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Nguyễn Tú
Giới thiệu về bản thân
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Số gạo đã lấy bằng:
\(\dfrac{1}{2}+\dfrac{1}{8}=\dfrac{5}{8}\) (bao)
Số gạo còn lại bằng:
\(1-\dfrac{5}{8}=\dfrac{3}{8}\) (bao)
Số gạo trong bao ban đầu là:
\(18:\dfrac{3}{8}=48\) (kg)
Đáp số: 48 kg
\(3x^3-14x^2+4x+3\)
\(=\left(3x^3-15x^2+9x\right)+\left(x^2-5x+3\right)\)
\(=3x\left(x^2-5x+3\right)+\left(x^2-5x+3\right)\)
\(=\left(3x+1\right)\left(x^2-5x+3\right)\)
Ta có: \(\dfrac{A+C+E}{3}+\dfrac{A+B+D}{3}=40+28\)
\(\Rightarrow\dfrac{2A+B+C+D+E}{3}=68\)
\(\Rightarrow\dfrac{2A}{3}+\dfrac{B+C+D+E}{3}=68\)
Thay \(\dfrac{B+C+D+E}{3}=33\) được:
\(\dfrac{2A}{3}+33=68\)
\(\Rightarrow\dfrac{2}{3}A=68-33\)
\(\Rightarrow\dfrac{2}{3}A=35\)
\(\Rightarrow A=35:\dfrac{2}{3}\)
\(\Rightarrow A=\dfrac{105}{2}=52,5\)
Vậy \(A=52,5\)
\(\dfrac{2^{17}\cdot9^4}{6^3\cdot8^3}\)
\(=\dfrac{2^{17}\cdot3^8}{2^3\cdot3^3\cdot2^9}\)
\(=\dfrac{2^{17}\cdot3^8}{2^{12}\cdot3^3}\)
\(=2^5\cdot3^5\)
\(=6^5=7776\)
\(\left(x-1\right)^2+\left(y+1\right)^2=0\)
Nhận xét:
\(\left\{{}\begin{matrix}\left(x-1\right)^2\ge0\\\left(y+1\right)^2\ge0\end{matrix}\right.\)
\(\Rightarrow\left(x-1\right)^2+\left(y+1\right)^2\ge0\)
Do đó: Dấu "=" xảy ra khi:
\(\left\{{}\begin{matrix}x-1=0\\y+1=0\end{matrix}\right.\)
\(\Rightarrow x=1;y=-1\)
Vậy \(x=1;y=-1\)
Từ (1) \(\Rightarrow A+C+E=40\cdot3=120\)
Từ (2) \(\Rightarrow A+B+D=28\cdot3=84\)
Từ (3) \(\Rightarrow B+C+D+E=33\cdot3=99\)
Suy ra:
\(\left(A+C+E+A+B+D\right)-\left(B+C+D+E\right)=\left(120+84\right)-99\)
\(2A+\left(B+C+D+E\right)-\left(B+C+D+E\right)=105\)
\(2A=105\)
\(A=52,5\)
Vậy \(A=52,5\)
\(2,6-1,7-1,6+5,5+3,2\)
\(=2,6-1,7-1,6+\left(5,5+3,2\right)\)
\(=2,6-1,7-1,6+8,7\)
\(=\left(2,6-1,6\right)+\left(8,7-1,7\right)\)
\(=1+7\)
\(=8\)
Phần định lý kia nếu muốn đầy đủ thì bạn ghi là "quan hệ giữa góc và cạnh đối diện" nhé
Ta có:
\(\widehat{M}=30^o< \widehat{N}=50^o< \widehat{P}=100^o\) (gt)
\(\Rightarrow NP< MP< MN\) (định lý)
Vậy...
Số học sinh của lớp 6A là:
\(15+20-8+10=37\) (học sinh)
Đáp số: 37 học sinh