giúp mình nha mn! Tks
Tính;
\(\frac{3^6.15^8}{9^7.5^9}\)
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Bài 7:
Ta thấy: $\widehat{xOy}+\widehat{yOx'}=\widehat{xOx'}=180^0$
$\widehat{xOy}-\widehat{yOx'}=30^0$
$\Rightarrow \widehat{yOx'}=\frac{180^0-30^0}{2}=75^0$
$\widehat{xOy'}=\widehat{yOx'}=75^0$ (hai góc đối đỉnh)
Bài 8:
$\widehat{AOC}+\widehat{BOD}=140^0$
$\widehat{AOC}=\widehat{BOD}$ (hai góc đối đỉnh)
$\Rightarrow \widehat{AOC}=\widehat{BOD}=\frac{140^0}{2}=70^0$
$\widehat{COB}=180^0-\widehat{AOC}=180^0-70^0=110^0$
$\widehat{DOA}=\widehat{COB}=110^0$ (hai góc đối đỉnh)
Câu 12 D là cửa hàng tạp hóa mà. nên câu đó chọn A
15 thì insisted that + V nguyên ko to nên câu này chọn A
2 she had got the price of the contest, she would be happy now
3 I had gone to the concert last night ,I would be tired
4 John had finished all his homework, he could see a movie now
5 we had listened to my mother, we wouldn't be in trouble right now
Ex5
1 you arrive at the offcie earlier than I do, please turn on the air-conditioner
2 Were it not snowy, the children would go to school
3 he not died so young, he would be a famous musician
4 Were I you, I would take care of my car
5 not very bad-tempered, his wife would give him soon are marrage
6 Had she not got married at such an early age, he would be at university
1: A=-1/2*xy^3*4x^2y^2=-2x^3y^5
Bậc là 8
Phần biến là x^3;y^5
Hệ số là -2
2:
a: P(x)=3x+4x^4-2x^3+4x^2-x^4-6
=3x^4-2x^3+4x^2+3x-6
Q(x)=2x^4+4x^2-2x^3+x^4+3
=3x^4-2x^3+4x^2+3
b: A(x)=P(x)-Q(x)
=3x^4-2x^3+4x^2+3x-6-3x^4+2x^3-4x^2-3
=3x-9
A(x)=0
=>3x-9=0
=>x=3
ta có
\(\frac{3^6.\left(3.5\right)^8}{\left(3.3\right)^7.5^9}\)= \(\frac{3^6.3^8.5^8}{3^7.3^7.5^9}\)= \(\frac{1}{5}\)
Đặt \(A=\frac{3^6.15^8}{9^7.5^9}\)
\(\Leftrightarrow A=\frac{3^6.\left(3.5\right)^8}{\left(3^2\right)^7.5^9}\)
\(\Leftrightarrow A=\frac{3^6.3^8.5^8}{3^{14}.5^9}\)
\(\Leftrightarrow A=\frac{3^{14}.5^8}{3^{14}.5^9}\)
\(\Leftrightarrow A=\frac{3^{14}}{3^{14}}.\frac{5^8}{5^9}\)
\(\Leftrightarrow A=1.\frac{1}{5}=\frac{1}{5}\)