Question

Bài 1:

 Mn giúp với cần gấp !!!

Answer 1

Bài 2:

a: xy=z; yz=4x; xz=9y

Do đó: \(xy\cdot yz\cdot xz=z\cdot4x\cdot9y=36xyz\)

=>\(\left(xyz\right)^2=36xyz\)

=>xyz(xyz-36)=0

TH1: xyz=0

=>x=y=z=0

TH2: xyz-36=0

=>xyz=36

Ta có: xyz=36

=>\(z\cdot z=36\)

=>\(z^2=36\)

=>\(\left[\begin{array}{l}z=6\\ z=-6\end{array}\right.\)

Ta có: xyz=36

=>\(4x\cdot x=36\)

=>\(4x^2=36\)

=>\(x^2=9\)

=>\(\left[\begin{array}{l}x=3\\ x=-3\end{array}\right.\)

Ta có: xyz=36

=>\(9y\cdot y=36\)

=>\(9y^2=36\)

=>\(y^2=4\)

=>\(\left[\begin{array}{l}y=2\\ y=-2\end{array}\right.\)

vậy: (x;y;z)∈{(0;0;0);(3;2;6);(-3;-2;-6)}

b: \(3\left(x-2022\right)^2=21-y^2\)

=>\(21-y^2\) ⋮3

=>\(y^2\) ⋮3

=>y⋮3(1)

Ta có: \(3\left(x-2022\right)^2>=0\forall x\)

=>\(21-y^2\ge0\)

=>\(y^2\le21\)

mà y⋮3

và y nguyên

nên y∈{0;3;-3}

TH1: y=0

\(3\left(x-2022\right)^2=21-y^2\)

=>\(3\left(x-2022\right)^2=21\)

=>\(\left(x-2022\right)^2=7\)

=>\(\left[\begin{array}{l}x-2022=\sqrt7\\ x-2022=-\sqrt7\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\sqrt7+2022\left(loại\right)\\ x=-\sqrt7+2022\left(loại\right)\end{array}\right.\)

TH2: y∈{3;-3}

=>\(y^2=9\)

=>\(3\left(x-2022\right)^2=21-9=12\)

=>\(\left(x-2022\right)^2=4\)

=>\(\left[\begin{array}{l}x-2022=2\\ x-2022=-2\end{array}\right.\Rightarrow\left[\begin{array}{l}x=2024\\ x=2020\end{array}\right.\)

vậy: (x;y)∈{(2020;3);(2020;-3);(2024;3);(2024;-3)}

Bài 1:

a: \(\frac{2a-b}{a-b}=\frac23\)

=>3(2a-b)=2(a-b)

=>6a-3b=2a-2b

=>6a-2a=-2b+3b

=>4a=b

=>\(a=\frac{b}{4}\)

\(M=\frac{4ab+2b^2}{8a^2-ab}\)

\(=\frac{b\cdot b+2b^2}{8\cdot\left(\frac{b}{4}\right)^2-b\cdot\frac{b}{4}}\)

\(=\frac{3b^2}{8\cdot\frac{b^2}{16}-\frac{b^2}{4}}=\frac{3b^2}{\frac12b^2-\frac14b^2}=\frac{3b^2}{\frac14b^2}=3:\frac14=12\)

b: ta có công thức tổng quát:

\(1-\frac{1}{1+2+\cdots+n}\)

\(=1-\frac{1}{\frac{n\left(n+1\right)}{2}}=1-\frac{2}{n\left(n+1\right)}\)

\(=\frac{n\left(n+1\right)-2}{n\left(n+1\right)}=\frac{n^2+n-2}{n\left(n+1\right)}=\frac{\left(n+2\right)\left(n-1\right)}{n\left(n+1\right)}\)

Do đó,ta có:

\(1-\frac{1}{1+2}=\frac{\left(2+2\right)\left(2-1\right)}{2\cdot\left(2+1\right)}=\frac{4\cdot1}{2\cdot3}\)

\(1-\frac{1}{1+2+3}=\frac{\left(3+2\right)\left(3-1\right)}{3\left(3+1\right)}=\frac{5\cdot2}{3\cdot4}\)

....

\(1-\frac{1}{1+2+3+\cdots+2024}=\frac{\left(2024+2\right)\left(2024-1\right)}{2024\left(2024+1\right)}=\frac{2026\cdot2023}{2024\cdot2025}\)

Do đó: \(A=\frac{4\cdot1}{2\cdot3}\cdot\frac{5\cdot2}{3\cdot4}\cdot\ldots\cdot\frac{2026\cdot2023}{2024\cdot2025}\)

\(=\frac{1\cdot2\cdot\ldots\cdot2023}{2\cdot3\cdot\ldots\cdot2024}\cdot\frac{4\cdot5\cdot\ldots\cdot2026}{3\cdot4\cdot\ldots\cdot2025}=\frac{1}{2024}\cdot\frac{2026}{3}=\frac{1013}{3\cdot1012}=\frac{1013}{3036}\)