đùa người

a) x=7-3

x=4

b) x=0-5

x=-5

c) x=2-9

x=-7

1.Tim x:

a)| x + 1 | = 5 -> Th1: x+1=5-> x= 5-1=4

Th2: x+1=-5-> x= (-5) -1=-6(Loại. vì x lớn hơn hoặc bằng 0)

Vậy x= 4

b)| x - 3 | = 7 -> TH1: x-3=7-> x=7+3=10(Loại. Vì x<3)

TH2: x-3=-7-> x=-7+3=-4

Vậy x= -4

c) x + | 2 - x | = 6

-> | 2 - x | =6 -x

-> TH1: 2-x = 6-x

-> -x+ x= 2-6

-> 0x =-4(LOẠI)

TH2: 2-x= -6+x

->(-x)-x= 2+6

-> -2.x=8

-> x=8: -2=-4

Vậy x=-4

Tick cho mik nha!!!

2. Tìm x

a) | x | = 7-> x=-7 hoặc x=7

b) | x | < 7.Vì| x | lớn hơn hoặc bằng 0

-> | x | =(0;1;2;3;4;5;6)

-> x= (-6;-5;-4;-3;-2;-1;0;1;2;3;4;5;6)

c) | x | > 7

-> | x | =(8;9;10;11;12;13.............)

-> x= (...............;-9;-8;8;9;10;.............)

Câu a:

\(\frac{-8}{3x-1}\) = \(\frac{4}{-7}\)

-8.(-7) = 4.(3\(x\) - 1)

56 = 12\(x\) - 4

12\(x\) = 56+ 4

12\(x\) = 60

\(x\) = 60 : 12

\(x\) = 5

Vậy \(x\) = 5

Câu b:

\(\frac{x}{-3}\) = \(\frac{-3}{x}\)

\(x^2\) = (-3)\(^2\)

\(\left[\begin{array}{l}x=-3\\ x=3\end{array}\right.\)

Vậy \(x\in\left\lbrace-3;3\right\rbrace\)

Câu c:

\(-\frac{4}{y}=\frac{x}{2}\)

-4.2 = \(x.y\)

\(xy=-8\)

Ư(8) = (-8; -4; -2; -1; 1; 2; 4; 8}

Vậy (\(x;y\)) = (-8; 1); (-4; 2); (-2; 4); (-1; 8); (1; -8); (2; -4); (4; -2); (8; -1)

Câu 2:

(\(x-1)\)(y + 2) = 7

Ư(7) = {-7; -1; 1; 7}

Lập bảng ta có:

Theo bảng trên ta có:

(\(x;y\)) = (-6; -3); (0; -9); (2; 5); (8; - 1)

Vậy (\(x;y\)) = (-6; -3); (0; -9); (2; 5); (8; -1)

|\(\frac32x\) + \(\frac12\)| = |4\(x\) - 1|

\(\left[\begin{array}{l}\frac32x+\frac12=-4x+1\\ \frac32x+\frac12=4x-1\end{array}\right.\)

\(\left[\begin{array}{l}\frac32x+4x=1-\frac12\\ \frac32x-4x=-1-\frac12\end{array}\right.\)

\(\left[\begin{array}{l}\frac{11}{2}x=\frac12\\ -\frac52x=-\frac32\end{array}\right.\)

\(\left[\begin{array}{l}x=\frac12:\frac{11}{2}\\ x=-\frac32:\frac{-5}{2}\end{array}\right.\)

\(\left[\begin{array}{l}x=\frac12\times\frac{2}{11}\\ x=-\frac32\times\frac{-2}{5}\end{array}\right.\)

\(\left[\begin{array}{l}x=\frac{1}{11}\\ x=\frac35\end{array}\right.\)

Vậy \(x\in\) {\(\frac{1}{11};\frac35\)}

|\(\frac54x\) - \(\frac72\)| - |\(\frac58x\) + \(\frac35\)| = 0

|\(\frac54x\) - \(\frac72\)| = |\(\frac58x\) + \(\frac35\)|

\(\left[\begin{array}{l}\frac54x-\frac72=-\frac58x-\frac35\\ \frac54x-\frac72=\frac58x+\frac35\end{array}\right.\)

\(\left[\begin{array}{l}\frac54x+\frac58x=\frac72-\frac35\\ \frac54x-\frac58x=\frac72+\frac35\end{array}\right.\)

\(\left[\begin{array}{l}\frac{15}{8}x=\frac{29}{20}\\ \frac58x=\frac{41}{10}\end{array}\right.\)

\(\left[\begin{array}{l}x=\frac{29}{10}:\frac{15}{8}\\ x=\frac{41}{10}:\frac58\end{array}\right.\)

\(\left[\begin{array}{l}x=\frac{116}{75}\\ x=\frac{164}{25}\end{array}\right.\)

Vậy \(x\in\) {\(\frac{116}{75}\); \(\frac{164}{25}\)}

a) \(\frac{9}{20}\)                                                   c) \(\frac{-55}{4}\)                                                                                                                                 

b) \(\frac{116}{75}\)                                                d) \(\frac{-76}{45}\)

đúng hết đấy nhé mình tính kĩ lắm ko sai đâu

       chúc may mắn

a) Ta có : \(\left|x-2\right|\ge0\forall x\)

                 \(\left|x+y-10\right|\ge0\forall x\)

Nên : \(\left|x-2\right|+\left|x+y-10\right|\ge0\forall x\)

Mà đề bài cho \(\left|x-2\right|+\left|x+y-10\right|\le0\)

Nên : \(\hept{\begin{cases}x-2=0\\x+y-10=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=2\\2+y-10=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x=2\\y=8\end{cases}}}\)

Vậy x = 2 ; y = 8 

Ta có : \(\left|x-2\right|\ge0\forall x\)

               \(\left|x.y-6\right|\ge0\forall x,y\)

Mà : \(\left|x-2\right|+\left|x.y-6\right|=0\)

Nên : pt \(\Leftrightarrow\hept{\begin{cases}x-2=0\\x.y-6=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=2\\x.y=6\end{cases}\Leftrightarrow}\hept{\begin{cases}x=2\\2y=6\end{cases}\Leftrightarrow}\hept{\begin{cases}x=2\\y=3\end{cases}}}\)