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\(a,-3x^2+7x-9+\left(x-1\right)\left(x+2\right)\\ =-3x^2+7x-9+x^2-x+2x-2\\ =\left(-3x^2+x^2\right)+\left(7x-x+2x\right)-\left(9+2\right)\\ =-2x^2+8x-11\\ b,x\left(x-5\right)-2x\left(x+1\right)\\ =x^2-5x-2x^2-2x\\ =\left(x^2-2x^2\right)-\left(5x+2x\right)\\ =-3x^2-7x\\ c,4x\left(x^2-x+1\right)-\left(x-1\right)\left(x^2-x\right)\\ =4x^3-4x^2+4x-x\left(x^2-x\right)+x^2-x\\ =4x^3-4x^2+4x-x^3+x^2+x^2-x\\ =\left(4x^3-x^3\right)+\left(-4x^2+x^2+x^2\right)+\left(4x-x\right)\\ =3x^3-2x^2+3x\\ =x\left(3x^2-2x+3\right)\)
\(d,-5x\left(x-5\right)+\left(x-3\right)\left(x^2-7\right)\\ =-5x^2+25x+x\left(x^2-7\right)-3\left(x^2-7\right)\\ =-5x^2+25x+x^3-7x-3x^2+21\\ =\left(-5x^2-3x^2\right)+\left(25x-7x\right)+x^3+21\\ =-8x^2+x^3+18x+21\)
a, \(-4x+5+2x-1=3\Leftrightarrow-2x=-1\Leftrightarrow x=\dfrac{1}{2}\)
b, \(-2x+2=2\Leftrightarrow x=0\)
c, \(-2x-6=-8\Leftrightarrow x=1\)
Bạn nên viết lại đề bài cho sáng sủa, rõ ràng để người đọc dễ hiểu hơn.
f: =>4(x^2+4x-5)-x^2-7x-10=3(x^2+x-2)
=>4x^2+16x-20-x^2-7x-10-3x^2-3x+6=0
=>6x-24=0
=>x=4
e: =>8x+16-5x^2-10x+4(x^2-x-2)=4-x^2
=>-5x^2-2x+16+4x^2-4x-8=4-x^2
=>-6x+8=4
=>-6x=-4
=>x=2/3
d: =>2x^2+3x^2-3=5x^2+5x
=>5x=-3
=>x=-3/5
b: =>2x^2-8x+3x-12+x^2-7x+10=3x^2-12x-5x+20
=>-12x-2=-17x+20
=>5x=22
=>x=22/5
b: =>2x^2-8x+3x-12+x^2-7x+10=3x^2-17x+20
=>-12x-2=-17x+20
=>5x=22
=>x=22/5
c: =>24x^2+16x-9x-6-4x^2-16x-7x-28=20x^2-4x+5x-1
=>-16x-34=x-1
=>-17x=33
=>x=-33/17
d: =>2x^2+3x^2-3=5x^2+5x
=>5x=-3
=>x=-3/5
e: =>8x+16-5x^2-10x+4x^2-4x-8=4-x^2
=>-6x+8=4
=>-6x=-4
=>x=2/3
f: =>4(x^2+4x-5)-x^2-7x-10=3x^2+3x-6
=>4x^2+16x-20-4x^2-10x+4=0
=>6x=16
=>x=8/3
Noob ơi, bạn phải đưa vào máy tính ý solve cái là ra x luôn, chỉ tội là đợi hơi lâu
a, 4.(18 - 5x) - 12(3x - 7) = 15(2x - 16) - 6(x + 14)
=> 72 - 20x - 36x + 84 = 30x - 240 - 6x - 84
=> (72 + 84) + (-20x - 36x) = (30x - 6x) + (-240 - 84)
=> 156 - 56x = 24x - 324
=> 24x + 56x = 324 + 156
=> 80x = 480
=> x = 480 : 80 = 6
Vậy x = 6
a: \(\Leftrightarrow12x^2-10x-12x^2-28x=7\)
=>-38x=7
hay x=-7/38
b: \(\Leftrightarrow-10x^2-5x+9x^2+6x+x^2-\dfrac{1}{2}x=0\)
=>1/2x=0
hay x=0
c: \(\Leftrightarrow18x^2-15x-18x^2-14x=15\)
=>-29x=15
hay x=-15/29
d: \(\Leftrightarrow x^2+2x-x-3=5\)
\(\Leftrightarrow x^2+x-8=0\)
\(\text{Δ}=1^2-4\cdot1\cdot\left(-8\right)=33>0\)
Do đó: Phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}x_1=\dfrac{-1-\sqrt{33}}{2}\\x_2=\dfrac{-1+\sqrt{33}}{2}\end{matrix}\right.\)
e: \(\Leftrightarrow-15x^2+10x-10x^2-5x-5x=4\)
\(\Leftrightarrow-25x^2=4\)
\(\Leftrightarrow x^2=-\dfrac{4}{25}\left(loại\right)\)
1: (3x+2)(x+2)(2x-1)
=(3x^2+6x+2x+4)(2x-1)
=(3x^2+8x+4)(2x-1)
=6x^3-3x^2+16x^2-8x+8x-4
=6x^3+13x^2-4
2: (5x+1)(x-1)+3x(2x+2)
=5x^2-5x+x-1+6x^2+6x
=11x^2+10x-1
3: 4x(2x+1)(x-1)+(x+5)(x-3)
=4x(2x^2-2x+x-1)+x^2+2x-15
=8x^3-4x^2-4x+x^2+2x-15
=8x^3-3x^2-2x-15
4: (2x-1)(x+2)(x-2)+(3x-1)(x-1)
=(2x-1)(x^2-4)+3x^2-4x+1
=2x^3-8x-x^2+4+3x^2-4x+1
=2x^3+2x^2-12x+5
a: 4x(x-5)-(x-1)(4x-3)=5
=>\(4x^2-20x-\left(4x^2-3x-4x+3\right)=5\)
=>\(4x^2-20x-4x^2+7x-3=5\)
=>-13x=8
=>\(x=-\dfrac{8}{13}\)
b: \(\left(x-5\right)\left(x-4\right)-\left(x+1\right)\left(x-2\right)=7\)
=>\(x^2-9x+20-\left(x^2-x-2\right)=7\)
=>\(x^2-9x+20-x^2+x+2=7\)
=>-8x+22=7
=>-8x=-15
=>\(x=\dfrac{15}{8}\)
c: \(5x\left(x-3\right)=\left(x-2\right)\left(5x-1\right)-5\)
=>\(5x^2-15x=5x^2-x-10x+2-5\)
=>-15x=-11x-3
=>-4x=-3
=>\(x=\dfrac{3}{4}\)
d: \(\left(x+1\right)\left(x^2+3x+9\right)+x\left(5-x^2\right)=6x\)
=>\(x^3+3x^2+9x+x^2+3x+9+5x-x^3-6x=0\)
=>\(4x^2+11x+9=0\)
=>\(x\in\varnothing\)