Bài 2: 

\(a,x-\dfrac{3}{10}=\dfrac{7}{15}\cdot\dfrac{3}{5}\\ =>x-\dfrac{3}{10}=\dfrac{7}{25}\\ =>x=\dfrac{7}{25}+\dfrac{3}{10}\\ =>x=\dfrac{29}{50}\\ b.2x+\dfrac{3}{2}=\dfrac{-2}{5}\\ =>2x=\dfrac{-2}{5}-\dfrac{3}{2}\\ =>x=\dfrac{-4}{10}-\dfrac{15}{10}=\dfrac{-19}{10}\\ =>x=\dfrac{-19}{10}:2=-\dfrac{19}{20}\\ c,\left(x-\dfrac{1}{2}\right)^3=-8\\ =>\left(x-\dfrac{1}{2}\right)^3=\left(-2\right)^3\\ =>x-\dfrac{1}{2}=-2\\ =>x=-2+\dfrac{1}{2}\\ =>x=\dfrac{-3}{2}\\ d,\left(\dfrac{7}{5}\right)^x=\dfrac{49}{25}\\ =>\left(\dfrac{7}{5}\right)^x=\left(\dfrac{7}{5}\right)^2\\=>x=2\)

Câu 3:

a: \(\widehat{xOz}+\widehat{yOz}=180^0\)(hai góc kề bù)

=>\(\widehat{yOz}=180^0-\widehat{xOz}=180^0-60^0=120^0\)

b: Ot là phân giác của góc yOz

=>\(\widehat{yOt}=\widehat{zOt}=\dfrac{\widehat{yOz}}{2}=\dfrac{120^0}{2}=60^0\)

Ta có: \(\widehat{xOt}+\widehat{yOt}=180^0\)(hai góc kề bù)

=>\(\widehat{xOt}+60^0=180^0\)

=>\(\widehat{xOt}=120^0\)

c: Ta có: \(\widehat{xOm}=\widehat{yOt}\)(hai góc đối đỉnh)

mà \(\widehat{yOt}=60^0\)

nên \(\widehat{xOm}=60^0\)

Ta có: \(\widehat{xOm}=\widehat{xOz}\left(=60^0\right)\)

=>Ox là phân giác của góc mOz

Câu 1:

b: \(\dfrac{11}{2}\cdot4\dfrac{5}{3}-2\dfrac{5}{3}\cdot\dfrac{11}{2}\)

\(=\dfrac{11}{2}\left(4+\dfrac{5}{3}-2-\dfrac{5}{3}\right)\)

\(=\dfrac{11}{2}\cdot2=11\)

d: \(\left(\dfrac{3}{7}\right)^0\cdot1^{15}+\dfrac{7}{9}:\left(\dfrac{2}{3}\right)^2-\dfrac{4}{5}\)

\(=1\cdot1+\dfrac{7}{9}:\dfrac{4}{9}-\dfrac{4}{5}\)

\(=1-\dfrac{4}{5}+\dfrac{7}{4}=\dfrac{1}{5}+\dfrac{7}{4}=\dfrac{4}{20}+\dfrac{35}{20}=\dfrac{39}{20}\)

\(\left[\left(-\dfrac{8}{3}\right)^2\right]^{1010}=\left(-\dfrac{8}{3}\right)^{2020}\)

\(\left[\left(\dfrac{7}{3}\right)^{505}\right]^3=\left(\dfrac{7}{3}\right)^{1515}\)

`-1,25 . (3/2 - 0,75) + 3,5`

`= -1,25 . (1,5 - 0,75) + 3,5`

`= -1,25 . 0,75 + 3,5`

`= -0,9375 + 3,5`

`= 2,5625`

\(-1,25\cdot\left(\dfrac{3}{2}-0,75\right)+3,5\\ =-\dfrac{5}{4}.\left(\dfrac{6}{4}-\dfrac{3}{4}\right)+\dfrac{7}{2}\\ =-\dfrac{5}{4}\cdot\dfrac{3}{4}+\dfrac{7}{2}\\ =-\dfrac{15}{16}+\dfrac{56}{16}\\ =\dfrac{41}{16}\)

TH1: \(a+b+c=0\Rightarrow\left\{{}\begin{matrix}a+b=-c\\b+c=-a\\c+a=-b\end{matrix}\right.\)

\(\Rightarrow P=\left(1+\dfrac{a}{-a}\right)\left(1+\dfrac{b}{-b}\right)\left(1+\dfrac{c}{-c}\right)=0\)

Th2: \(a+b+c\ne0\)

Áp dụng t/c dãy tỉ số bằng nhau:

\(\dfrac{a}{b+c}=\dfrac{b}{c+a}=\dfrac{c}{a+b}=\dfrac{a+b+c}{b+c+c+a+a+b}=\dfrac{a+b+c}{2\left(a+b+c\right)}=\dfrac{1}{2}\)

\(\Rightarrow P=\left(1+\dfrac{1}{2}\right)\left(1+\dfrac{1}{2}\right)\left(1+\dfrac{1}{2}\right)=\dfrac{3}{2}.\dfrac{3}{2}.\dfrac{3}{2}=\dfrac{27}{8}\)

TH1:

5x+10=0

5x=-10

x=-2

Th2:

2x-10=0

2x=10

x=5

Vậy x thuộc tập hợp -2 và 5

a: Vì OA và OB là hai tia đối nhau

nên O nằm giữa A và B

=>AB=OA+OB=6+2=8(cm)

b: I là trung điểm của AB

=>\(IA=IB=\dfrac{AB}{2}=4\left(cm\right)\)

Vì AI<AO

nên I nằm giữa A và O

=>AI+IO=AO

=>IO+4=6

=>IO=2(cm)

=>OA=3IO

c: Các góc đỉnh O có trên hình là \(\widehat{xOt};\widehat{xOz};\widehat{xOy};\widehat{tOz};\widehat{tOy};\widehat{zOy}\)

Ta có; ΔABC=ΔDEF

=>AB=DE; BC=EF; AC=DF; \(\widehat{BAC}=\widehat{EDF};\widehat{ABC}=\widehat{DEF};\widehat{ACB}=\widehat{DFE}\)

Xét ΔBAM và ΔEDN có

AB=DE

\(\widehat{ABM}=\widehat{DEN}\)

BM=EN

Do đó: ΔBAM=ΔEDN

=>AM=DN và \(\widehat{BAM}=\widehat{EDN}\)

3: \(564\left(\dfrac{12+\dfrac{12}{7}-\dfrac{12}{25}-\dfrac{12}{71}}{4+\dfrac{4}{7}-\dfrac{4}{25}-\dfrac{4}{71}}:\dfrac{3+\dfrac{3}{13}+\dfrac{3}{19}+\dfrac{3}{101}}{5+\dfrac{5}{13}+\dfrac{5}{19}+\dfrac{5}{101}}\right)\)

\(=564\left(\dfrac{12\left(1+\dfrac{1}{7}-\dfrac{1}{25}-\dfrac{1}{71}\right)}{4\left(1+\dfrac{1}{7}-\dfrac{1}{25}-\dfrac{1}{71}\right)}:\dfrac{3\left(1+\dfrac{1}{13}+\dfrac{1}{19}+\dfrac{1}{101}\right)}{5\left(1+\dfrac{1}{13}+\dfrac{1}{19}+\dfrac{1}{101}\right)}\right)\)

\(=564:\left(3\cdot\dfrac{5}{3}\right)=564\cdot5=2820\)

4: \(\dfrac{155-\dfrac{10}{7}-\dfrac{5}{11}+\dfrac{5}{23}}{402-\dfrac{26}{7}-\dfrac{13}{11}+\dfrac{13}{23}}+\dfrac{\dfrac{3}{5}+\dfrac{3}{13}-0,9}{\dfrac{7}{91}+0,2-\dfrac{3}{10}}\)

\(=\dfrac{5\left(31-\dfrac{2}{7}-\dfrac{1}{11}+\dfrac{1}{23}\right)}{13\left(31-\dfrac{2}{7}-\dfrac{1}{11}+\dfrac{1}{23}\right)}+\dfrac{\dfrac{3}{5}+\dfrac{3}{13}-\dfrac{9}{10}}{\dfrac{1}{13}+\dfrac{1}{5}-\dfrac{1}{10}}\)

\(=\dfrac{5}{13}+\dfrac{3\left(\dfrac{1}{5}+\dfrac{1}{13}-\dfrac{1}{10}\right)}{\dfrac{1}{5}+\dfrac{1}{13}-\dfrac{1}{10}}=\dfrac{5}{13}+3=\dfrac{44}{13}\)

5: \(\dfrac{0,375-0,3+\dfrac{3}{11}+\dfrac{3}{12}}{-0,625+0,5-\dfrac{5}{11}-\dfrac{5}{12}}+\dfrac{1,5+1-0,75}{2,5+\dfrac{5}{3}-1,25}\)

\(=-\dfrac{\dfrac{3}{8}-\dfrac{3}{10}+\dfrac{3}{11}+\dfrac{3}{12}}{\dfrac{5}{8}-\dfrac{5}{10}+\dfrac{5}{11}+\dfrac{5}{12}}+\dfrac{\dfrac{3}{2}+\dfrac{3}{3}-\dfrac{3}{4}}{\dfrac{5}{2}+\dfrac{5}{3}-\dfrac{5}{4}}\)

\(=-\dfrac{3\left(\dfrac{1}{8}-\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}\right)}{5\left(\dfrac{1}{8}-\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}\right)}+\dfrac{3\left(\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}\right)}{5\left(\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}\right)}\)

\(=-\dfrac{3}{5}+\dfrac{3}{5}=0\)

1: \(\left(\dfrac{1,5+1-0,75}{2,5+\dfrac{5}{3}-1,25}+\dfrac{0,375-0,3+\dfrac{3}{11}+\dfrac{3}{12}}{-0,625+0,5-\dfrac{5}{11}-\dfrac{5}{12}}\right):\dfrac{1890}{2005}+115\)

\(=\left(\dfrac{3\left(\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}\right)}{5\left(\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}\right)}+\dfrac{3\left(\dfrac{1}{8}-\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}\right)}{-5\left(\dfrac{1}{8}-\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}\right)}\right)\cdot\dfrac{2005}{1890}+115\)

\(=0\cdot\dfrac{2005}{1890}+115=115\)

2: \(\dfrac{\dfrac{1}{9}-\dfrac{1}{7}-\dfrac{1}{11}}{\dfrac{4}{9}-\dfrac{4}{7}-\dfrac{4}{11}}+\dfrac{0,6-\dfrac{3}{25}-\dfrac{3}{125}-\dfrac{3}{625}}{\dfrac{4}{5}-0,16-\dfrac{4}{125}-\dfrac{4}{625}}\)

\(=\dfrac{\dfrac{1}{9}-\dfrac{1}{7}-\dfrac{1}{11}}{4\left(\dfrac{1}{9}-\dfrac{1}{7}-\dfrac{1}{11}\right)}+\dfrac{3\left(\dfrac{1}{5}-\dfrac{1}{25}-\dfrac{1}{125}-\dfrac{1}{625}\right)}{4\left(\dfrac{1}{5}-\dfrac{1}{25}-\dfrac{1}{125}-\dfrac{1}{625}\right)}\)

\(=\dfrac{1}{4}+\dfrac{3}{4}=1\)