a) \(3^2.5+2^3.10-81:3\\ =9.5+8.10-27\\ =45+80-27=98\)

b) \(5^{13}:5^{10}-25.2^2\\ =5^3-25.4\\ =125-100=25\)

c) \(20:2^2+5^9:5^8\\ =20:4+5^1\\ =5+5=10\)

d) \(100:5^2+7.3^2\\ =100:25+7.9\\ =4+63=67\)

e) \(84:4+3^9:3^7+5^0\\ =21+3^2+1\\ =21+9+1=31\)

98

25

10

67

31

`#3107.101107`

\(\dfrac{4^5\cdot9^4-2\cdot6^9}{2^{10}\cdot3^8+6^8\cdot20}\\ =\dfrac{\left(2^2\right)^5\cdot\left(3^2\right)^4-2\cdot2^9\cdot3^9}{2^{10}\cdot3^8+3^8\cdot2^8\cdot2^2\cdot5}\\ =\dfrac{2^{10}\cdot3^8-2^{10}\cdot3^9}{2^{10}\cdot3^8+3^8\cdot2^{10}\cdot5}\\ =\dfrac{2^{10}\cdot3^8\cdot\left(1-3\right)}{2^{10}\cdot3^8\cdot\left(1+5\right)}\\ =\dfrac{-2}{6}=-\dfrac{1}{3}\)

\(\dfrac{4^5.9^4-2.6^9}{2^{10}.3^8+6^8.20}=\dfrac{\left(2^2\right)^5.\left(3^2\right)^4-2.6^9}{2^8.3^8.2^2+6^8.20}\\ =\dfrac{2^{10}.3^8-2.6^9}{\left(2.3\right)^8.2^2+6^8.20}=\dfrac{2^8.3^8.2^2-2.6^9}{6^8.4+6^8.20}\\ =\dfrac{6^8.4-2.6.6^8}{6^8.\left(4+20\right)}=\dfrac{6^8.\left(4-2.6\right)}{6^8.24}\\ =\dfrac{4-12}{24}=\dfrac{-8}{24}=-\dfrac{1}{3}\)

`#3107.101107`

\(\dfrac{2^8\cdot2^{18}}{8^5\cdot4^6}=\dfrac{2^{8+18}}{\left(2^3\right)^5\cdot\left(2^2\right)^6}=\dfrac{2^{26}}{2^{15}\cdot2^{12}}=\dfrac{2^{26}}{2^{15+12}}=\dfrac{2^{26}}{2^{27}}=\dfrac{1}{2}\)

\(\dfrac{2^8.2^{18}}{8^5.4^6}=\dfrac{2^{8+18}}{\left(2^3\right)^5.\left(2^2\right)^6}\\ =\dfrac{2^{26}}{2^{15}.2^{12}}=\dfrac{2^{26}}{2^{15+12}}\\ =\dfrac{2^{26}}{2^{27}}=\dfrac{1}{2^1}=\dfrac{1}{2}\)

Bài 3:

a: \(\left\{{}\begin{matrix}4x+y=2\\\dfrac{4}{3}x+\dfrac{1}{3}y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=2-4x\\\dfrac{4}{3}x+\dfrac{1}{3}\left(2-4x\right)=1\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=2-4x\\\dfrac{4}{3}x+\dfrac{2}{3}-\dfrac{4}{3}x=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=2-4x\\\dfrac{2}{3}=1\left(vôlý\right)\end{matrix}\right.\)

=>Hệ phương trình vô nghiệm

b: \(\left\{{}\begin{matrix}x-y\sqrt{2}=0\\2x+y\sqrt{2}=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=y\sqrt{2}\\2y\sqrt{2}+y\sqrt{2}=3\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}3y\sqrt{2}=3\\x=y\sqrt{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{1}{\sqrt{2}}=\dfrac{\sqrt{2}}{2}\\x=\dfrac{y\sqrt{2}}{2}=\dfrac{\sqrt{2}}{2}\cdot\sqrt{2}=1\end{matrix}\right.\)

c: \(\left\{{}\begin{matrix}5x\sqrt{3}+y=2\sqrt{2}\\x\sqrt{6}-y\sqrt{2}=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=2\sqrt{2}-5x\sqrt{3}\\x\sqrt{6}-\sqrt{2}\left(2\sqrt{2}-5x\sqrt{3}\right)=2\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x\sqrt{6}-4+5x\sqrt{6}=2\\y=2\sqrt{2}-5x\sqrt{3}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}6x\sqrt{6}=6\\y=2\sqrt{2}-5\sqrt{3}\cdot x\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=\dfrac{1}{\sqrt{6}}=\dfrac{\sqrt{6}}{6}\\y=2\sqrt{2}-5\sqrt{3}\cdot\dfrac{\sqrt{6}}{6}=-\dfrac{\sqrt{2}}{2}\end{matrix}\right.\)

d: \(\left\{{}\begin{matrix}2\left(x+y\right)+3\left(x-y\right)=4\\\left(x+y\right)+2\left(x-y\right)=5\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}2x+2y+3x-3y=4\\x+y+2x-2y=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}5x-y=4\\3x-y=5\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=5x-4\\3x-\left(5x-4\right)=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=5x-4\\-2x+4=5\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=-\dfrac{1}{2}\\y=5\cdot\dfrac{-1}{2}-4=-\dfrac{5}{2}-4=-\dfrac{13}{2}\end{matrix}\right.\)

\(\dfrac{5}{11}.\left(\dfrac{-3}{7}\right)+\dfrac{5}{11}.\left(\dfrac{-5}{7}\right)+\left(\dfrac{-8}{7}\right).\dfrac{6}{11}\\ =\dfrac{5}{11}.\left(\dfrac{-3}{7}+\dfrac{-5}{7}\right)+\left(\dfrac{-8}{7}\right).\dfrac{6}{11}\\ =\dfrac{5}{11}.\dfrac{-8}{7}+\dfrac{-8}{7}.\dfrac{6}{11}\\ =\dfrac{-8}{7}.\left(\dfrac{5}{11}+\dfrac{6}{11}\right)\\ =\dfrac{-8}{7}.\dfrac{11}{11}\\ =\dfrac{-8}{7}.1=-\dfrac{8}{7}\)

3x-7=x-7

=> 3x-x=7-7

=> 2x=0

=> x=0:2

=> x=0

x=0

Ta có: M1 đối đỉnh với M3 

⇒ M1 kề bù với M2 

⇒ \(M1+M2=180^o\) 

Mà: \(M1=3\cdot M2\)

\(\Rightarrow3\cdot M2+M2=180^o\)

\(\Rightarrow4\cdot M2=180^o\)

\(\Rightarrow M2=\dfrac{180^o}{4}=45^o\)

Mà: M4 = M2 = `45^o` 

⇒ M1 = 3.M2 = 3.45 = `135^o` 

Mà: M1 = M3 

⇒ M3 = `135^o`  

lạ vải vậy mấy tụi nhóc con !

`5x^3 : x + 2x - 5x^2 = 1`

`<=> 5x^2 + 2x - 5x^2 = 1`

`<=> 2x = 1`.

`<=> x = 1/2`

Không dùng dấu ⇔ nhé em!

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