Rút gọn biểu thức
a)(x - 5).(2x +3) - (2x -1).(x +7) - (x -1).(x+2)
b)(6x +1).(x +5) - (3x + 5).(2x - 10)
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\(D=\dfrac{5}{2x^2+6x}-\dfrac{4-3x^2}{x^2-9}-3\) (đk:\(x\ne3;x\ne-3\))
\(=\dfrac{5}{2x\left(x+3\right)}-\dfrac{4-3x^2}{\left(x-3\right)\left(x+3\right)}-3\)
\(=\dfrac{5\left(x-3\right)}{2x\left(x-3\right)\left(x+3\right)}-\dfrac{\left(4-3x^2\right).2x}{2x\left(x-3\right)\left(x+3\right)}-\dfrac{3.2x\left(x-3\right)\left(x+3\right)}{2x\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{5x-15-8x+6x^3-6x\left(x^2-9\right)}{2x\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{51x-15}{2x\left(x-3\right)\left(x+3\right)}\)
Bài 1:
\(D=-3x^2+x+15x-5-3\left(2x^2-5x+2\right)\)
\(=-3x^2+16x-5-6x^2+15x-6\)
\(=-9x^2+31x-11\)
\(=-9\cdot\dfrac{1}{9}+\dfrac{31}{3}-11\)
=-11-1+31/3=-12+31/3=-5/3
b: \(E=x^2+x-56-x^2+7x-10=8x-66\)
\(=-\dfrac{8}{5}-66=-\dfrac{338}{5}\)
c: \(F=-3\left(2x^2+x-16x-8\right)-\left(-3x^2+2x-15x+10\right)-4x^2+24x\)
\(=-6x^2+45x+24+3x^2+13x-10-4x^2+24x\)
\(=-4x^2+82x+14\)
\(=-4\cdot9-82\cdot3+14=-268\)
(6x + 1)(x + 5) - (3x + 5)(2x - 10)
= (6x + 1)(x + 5) + 5(3x + 5)(x + 5)
= 5(x + 5)(6x + 1 + 3x + 5)
= 5(x + 5)(9x + 6)
Ta có: (6x +1).(x +5) - (3x + 5).(2x - 10)
= 6x2 + 30x + x + 5 - 6x2 + 30x - 10x + 50
= 51x + 55
Mk nhầm nha câu đầu chỉ có 1 cái x-1 + x -2 thôi ko có cái đằng sau nhé ! giá trị tuyệt đối thì vẫn giữ nguyên !
\(A=\dfrac{2x^2\left(3x-4y+2\right)}{x\left(3x+y\right)\left(3x-y\right)}=\dfrac{2x\left(3x-4y+2\right)}{\left(3x+y\right)\left(3x-y\right)}\\ A=\dfrac{2\left(3-8+2\right)}{\left(3+2\right)\left(3-2\right)}=\dfrac{2\left(-3\right)}{5}=\dfrac{-6}{5}\)
Ta có: a)(x - 5).(2x +3) - (2x -1).(x +7) - (x -1).(x+2)
= 2x2 + 3x - 10x - 15 - 2x2 - 14x + x + 7 - x2 - 2x + x + 2
= -x2 - 21x - 6
Ta có: \(\left(x+5\right)\left(x^2-5x+25\right)-\left(x+3\right)^3+\left(x-2\right)\left(x^2+2x+4\right)-\left(x-1\right)^3\)
\(=x^3+125-x^3-9x^2-27x-27+x^3-8-x^3+3x^2-3x+1\)
\(=-6x^2-30x+91\)
a, Với x khác 1
\(A=\dfrac{x^2+x+1-3x^2+2x\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{1-x}{\left(x-1\right)\left(x^2+x+1\right)}=-\dfrac{1}{x^2+x+1}\)
b, Ta có \(x^2+x+1=\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\Rightarrow\dfrac{-1}{\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}}< 0\)
Vậy với x khác 1 thì bth A luôn nhận gtri âm
a: \(=2x\left(4x^2-4x+1\right)-3x\left(x^2-9\right)-4x\left(x^2+2x+1\right)\)
\(=8x^3-8x^2+2x-3x^3+27x-4x^3-8x^2-4x\)
\(=x^3-16x^2+25x\)
b: \(=\left(a-b+c-b+c\right)\left(a-b+c+b-c\right)+2ab-2ac\)
\(=\left(a-2b+2c\right)\cdot a+2ab-2ac\)
\(=a^2-2ab+2ac+2ab-2ac=a^2\)
c: \(\left(3x+1\right)^2-2\left(3x+1\right)\left(3x+5\right)+\left(3x+5\right)^2\)
\(=\left(3x+1-3x-5\right)^2=\left(-4\right)^2=16\)