giúp mình vs

Bài a

a: \(49-y^2=6\left(x-2021\right)^2\)

=>\(49-y^2\ge0\) và \(49-y^2\) ⋮6

=>\(y^2\in\left\lbrace1;16;25;49\right\rbrace\)

TH1: \(y^2=1\)

Ta có: \(49-y^2=6\left(x-2021\right)^2\)

=>\(6\left(x-2021\right)^2=49-1=48\)

=>\(\left(x-2021\right)^2=8\)

mà x nguyên

nên x∈∅

TH2: \(y^2=16\)

Ta có: \(49-y^2=6\left(x-2021\right)^2\)

=>\(6\left(x-2021\right)^2=49-16=33\)

=>\(\left(x-2021\right)^2=5,5\)

mà x nguyên

nên x∈∅

TH3: \(y^2=25\)

Ta có: \(49-y^2=6\left(x-2021\right)^2\)

=>\(6\left(x-2021\right)^2=49-25=24\)

=>\(\left(x-2021\right)^2=4\)

=>x-2021=2 hoặc x-2021=-2

=>x=2023(nhận) hoặc x=2019(nhận)

\(y^2=25\)

=>y=5(nhận) hoặc y=-5(nhận)

TH4: \(y^2=49\)

Ta có: \(49-y^2=6\left(x-2021\right)^2\)

=>\(6\left(x-2021\right)^2=49-49=0\)

=>\(\left(x-2021\right)^2=0\)

=>x-2021=0

=>x=2021(nhận)

\(y^2=49\)

=>y=7(nhận) hoặc y=-7(nhận)

theo bai co 1/3+1/6+...+2/x.x+1=499/1000

(=)2/2.3+2/3.4+...+2/x.(x+1)=499/1000

(=)2(1/2-...-1/x+1)=499/1000

(=)1/2-1/x+1=499/2000

(=)1/x+1=501/2000

den day thi minh chiu roi!

Ta có: \(\left|x+\frac{1}{2}\right|\ge0;\left|x+\frac{1}{6}\right|\ge0;...;\left|x+\frac{1}{110}\ge0\right|\)

=> \(\left|x+\frac{1}{2}\right|+\left|x+\frac{1}{6}\right|+...+\left|x+\frac{1}{100}\right|\ge0\)

=> 11x \(\ge\)0 => x\(\ge\)0

=> \(x+\frac{1}{2}>0;x+\frac{1}{6}>0;...;x+\frac{1}{110}>0\)

=> \(\left|x+\frac{1}{2}\right|=x+\frac{1}{2};\left|x+\frac{1}{6}\right|=x+\frac{1}{6};...;\left|x+\frac{1}{110}\right|=x+\frac{1}{110}\)

=> \(\left(x+\frac{1}{2}\right)+\left(x+\frac{1}{6}\right)+...+\left(x+\frac{1}{110}\right)=11x\)

=> 10x + \(\left(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{110}\right)=11x\)

=> 10x + \(\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{10.11}\right)\)= 11x

=> 10x  + \(\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{10}-\frac{1}{11}\right)\)= 11x

=> 10x + \(\frac{10}{11}\)= 11x

=> x = \(\frac{10}{11}\)

Vậy x = \(\frac{10}{11}\)

bạn ơi sao lại là 10x ?

với x≥-1 ta có:

x+1+x+7=6-(x+2)²

⇔2x+8=6-x²-4x-4

⇔x²+6x+6=0

với x≤-7 ta có

 -x-1-x-7=6-(x+2)²

⇔-2x-8=6-x²-4x-4

⇒x²+2x-10=0

⇒x=-1±√11

với -7<x<-1 ta có 

-x-1+x+7=6-(x+2)²

⇔6=6-(x+2)²

⇒x=-2