a) Ta có: \(x+\dfrac{1}{3}=\dfrac{2}{6}\)

\(\Leftrightarrow x+\dfrac{1}{3}=\dfrac{1}{3}\)

hay x=0

Vậy: x=0

b) Ta có: \(x-\dfrac{1}{4}=\dfrac{1}{-2}\)

\(\Leftrightarrow x-\dfrac{1}{4}=\dfrac{-1}{2}\)

\(\Leftrightarrow x=\dfrac{-1}{2}+\dfrac{1}{4}=\dfrac{-2}{4}+\dfrac{1}{4}=\dfrac{-1}{4}\)

Vậy: \(x=-\dfrac{1}{4}\)

c) Ta có: \(\dfrac{-1}{6}=\dfrac{3}{2}x\)

\(\Leftrightarrow x=\dfrac{-1}{6}:\dfrac{3}{2}=\dfrac{-1}{6}\cdot\dfrac{2}{3}\)

hay \(x=\dfrac{-1}{9}\)

Vậy: \(x=\dfrac{-1}{9}\)

a) \(x+1\dfrac{4}{7}=2\dfrac{2}{3}\)

\(x+\dfrac{11}{7}=\dfrac{8}{3}\)

\(x=\dfrac{8}{3}-\dfrac{11}{7}\)

\(x=\dfrac{23}{21}\)

b) \(3\dfrac{9}{12}:x=4\dfrac{1}{6}\)

\(\dfrac{15}{4}:x=\dfrac{25}{6}\)

\(x=\dfrac{15}{4}:\dfrac{25}{6}\)

\(x=\dfrac{9}{10}\)

a: Ta có: \(\left(x+2\right)\left(x^2-2x+4\right)-x\left(x^2+2\right)=15\)

\(\Leftrightarrow x^3+8-x^3-2x=15\)

\(\Leftrightarrow2x=-7\)

hay \(x=-\dfrac{7}{2}\)

b: Ta có: \(\left(x-2\right)^3-\left(x-4\right)\left(x^2+4x+16\right)+6\left(x+1\right)^2=49\)

\(\Leftrightarrow x^3-6x^2+12x-8-x^3+64+6\left(x+1\right)^2=49\)

\(\Leftrightarrow-6x^2+12x+56+6x^2+12x+6=49\)

\(\Leftrightarrow24x=-13\)

hay \(x=-\dfrac{13}{24}\)

a) \(6.8^{x-1}+8^{x+1}=6.8^{19}+8^{21}\)

\(\Rightarrow x-1+x+1=19+21\)

\(=2x=40\)

\(\Rightarrow x=20\)

b) \(4.3^{x-1}+2.3^{x+2}=4.3^6+2.3^9\)

\(\Rightarrow x-1+x+2=6+9\)

\(\Rightarrow2x+1=15\)

\(\Rightarrow2x=14\)

\(\Rightarrow x=7\)

rảnh à

100

a)\(\left(x-2\right)^2-\left(x-3\right)\left(x+3\right)=6.\)

\(\Leftrightarrow x^2-4x+4-x^2+9-6=0\)

\(\Leftrightarrow-4x+7=0\)

\(\Leftrightarrow4x=7\Leftrightarrow x=1,75\)

\(b,4\left(x-3\right)^2-\left(2x-1\right)\left(2x+1\right)=10.\)

\(\Leftrightarrow4\left(x^2-6x+9\right)-4x^2+1-10=0\)

\(\Leftrightarrow-24x+27=0\)

\(\Leftrightarrow24x=27\Leftrightarrow x=1,125\)

a: \(A=x^4+3x^3+2x^2-x-4\)

\(=x^4-2x^3+3x^2+5x^3-10x^2+15x+9x^2-18x+27\) +2x-31

\(=\left(x^2-2x+3\right)\left(x^2+5x+9\right)+2x-31=B\left(x^2+5x+9\right)+2x-31\)

=>Dư là R=2x-31; thương là \(Q=x^2+5x+9\)

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

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

\(=\left(x^2-x+3\right)\left(2x+1\right)+x-7=B\left(2x+1\right)+x-7\)

=>Thương là Q=2x+1; dư là R=x-7

c: \(A=2x^4+x^3+3x^2+4x+9\)

\(=2x^4+2x^2+x^3+x+x^2+1+3x+8\)

\(=\left(x^2+1\right)\left(2x^2+x+1\right)+3x+8=B\left(2x^2+x+1\right)+3x+8\)

=>Thương là \(Q=2x^2+x+1\) ; dư là R=3x+8

d: \(A=2x^3-11x^2+19x-6\)

\(=2x^3-6x^2+2x-5x^2+15x-5+2x-1\)

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

=>Thương là Q=2x-5; dư là R=2x-1

e: \(A=2x^4-x^3-x^2-x+1\)

\(=2x^4+2x^2-x^3-x-3x^2-3+4\)

\(=\left(x^2+1\right)\left(2x^2-x-3\right)+4=B\left(2x^2-x-3\right)+4\)

=>Thương là \(Q=2x^2-x-3\) và dư là R=4