32 - 42 * (-19) + 38 * 5

= 32 - (-798) + 190

= 32 + 798 + 190

= 830 + 190

= 1020

32 - 42 x (-19) + 38 x 5

= 32 - ( - 798 ) + 190

= 830 + 190

= 1020

\(A=\frac{35}{42}+\frac{1}{12}+\frac{8}{12}+\frac{12}{42}+\frac{1}{25}+\frac{3}{12}-\frac{5}{42}\)

\(A=\left(\frac{35}{42}+\frac{12}{42}-\frac{5}{42}\right)+\left(\frac{1}{12}+\frac{8}{12}+\frac{3}{12}\right)+\frac{1}{25}\)

\(A=1+1+\frac{1}{25}\)

\(A=2+\frac{1}{25}\)

\(A=\frac{51}{25}\)

\(\frac{3}{x}+\frac{4}{3}=\frac{5}{6}\)

\(\frac{3}{x}=\frac{5}{6}-\frac{4}{3}\)

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

\(\Rightarrow3.2=\left(-1\right).x\)

\(\Rightarrow6=\left(-1\right).x\)

\(\Rightarrow x=6:\left(-1\right)\)

\(\Rightarrow x=-6\)

\(\frac{x}{2}-\frac{2}{y}=\frac{1}{2}\)

\(\Rightarrow\frac{x}{2}-\frac{1}{2}=\frac{2}{y}\)

\(\Rightarrow\frac{x-1}{2}=\frac{2}{y}\)

\(\Rightarrow\hept{\begin{cases}x-1=2\\2=y\end{cases}\Rightarrow}\hept{\begin{cases}x=3\\y=2\end{cases}}\)

\(b,\frac{3}{x}+\frac{4}{3}=\frac{5}{6}\)

\(\Rightarrow\frac{3}{x}=\frac{5}{6}-\frac{4}{3}\)

\(\Rightarrow\frac{3}{x}=\frac{5}{6}-\frac{8}{6}\)

\(\Rightarrow\frac{3}{x}=\frac{-3}{6}\)

\(\Rightarrow x\cdot(-3)=18\Rightarrow x=-6\)

1)

a)

\(\frac{-5}{6}.\frac{120}{25}< x< \frac{-7}{15}.\frac{9}{14}\)

\(\frac{-1}{1}.\frac{20}{5}< x< \frac{-1}{5}.\frac{3}{2}\)

\(\frac{-20}{5}< x< \frac{-3}{10}\)

\(\frac{-40}{10}< x< \frac{-3}{10}\)

\(\Rightarrow Z\in\left\{-4;-5;-6;-7;-8;-9;-10;...;-39\right\}\)

\(\left(\frac{-5}{3}\right)^3< x< \frac{-24}{35}.\frac{-5}{6}\)

\(\frac{25}{3}< x< \frac{-4}{7}.\frac{1}{1}\)

\(\frac{-25}{3}< x< \frac{-4}{7}\)

\(\frac{-175}{21}< x< \frac{-12}{21}\)

\(\Rightarrow Z\in\left\{-13;-14;-15;-16;...;-174\right\}\)

TH1 x=0

TH2 x=2

x.(2-x)=0

x=0 hoặc 2-x=0

suy ra x=0 hoặc x=2

Ta có: a=18+24-36+x

a=6+x

a) Để a chia hết cho 6 thì 6+x chia hết cho 6

Vì 6 chia hết cho 6 nên x chia hết cho 6.

Vậy x=6k

b) Để a ko chia hết cho 6 thì 6+x ko chia hết cho 6

Vì 6 chia hết cho 6 nên x ko chia hết cho 6.

Vậy x ≠ 6k

a: a=18+24-36+x

=>a=x+6

Để a⋮6 thì x+6⋮6

=>x⋮6

b: Để a không chia hết cho 6 thì x+6 không chia hết cho 6

=>x không chia hết cho 6

Đề bài:

\(A = \frac{1 2^{3} \cdot 12 1^{2} \cdot 5 - 2 2^{4} \cdot 3^{3}}{7 5^{2} \cdot 1 1^{4} - 3 0^{2} \cdot 1 1^{5}}\)

Bước 1: Biến đổi các lũy thừa

Ta rút gọn từng số:

• \(1 2^{3} = \left(\right. 3 \cdot 4 \left.\right)^{3} = 3^{3} \cdot 4^{3} = 3^{3} \cdot \left(\right. 2^{2} \left.\right)^{3} = 3^{3} \cdot 2^{6}\)
• \(12 1^{2} = \left(\right. 1 1^{2} \left.\right)^{2} = 1 1^{4}\)
• \(2 2^{4} = \left(\right. 2 \cdot 11 \left.\right)^{4} = 2^{4} \cdot 1 1^{4}\)
• \(3^{3} = 3^{3}\)
• \(7 5^{2} = \left(\right. 3 \cdot 5^{2} \left.\right)^{2} = 3^{2} \cdot 5^{4}\)
• \(1 1^{4} = 1 1^{4}\)
• \(3 0^{2} = \left(\right. 2 \cdot 3 \cdot 5 \left.\right)^{2} = 2^{2} \cdot 3^{2} \cdot 5^{2}\)
• \(1 1^{5} = 1 1^{5}\)

Bước 2: Thay vào biểu thức

Tử số:

\(1 2^{3} \cdot 12 1^{2} \cdot 5 - 2 2^{4} \cdot 3^{3} = \left(\right. 3^{3} \cdot 2^{6} \cdot 1 1^{4} \cdot 5 \left.\right) - \left(\right. 2^{4} \cdot 1 1^{4} \cdot 3^{3} \left.\right)\)

Rút chung:

\(= 3^{3} \cdot 2^{4} \cdot 1 1^{4} \cdot \left(\right. 2^{2} \cdot 5 - 1 \left.\right) = 3^{3} \cdot 2^{4} \cdot 1 1^{4} \cdot \left(\right. 4 \cdot 5 - 1 \left.\right) = 3^{3} \cdot 2^{4} \cdot 1 1^{4} \cdot \left(\right. 20 - 1 \left.\right) = 3^{3} \cdot 2^{4} \cdot 1 1^{4} \cdot 19\)

Mẫu số:

\(7 5^{2} \cdot 1 1^{4} - 3 0^{2} \cdot 1 1^{5} = \left(\right. 3^{2} \cdot 5^{4} \cdot 1 1^{4} \left.\right) - \left(\right. 2^{2} \cdot 3^{2} \cdot 5^{2} \cdot 1 1^{5} \left.\right)\)

Rút chung \(3^{2} \cdot 5^{2} \cdot 1 1^{4}\):

\(= 3^{2} \cdot 5^{2} \cdot 1 1^{4} \cdot \left(\right. 5^{2} - 2^{2} \cdot 11 \left.\right) = 3^{2} \cdot 5^{2} \cdot 1 1^{4} \cdot \left(\right. 25 - 4 \cdot 11 \left.\right) = 3^{2} \cdot 5^{2} \cdot 1 1^{4} \cdot \left(\right. 25 - 44 \left.\right) = 3^{2} \cdot 5^{2} \cdot 1 1^{4} \cdot \left(\right. - 19 \left.\right)\)

Bước 3: Viết lại biểu thức đầy đủ

\(A = \frac{3^{3} \cdot 2^{4} \cdot 1 1^{4} \cdot 19}{3^{2} \cdot 5^{2} \cdot 1 1^{4} \cdot \left(\right. - 19 \left.\right)}\)

Rút gọn:

• \(3^{3} / 3^{2} = 3\)
• \(1 1^{4}\) triệt tiêu
• \(19 / \left(\right. - 19 \left.\right) = - 1\)

Còn lại:

\(A = \frac{3 \cdot 2^{4}}{5^{2}} \cdot \left(\right. - 1 \left.\right) = \frac{3 \cdot 16}{25} \cdot \left(\right. - 1 \left.\right) = \frac{48}{25} \cdot \left(\right. - 1 \left.\right) = \boxed{- \frac{48}{25}}\)

✅ Kết quả cuối cùng:

\(\boxed{A = - \frac{48}{25}}\)

Ta có: \(12^3\cdot121^2\cdot5-22^4\cdot3^3\)

\(=\left(2^2\cdot3\right)^3\cdot\left(11^2\right)^2\cdot5-11^4\cdot2^4\cdot3^3\)

\(=2^6\cdot3^3\cdot11^4\cdot5-11^4\cdot2^4\cdot3^3=11^4\cdot3^3\cdot2^4\left(2^2\cdot5-1\right)\)

\(=11^4\cdot3^3\cdot2^4\cdot19\)

Ta có: \(75^2\cdot11^4-30^2\cdot11^5\)

\(=\left(3\cdot5^2\right)^2\cdot11^4-\left(2\cdot3\cdot5\right)^2\cdot11^5\)

\(=3^2\cdot5^4\cdot11^4-2^2\cdot3^2\cdot5^2\cdot11^5\)

\(=3^2\cdot5^2\cdot11^4\left(5^2-2^2\cdot11\right)=3^2\cdot5^2\cdot11^4\cdot\left(-19\right)\)

Ta có: \(A=\frac{12^3\cdot121^2\cdot5-22^4\cdot3^3}{75^2\cdot11^4-30^2\cdot11^5}\)

\(=\frac{2^4\cdot3^3\cdot11^4\cdot19}{3^2\cdot5^2\cdot11^4\cdot\left(-19\right)}=\frac{2^4\cdot3}{5^2\cdot\left(-1\right)}=\frac{48}{-25}\)

(3x - 1) * 2 - 21 = 43

(3x - 1) * 2 = 43 + 21

(3x - 1) * 2 = 64

3x - 1 = 64 : 2

3x - 1 = 32

3x = 32 + 1

3x = 33

x = 33 : 3

x = 11

Vậy x = 11

\(\left(3x-1\right)^2-21=43\)

=>\(\left(3x-1\right)^2=21+43=64\)

=>\(\left[\begin{array}{l}3x-1=8\\ 3x-1=-8\end{array}\right.\Rightarrow\left[\begin{array}{l}3x=9\\ 3x=-7\end{array}\right.=>\left[\begin{array}{l}x=3\\ x=-\frac73\end{array}\right.\)