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\(a,=\dfrac{3}{4}-\dfrac{7}{2}-5=-\dfrac{31}{4}\\ b,=-\dfrac{1}{15}+\dfrac{4}{9}\cdot\dfrac{3}{8}-\dfrac{5}{6}=-\dfrac{9}{10}+\dfrac{1}{6}=-\dfrac{11}{15}\\ c,=\dfrac{1}{12}-\dfrac{4}{15}\cdot\dfrac{5}{6}+\left(-\dfrac{2}{3}\right)^3=\dfrac{1}{12}-\dfrac{2}{9}-\dfrac{8}{27}=-\dfrac{47}{108}\\ d,=\left[2\left(-\dfrac{1}{2}\right)\right]^5-\left[3\cdot\left(-\dfrac{1}{3}\right)\right]^3+\dfrac{2}{3}:\left(\dfrac{5}{3}-\dfrac{13}{6}\right)=-1-\left(-1\right)+\dfrac{2}{3}:\left(-\dfrac{1}{2}\right)=-\dfrac{4}{3}\)
a: \(=\dfrac{5}{6}\cdot10=\dfrac{50}{6}=\dfrac{25}{3}\)
g: \(\Leftrightarrow\left[{}\begin{matrix}x-3=-6\\x-3=6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=9\end{matrix}\right.\)
\(a,=\dfrac{9}{4}+\dfrac{5}{6}-\dfrac{3}{2}\cdot\dfrac{1}{6}=\dfrac{37}{12}-\dfrac{1}{4}=\dfrac{17}{6}\\ b,=-\dfrac{5}{3}\left(16\dfrac{2}{7}-28\dfrac{2}{7}\right)=-\dfrac{5}{3}\left(-12\right)=20\\ c,M=\dfrac{2^{60}+2^{40}}{2^{50}+2^{30}}=\dfrac{2^{40}\left(2^{20}+1\right)}{2^{30}\left(2^{20}+1\right)}=2^{10}=1024\\ d,=\left(\dfrac{3}{2}-\dfrac{1}{2}\right)\cdot\dfrac{1}{9}+\dfrac{1}{6}=\dfrac{1}{9}+\dfrac{1}{6}=\dfrac{5}{18}\)
Bài 5A:
a) Ta có: \(\widehat{aAe}=\widehat{bBA}=100^o\)
\(\Rightarrow a//b\)
\(\Rightarrow\widehat{gCc}=\widehat{CDd}\) (đồng vị)
\(\widehat{CDd}=135^o\)
Mà: \(x+\widehat{CDd}=180^o\) (kề bù)
\(\Rightarrow x=180^o-\widehat{CDd}=180^o-135^o=45^o\)
b) Xét tứ giác MNPQ có:
\(\widehat{QPM}+\widehat{PQN}+\widehat{PMN}+\widehat{QNM}=360^o\)
\(\Rightarrow2y+y+90^o+90^o=360^o\)
\(\Rightarrow3y=180^o\)
\(\Rightarrow y=\dfrac{180^o}{3}=60^o\)
Bài 3A:
Ta có: \(\widehat{A_1}+\widehat{B_3}=80^o+100^o=180^o\)
Mà hai góc này ở vị trí trong cùng phía
\(\Rightarrow a//b\)
Bài 4A:
\(a//b\Rightarrow\widehat{A_2}=\widehat{bBA}\)
\(\Rightarrow\widehat{bBA}=75^o\)
Mà: \(\widehat{bBA}+\widehat{B_3}=180^o\) (kề bù)
\(\Rightarrow\widehat{B_3}=180^o-\widehat{bBA}=180^o-75^o=105^o\)