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\(a,4y\left(x-1\right)-\left(1-x\right)\)
\(=4y\left(x-1\right)+\left(x-1\right)\)
\(=\left(4y+1\right)\left(x-1\right)\)
\(b,18x^2\left(3+x\right)+3\left(x+3\right)\)
\(=\left(18x^2+3\right)\left(3+x\right)\)
\(a.\left(2x-3\right)\left(4x^2+6x+9\right)-\left(2x+3\right)\left(4x^2-6x+9\right)\\ =\left(2x\right)^3-3^3-\left[\left(2x\right)^3+3^3\right]\\ =8x^3-9-\left(8x^3+9\right)\\ =8x^3-9-8x^3-9=-18\)
\(b.\left(x+1\right)\left(x^2-x+1\right)-\left(x-1\right)\left(x^2+x+1\right)\\ =x^3+1-\left(x^3-1\right)\\ =x^3+1-x^3+1=2\)
\(c.\left(3x-1\right)\left(3x+1\right)-\left(3x-2\right)^2\\ =9x^2-1-\left(9x^2-12x+4\right)\\ =9x^2-1-9x^2+12x-4\\ =12x-5\)
\(d.\left(2x-3\right)^2-\left(2x+3\right)\left(2x-3\right)\\ =\left(2x-3\right)\cdot\left[\left(2x-3\right)-\left(2x+3\right)\right]\\ =\left(2x-3\right)\cdot\left(2x-3-2x-3\right)\\ =\left(2x-3\right)\cdot\left(-6\right)\\ =-12x\cdot18\)
\(e.\left(3x-4\right)^2-\left(2x+4\right)^2\\ =9x^2-24x+16-\left(4x^2+16x+16\right)\\ =9x^2-24x+16-4x^2-16x-16\\ =5x^2-40x\)
\(f.\left(3x-5\right)^3-\left(3x+5\right)^3\\ =27x^3-135x^2+225x-125-\left(27x^3+135x^2+225x+125\right)\\ =27x^3-135x^2+225x-125-27x^3-135x^2-225x-125\\ =-270x^2-250\)
\(g.\left(2x-1\right)^2-\left(3x-1\right)^2\\ =4x^2-4x+1-\left(9x^2-6x+1\right)\\ =4x^2-4x+1-9x^2+6x-1\\ =-5x^2+2x\)
\(h.\left(x-2y\right)\left(x^2+2xy+4y^2\right)+\left(x^3-6y^3\right)\\ =x^3-8y^3+x^3-6y^3\\ =2x^3-14y^3\)
\(\frac{3}{x+1}+\frac{2}{x+2}=\frac{5x+4}{x^2+3x+2}.\)ĐKXĐ: \(x\ne-1;-2\)
\(\Leftrightarrow\frac{3\left(x+2\right)}{\left(x+1\right)\left(x+2\right)}+\frac{2\left(x+1\right)}{\left(x+1\right)\left(x+2\right)}=\frac{5x+4}{\left(x+1\right)\left(x+2\right)}\)
\(\Leftrightarrow3x+6+2x+2=5x+4\)
\(\Leftrightarrow3x+2x-5x=-6-2+4\)
\(\Leftrightarrow0x=-4\)
=> PT vô nghiệm
\(2;\frac{2}{3x-1}-\frac{15}{6x^2-x-1}=\frac{3}{2x-1}\)
\(\Leftrightarrow\frac{2\left(2x-1\right)}{\left(2x-1\right)\left(3x-1\right)}-\frac{15}{6x^2+3x-2x-1}=\frac{3\left(3x-1\right)}{\left(2x-1\right)\left(3x-1\right)}\)
\(\Leftrightarrow\frac{4x-2-15}{\left(2x-1\right)\left(3x-1\right)}=\frac{9x-3}{\left(2x-1\right)\left(3x-1\right)}\)
\(\Leftrightarrow4x-2-15=9x-3\)
\(\Leftrightarrow4x-9x=2+15-3\)
\(\Leftrightarrow-5x=14\)
.....
tìm x:
(x+3)3 - (x(3x+1)2 + (2x+1)4x2 - (2x+1) - 3x2=54
(x-3)3 - (x-3)(x2+3x+9)+ 6(x+1)2 + 3x2 = -33
1.
(x + 3)3 - x(3x + 1)2 + (2x + 1)(4x2 - 2x + 1) - 3x2 = 54
x3 + 9x2 + 27x + 27 - x(9x2 + 6x + 1) + 8x3 + 1 - 3x2 = 54
9x3 + 6x2 + 27x - 9x3 - 6x2 - x
= 54 - 27 - 1
26x = 26
x = 1
a) (x-3)(x+3)-(x-1)^2=0
=> (x^2-9)-(x^2-2x+1)=0
=>x^2-9-x^2+2x-1=0
=>(x^2-x^2)-9-1+2x=0
=>-10+2x=0
=>-2.(-5-x)=0
=>-5-x=0
=>-x=0+5
=>x=-5
vậy x=-5
b) x^3-3x^2+3x-1=0
=>(x-1)^3=0
=>x-1=0
=>x=0+1
=>x=1
vậy x=1
c) 4x^2-28x=0
=>4x.(x-7)=0
=> 2 TH
* 4x=0=>x=0
*x-7=0=>x=0+7=>x=7
vậy x=0 hoặc x=7
\(B=\left(x-3\right)^3-\left(x+3\right)\left(x^2-3x+9\right)+\left(3x-1\right)\left(3x+1\right)\)
\(B=x^3-9x^2+27x-27-\left(x^3-3x^2+9x+3x^2-9x+27\right)+\left(9x^2-1\right)\)
\(B=x^3-9x^2+27x-27-\left(x^3+27\right)+9x^2-1\)
\(B=x^3-9x^2+27x-27-x^3-27+9x^2-1\)
\(B=27x-55\)
giải cho tui cả phần c nx đi