end{matrix}\right.\)

\(\left[{}\begin{matrix}x=0\\x=\dfrac{1}{5}\end{matrix}\right.\)

Vậy \(x\) \(\in\) {0; \(\dfrac{1}{5}\)}

\(x\left(2x+\dfrac{-4}{10}\right)=0\\ =>x\left(2x+\dfrac{-2}{5}\right)=0\\ =>2x\left(x-\dfrac{1}{5}\right)=0\\ TH1:2x=0\\ =>x=0\\ TH2:x-\dfrac{1}{5}=0\\ =>x=\dfrac{1}{5}\)

Số đối của 5/6 là -5/6

Số đối của \(-\dfrac{5}{2}\) là \(\dfrac{5}{2}\)

Số đối của 0,8 là -0,8

Biểu diễn:

\(a.\left(4x+1\right)\left(-2x+\dfrac{1}{3}\right)=0\\ TH1:4x+1=0\\ =>4x=-1\\ =>x=-\dfrac{1}{4}\\ TH2:-2x+\dfrac{1}{3}=0\\ =>2x=\dfrac{1}{3}\\ =>x=\dfrac{1}{3}:2=\dfrac{1}{6}\\ b.\left(x-\dfrac{5}{2}\right)^3=\dfrac{-1}{8}\\ =>\left(x-\dfrac{5}{2}\right)^3=\left(-\dfrac{1}{2}\right)^2\\ =>x-\dfrac{5}{2}=-\dfrac{1}{2}\\ =>x=-\dfrac{1}{2}+\dfrac{5}{2}\\ =>x=\dfrac{4}{2}=2\\ c.\left(\dfrac{2}{5}-3x\right)^2-\dfrac{1}{5}=\dfrac{4}{25}\\ =>\left(\dfrac{2}{5}-3x\right)^2=\dfrac{4}{25}+\dfrac{1}{5}=\dfrac{9}{25}=\left(\dfrac{3}{5}\right)^2\\TH1:\dfrac{2}{5}-3x=\dfrac{3}{5}\\ =>3x=\dfrac{2}{5}-\dfrac{3}{5}=-\dfrac{1}{5}\\ =>x=\dfrac{-1}{5}:3=-\dfrac{1}{15}\\ TH2:\dfrac{2}{5}-3x=-\dfrac{3}{5}=>3x=\dfrac{2}{5}+\dfrac{3}{5}=1\\ =>x=1:3=\dfrac{1}{3}\) 

\(d.\left(\dfrac{2}{3}\right)^{x+2}+\left(\dfrac{2}{3}\right)^{x+1}=\dfrac{20}{27}\\ =>\left(\dfrac{2}{3}\right)^{x+1}\cdot\left(\dfrac{2}{3}+1\right)=\dfrac{20}{27}\\ =>\left(\dfrac{2}{3}\right)^{x+1}\cdot\dfrac{5}{3}=\dfrac{20}{27}\\ =>\left(\dfrac{2}{3}\right)^{x+1}=\dfrac{20}{27}:\dfrac{5}{3}=\dfrac{4}{9}=\left(\dfrac{2}{3}\right)^2\\ =>x+1=2\\ =>x=2-1\\ =>x=1\)

Bài 1:

a: OM là phân giác của góc AOB

=>\(\widehat{AOM}=\dfrac{\widehat{AOB}}{2}=\dfrac{120^0}{2}=60^0\)

b: Vì OM và OM là hai tia trùng nhau

nên \(\widehat{MOM}=0^0\)

a: Ta có: \(\widehat{A'OC}+\widehat{AOC}=180^0\)(kề bù)

=>\(\widehat{AOC}+90^0=180^0\)

=>\(\widehat{AOC}=90^0\)

Trên cùng một nửa mặt phẳng bờ chứa tia OA, ta có: \(\widehat{AOB}< \widehat{AOC}\)

nên tia OB nằm giữa hai tia OA và OC

=>\(\widehat{AOB}+\widehat{BOC}=\widehat{AOC}\)

=>\(\widehat{BOC}=90^0-45^0=45^0\)

Ta có: \(\widehat{AOB}=\widehat{BOC}\)

mà tia OB nằm giữa hai tia OA và OC

nên OB là phân giác của góc AOC

b: Ta có: \(\widehat{COB}+\widehat{COE}=180^0\)(hai góc kề bù)

=>\(\widehat{COE}+45^0=180^0\)

=>\(\widehat{COE}=135^0\)