a)\(\left[{}\begin{matrix}\dfrac{5}{2}-x=\dfrac{1}{3}\\\dfrac{5}{2}-x=-\dfrac{1}{3}\end{matrix}\right.\left[{}\begin{matrix}x=\dfrac{13}{6}\\x=\dfrac{17}{6}\end{matrix}\right.\)

b) 8/6-x-1/5=0

9/6-x=1/5

x=13/10

a = 1,2

b = 1,8

a) \(\left|2,5-x\right|=1,3\)

\(\Leftrightarrow\left[{}\begin{matrix}2,5-x=1,3\\2,5-x=-1,3\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1,2\\x=3,8\end{matrix}\right.\)

b) \(1,6-\left|x-0,2\right|=0\)

\(\Leftrightarrow\left|x-0,2\right|=1,6\)

\(\Leftrightarrow\left[{}\begin{matrix}x-0,2=1,6\\x-0,2=-1,6\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1,8\\x=-1,4\end{matrix}\right.\)

a) \(\left(x-1,3\right)^2=9\Leftrightarrow\left[{}\begin{matrix}x-1,3=3\\x-1,3=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4,3\\x=-1,7\end{matrix}\right.\)

b) 2^4-x = 32

⇔ 2^4-x = 2⁵

⇔ 4-x=5

⇔ x=-1

c) (x+1,5)²+(y-2,5)¹⁰=0

\(\Leftrightarrow\left\{{}\begin{matrix}x+1,5=0\\y-2,5=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-1,5\\y=2,5\end{matrix}\right.\)

\(a,\left(x-1,3\right)^2=9\\ \Leftrightarrow\left(x-1,3+9\right)\left(x-1,3-9\right)=0\\ \Leftrightarrow\left(x-7,7\right)\left(x-10,3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=7,7=\dfrac{77}{10}\\x=10,3=\dfrac{103}{10}\end{matrix}\right.\)

\(b,2^{4-x}=32=2^5\\ \Leftrightarrow4-x=5\\ \Leftrightarrow x=-1\)

\(c,\left(x+1,5\right)^2+\left(y-2,5\right)^{10}=0\\ \Leftrightarrow\left\{{}\begin{matrix}x+1,5=0\\y-2,5=0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=-1,5=-\dfrac{3}{2}\\y=2,5=\dfrac{5}{2}\end{matrix}\right.\)

a. (x - 1,3)² = 9

<=> (x - 1,3)² - 9 = 0

<=> (x - 1,3)² - 3² = 0

<=> (x - 1,3 - 3)(x - 1,3 + 3) = 0

<=> (x - 4,3)(x + 1,7) = 0

<=> \(\left[{}\begin{matrix}x-4,3=0\\x+1,7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4,3\\x=-1,7\end{matrix}\right.\) 

\(\Rightarrow x-2,2=0,2+x \)

\(\Leftrightarrow x\in rổng\)

\(Do\left|x+\frac{8}{5}\right|\ge0;\left|2,2-2y\right|\ge0=>\left|x+\frac{8}{5}\right|+\left|2,2-2y\right|\ge0\)

Mà \(\left|x+\frac{8}{5}\right|+\left|2,2-2y\right|\le0=>\left|x+\frac{8}{5}\right|+\left|2,2-2y\right|=0\)

\(=>\hept{\begin{cases}\left|x+\frac{8}{5}\right|=0\\\left|2,2-2y\right|=0\end{cases}=>\hept{\begin{cases}x+\frac{8}{5}=0\\2,2-2y=0\end{cases}=>\hept{\begin{cases}x=-\frac{8}{5}\\2y=2,2\end{cases}=>\hept{\begin{cases}x=-1,6\\y=1,1\end{cases}}}}}\)

Vậy x = -1,6; y = 1,1

Ủng hộ mk nha ^_-

trả lời nhanh nha!!!

Hồng Phúc Nguyễn bn hk cx giỏi nhỉ bn giúp mk bài này nha

xin bn đấy!!!

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

\(\left|x\right|=1,3+2,2\)

\(\left|x\right|=3,5\)

Vậy x = 3,5 hoặc x = -3,5

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

                \(x=1,3+2,2\)

                 \(x=3,5\)

hoặc : |x| - 2,2 = 1,3

         -x - 2,2 = 1,3

                 - x= 1,3 + 2,2

                  -x = 3,5

                   x = -3,5

Vậy x = 3,5 hoặc x = -3,5

mik nhah nhất đó

/x/-2,2=1,3

/x/=1,3+2,2

/x/=3,5

Vậy x=3,5 hoặc x=-3,5

Ta luôn có : \(\left|x+\frac{8}{5}\right|\ge0\) , \(\left|2,2-2y\right|\ge0\)

Suy ra \(\left|x+\frac{8}{5}\right|+\left|2,2-2y\right|\ge0\)

mà \(\left|x+\frac{8}{5}\right|+\left|2,2-2y\right|\le0\)

Do đó : \(\left|x+\frac{8}{5}\right|+\left|2,2-2y\right|=0\)

\(\Rightarrow\begin{cases}\left|x+\frac{8}{5}\right|=0\\\left|2,2-2y\right|=0\end{cases}\) \(\Rightarrow\begin{cases}x=-\frac{8}{5}\\y=\frac{11}{10}\end{cases}\)

Ta có

\(\begin{cases}\left|x+\frac{8}{5}\right|\ge0\\\left|2,3-2y\right|\ge0\end{cases}\)

=> \(\left|x+\frac{8}{5}\right|+\left|2,3-2y\right|\ge0\)

=> \(x,y\in\varnothing\)

Vì : \(\left|x+\frac{8}{5}\right|\ge0;\left|2,2-2y\right|\ge0\)

\(\Rightarrow\left|x+\frac{8}{5}\right|+\left|2,2-2y\right|\ge0\)

Mà theo đề bài : \(\left|x+\frac{8}{5}\right|+\left|2,2-2y\right|\le0\)

\(\Rightarrow\left|x+\frac{8}{5}\right|+\left|2,2-2y\right|=0\)

\(\Rightarrow\begin{cases}\left|x+\frac{8}{5}\right|=0\\\left|2,2-2y\right|=0\end{cases}\)

\(\Rightarrow\begin{cases}x+\frac{8}{5}=0\\2,2-2y=0\end{cases}\)

\(\Rightarrow\begin{cases}x=-\frac{8}{5}\\2y=2,2\end{cases}\)

\(\Rightarrow\begin{cases}x=\frac{-8}{5}\\y=1,1\end{cases}\)

\(\Rightarrow\begin{cases}x=\frac{-8}{5}\\y=\frac{11}{10}\end{cases}\)

ta có:

Thầy làm chi tiết giúp em được ko ạ ?

Tl

a =x-1,3-2,5+x

=-3,8

b,D=x+3 và 1/2+x - 3 và 1/2

=2x

c,A=-x +1/7-x-5/5+4/5=1/7+1/5=12/35

Hok tốt