9 x 10 x 2 =
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9 x 2 = 18
9 x 5 = 45
9 x 8 = 72
9 x 10 = 90
2 x 9 = 18
5 x 9 = 45
8 x 9 = 72
10 x 9 = 90
a: \(\dfrac{x-1}{x^2-x+1}-\dfrac{x+1}{x^2+x+1}=\dfrac{10}{x\left(x^4+x^2+1\right)}\)
\(\Leftrightarrow x\left(x-1\right)\left(x^2+x+1\right)-x\left(x+1\right)\left(x^2-x+1\right)=10\)
\(\Leftrightarrow x\left(x^3-1\right)-x\left(x^3+1\right)=10\)
=>-2x=10
hay x=-5
d: \(\Leftrightarrow\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}+...+\dfrac{1}{\left(x+7\right)\left(x+8\right)}=\dfrac{1}{14}\)
\(\Leftrightarrow\dfrac{1}{x+1}-\dfrac{1}{x+8}=\dfrac{1}{14}\)
\(\Leftrightarrow\left(x+1\right)\left(x+8\right)=14\left(x+8\right)-14\left(x+1\right)\)
\(\Leftrightarrow x^2+9x+8=14x+112-14x-14=98\)
\(\Leftrightarrow x^2+9x-90=0\)
\(\Leftrightarrow x\in\left\{6;-15\right\}\)
Tử số= 2^19.9^3.3^3 + 5.3.2^9.2^9.9^4
= 2^19.9^3.3^3 + 5.3.2^18.9.9^3
= 2^19.9^3.3.3^2 + 5.3.2^18.3^2.9^3
= 2^18.9^3.3^2(2 + 5.) (đặt nhân tử chung)
=7.2^18.9^3.3^2
=7.2^18.9.9.9.3^2
=7.2^18.3^2.3^2.3^2.3^2
=7.2^18.3^8
Mẫu số= 6^9.2^10 + 6^10.2^10
= 6^9.2^10 + 6^10.2^10
=6^9.2^10(1+6)
=7.6^9.2^10.
=7.2^9.3^9.2^10
=7.2^19.3^9
Lấy tử số chia mẫu số ta được : 1/2.3 = 1/6
-9/10x(5/14+1/7)+1/10x-9/2
=-9/10x7/14+1/10x-9/2
-9/10x1/2+1/10x-9/2
=-9/20+--9/20
=-18/20=-9/10
Mình xin trình bày 2 cách, một là phân tích bình thường, 2 là xài L'Hospital. Bởi c3 ko ai cho xài L'Hospital để hack tự luận cả
C1: Normal
\(\left(2-x\right)+\left(2-x\right)^2+...+\left(2-x\right)^9-9\)
\(=\left[\left(2-x\right)-1\right]+\left[\left(2-x\right)^2-1\right]+...+\left[\left(2-x\right)^9-1\right]\)
\(=\left(2-x-1\right)+\left(2-x-1\right)\left(2-x+1\right)+\left(2-x-1\right)\left[\left(2-x\right)^2+\left(2-x\right)+1\right]+...+\left(2-x-1\right)\left[\left(2-x\right)^8+\left(2-x\right)^7+...+1\right]\)
\(=-\left(x-1\right)\left(1+2-x+1+\left(2-x\right)^2+\left(2-x\right)+1+....+\left(2-x\right)^8+\left(2-x\right)^7+...+1\right)\)
Lai co:
\(x+x^2+...+x^{10}-10=\left(x-1\right)+\left(x^2-1\right)+...+\left(x^{10}-1\right)\)
\(=\left(x-1\right)+\left(x-1\right)\left(x+1\right)+....+\left(x-1\right)\left(x^9+x^8+...+1\right)\)
\(=\left(x-1\right)\left[1+x+1+x^2+x+1+....+x^9+x^8+...+1\right]\)
\(\Rightarrow\lim\limits_{x\rightarrow1}....=\lim\limits_{x\rightarrow1}\dfrac{-[1+2-x+1+\left(2-x\right)^2+\left(2-x\right)+1+...+\left(2-x\right)^8+\left(2-x\right)^7+...+1]}{1+x+1+x^2+x+1+...+x^9+x^8+...+1}\)
\(=\lim\limits_{x\rightarrow1}\dfrac{-[9.1+8.\left(2-x\right)+7\left(2-x\right)^2+6\left(2-x\right)^3+5\left(2-x\right)^4+4\left(2-x\right)^5+3\left(2-x\right)^6+2\left(2-x\right)^7+\left(2-x\right)^8]}{10.1+9x^2+8x^3+7x^4+6x^5+5x^6+4x^7+3x^8+2x^9+x^{10}}\)
\(=\dfrac{-[1+2+3+...+9]}{1+2+3+...+10}=\dfrac{-45}{55}\)
C2: L'Hospital
\(=\lim\limits_{x\rightarrow1}\dfrac{-1-2\left(2-x\right)-3\left(2-x\right)^2-...-9\left(2-x\right)^8}{1+2x+3x^2+...+10x^9}\)
\(=\lim\limits_{x\rightarrow1}\dfrac{-1-2-3-...-9}{1+2+3+...+10}=-\dfrac{45}{55}\)
Đáp án là 180
9 x 10 x 2 = 180
k mình,mình k lại