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)

A=7 B=1 OR =0 C=2 D=4.5.6

a: \(2^{x}\cdot4=128\)

=>\(2^{x}=\frac{128}{4}=32=2^5\)

=>x=5

b: \(x^{15}=x\)

=>\(x^{15}-x=0\)

=>\(x\left(x^{14}-1\right)=0\)

=>\(\left[\begin{array}{l}x=0\\ x^{14}-1=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=0\\ x^{14}=1\end{array}\right.\Rightarrow\left[\begin{array}{l}x=0\\ x=1\end{array}\right.\)

c: \(\left(2x+1\right)^3=125\)

=>\(\left(2x+1\right)^3=5^3\)

=>2x+1=5

=>2x=5-1=4

=>\(x=\frac42=2\)

d: \(\left(x-5\right)^4=\left(x-5\right)^6\)

=>\(\left(x-5\right)^6-\left(x-5\right)^4=0\)

=>\(\left(x-5\right)^4\cdot\left\lbrack\left(x-5\right)^2-1\right\rbrack=0\)

=>\(\left(x-5\right)^4\cdot\left(x-5-1\right)\left(x-5+1\right)=0\)

=>\(\left(x-5\right)^4\cdot\left(x-6\right)\left(x-4\right)=0\)

=>\(\left[\begin{array}{l}x-5=0\\ x-6=0\\ x-4=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=5\\ x=6\\ x=4\end{array}\right.\)

|\(\frac32x\) + \(\frac12\)| = |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) \(\frac{x-7}{160}=\frac{9}{24}\)=> \(\frac{x-7}{160}=\frac{3}{8}\)=> \(\frac{\left(x-7\right):20}{160:20}=\frac{3}{8}\)

=> \(\left(x-7\right):20=3\)

=> \(x-7=60\)

=> \(x=67\)

b) \(\frac{x-3}{8}=\frac{23-5x}{24}\)

=> \(\frac{3\left(x-3\right)}{3\cdot8}=\frac{23-5x}{24}\)

=> \(\frac{3x-9}{24}=\frac{23-5x}{24}\)

=> \(3x-9=23-5x\)

=> \(3x+5x=23+9\)

=> \(8x=32\)

=> \(x=4\)

c) * Suy nghĩ các thứ * 

a) x-7= 9/24 .160=60

=>x=67

b)\(\frac{3x-9}{24}-\frac{23-5x}{24}=0.\)

<=>3x-9-23+5x=0

<=>8x-32=0

<=>x=4

c)xy=-10

mà x,y thuộc Z,x<0<y

=>x=-5,y=2

học tốt

Bài 1:

a; \(\dfrac{x}{3}\) = \(\dfrac{4}{y}\)

     \(xy\) = 12

12 = 2².3; Ư(12) = {-12; -6; -4; -3; -2; -1; 1; 2; 3; 4; 6;12}

Lập bảng ta có:

Theo bảng trên ta có các cặp \(x;y\) nguyên thỏa mãn đề bài là:

(\(x\)\(;y\)) =(-12; -1);(-6; -2);(-4; -3);(-2; -6);(-1; 12);(1; 12);(2;6);(3;4);(4;3);(6;2);(12;1)

b; \(\dfrac{x}{y}\) = \(\dfrac{2}{7}\)

    \(x\) = \(\dfrac{2}{7}\).y 

     \(x\) \(\in\)z ⇔ y ⋮ 7

   y = 7k;

   \(x\) = 2k 

Vậy \(\left\{{}\begin{matrix}x=2k\\y=7k;k\in z\end{matrix}\right.\) 

3 + (-2) + x = 5

1 + x = 5 x

= 5 - 1 (chuyển 1 sang vế phải)

x = 4

Sách Giáo Khoa

Bài giải:

Ta có đẳng thức tìm x như sau:

3 + (-2) + x = 5

1 + x = 5

x = 5 - 1

x = 4

Đáp số: x = 4.

Bài 1

a) \(\frac{5}{6}=\frac{x-1}{x}\)

<=> 5x=6x-6

<=> 5x-6x=-6

<=> -11x=-6

<=> \(x=\frac{6}{11}\)

b)c)d) nhân chéo làm tương tự

Câu 1

b; 1/2=x+1/3x

x.(1+ 1/3) = 1/2

x.4/3 = 1/2

x = 1/2 : 4/3

x = 3/8

Vậy x = 3/8

c; 3/2+x=5/2x+1

5/2x - x = 3/2 - 1

x(5/2 - 1) = 1/2

x.3/2 = 1/2

x = 1/2 : 3/2

x = 1/2 x 2/3

x = 1/3

Vậy x = 1/3

d/5/8x−2=−4/7−x