a.

\(10⋮\left(x-1\right)\)

\(\Rightarrow x-1=Ư\left(10\right)\)

\(\Rightarrow x-1=\left\{-10;-5;-2;-1;1;2;5;10\right\}\)

\(\Rightarrow x=\left\{-9;-4;-1;0;2;3;6;11\right\}\)

b.

\(\left(x+5\right)⋮\left(x-2\right)\Rightarrow\left(x-2\right)+7⋮x-2\)

\(\Rightarrow7⋮x-2\)

\(\Rightarrow x-2=Ư\left(7\right)=\left\{-7;-1;1;7\right\}\)

\(\Rightarrow x=\left\{-5;1;3;9\right\}\)

c.

\(\left(3x+8\right)⋮\left(x-1\right)\)

\(\Rightarrow\left(3x-3+11\right)⋮\left(x-1\right)\)

\(\Rightarrow3\left(x-1\right)+11⋮x-1\)

\(\Rightarrow11⋮\left(x-1\right)\)

\(\Rightarrow x-1=Ư\left(11\right)=\left\{-11;-1;1;11\right\}\)

\(\Rightarrow x=\left\{-10;0;2;12\right\}\)

\(1)\)

\(VT=\left(\left|x-6\right|+\left|2022-x\right|\right)+\left|x-10\right|+\left|y-2014\right|+\left|z-2015\right|\)

\(\ge\left|x-6+2022-x\right|+\left|0\right|+\left|0\right|+\left|0\right|=2016\)

Dấu "=" xảy ra \(\Leftrightarrow\)\(\hept{\begin{cases}\left(x-6\right)\left(2022-x\right)\ge0\left(1\right)\\x-10=y-2014=z-2015=0\left(2\right)\end{cases}}\)

\(\left(2\right)\)\(\Leftrightarrow\)\(\hept{\begin{cases}x=10\\y=2014\\z=2015\end{cases}}\)

\(\left(1\right)\)

TH1 : \(\hept{\begin{cases}x-6\ge0\\2022-x\ge0\end{cases}\Leftrightarrow\hept{\begin{cases}x\ge6\\x\le2022\end{cases}\Leftrightarrow}6\le x\le2022}\) ( nhận ) 

TH2 : \(\hept{\begin{cases}x-6\le0\\2022-x\le0\end{cases}\Leftrightarrow\hept{\begin{cases}x\le6\\x\ge2022\end{cases}}}\) ( loại ) 

Vậy \(x=10\)\(;\)\(y=2014\) và \(z=2015\)

\(2)\)

\(VT=\left|x-5\right|+\left|1-x\right|\ge\left|x-5+1-x\right|=\left|-4\right|=4\)

\(VP=\frac{12}{\left|y+1\right|+3}\le\frac{12}{3}=4\)

\(\Rightarrow\)\(VT\ge VP\)

Dấu "=" xảy ra \(\Leftrightarrow\)\(\hept{\begin{cases}\left(x-5\right)\left(1-x\right)\ge0\left(1\right)\\\left|y+1\right|=0\left(2\right)\end{cases}}\)

\(\left(1\right)\)

TH1 : \(\hept{\begin{cases}x-5\ge0\\1-x\ge0\end{cases}\Leftrightarrow\hept{\begin{cases}x\ge5\\x\le1\end{cases}}}\) ( loại ) 

TH2 : \(\hept{\begin{cases}x-5\le0\\1-x\le0\end{cases}\Leftrightarrow\hept{\begin{cases}x\le5\\x\ge1\end{cases}\Leftrightarrow}1\le x\le5}\) ( nhận ) 

\(\left(2\right)\)\(\Leftrightarrow\)\(y=-1\)

Vậy \(1\le x\le5\) và \(y=-1\)

\(\left|x+1\right|+\left|x+2\right|+...+\left|x+100\right|=605x\)(1)

Vì \(VT>0\forall x\)

\(\Rightarrow VP>0\Leftrightarrow605x>0\Leftrightarrow x>0\)

Khi đó \(\left(1\right)\Leftrightarrow x+1+x+2+...+x+100=605x\)

\(\Leftrightarrow100x+5050=605x\)

\(\Leftrightarrow505x=5050\)

\(\Leftrightarrow x=10\)( thỏa mãn )

Vậy....

a) Quy luật là gì ??

b) 

Đặt

 \(A=\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2020}}\\\Rightarrow2A=1+\dfrac{1}{2}+...+\dfrac{1}{2^{2019}}\\ \Rightarrow2A-A=1-\dfrac{1}{2^{2020}}\Rightarrow A=1-\dfrac{1}{2^{2020}}\)

Suy ra , phương trình trở thành :

213 -x  =13

<=> x=200

a/ \(x=\dfrac{-5}{12}\)

b/ \(x\approx-1,9526\)

c/ \(x=\dfrac{21-i\sqrt{199}}{10}\)

d/ \(x=\dfrac{-20}{13}\)

a) (x-2)³+6(x+1)²-x³+12=0

⇒ x³-6x²+12x-8+6(x²+2x+1)-x³+12=0

⇒ x³-6x²+12x-8+6x²+12x+6-x³+12=0

⇒ 24x+10=0

⇒ 24x=-10

⇒ x=-5/12

a.

PT \(\Leftrightarrow x^3-6x^2+12x-8+6(x^2+2x+1)-x^3+12=0\)

\(\Leftrightarrow x^3-6x^2+12x-8+6x^2+12x+6-x^3+12=0\)

\(\Leftrightarrow 24x+10=0\Leftrightarrow x=\frac{-5}{12}\)

b. Bạn xem lại đề, nghiệm khá xấu không phù hợp với mức độ tổng thể của bài.

c.

PT $\Leftrightarrow (4x^2+12x+9)+(x^2-1)=5(x^2+4x+4)+(x^2-4x-5)+9(x^2+6x+9)$
$\Leftrightarrow 10x^2+42x+64=0$

$\Leftrightarrow x^2+(3x+7)^2=-15< 0$ (vô lý) 

Do đó pt vô nghiệm.

d.

PT $\Leftrightarrow (1-6x+9x^2)-(9x^2-17x-2)=(9x^2-16)-9(x^2+6x+9)$

$\Leftrightarrow 11x+3=-54x-97$

$\Leftrightarrow 65x=-100$

$\Leftrightarrow x=\frac{-20}{13}$

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

\(\Rightarrow f\left(1\right)+f\left(2\right)+....+f\left(x\right)=1-\frac{1}{2^2}+\frac{1}{2^2}-....-\frac{1}{\left(x+1\right)^2}\)

\(\Rightarrow\frac{2y\left(x+1\right)^3-1}{\left(x+1\right)^2}-19+x=\frac{x\left(x+2\right)}{\left(x+1\right)^2}\)

\(\Leftrightarrow\frac{2y\left(x+1\right)^3-1}{\left(x+1\right)^2}-19+x=\frac{2y\left(x+1\right)^3-1}{\left(x+1\right)^2}-20+\left(x+1\right)=\frac{x\left(x+2\right)}{\left(x+1\right)^2}\)

Dat:\(x+1=a\Rightarrow\frac{\left(2y+1\right)a^3-20a^2-1}{a^2}=\frac{a^2-1}{a^2}\Leftrightarrow\left(2y+1\right)a^3-20a^2-1=a^2-1\)

\(\Leftrightarrow\left(2y+1\right)a^3-20a^2=a^2\Leftrightarrow\left(2ay+a\right)-20=1\left(coi:x=-1cophailanghiemko\right)\)

\(\Leftrightarrow2ay+a=21\Leftrightarrow a\left(2y+1\right)=21\Leftrightarrow\left(x+1\right)\left(2y+1\right)=21\)

\(\left(x+1\right)+\left(x+2\right)+...+\left(x+100\right)=5750\)

\(\left(x\cdot100\right)+\left(1+2+...+100\right)=5750\)

\(\left(x\cdot100\right)+\left(100+1\right)\cdot\frac{100}{2}=5750\)

\(\left(x\cdot100\right)+101\cdot50=5750\)

\(\left(x\cdot100\right)+5050=5750\)

\(x\cdot100=5750-5050\)

\(x\cdot100=700\)

\(x=700\div100\)

\(x=7\)

Ta có: ( x+1)+(x+2)+(x+3)+.....+(x+99)+(x+100)=5750

<=>(x+x+x+....+x+x)+(1+2+3+..+99+100)=5750

<=> 100x+5050=5750

=>100x=5750-5050

=>100x=700

=>x=700:100

=>x=7

Vậy x=7

 hoặc mở câu hỏi tương tự tham khảo.

thank you

chỉ cách giải luôn nha bạn chỉ rõ mới k