\(3x^2-3+2x^4+6x^2-4-5x-8x^3=2x^4-8x^3-3x^2-5x-7\)

=>36/18>x>8/15

=>2>x>8/15

mà x nguyên

nên x=1

a: \(B=\dfrac{x^2+5x+5x+25}{x\left(x+5\right)}=\dfrac{x+5}{x}\)

a = 12
7a6b = 7161

ta có 7+a+6+b =13+a+b chia hết cho 3=>a+b ko chia hết cho 3 mà a-b=4(a,b là số có 1 chữ số)

=>a=8;b=4

ta có 7+a+6+b =13+a+b chia hết cho 3=>a+b ko chia hết cho 3 mà a-b=4(a,b là số có 1 chữ số)

=>a=9;b=5

\(4A-3x^2+7-6x=x^2+3A-4x-3\)

\(\Rightarrow4A-3A=\left(x^2+3x^2\right)-\left(4x-6x\right)-\left(3+7\right)\)

\(\Rightarrow A=4x^2-\left(-2x\right)-10\)

\(\Rightarrow A=4x^2+2x-10\)

\(=-7\cdot\left[16+\left(-36\right):\left(-9\right)\right]+125=-77\cdot20+125=-140+125=-15\)

\(-7.\left[\left(-2\right)^4+\left(-36\right):\left(-3\right)^2\right]-\left(-5\right)^3\\ =-7.\left[16+\left(-36\right):9\right]+125\\ =-7.\left[16+\left(-4\right)\right]+125\\ =-7.12+125\\ =-84+125\\ =41\)

x=0

~~~~~~~

^_^

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

\(3^x\cdot3-3^x=2\)

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

\(3^x\cdot2=2\)

\(3^x=1\)

\(\Rightarrow x=0\)

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

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

\(3^x.2=2=>3^x=1=>x=0\)

\(\left(x-5\right)^2=\left(18\dfrac{1}{3}:5\right).\dfrac{11}{3}\)

\(\Leftrightarrow\left(x-5\right)^2=\dfrac{55}{3}.\dfrac{1}{5}.\dfrac{11}{3}\)

\(\Leftrightarrow\left(x-5\right)^2=\dfrac{121}{9}\)

\(\Leftrightarrow\left[{}\begin{matrix}x-5=\dfrac{11}{3}\\x-5=-\dfrac{11}{3}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{26}{3}\\x=\dfrac{4}{3}\end{matrix}\right.\)