3 + | x + 2 | = 2

| x + 2 | = 2 - 3

| x + 2 | = - 1

\(\Rightarrow\)x + 2 = 1 hoặc - 1

Ta xét 2 trường hợp :

TH1 : x + 2 = 1

          x = 1 - 2

          x = - 1

TH2 : x + 2 = - 1

          x = - 1 - 2

          x = - 3

Vậy x \(\in\){ - 1 ; - 3 }

3 + | x + 2 | = 2

| x + 2 | = 2 - 3

| x + 2 | = - 1

\(\Rightarrow\)x + 2 = 1 hoặc - 1

Ta xét 2 trường hợp :

Th 1 :

x + 2 = 1

x = 1 - 2

x = - 1

Th 2 :

x + 2 = - 1

x = - 1 - 2

x = - 3

Vậy x \(\in\){ - 1 ; - 3 }

Bài 2. 

a) \(\left(x+2\right)\left(x^2-2x+4\right)-x\left(x+3\right)\left(x-3\right)=26\)

\(\Leftrightarrow x^3+8-x\left(x^2-9\right)=26\)

\(\Leftrightarrow9x=18\)

\(\Leftrightarrow x=2\).

b) \(\left(4x+1\right)\left(1-4x+16x^2\right)-16x\left(4x^2-5\right)=17\)

\(\Leftrightarrow64x^3+1-64x^3+80x=17\)

\(\Leftrightarrow80x=16\)

\(\Leftrightarrow x=\frac{1}{5}\)

c) \(\left(x+3\right)^3-x\left(3x+1\right)^2+\left(2x+1\right)\left(4x^2-2x+1\right)-3x^2=54\)

\(\Leftrightarrow x^3+9x^2+27x+27-9x^3-6x^2-x+8x^3+1-3x^2=54\)

\(\Leftrightarrow26x=26\)

\(\Leftrightarrow x=1\)

d) \(7x^3+3x^2-3x+1=0\)

\(\Leftrightarrow8x^3=x^3-3x^2+3x-1\)

\(\Leftrightarrow\left(2x\right)^3=\left(x-1\right)^3\)

\(\Leftrightarrow2x=x-1\)

\(\Leftrightarrow x=-1\)

e) \(x^2-6x-7=0\)

\(\Leftrightarrow\left(x-7\right)\left(x+1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=7\\x=-1\end{cases}}\)

f) \(4x^2-8x+4y^2+4y+5=0\)

\(\Leftrightarrow4x^2-8x+4+4y^2+4y+1=0\)

\(\Leftrightarrow4\left(x-1\right)^2+\left(2y+1\right)^2=0\)

\(\Leftrightarrow\hept{\begin{cases}x-1=0\\2y+1=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=1\\y=-\frac{1}{2}\end{cases}}\)

g) \(x^2+5y^2-4xy-2y+1=0\)

\(\Leftrightarrow x^2-4xy+4y^2+y^2-2y+1=0\)

\(\Leftrightarrow\left(x-2y\right)^2+\left(y-1\right)^2=0\)

\(\Leftrightarrow\hept{\begin{cases}x-2y=0\\y-1=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=2\\y=1\end{cases}}\)

Đk: x \(\ne\)0; x \(\ne\)-2; x \(\ne\)-4; x \(\ne\)-6

Ta có: \(\frac{1}{x\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+4\right)}+\frac{1}{\left(x+4\right)\left(x+6\right)}+\frac{1}{2\left(x+6\right)}=\frac{1}{4042}\)

<=> \(\frac{2}{x\left(x+2\right)}+\frac{2}{\left(x+2\right)\left(x+4\right)}+\frac{2}{\left(x+4\right)\left(x+6\right)}+\frac{1}{x+6}=\frac{1}{2021}\)

<=> \(\frac{1}{x}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+4}+\frac{1}{x+4}-\frac{1}{x+6}+\frac{1}{x+6}=\frac{1}{2021}\)

<=> \(\frac{1}{x}=\frac{1}{2021}\) <=> x = 2021 (tm)

Vậy x = 2021

Cảm ơn bạn nhìu!

13+16+110+...+2x(x+1)=2019202113+16+110+...+2x(x+1)=20192021

⇒16+112+120+...+1x(x+1)=20194042⇒16+112+120+...+1x(x+1)=20194042

12.3+13.4+14.5+...+1x(x+1)=2019404212.3+13.4+14.5+...+1x(x+1)=20194042

12−13+13−14+14−15+...+1x−1x+1=2019404212−13+13−14+14−15+...+1x−1x+1=20194042

12−1x+1=2019202112−1x+1=20192021

⇒1x+1=12021⇒1x+1=12021

⇒x+1=2021⇒x+1=2021

⇒x=2020⇒x=2020

Vậy x=2020x=2020.

  Chúc em Hok Tốt ~

ĐK : 11x \(\ge0\Leftrightarrow x\ge0\)

Với x \(\ge0\Rightarrow x+\frac{1}{2}\ge0;...;x+\frac{1}{110}\ge0\)

Khi đó \(\left|x+\frac{1}{2}\right|+\left|x+\frac{1}{6}\right|+\left|x+\frac{1}{12}\right|+...+\left|x+\frac{1}{110}\right|=11x\)(10 hạng tử ở vế trái)

<=> \(x+\frac{1}{2}+x+\frac{1}{6}+x+\frac{1}{12}+...+x+\frac{1}{110}=11x\)

<=> \(10x+\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{110}\right)=11x\)

<=> \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{10.11}=x\)

<=> \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{10}-\frac{1}{11}=x\)

<=> \(1-\frac{1}{11}=x\)

<=> \(x=\frac{10}{11}\)

Giúp mình với mình cần gấp hôm nay nộp rồi 

13+16+110+...+2x(x+1)=2019202113+16+110+...+2x(x+1)=20192021

⇒16+112+120+...+1x(x+1)=20194042⇒16+112+120+...+1x(x+1)=20194042

12.3+13.4+14.5+...+1x(x+1)=2019404212.3+13.4+14.5+...+1x(x+1)=20194042

12−13+13−14+14−15+...+1x−1x+1=2019404212−13+13−14+14−15+...+1x−1x+1=20194042

12−1x+1=2019202112−1x+1=20192021

⇒1x+1=12021⇒1x+1=12021

⇒x+1=2021⇒x+1=2021

⇒x=2020⇒x=2020

Vậy x=2020x=2020.

    Chúc em Hok Tốt~

Do vai trò \(x,y,z\)như nhau nên không mất tính tổng quát, giả sử \(x\ge y\ge z\ge1\).

Ta có: \(xyz=x+y+z\le3x\Leftrightarrow yz\le3\)

- \(yz=3\Rightarrow y=3,z=1\)

\(x+3+1=3x\Leftrightarrow x=2\)không thỏa do \(x< y\).

- \(yz=2\Rightarrow y=2,z=1\)

\(x+2+1=2x\Leftrightarrow x=3\)(thỏa mãn) 

- \(yz=1\Rightarrow y=z=1\)

\(x+1+1=x\Leftrightarrow0x=2\)(vô nghiệm).

Vậy phương trình đã cho có nghiệm là \(\left(3,2,1\right)\)và các hoán vị. 

\(A=\frac{5x-3}{x+1}=\frac{5x+5-8}{x+1}=5-\frac{8}{x+1}\inℤ\Leftrightarrow\frac{8}{x+1}\inℤ\)

mà \(x\inℤ\)nên \(x+1\inƯ\left(8\right)=\left\{-8,-4,-2,-1,1,2,4,8\right\}\)

\(\Leftrightarrow x\in\left\{-9,-5,-3,-2,0,1,3,7\right\}\).

Ta có: \(\left|x+1\right|+\left|x+7\right|=\left|x+1\right|+\left|-x-7\right|\ge\left|x+1-x-7\right|=\left|-6\right|=6\)

Lại có: \(6-\left(x+2\right)^2\le6\)

Dấu = xảy ra

\(\Leftrightarrow\hept{\begin{cases}\left(x+1\right)\left(-x-7\right)>0\\x+2=0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}-7< x< -1\\x=-2\left(tm\right)\end{cases}}\)

Vậy ...