\(4^5:4^2-x=12\)

\(\Rightarrow4^{5-2}-x=12\)

\(\Rightarrow4^3-x=12\)

\(\Rightarrow64-x=12\)

\(\Rightarrow x=64-12\)

\(\Rightarrow x=52\)

Vậy: x=52

4^5 : 4^2 - x = 12

4^3 - x = 12

64 - x = 12

x = 64 - 12

x = 52

1) \(-4< x< 3\)

\(\Rightarrow x\in\left\{-3;-2;-1;0;1;2\right\}\)

Tổng:

\(\left(-3\right)+\left(-2\right)+\left(-1\right)+0+1+2\)

\(=\left(-2+2\right)+\left(-1+1\right)+0-3\)

\(=-3\)

2) \(-5< x< 5\)

\(\Rightarrow x\in\left\{-4;-3;-2;-1;0;1;2;3;4\right\}\)

Tổng:

\(\left(-4\right)+\left(-3\right)+\left(-2\right)+\left(-1\right)+0+1+2+3+3\)

\(=\left(-4+4\right)+\left(-3+3\right)+\left(-2+2\right)+\left(-1+1\right)+0\)

\(=0\)

3) \(-10< x< 6\)

\(\Rightarrow x\in\left\{-9;-8;-7;-6;-5;-4;-3;-2;-1;0;1;2;3;4;5\right\}\)

Tổng:

\(\left(-9\right)+\left(-8\right)+\left(-7\right)++\left(-6\right)+\left(-5\right)+\left(-4\right)+\left(-3\right)+\left(-2\right)+\left(-1\right)+0+1+2+3+4+5\)

\(=-24\)

4) \(-6< x< 5\)

\(\Rightarrow x\in\left\{-5;-4;-3;-2;-1;0;1;2;3;4\right\}\)

Tổng:

\(\left(-5\right)+\left(-4\right)+\left(-3\right)+\left(-2\right)+\left(-1\right)+0+1+2+3+4\)

\(=\left(-4+4\right)+\left(-3+3\right)+\left(-2+2\right)+\left(-1+1\right)+0-5\)

\(=-5\)

5) \(-5< x< 2\)

\(\Rightarrow x\in\left\{-4;-3;-2;-1;0;1\right\}\)

Tổng:

\(\left(-4\right)+\left(-3\right)+\left(-2\right)+\left(-1\right)+0+1\)

\(=\left(-1+1\right)+0+\left(-4-3-2\right)\)

\(=-6\)

\(\overline{abc}+\overline{bca}+\overline{cab}\)

\(=100a+10b+c+100b+10c+a+100c+10a+b\)

\(=111a+111b+111c\)

\(=111\left(a+b+c\right)⋮3\) (vì \(111⋮3\))

\(\Rightarrow\overline{abc}+\overline{bca}+\overline{cab}⋮3\left(dpcm\right)\)

\(x\inƯ\left(6\right)=\left\{1;2;3;6\right\}\)

\(x\in B\left(2\right)=\left\{2;4;6;8;10;.....\right\}\)

\(\left\{{}\begin{matrix}x\in U\left(6\right)=\left\{1;2;3;6\right\}\\x\in B\left(2\right)=\left\{2;4;6;8...\right\}\end{matrix}\right.\)

\(\Rightarrow x\in\left\{2;4;6\right\}\)

Gọi số cần tìm có dạng \(\overline{abc}\)

Chọn a từ tập \(\left\{1;2;3;...;9\right\}\): có 9 cách chọn

Chọn b từ tập \(\left\{0;1;2;3;...;9\right\}\): có 10 cách chọn

Chọn c từ tập \(\left\{0;5\right\}\): có 2 cách chọn.

Vậy ta có: \(9.10.2=180\left(tnđb\right)\)

Thoả mãn đề bài nhé

Lời giải:

$7^{a-2}=5^{2a-4}$

$\Rightarrow a-2=2a-4$

$\Rightarrow a-2=0$

$\Rightarrow a=2$

28 < x < 29

29< x < 35

vì a : 18 dư 9 nên a có dạng: a = 18k + 9 = 9.(2k + 1)

9⋮ 3 ⇒ a ⋮ 3; 

a = 18k + 9 = 6.(3k + 1) + 3 vì 6.(3k + 1) ⋮ 6 và 3 không chia hết cho 6 nên a không chia hết cho 6

gấp mn

a) \(\left(x+1\right)\left(x+2\right)=272\)

\(\Rightarrow x^2+3x+2=272\)

\(\Rightarrow x^2+3x-270=0\)

\(\Rightarrow x^2+18x-15x-270=0\)

\(\Rightarrow x\left(x+18\right)-15\left(x+18\right)=0\)

\(\Rightarrow\left(x+18\right)\left(x-15\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x+18=0\\x-15=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-18\\x=15\end{matrix}\right.\)

d) \(\left(x+4\right)\left(x+5\right)=552\)

\(\Rightarrow x^2+9x+20=552\)

\(\Rightarrow x^2+9x-532=0\)

\(\Rightarrow x^2+28x-19x-532=0\)

\(\Rightarrow x\left(x+28\right)-19\left(x+28\right)=0\)

\(\Rightarrow\left(x+28\right)\left(x-19\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x+28=0\\x-19=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-28\\x=19\end{matrix}\right.\)

\(\text{#040911}\)

\(\left[x\cdot\left(x+1\right)\right]\div2=153\\ \Rightarrow x\cdot\left(x+1\right)=153\cdot2\\ \Rightarrow x\cdot\left(x+1\right)=306\\ \Rightarrow x^2+x=306\\ \Rightarrow x^2+x-306=0\\ \Rightarrow x^2+18x-17x-306=0\\ \Rightarrow\left(x^2+18x\right)-\left(17x+306\right)=0\\ \Rightarrow x\left(x+18\right)-17\left(x+18\right)=0\\ \Rightarrow\left(x-17\right)\left(x+18\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x-17=0\\x+18=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=17\\x=-18\end{matrix}\right.\\ \text{Vậy, x }\in\left\{-18;17\right\}.\)