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\(a.\left(2x-3\right)\left(4x^2+6x+9\right)-\left(2x+3\right)\left(4x^2-6x+9\right)\\ =\left(2x\right)^3-3^3-\left[\left(2x\right)^3+3^3\right]\\ =8x^3-9-\left(8x^3+9\right)\\ =8x^3-9-8x^3-9=-18\)
\(b.\left(x+1\right)\left(x^2-x+1\right)-\left(x-1\right)\left(x^2+x+1\right)\\ =x^3+1-\left(x^3-1\right)\\ =x^3+1-x^3+1=2\)
\(c.\left(3x-1\right)\left(3x+1\right)-\left(3x-2\right)^2\\ =9x^2-1-\left(9x^2-12x+4\right)\\ =9x^2-1-9x^2+12x-4\\ =12x-5\)
\(d.\left(2x-3\right)^2-\left(2x+3\right)\left(2x-3\right)\\ =\left(2x-3\right)\cdot\left[\left(2x-3\right)-\left(2x+3\right)\right]\\ =\left(2x-3\right)\cdot\left(2x-3-2x-3\right)\\ =\left(2x-3\right)\cdot\left(-6\right)\\ =-12x\cdot18\)
\(e.\left(3x-4\right)^2-\left(2x+4\right)^2\\ =9x^2-24x+16-\left(4x^2+16x+16\right)\\ =9x^2-24x+16-4x^2-16x-16\\ =5x^2-40x\)
\(f.\left(3x-5\right)^3-\left(3x+5\right)^3\\ =27x^3-135x^2+225x-125-\left(27x^3+135x^2+225x+125\right)\\ =27x^3-135x^2+225x-125-27x^3-135x^2-225x-125\\ =-270x^2-250\)
\(g.\left(2x-1\right)^2-\left(3x-1\right)^2\\ =4x^2-4x+1-\left(9x^2-6x+1\right)\\ =4x^2-4x+1-9x^2+6x-1\\ =-5x^2+2x\)
\(h.\left(x-2y\right)\left(x^2+2xy+4y^2\right)+\left(x^3-6y^3\right)\\ =x^3-8y^3+x^3-6y^3\\ =2x^3-14y^3\)
Bài 2
a) 4x(x-3)-3x+9
=4x(x-3)-3(x-3)
= (x-3)(4x-3)
b) x3+2x2-2x-4
=(x3+2x2)-(2x+4)
=x2(x+2)-2(x+2)
=(x+2)(x2-2)
c) 4x2-4y+4y-1
=4x2-1
=(2x-1)(2x+1)
d) x5-x
=x(x4-1)
=x(x2-1)(x2+1)
a) 4x(x-3)-3x+9
= 4x(x-3) - 3(x-3)
= (x-3)(4x-3)
b)x3 + 2x2 - 2x - 4
= x2(x + 2) - 2(x + 2)
= (x+2)(x2-2)
c) 4x2 - 4y +4y -1
= [(2x)2-12] + (-4y+4y)
= (2x+1)(2x-1)
d) x5-x
= x(x4 - 1)
Bài 1:tìm x ,biết:
a) (2x - 1)(3x + 2) - 6x(x + 1) = 0
\(\Leftrightarrow6x^2+x-2-6x^2-6x=0\)
\(\Leftrightarrow-5x=2\)
\(\Leftrightarrow x=\frac{-2}{5}\)
b) \(\left(4x-1\right)^2-\left(2x+1\right)\left(8x-3\right)=0\)
\(\Leftrightarrow16x^2-8x+1-16x^2-2x+3=0\)
\(\Leftrightarrow-10x=-4\)
\(\Leftrightarrow x=\frac{2}{5}\)
c) \(4x^2-1=2\left(2x+1\right)\)
\(\Leftrightarrow\left(2x+1\right)\left(2x-1\right)-2\left(2x+1\right)=0\)
\(\Leftrightarrow\left(2x+1\right)\left(2x-3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=-\frac{1}{2}\\x=\frac{3}{2}\end{cases}}\)
2a) \(4x^2-9y^2-6y-1=4x^2-\left(3y+1\right)^2\)
\(=\left(2x-3y-1\right)\left(2x+3y+1\right)\)
b) \(4x^2-1-2x\left(2x-1\right)=\left(2x-1\right)\left(2x+1\right)-2x\left(2x-1\right)\)
\(=1.\left(2x-1\right)\)
c) \(x^2-8x-4y^2+16=\left(x-4\right)^2-4y^2\)
\(=\left(x-4-2y\right)\left(x-4+2y\right)\)
d) \(9x^2-12x-y^2+4=\left(3x-2\right)^2-y^2\)
\(=\left(3x-2-y\right)\left(3x-2+y\right)\)
e) \(4x^2+10x-5=4x^2+2.2.\frac{5}{2}x+\frac{25}{4}-\frac{25}{4}-5\)
\(=\left(2x+\frac{5}{2}\right)^2-\frac{45}{4}\)
\(=\left(2x+\frac{5+3\sqrt{5}}{2}\right)\left(2x+\frac{5-3\sqrt{5}}{2}\right)\)
\(a,\left(6x+1\right)\left(x+2\right)-2x\left(3x-5\right)\)
\(=6x^2+12x+x+2-6x^2+10x\)
\(=23x+2\)
a) (6x + 1)(x + 2) - 2x(3x - 5)
= 6x2 + 12x + x + 2 - 6x2 + 10x
= (6x2 - 6x2) + (12x + x + 10x) + 2
= 23x + 2
b) (2x - 1)2 - (2x - 3)(2x + 3)
= 4x2 - 4x + 1 - 4x2 + 9
= (4x2 - 4x2) - 4x + (1 + 9)
= -4x + 10
c) (2x - 3)3 - (3x + 1)(5 - 4x) - 16x2
= 8x3 - 36x2 + 54x - 15x + 12x2 - 5 + 4x - 16x2
= 8x3 - (36x2 - 12x2 + 16x2) + (54x - 15x + 4x) - 5
= 8x3 - 40x2 + 43x - 5
d) (3x + 2) - (x - 5) - x(3x - 13)
= 3x + 2 - x + 5 - 3x2 + 13x
= (3x - x + 13x) + (2 + 5) - 3x2
= 15x + 7 - 3x2
mk giải bài 2 cho bạn thôi nhak vì bài 1 mk k bt cách bấm bình phương mũ 2
bình phương là nhấn x2
bạn giải nhanh lên giùm mình sắp đến giờ nộp bài rồi bạn
x3 + 2x2y + xy2
= x(x2 + 2xy + y2)
= x(x + y)2
x2 - xy - 4x + 4y
= x(x - y) - 4(x - y)
= (x - y)(x - 4)
1.
\(x^3+2x^2y+xy^2\\ =\left(x^3+x^2y\right)+\left(x^2y+xy^2\right)\\ =x^2\left(x+y\right)+xy\left(x+y\right)\\ =\left(x+y\right)\left(x^2+xy\right)\\ =\left(x+y\right)^2.x\)
\(x^2-xy-4x+4y\\ =\left(x^2-xy\right)-\left(4x-4y\right)\\ =x\left(x-y\right)-4\left(x-y\right)=\left(x-y\right)\left(x-4\right)\)
\(\dfrac{x-1}{x-2}+\dfrac{2x-3}{x-2}+\dfrac{x-4}{x-2}\\ =\dfrac{4x-8}{x-2}=4\)
\(a,VT=\left(a+b+c\right)\left(a-b+c\right)\)
\(=\left(a+c+b\right)\left(a+c-b\right)\)
\(=\left(a+c\right)^2-b^2\)
\(=a^2+2ac+c^2-b^2=VP\)
\(b,VT=\left(3x+2y\right)\left(3x-2y\right)-\left(4x-2y\right)\left(4x+2y\right)\)
\(=9x^2-4y^2-16x^2+4y^2=-7x^2=VP\)
\(c,VT=x^3-1-x^3-1=-2=VP\)
\(d,VT=8x^3+1-8x^3+1=2=VP\)
\(e,VT=\left(x^2+2xy+4y^2\right)\left(x-2y-2x+1\right)\)
\(=\left(x^2+2xy+4y^2\right)\left(-x-2y+1\right)\)
\(=-x^3-2x^2y+x^2-2x^2y-4xy^2+2xy-4xy^2-8y^3+4y^2\)
( bn kiểm tra lại đề nhé)
Bài 1:
a) \(\left(2x-1\right)\left(2x+1\right)=\left(2x\right)^2-1^2=4x^2-1\)
b) \(-\left(5+4y\right)\left(5-4y\right)=-\left[\left(5+4y\right)\left(5-4y\right)\right]=-\left[5^2-\left(4y\right)^2\right]=-\left(25-16y^2\right)=-25+16y^2\)