\(a.\)

\(x+5=-10\)

\(\Rightarrow x=-10-5=-15\)

a ) x +5 = -10

x = -10 -5

x = - 15

b) x - ( - 10 ) = 5

x = 5+(-10)

x = -5

c) \(\left|x\right|\) -5 = 3

\(\left|x\right|=8\)

x ϵ { -8 ; 8 }

d) 15 - ( - x ) = 20

Không có số tự nhiên x nào mà 15 ( - x ) = 20

e ) \(\left|x-4\right|=3-\left(-7\right)\\ \left|x-4\right|=10\\ \left|x\right|=14\\ x\in\left\{\pm14\right\}\)

f ) \(\left|x+5\right|=10-\left(-20\right)\\ \left|x+5\right|=30\\ \left|x\right|=25\\ x\in\left\{\pm25\right\}\)

a) \(\left(x-1\right):3=2^3\) \(\Leftrightarrow\) \(\left(x-1\right):3=8\) \(x+1=24\) \(\Leftrightarrow\) \(x=23\) vậy \(x=23\)

b) \(12-2\left(x+5\right)=-10\) \(\Leftrightarrow\) \(12-2x-10=-10\)

\(\Leftrightarrow\) \(-2x=-12\) \(\Leftrightarrow\) \(x=6\) vậy \(x=6\)

c) \(x-12\left(x+5\right)=-10\) \(\Leftrightarrow\) \(x-12x-60=-10\)

\(\Leftrightarrow\) \(-11x=50\) \(\Leftrightarrow\) \(x=\dfrac{50}{-11}\) vậy \(x=\dfrac{50}{-11}\)

e) \(13-x:2=10\Leftrightarrow-x:2=-3\Leftrightarrow x=\dfrac{3}{2}\)

f) \(\left|12-x\right|-7=5\)

th1 : \(x\le12\) thì \(\left|12-x\right|-7=5\) \(\Leftrightarrow\) \(12-x-7=5\) \(\Leftrightarrow\) \(-x=0\Leftrightarrow x=0\)

th2 : \(x>12\) thì \(\left|12-x\right|-7=5\) \(\Leftrightarrow\) \(x-12-7=5\) \(\Leftrightarrow\) \(x=24\) vậy \(x=0;x=24\)

i) \(x^2-7=2\Leftrightarrow x^2=9\Leftrightarrow x=3\) vậy \(x=3\)

k) \(x^3-4=-12\) \(\Leftrightarrow\) \(x^3=-8\) \(\Leftrightarrow x=-2\) vậy \(x=-2\)

a)\(\left(x-1\right):3=2^3\Rightarrow x-1=2^3.3=24\Rightarrow x=25\)

b)\(12-2\left(x+5\right)=-10\Leftrightarrow12-2x-10=-10\Rightarrow2-2x=-10\Rightarrow2x=12\Rightarrow x=6\)c)\(x-12\left(x+5\right)=-10\Rightarrow x-12x-60=-10\Rightarrow-11x-60=-10\Rightarrow-11x=-70\Rightarrow x=\dfrac{70}{-11}\)d)\(6-\left|x\right|=5\Rightarrow\left|x\right|=1\Rightarrow x=\left\{\pm1\right\}\)

Làm nốt nha

\(a,\frac{1}{2}+\frac{2}{3}x=\frac{4}{5}\)

=> \(\frac{2}{3}x=\frac{4}{5}-\frac{1}{2}=\frac{3}{10}\)

=> \(x=\frac{3}{10}:\frac{2}{3}=\frac{9}{20}\)

Vậy \(x\in\left\{\frac{9}{20}\right\}\)

\(b,x+\frac{1}{4}=\frac{4}{3}\)

=> \(x=\frac{4}{3}-\frac{1}{4}=\frac{13}{12}\)

Vậy \(x\in\left\{\frac{13}{12}\right\}\)

\(c,\frac{3}{5}x-\frac{1}{2}=-\frac{1}{7}\)

=> \(\frac{3}{5}x=-\frac{1}{7}+\frac{1}{2}=\frac{5}{14}\)

=> \(x=\frac{5}{14}:\frac{3}{5}=\frac{25}{42}\)

Vậy \(x\in\left\{\frac{25}{42}\right\}\)

\(d,\left|x+5\right|-6=9\)

=> \(\left|x+5\right|=9+6=15\)

=> \(\left[{}\begin{matrix}x+5=15\\x+5=-15\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}x=15-5=10\\x=-15-5=-20\end{matrix}\right.\)

Vậy \(x\in\left\{10;-20\right\}\)

\(e,\left|x-\frac{4}{5}\right|=\frac{3}{4}\)

=> \(\left[{}\begin{matrix}x-\frac{4}{5}=\frac{3}{4}\\x-\frac{4}{5}=-\frac{3}{4}\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}x=\frac{3}{4}+\frac{4}{5}=\frac{31}{20}\\x=-\frac{3}{4}+\frac{4}{5}=\frac{1}{20}\end{matrix}\right.\)

Vậy \(x\in\left\{\frac{31}{20};\frac{1}{20}\right\}\)

\(f,\frac{1}{2}-\left|x\right|=\frac{1}{3}\)

=> \(\left|x\right|=\frac{1}{2}-\frac{1}{3}\)

=> \(\left|x\right|=\frac{1}{6}\)

=> \(\left[{}\begin{matrix}x=\frac{1}{6}\\x=-\frac{1}{6}\end{matrix}\right.\)

Vậy \(x\in\left\{\frac{1}{6};-\frac{1}{6}\right\}\)

\(g,x^2=16\)

=> \(\left|x\right|=\sqrt{16}=4\)

=> \(\left[{}\begin{matrix}x=4\\x=-4\end{matrix}\right.\)

vậy \(x\in\left\{4;-4\right\}\)

\(h,\left(x-\frac{1}{2}\right)^3=\frac{1}{27}\)

=> \(x-\frac{1}{2}=\sqrt[3]{\frac{1}{27}}=\frac{1}{3}\)

=> \(x=\frac{1}{3}+\frac{1}{2}=\frac{5}{6}\)

Vậy \(x\in\left\{\frac{5}{6}\right\}\)

\(i,3^3.x=3^6\)

\(x=3^6:3^3=3^3=27\)

Vậy \(x\in\left\{27\right\}\)

\(J,\frac{1,35}{0,2}=\frac{1,25}{x}\)

=> \(x=\frac{1,25.0,2}{1,35}=\frac{5}{27}\)

Vậy \(x\in\left\{\frac{5}{27}\right\}\)

\(k,1\frac{2}{3}:x=6:0,3\)

=> \(\frac{5}{3}:x=20\)

=> \(x=\frac{5}{3}:20=\frac{1}{12}\)

Vậy \(x\in\left\{\frac{1}{12}\right\}\)

\(\frac{1}{3}x+\frac{2}{5}\left(x-1\right)=0\)

\(\Leftrightarrow\frac{1}{3}x+\frac{2}{5}x-\frac{2}{5}=0\)

\(\Leftrightarrow\frac{11}{15}x=\frac{2}{5}\)

\(\Leftrightarrow x=\frac{2}{5}\div\frac{11}{15}=\frac{2.15}{5.11}=\frac{6}{11}\)

Vậy x = 6/11 

a) \(\frac{1}{3}.x+\frac{2}{5}.\left(x-1\right)=0\)

\(\frac{1}{3}.x+\frac{2}{5}.x-\frac{2}{5}=0\)

\(x.\left(\frac{1}{3}+\frac{2}{5}\right)-\frac{2}{5}=0\)

\(x.\frac{11}{15}-\frac{2}{5}=0\)

\(x.\frac{11}{15}=\frac{2}{5}\)

\(x=\frac{2}{5}:\frac{11}{15}\)

\(x=\frac{6}{11}\)

b) \(3.\left(x-\frac{1}{2}\right)-5.\left(x+\frac{3}{5}\right)=x+\frac{1}{5}\)

\(3x-\frac{3}{2}-5x-3=x+\frac{1}{5}\)

\(3x-5x-\left(\frac{3}{2}+3\right)=x+\frac{1}{5}\)

\(-2x-\frac{9}{2}=x+\frac{1}{5}\)

\(\Rightarrow-2x-x=\frac{1}{5}+\frac{9}{2}\)

\(-3x=\frac{47}{10}\)

\(x=\frac{47}{10}:\left(-3\right)\)

\(x=\frac{-47}{30}\)

Giải:

a) \(\dfrac{1}{3}x+\dfrac{2}{5}\left(x-1\right)=0\)

\(\Leftrightarrow\dfrac{1}{3}x+\dfrac{2}{5}x-\dfrac{2}{5}=0\)

\(\Leftrightarrow\dfrac{11}{15}x-\dfrac{2}{5}=0\)

\(\Leftrightarrow\dfrac{11}{15}x=\dfrac{2}{5}\)

\(\Leftrightarrow x=\dfrac{6}{11}\)

Vậy ...

b) \(3\left(x-\dfrac{1}{2}\right)-5\left(x+\dfrac{3}{5}\right)=x+\dfrac{1}{5}\)

\(\Leftrightarrow3x-\dfrac{3}{2}-5x-3=x+\dfrac{1}{5}\)

\(\Leftrightarrow-2x-\dfrac{9}{2}=x+\dfrac{1}{5}\)

\(\Leftrightarrow-3x=\dfrac{47}{10}\)

\(\Leftrightarrow x=\dfrac{-47}{30}\)

Vậy ...

a, 1/3 . x + 2/5 . ( x - 1 ) = 0

1/3 . x + 2/5 . x - 2/5 = 0

x . ( 1/3 + 2/5 ) = 0 + 2/5

x . 11/15 = 2/5

x = 2/5 : 11/15

x = 6/11

b, 3 . ( x - 1/2 ) - 5 . ( x + 3/5 ) = x + 1/5

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

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

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

-3x = -13/10

x = -13/10 : -1

x = -13/10

a: Ta có: \(A=2+2^2+2^3+\cdots+2^{2025}\)

=>\(2A=2^2+2^3+2^4+\cdots+2^{2026}\)

=>\(2A-A=2^2+2^3+2^4+\cdots+2^{2026}-2-2^2-\cdots-2^{2025}\)

=>\(A=2^{2026}-2\)

b:Sửa đề: \(B=1+5+5^2+\cdots+5^{150}\)

=>\(5B=5+5^2+5^3+\cdots+5^{151}\)

=>\(5B-B=5+5^2+5^3+\cdots+5^{151}-1-5-5^2-\cdots-5^{150}\)

=>\(4B=5^{151}-1\)

=>\(B=\frac{5^{151}-1}{4}\)

c: Ta có: \(C=3+3^2+3^3+\ldots+3^{1000}\)

=>\(3C=3^2+3^3+3^4+\cdots+3^{1001}\)

=>\(3C-C=3^2+3^3+\cdots+3^{1001}-3-3^2-\cdots-3^{1000}\)

=>\(2C=3^{1001}-3\)

=>\(C=\frac{3^{1001}-3}{2}\)

\(a,234-\left(x-56\right)=789\)

\(\Leftrightarrow x-56=234-789\)

\(\Leftrightarrow x-56=-555\)

\(\Leftrightarrow x=\left(-555\right)+56=-499\)

Vậy x = -499

b) \(\frac{x+3}{-5}=\frac{x-15}{4}\)

\(\Leftrightarrow4\left(x+3\right)=-5\left(x-15\right)\)

\(\Leftrightarrow4x+12=-5x+75\)

\(\Leftrightarrow4x+12-\left(-5x\right)=75\)

\(\Leftrightarrow4x-\left(-5x\right)+12=75\)

\(\Leftrightarrow4x+5x=63\)

\(\Leftrightarrow9x=63\)

\(\Leftrightarrow x=7\)

Vậy x = 7

c) \(8\left(x-1\right)-7=2\left(x+2\right)+5\)

\(\Leftrightarrow8x-8-7=2x+4+5\)

\(\Leftrightarrow8x-8-7-2x+4=5\)

\(\Leftrightarrow8x-2x-8-7+4=5\)

\(\Leftrightarrow8x-2x=5-4+7+8\)

\(\Leftrightarrow4x=16\)

\(\Leftrightarrow x=4\)

Vậy x = 4

d) Đặt \(D=\frac{2x+3}{x-1}=\frac{2x-2+5}{x-1}=\frac{2\left(x-1\right)+5}{x-1}=2+\frac{5}{x-1}\)

=> \(5⋮x-1\)

=> \(x-1\inƯ\left(5\right)\)

=> \(x-1\in\left\{\pm1;\pm5\right\}\)

=> \(x\in\left\{2;0;6;-4\right\}\)

Câu 5

Nếu p lẻ thì 3p lẻ nên 3p+7 chẵn,mà 3p+7 lầ số nguyên tố

Suy ra 3p+7=2(L)

Khí đó p chẵn,mà p là số nguyên tố nên p=2

Vậy p=2

Câu 3

Ta có:\(\overline{ab}-\overline{ba}=9\times\left(a-b\right)=3^2\times\left(a-b\right)\)

Mà ab-ba là số chính phương nên 3^2X(a-b) là số chính phương

Suy ra a-b là số chính phương

Mà 0<a-b<9 nên \(a-b\in\left\{1;4\right\}\)

Với a-b=1 mà 0<b<a nên ta có bảng sau:

Với a-b=4 mà a>b>0 nên ta có bảng sau:

Vậy ..............

Bài 1

a.\(\frac{-3}{4}\)-y:\(\frac{1}{5}\)=\(\frac{9}{28}\)

                y:\(\frac{1}{5}\)=\(\frac{-15}{14}\)

                          y= \(\frac{-3}{14}\)

b.5^x + 5^x+2=650

5^x . 1 + 5^x + 5²=650

5^x(1+25)=650

5^x.26=650

5^x=25

x=2

1:2:3:4:5:7:9:10