\(\Leftrightarrow\left[{}\begin{matrix}x+1=8\\x+1=-8\end{matrix}\right.\Leftrightarrow x\in\left\{7;-9\right\}\)

Lời giải:

$N=x^2-2xy+2y^2-x=(2y^2-2xy+\frac{x^2}{2})+(\frac{x^2}{2}-x+\frac{1}{2})-\frac{1}{2}$

$=2(y-\frac{x}{2})^2+\frac{1}{2}(x-1)^2-\frac{1}{2}\geq \frac{-1}{2}$

Vậy GTNN của $N$ là $\frac{-1}{2}$

Giá trị này đạt tại $y-\frac{x}{2}=x-1=0$

$\Leftrightarrow x=1; y=\frac{1}{2}$

Ta có: N = x^2 -2xy +2y^2 -x

         2N = 2x^2 - 4xy + 4y^2 - 2x

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

              = (x-2y)^2 + ( x-1)^2 -1 

=> 2N lớn hơn hoặc bằng -1

=> N lớn hơn hoặc bằng -1/2

   Dấu "=" xảy ra <=> ( x-2y )^2 = 0 và ( x-1 )^2 = 0

                             => x-2y=0 và x-1=0

                             => x=1 và y=1/2

Vậy tại x=1 và y=1/2 thì biểu thức N đạt GTNN là -1/2

\(a,\Leftrightarrow\left|x+\dfrac{2}{5}\right|=\dfrac{7}{4}\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{2}{5}=\dfrac{7}{4}\left(x\ge-\dfrac{2}{5}\right)\\x+\dfrac{2}{5}=-\dfrac{7}{4}\left(x< -\dfrac{2}{5}\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{27}{20}\left(tm\right)\\x=-\dfrac{43}{20}\left(tm\right)\end{matrix}\right.\)

\(b,\Leftrightarrow\left|x-\dfrac{13}{10}\right|=\dfrac{13}{10}\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{13}{10}=\dfrac{13}{10}\left(x\ge\dfrac{13}{10}\right)\\x-\dfrac{13}{10}=-\dfrac{13}{10}\left(x< \dfrac{13}{10}\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{13}{5}\left(tm\right)\\x=0\left(tm\right)\end{matrix}\right.\)

\(c,\Leftrightarrow\left|\dfrac{3}{4}-\dfrac{1}{2}x\right|=\dfrac{1}{2}\Leftrightarrow\left[{}\begin{matrix}\dfrac{3}{4}-\dfrac{1}{2}x=\dfrac{1}{2}\left(x\le\dfrac{3}{2}\right)\\\dfrac{1}{2}x-\dfrac{3}{4}=\dfrac{1}{2}\left(x>\dfrac{3}{2}\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\left(tm\right)\\x=\dfrac{5}{2}\left(tm\right)\end{matrix}\right.\)

\(d,\Leftrightarrow\left|5-2x\right|=4\Leftrightarrow\left[{}\begin{matrix}5-2x=4\left(x\le\dfrac{5}{2}\right)\\2x-5=4\left(x>\dfrac{5}{2}\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\left(tm\right)\\x=\dfrac{9}{2}\left(tm\right)\end{matrix}\right.\)

\(đ,\Leftrightarrow\left\{{}\begin{matrix}x-3,5=0\\x-1,3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3,5\\x=1,3\end{matrix}\right.\left(vô.lí\right)\Leftrightarrow x\in\varnothing\)

\(e,\Leftrightarrow\left\{{}\begin{matrix}x-2021=0\\x-2022=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2021\\x=2022\end{matrix}\right.\left(vô.lí\right)\Leftrightarrow x\in\varnothing\)

\(f,\Leftrightarrow\left|x\right|=\dfrac{1}{3}-x\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}-x\left(x\ge0\right)\\x=x-\dfrac{1}{3}\left(x< 0\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{6}\left(tm\right)\\0x=-\dfrac{1}{3}\left(vô.lí\right)\end{matrix}\right.\Leftrightarrow x=\dfrac{1}{6}\)

\(g,\Leftrightarrow\left[{}\begin{matrix}x-2=x\left(x\ge2\right)\\2-x=x\left(x< 2\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}0x=2\left(vô.lí\right)\\x=1\left(tm\right)\end{matrix}\right.\Leftrightarrow x=1\)

x + (-15) - (-12) =0
x -3 =0
x = 3

\(x-3=0\Leftrightarrow x=3\)

x +(-15)-(-12)=0

x +(-3)            =0

x                      =0 -(-3)

x                      = 3

x

vì \(\left(x+1\right)< \left(x+2\right)\)

để \(\left(x+1\right).\left(x+2\right)>0\)

=> \(\hept{\begin{cases}x+1< 0\\x+2>0\end{cases}\Rightarrow\hept{\begin{cases}x< -1\\x>-2\end{cases}}}\)

=> ko có giá trị x t/mãn

b) 

để \(\left(x-2\right).\left(x+\frac{2}{3}\right)>0\)

=> \(\hept{\begin{cases}x-2>0\\\left(x+\frac{2}{3}\right)\end{cases}>0}hay\hept{\begin{cases}x-2< 0\\x+\frac{2}{3}< 0\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}x>2\\x>-\frac{2}{3}\end{cases}}hay\hept{\begin{cases}x< 2\\x< -\frac{2}{3}\end{cases}}\)

vậy \(x>2,x< -\frac{2}{3}\)

eei dòng thứ hai ấy tớ viết lộn nha :))

\(\left(x+1\right).\left(x+2\right)< 0\)

x + 1 < x + 2 ? 

nếu x âm thì sao ?????????????????????????

\(\left(x+1\right)\left(x+2\right)< 0\)

th1 :

\(\hept{\begin{cases}x+1< 0\\x+2>0\end{cases}\Rightarrow\hept{\begin{cases}x< -1\\x>-2\end{cases}\left(vl\right)}}\)

th2 : 

\(\hept{\begin{cases}x+1>0\\x+2< 0\end{cases}\Rightarrow\hept{\begin{cases}x>-1\\x< -2\end{cases}\left(vl\right)}}\)

... 

b tương tự

a: Ta có: x+17<10

nên x<-7

b: Ta có: 9-2x<0

\(\Leftrightarrow2x>9\)

hay \(x>\dfrac{9}{2}\)

c: Ta có: \(-3x-11\ge0\)

\(\Leftrightarrow-3x\ge11\)

hay \(x\le-\dfrac{11}{3}\)

x-5/2<0

=>x>5/2

7-x/3>

=>x/3<7

(14,78-a)/(2,87+a)=4/1

14,78+2,87=17,65

Tổng số phần bằng nhau là 4+1=5

Mỗi phần có giá trị bằng 17,65/5=3,53

=>2,87+a=3,53

=>a=0,66.

bexiu đang trả lời câu nào vậy ghi giúp mik cái đề