3) a) \(\left|x\right|=\dfrac{1}{2}\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{-1}{2}\end{matrix}\right.\) vậy \(x=\dfrac{1}{2};x=\dfrac{-1}{2}\)

b) \(\left|x\right|=3,12\Leftrightarrow\left[{}\begin{matrix}x=3,12\\x=-3,12\end{matrix}\right.\) vậy \(x=3,12;x=-3,12\)

c) \(\left|x\right|=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-0\end{matrix}\right.\) \(\Rightarrow x=0\) vậy \(x=0\)

d) th1: \(\left|x\right|=2\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)

th2: \(\left|x\right|=\dfrac{1}{7}\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{7}\\x=\dfrac{-1}{7}\end{matrix}\right.\)

vậy \(x=2;x=-2;x=\dfrac{1}{7};x=\dfrac{-1}{7}\)

Bạn viết đề cho rõ đi

vừa viết tắt ; vừa viết đề sai

a)|x|=\(\frac{1}{2}\)

=>x=   +-\(\frac{1}{2}\)

|x|=3,12

=>x= +-3,12

c,|x|=0

x=0

d, |x|=\(2\frac{1}{7}\)

=>x=+-\(2\frac{1}{7}\)

                                      CHÚC BẠN HỌC TỐT

A) |x|=\(\frac{1}{2}\)                          b)|x|=3,12

=>x= +- \(\frac{1}{2}\)                           => x= +- 3,12

c)|x|=0                                  d) |x|=\(2\frac{1}{7}\)

=>x =0                                 =>x =+-\(2\frac{1}{7}\)                 

                                   HỌC TỐT NHÉ

/-3/ = /3/

/1,3/ > /0,5/

/-100/ > /20/

bài 2: tìm x

/x/ = \(\frac{1}{2}\)

\(x=\orbr{\begin{cases}\frac{1}{2}\\\frac{-1}{2}\end{cases}}\)

/x/ = 3,12

\(x=\orbr{\begin{cases}3,12\\-3,12\end{cases}}\)

/x/ = 0

=> x = 0

/x/ = \(2\frac{1}{7}\)

=> /x/ = \(\frac{15}{7}\)

=> \(x=\orbr{\begin{cases}\frac{15}{7}\\\frac{-15}{7}\end{cases}}\)

chúc bn học tốt

bạn làm đúng thật mẹ tôi bảo là chuẩn

\(\left|x\right|=\frac{1}{2}\)

\(\Rightarrow x=\pm\frac{1}{2}\)

\(\left|x\right|=3,12\)

\(\Rightarrow x=\pm3,12\)

\(\left|x\right|=0\)

\(\Rightarrow x=0\)

\(\left|x\right|=2\frac{1}{7}\)

\(\Rightarrow x=\pm2\frac{1}{7}\)

1, a/ \(\left|x\right|=\dfrac{1}{2}\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{-1}{2}\end{matrix}\right.\)

Vậy .............

b/ \(\left|x\right|=3,12\Leftrightarrow\left[{}\begin{matrix}x=3,12\\x=-3,12\end{matrix}\right.\)

Vậy ...........

c/ \(\left|x\right|=0\Leftrightarrow x=0\)

Vậy ..........

d/ \(\left|x\right|=2\dfrac{1}{7}\) \(\Leftrightarrow\left[{}\begin{matrix}x=2\dfrac{1}{7}\\x=-2\dfrac{1}{7}\end{matrix}\right.\)

Vậy ..............

2, a/ \(\left|x\right|=2,1\Leftrightarrow\left[{}\begin{matrix}x=2,1\\x=-2,1\end{matrix}\right.\)

Vậy ...........

b/ \(\left|x\right|=\dfrac{17}{9}\) ; \(x< 0\)

\(\Leftrightarrow x=-\dfrac{17}{9}\)

Vậy ..........

c/ \(\left|x\right|=1\dfrac{2}{5}\Leftrightarrow\left[{}\begin{matrix}x=1\dfrac{2}{5}\\x=-1\dfrac{2}{5}\end{matrix}\right.\)

Vậy ...........

d/ \(\left|x\right|=0,35\) ; \(x>0\Leftrightarrow x=0,35\)

3, a/ \(\left|x-1,7\right|=2,3\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1,7=2,3\\x-1,7=-2,3\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-0,6\end{matrix}\right.\)

Vậy ...........

b/ \(\left|x+\dfrac{3}{4}\right|-\dfrac{1}{3}=0\)

\(\Leftrightarrow\left|x+\dfrac{3}{4}\right|=\dfrac{1}{3}\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{3}{4}=\dfrac{1}{3}\\x+\dfrac{3}{4}=-\dfrac{1}{3}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-5}{12}\\x=-\dfrac{13}{12}\end{matrix}\right.\)

Vậy ...........

Đề dễ lắm sao ko tự làm đi

Bài1:

a/ |x|=

a/|x|=2,1

=>\(x=-2,1\)hoặc \(x=2,1\)

Vậy...

Bài3:

a/ | x - 1,7| =2,3

=>\(x-1,7=2,3\)hoặc\(x-1,7=-2,3\)

=>x=4 hoặc x=-0,6

Vậy ...

Các bạn giúp mình giải với nhé! Đúng thì mình k đúng nhé. Cảm ơn các bạn nhiều lắm. Yêu cả nhà.

\(1.\left(x-5\right)^{23}.\left(y+2\right)^7=0\)

\(\Rightarrow\hept{\begin{cases}\left(x-5\right)^{23}=0\\\left(y+2\right)^7=0\end{cases}\Rightarrow\hept{\begin{cases}\left(x-5\right)^{23}=0^{23}\\\left(y+2\right)^7=0^7\end{cases}}}\)\(\Rightarrow\hept{\begin{cases}x-5=0\\y+2=0\end{cases}\Rightarrow\hept{\begin{cases}x=0+5\\y=0-2\end{cases}}}\)\(\Rightarrow\hept{\begin{cases}x=5\\y=-2\end{cases}}\)

Vậy \(\left(x;y\right)=\left(5;-2\right)\)

2. \(A=\left(x-2\right)^2+|y+3|+7\)

Ta có :

\(\hept{\begin{cases}\left(x-2\right)^2\ge0\forall x\\|y+3|\ge0\forall y\end{cases}}\)

\(\Rightarrow\left(x-2\right)^2+|y+3|\ge0\forall x;y\)

\(\Rightarrow\left(x-2\right)^2+|y+3|+7\ge7\forall x;y\)

\(\Rightarrow A\ge7\forall x;y\)

Dấu bằng xảy ra

\(\Leftrightarrow\hept{\begin{cases}\left(x-2\right)^2=0\\|y+3|=0\end{cases}\Leftrightarrow\hept{\begin{cases}x-2=0\\y+3=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x=2\\y=-3\end{cases}}}\)

Vậy GTNN của A là 7 khi \(\left(x;y\right)=\left(2;-3\right)\)

a)|x|=1/2

=>x=\(\pm\dfrac{1}{2}\)

b)|x|=3,12

=>x=\(\pm3,12\)

a)

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

\(\Rightarrow\left(x+2\right)^2=9=3^2\)

\(\Rightarrow x+2=\pm3\)

\(\Rightarrow x=-5;1\)

b)

\(25x^2-10x+1=0\)

\(\left(5x\right)^2-2\cdot5x+1^2=0\)

\(\Rightarrow\left(5x+1\right)^2=0\)

\(\Rightarrow5x+1=0\)

\(\Rightarrow5x=-1;x=\dfrac{-1}{5}\)

c)

\(x^2+14x+49=0\)

\(\Rightarrow x^2+2\cdot7x+7^2=0\)

\(\Rightarrow\left(x+7\right)^2=0;x+7=0\)

\(\Rightarrow x=-7\)

d)

\(\left(2x-1\right)^2+\left(x+3\right)^2-5\left(x+7\right)\left(x-7\right)=0\)

\(4x^2-4x+1+x^2+6x+9-5x^2+5\cdot49=0\)

\(\Rightarrow5x^2-5x^2-4x+6x+10+245=0\)

\(\Rightarrow2x+255=0\)

\(\Rightarrow2x=-255\)

\(\Rightarrow x=\dfrac{-255}{2}\)

1) \(\Rightarrow x^2+4x+4-x^2+1=9\)

\(\Rightarrow4x=4\Rightarrow x=1\)

2) \(\Rightarrow x\left(2x+7\right)+2\left(2x+7\right)=0\)

\(\Rightarrow\left(2x+7\right)\left(x+2\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{7}{2}\\x=-2\end{matrix}\right.\)

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

\(\Rightarrow3x=1\Rightarrow x=\dfrac{1}{3}\)