\(|\frac{1}{2}x+1|-4=0\)

\(\Rightarrow|\frac{1}{2}x+1|=4\)

\(\Rightarrow\orbr{\begin{cases}\frac{1}{2}x+1=4\\\frac{1}{2}x+1=-4\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}\frac{1}{2}x=3\\\frac{1}{2}x=-5\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x=6\\x=-10\end{cases}}\)

Vậy x = 6 hoặc x = -10

_Chúc bạn học tốt_

+) Min: \(A=\frac{x^2}{x^4+x^2+1}\ge0\forall x\) 

Dấu "=" <=> x=0

+) Max: \(1-3A=\frac{x^4-2x^2+1}{x^4+x^2+1}=\frac{\left(x^2-1\right)^2}{x^4+x^2+1}\ge0\)

\(\Rightarrow A\le\frac{1}{3}\)Dấu "=" <=> x= 1,-1

a,ĐKXĐ \(x^3-8\ne0\Leftrightarrow x^3\ne8\Leftrightarrow x\ne2\)

b,\(\Leftrightarrow3x^2+6x+12=0\)

    \(\Leftrightarrow3\left(x^2+2x+1\right)+9=0\)

   \(\Leftrightarrow3\left(x+1\right)^2+9=0\)(VÔ LÝ VÌ 3(x+1)²>=0 =>3(x+1)²+9>0)

vì vây ko có giá trị x để F =0

C, VỚI ĐKXĐ trên ,ta có 

\(F=\frac{3\left(x^2+2x+4\right)}{\left(x-2\right)\left(x^2+2x+4\right)}\)

    \(=\frac{3}{x-2}\)

ta có:   \(x+\frac{1}{x}=3\)

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

\(\Leftrightarrow x^2+2.x.\frac{1}{x}+\left(\frac{1}{x}\right)^2=9\)

\(\Leftrightarrow x^2+2+\frac{1}{x^2}=9\)

\(\Leftrightarrow x^2+\frac{1}{x^2}=7\)

\(\Rightarrow\left(x^2+\frac{1}{x^2}\right)^2=7^2\)

\(\Leftrightarrow\left(x^2\right)^2+2.x^2.\left(\frac{1}{x^2}\right)+\left(\frac{1}{x^2}\right)^2=49\)

\(\Leftrightarrow x^4+2+\frac{1}{x^4}=49\)

\(\Leftrightarrow x^4+\frac{1}{x^4}=47\)

\(x+\frac{1}{x}=3\Rightarrow\left(x+\frac{1}{x}\right)^2=x^2+2+\frac{1}{x^2}=9\Rightarrow x^2+\frac{1}{x^2}=7\)

\(\Rightarrow\left(x^2+\frac{1}{x^2}\right)^2=x^4+2+\frac{1}{x^4}=49\Rightarrow x^4+\frac{1}{x^4}=47\)

a) Ta có: \(|-5x|-16=3x\) 

Đk:  \(3x\ge0\Rightarrow x\ge0\)

\(\Rightarrow\orbr{\begin{cases}-5x-16=3x\\5x-16=3x\end{cases}}\Rightarrow\orbr{\begin{cases}-5x-3x=16\\5x-3x=16\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}-8x=16\\-2x=16\end{cases}\Rightarrow\orbr{\begin{cases}x=-2\\x=8\end{cases}}}\)

Mà x \(\ge0\)\(\Rightarrow x=8\)

b) \(|3x-2|=1-x\)

\(\Rightarrow\orbr{\begin{cases}3x-2=1-x\\3x-2=-1+x\end{cases}\Rightarrow}\orbr{\begin{cases}3x+x=1+2\\3x-x=-1+2\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}4x=3\\2x=1\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{3}{4}\\x=\frac{1}{2}\end{cases}}\)

Vậy: x = \(\frac{3}{4}\)hoặc x\(=\frac{1}{2}\)

c) Ta có: \(|-2x|=4x-10\)

Đk: \(4x-10\ge0\Rightarrow4x\ge10\Rightarrow x\ge\frac{5}{2}\)

\(\Rightarrow\orbr{\begin{cases}-2x=4x-10\\2x=4x-10\end{cases}}\Rightarrow\orbr{\begin{cases}-2x-4x=-10\\2x-4x=-10\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}-6x=-10\\-2x=-10\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{5}{3}\\x=5\end{cases}}\)

mà x\(\ge\frac{5}{2}\)\(\Rightarrow x=5\)

pt <=> x^4+x^3+x^2+x^2+x+1=0 
<=> x^4+x^2+x^3+x+x^2+1=0 
<=> x^2(x^2+1)+x(x^2+1)+(x^2+1)=0 
<=>(x^2+x+1)(x^2+1)=0 
<=> x^2+x+1=0 (Vô nghiệm) 
hoặc x^2+1=0 (vô lý) 
=>pt vô nghiệm

tk mk nhé

b chép sai đề r híc-.-

a) Ta có: \(\frac{2}{x+1}-\frac{1}{x-2}=\frac{3x-11}{\left(x+1\right)\left(x-2\right)}\)

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

\(\Rightarrow\frac{x-5}{\left(x+1\right)\left(x-2\right)}=\frac{3x-11}{\left(x+1\right)\left(x-2\right)}\)\(\Rightarrow x-5=3x-11\Rightarrow x-3x=-11+5\Rightarrow-2x=-6\Rightarrow x=3\)

b)Ta có:  \(\frac{15-6x}{3}>5\)

\(\Rightarrow15-6x>15\)

\(\Rightarrow6x< 0\)

\(\Rightarrow x< 0\).

Kb với mình nha!

a) x=3

b) x<0

\(\widehat{ABE}=\widehat{ABC}+\widehat{CBE}=\widehat{ABC}+60^0\)    (do tam giác BCE đều)

\(\widehat{FDA}=\widehat{ADC}+\widehat{CDF}=\widehat{ADC}+60^0\) (do tam giác DFC đều)

 ABCD là hình bình hành  =>   \(\widehat{ABC}=\widehat{AD}C\)

suy ra:   \(\widehat{ABE}=\widehat{FDA}\)

Xét  \(\Delta ABE\)và     \(\Delta FDA\)có:

    \(AB=FD\)  (cùng bằng DC)

   \(\widehat{ABE}=\widehat{FDA}\) (cmt)

  \(BE=DA\)  (cùng bằng BC)

suy ra:   \(\Delta ABE=\Delta FDA\)  (c.g.c)

\(\Rightarrow\)\(AE=AF\)   (1)

Ta có:   \(\widehat{FCE}=360^0-\widehat{DCF}-\widehat{BCE}-\widehat{BCD}\)

                         \(=360^0-60^0-60^0-\widehat{BCD}\)

                         \(=240^0-\widehat{BCD}\)

                        \(=240^0-\left(180^0-\widehat{ABC}\right)=60^0+\widehat{ABC}\)

suy ra:   \(\widehat{FCE}=\widehat{ABE}\)

dễ dàng c/m:  \(\Delta ABE=\Delta FCE\)  (c.g.c)

\(\Rightarrow\)\(AE=FE\)       (2)

Từ (1) và (2) suy ra:   \(AF=FE=EA\)

hay    \(\Delta AEF\)đều

\(\Rightarrow\)\(\widehat{EAF}=60^0\)

mk lm dc rui nhug dù sao cux thanks bn