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a) 6(x+1)-3=2(x+2)+11
6x + 6 - 3= 2x+ 4+11
6x-2x +6-3= 4+11
4x +6-3= 15
4x+6= 15+3
4x+6= 18
4x= 18-6
4x= 12
x= 12:4
x=3
Vậy...
\(\frac{-1}{6}+\frac{1}{6}+0=0\)
\(\frac{-1}{3}+\frac{1}{3}+0=0\)
\(\frac{-1}{2}+\frac{1}{2}+0=0\)
\(8-12x+6x^2-x^3\)
\(=\left(2-x\right)^3\)
\(125x^3-75x^2+15x-1\)
\(=\left(5x-1\right)^3\)
\(x^2-xz-9y^2+3yz\)
\(=\left(x-3y\right)\left(x+3y\right)-z\left(x-3y\right)\)
\(=\left(x-3y\right)\left(x+3y-z\right)\)
\(x^3-x^2-5x+125\)
\(=\left(x+5\right)\left(x^2-5x+25\right)-x\left(x+5\right)\)
\(=\left(x+5\right)\left(x^2-5x+25-x\right)\)
\(=\left(x+5\right)\left(x^2-6x+25\right)\)
\(x^3+2x^2-6x-27\)
\(=x^3+5x^2+9x-3x^2-15x-27\)
\(=x\left(x^2+5x+9\right)-3\left(x^2+5x+9\right)\)
\(=\left(x-3\right)\left(x^2+5x+9\right)\)
\(12x^3+4x^2-27x-9\)
\(=4x^2\left(3x+1\right)-9\left(3x+1\right)\)
\(=\left(3x+1\right)\left(4x^2-9\right)\)
\(=\left(3x+1\right)\left(2x-3\right)\left(2x+3\right)\)
\(4x^4+4x^3-x^2-x\)
\(=4x^3\left(x+1\right)-x\left(x+1\right)\)
\(=x\left(x+1\right)\left(4x^2-1\right)\)
\(=x\left(x+1\right)\left(2x-1\right)\left(2x+1\right)\)
1, Ta có :
a . 81 = 34 => 3x= 34 => x = 4 .
b. 125 = 53 => 5x+2 = 53 =>x + 2 = 3 => x = 1
c. 23 * 2x - 1 = 64
=> 23 + ( x - 1 ) = 64 = 26
=> 3 + ( x - 1 ) = 6
=> x - 1 = 6 - 3 = 3
x = 3 + 1
x = 4
\(1)A=a\frac{1}{3}+a\frac{1}{4}-a\frac{1}{6}=a\left(\frac{1}{3}+\frac{1}{4}-\frac{1}{6}\right)=a\frac{5}{12}\)
Thay \(a=-\frac{3}{5}\) vào A,ta đc:
\(A=-\frac{3}{5}.\frac{5}{12}=-\frac{1}{4}\)
\(2)B=b\frac{5}{6}+b\frac{3}{4}-b\frac{1}{2}=b\left(\frac{5}{6}+\frac{3}{4}-\frac{1}{2}\right)=b\frac{13}{12}\)
Thay \(b=\frac{12}{13}\) vào B, ta đc: \(B=b\frac{13}{12}=\frac{12}{13}.\frac{13}{12}=1\)
a
\(5\frac{4}{7}:x+=13\)
\(\frac{39}{7}:x=13\)
\(x=\frac{39}{7}:13\)
\(x=\frac{3}{7}\)
\(\frac{4}{7}x=\frac{9}{8}-0,125\)
\(\frac{4}{7}x=1\)
\(x=1:\frac{4}{7}\)
\(x=\frac{7}{4}=1\frac{3}{4}\)
(x+3)^3=10^2-6 mũ mấy í bn êi
thiếu đề bài à
z lm s mà giải đc