\(\left(x+2\right)^2+2\left(y-3\right)^2< 4\)

mà x,y nguyên

nên \(\left[\left(x+2\right)^2;2\left(y-3\right)^2\right]\in\left\{\left(1;2\right);\left(0;2\right)\right\}\)

=>\(\left(x+2;y-3\right)\in\left\{\left(1;1\right);\left(1;-1\right);\left(-1;1\right);\left(-1;-1\right);\left(0;1\right);\left(0;-1\right)\right\}\)

=>\(\left(x;y\right)\in\left\{\left(-1;4\right);\left(-1;2\right);\left(-3;4\right);\left(-3;2\right);\left(-2;4\right);\left(-2;2\right)\right\}\)

\(0,\left(4\right)+\dfrac{10}{3}+0,4\left(2\right)\)

\(=\dfrac{4}{9}+\dfrac{10}{3}+\dfrac{19}{45}\)

\(=\dfrac{4}{9}+\dfrac{30}{9}+\dfrac{19}{45}=\dfrac{34}{9}+\dfrac{19}{45}=\dfrac{170+19}{45}=\dfrac{189}{45}=\dfrac{21}{5}\)

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

\(\dfrac{2^8\cdot2^{18}}{8^5\cdot4^6}\\ =\dfrac{2^{8+18}}{\left(2^3\right)^5\cdot\left(2^2\right)^6}\\ =\dfrac{2^{26}}{2^{3\cdot5}\cdot2^{2\cdot6}}\\ =\dfrac{2^{26}}{2^{15}\cdot2^{12}}\\ =\dfrac{2^{26}}{2^{27}}\\ =\dfrac{1}{2}\)

`(3x + 1)^4 =` \(\dfrac{1}{16}\)

`=> (3x + 1)^4 =` \(\left(\dfrac{1}{2}\right)^4\)

`=> 3x + 1 =` \(\dfrac{1}{2}\) hoặc `3x + 1 =` \(-\dfrac{1}{2}\)

`=> 3x =` \(-\dfrac{1}{2}\) hoặc `3x =` \(-\dfrac{3}{2}\)

`=> x =` \(-\dfrac{1}{6}\) hoặc `x =` \(-\dfrac{1}{2}\)

Vậy `x =` \(-\dfrac{1}{6}\) hoặc `x =` \(-\dfrac{1}{2}\)

\(2\left(x-\dfrac{1}{3}\right)-3\left(x-1\right)=\dfrac{2}{3}\left(2-3x\right)\)

=> \(2x-\dfrac{2}{3}-3x+3=\dfrac{4}{3}-2x\)

=> \(2x-\dfrac{2}{3}-3x+3-\dfrac{4}{3}+2x=0\)

=> \(2x-3x+2x-\dfrac{2}{3}+3-\dfrac{4}{3}=0\)

=> \(x+3-\left(\dfrac{2}{3}+\dfrac{4}{3}\right)=0\)

=> \(x+3-\dfrac{6}{3}=0\)

=> \(x+3-2=0\)

=> \(x+1=0\)

=> ` x = 0 - 1`

=> `x = -1`

bằng -1 nha

\(\dfrac{3}{1^2.2^2}+\dfrac{5}{2^2.3^2}+...+\dfrac{19}{9^2.10^2}\)

= \(\dfrac{3}{1.4.}+\dfrac{5}{4.9}+...+\dfrac{19}{81.100}\)

= \(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{9}+...+\dfrac{1}{81}+\dfrac{1}{100}\)

= \(1-\dfrac{1}{100}< 1\) (đpcm)

--------------------------------

Cho các số: a;b;c thuộc `N`; `c,b` khác `0` ta luôn có:

Nếu: `c-b = a` thì: 

\(\dfrac{a}{b.c}=\dfrac{1}{b}-\dfrac{1}{c}\)

`a` là số tự nhiên không chia hết cho `3` nên a có dạng: 

`a = 3k + 1` hoặc `a = 3k + 2`

(`k` thuộc `N`*)

Mà a là số tự nhiên lẻ `=> a^2` là số tự nhiên lẻ `=> a^2 - 1` là số chẵn 

`=> a^2 ⋮ 2`

Để `a^2 - 1 ⋮ 6` thì  `a^2 - 1 ⋮ 3` (Vì `UCLN(2;3) = 1`)

- Xét `a = 3k + 1`

`=> a^2 -1 = (3k+1)^2 -1= 9k^2 + 6k + 1 - 1= 9k^2 + 6k^2 ⋮ 3` (Thỏa mãn)

- Xét `a = 3k + 2`

`=> a^2 -1 = (3k+2)^2 -1 = 9k^2 + 12k + 4 - 1= 9k^2 + 12k^2 + 3 ⋮ 3` (Thỏa mãn)

Vậy ...

\(\left(x+5\right)^2-4x^2\\=\left(x+5\right)^2-\left(2x\right)^2\\ =\left[\left(x+5\right)-2x\right]\left[\left(x+5\right)+2x\right]\\ =\left(x+5-2x\right)\left(x+5+2x\right)\\ =\left(-x+5\right)\left(3x+5\right)\)

\(x^2-9x-26=0\)

\(\text{Δ}=\left(-9\right)^2-4\cdot1\cdot\left(-26\right)=81+104=185>0\)

Do đó: Phương trình có hai nghiệm phân biệt là:

\(\left[{}\begin{matrix}x=\dfrac{9-\sqrt{185}}{2}\\x=\dfrac{9+\sqrt{185}}{2}\end{matrix}\right.\)

TY :>