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1) a) \(\left(3x-1\right)\left(9x^2+3x+1\right)-4x\left(x-5\right)\)
\(=27x^3+9x^2+3x-9x^2-3x-1-4x^2+20x\)
\(=27x^3+\left(9x^2-9x^2-4x^2\right)+\left(3x-3x+20x\right)+\left(-1\right)\)
\(=27x^3-4x^2+20x-1\)
b)\(\left(7x+2\right)\left(3-4x\right)-\left(x+3\right)\left(x^2-3x+9\right)\)
\(=21x-28x^2+6-8x-x^3+3x^2-9x-3x^2+9x-27\)
\(=\left(21x-8x-9x+9x\right)+\left(-28x^2+3x^2-3x^2\right)\)\(+\left(6-27\right)\)\(+\left(-x^3\right)\)
\(=13x-28x^2-21-x^3\)
c)\(\left(4x+3\right)\left(4x-3\right)-\left(2-x\right)\left(4+2x+x^2\right)\)
\(=16x^2-12x+12x-9-8-4x-2x^2+4x+2x^2+x^3\)
\(=\left(16x^2-2x^2+2x^2\right)+\left(-12x+12x-4x+4x\right)\)\(+\left(-9-8\right)\)\(+x^3\)
\(=16x^2-17+x^3\)
d)\(\left(3x-8\right)\left(-5x+6\right)-\left(4x+1\right)\left(3x-2\right)\)
\(=-15x^2+18x+40x-48-12x^2+8x-3x+2\)
\(=\left(-15x^2-12x^2\right)+\left(18x+40x+8x-3x\right)\)\(+\left(-48+2\right)\)
\(=-27x^2+63x-46\)
e)\(\left(3x-6\right)4x-2x\left(3x+5\right)-4x^2\)
\(=12x^2-24x-6x^2-10x-4x^2\)
\(=\left(12x^2-6x^2-4x^2\right)+\left(-24x-10x\right)\)
\(=2x^2-34x\)
f)\(\left(5x-6\right)\left(6x-5\right)-x\left(3x+10\right)\)
\(=30x^2-25x-36x+30-3x^2-10x\)
\(=\left(30x^2-3x^2\right)+\left(-25x-36x-10x\right)+30\)
\(=27x^2-71x+30\)
2) a)\(x\left(x+3\right)-x^2=6\)
\(\Rightarrow x^2+3x-x^2=6\)
\(\Rightarrow\left(x^2-x^2\right)+3x=6\)
\(\Rightarrow3x=6\)
\(\Rightarrow x=2\)
Vậy x=2
b) \(2x\left(x-5\right)+x\left(-2x-1\right)=6\)
\(\Rightarrow2x^2-10x-2x^2-x=6\)
\(\Rightarrow\left(2x^2-2x^2\right)+\left(-10x-x\right)=6\)
\(\Rightarrow-11x=6\)
\(\Rightarrow x=-\dfrac{6}{11}\)
\(\)Vậy \(x=-\dfrac{6}{11}\)
c) x(x+5)-(x+1)(x-2)=7
\(\Rightarrow x^2+5x-x^2+2x-x+2=7\)
\(\Rightarrow\left(x^2-x^2\right)+\left(5x+2x-x\right)=7-2\)
\(\Rightarrow6x=5\)
\(\Rightarrow x=\dfrac{5}{6}\)
Vậy x=\(\dfrac{5}{6}\)
d)\(\left(3x+4\right)\left(6x-3\right)-\left(2x+1\right)\left(9x-2\right)=10\)
\(\Rightarrow18x^2-9x+24x-12-18x^2+4x-9x+2=10\)
\(\Rightarrow\left(18x^2-18x^2\right)+\left(-9x+24x+4x-9x\right)+\left(-12+2\right)=10\)
\(\Rightarrow10x-10=10\)
\(\Rightarrow10x=20\)
\(\Rightarrow x=2\)
Vậy x=2
a) (12x-5)(4x-1)+(3x-7)(1-16x)
= (48x^2 - 12x - 20x + 5) + (3x - 48x^2 - 7 + 112x)
= 48x^2 - 12x - 20x + 5 +3x - 48x^2 -7 + 112x
= 83x-2
những phần sau bạn cứ làm tương tự theo cách nhân đa thức với đa thức và phá ngiawcj là ra nha :0))
Mình ko ghi lại đề , bạn ghi ra xong rồi suy ra như mình nha .
1) \(=>A=\left(6x^2+3x-10x-5\right)-\left(6x^2+14x-9x-21\right)\)
\(=>A=-12x+16\)
2) \(=>B=8x^3+27-8x^3+2=29\)
3)\(=>C=[\left(x-1\right)-\left(x+1\right)]^3=\left(-2\right)^3=-8\)
4)\(=>D=[\left(2x+5\right)-\left(2x\right)]^3=5^3=125\)
5)\(=>E=\left(3x+1\right)^2-\left(3x+5\right)^2+12x+2\left(6x+3\right)\)
\(=>E=\left(3x+1+3x+5\right)\left(3x+1-3x-5\right)+12x+12x+6\)
\(=>E=\left(6x+6\right)\left(-4\right)+24x+6=-24x-24+24x+6=-18\)
6)\(=>F=\left(2x^2+3x-10x-15\right)-\left(2x^2-6x\right)+x+7=-8\)
k cho mik nha ,
a, \(E=4x^2+6x+5=4\left(x^2+\frac{2.3}{4}x+\frac{9}{16}-\frac{9}{16}\right)+5\)
\(=4\left(x+\frac{3}{4}\right)^2+\frac{11}{4}\ge\frac{11}{4}>0\forall x\)
Vậy ta có đpcm
b, \(F=2x^2-3x+7=2\left(x^2-\frac{2.3}{4}x+\frac{9}{16}-\frac{9}{16}\right)+7\)
\(=2\left(x-\frac{3}{4}\right)^2+\frac{47}{8}\ge\frac{47}{8}>0\forall x\)
Vậy ta có đpcm
c, \(K=5x^2-4x+1=5\left(x^2-\frac{2.2}{5}x+\frac{4}{25}-\frac{4}{25}\right)+1\)
\(=5\left(x-\frac{2}{5}\right)^2+\frac{1}{5}\ge\frac{1}{5}>0\forall x\)
Vậy ta có đpcm
d, \(Q=3x^2+2x+5=3\left(x^2+\frac{2}{3}x+\frac{1}{9}-\frac{1}{9}\right)+5\)
\(=3\left(x+\frac{1}{3}\right)^2+\frac{14}{3}\ge\frac{14}{3}>0\forall x\)
Vậy ta có đpcm
\(\left(x^2+3\right)\left(3-x^2\right)\)
\(\left(x^2+3\right)\left(-x^2+3\right)\)
\(\left(-x^2+3\right).x^2+3\left(-x^2+3\right)\)
\(-x^2.x^2+3x^2+3\left(-x^2+3\right)\)
\(-x^2.x^2+3x^2-3x^2+9\)
\(-x^2.x^2+9\)
Ai giúp em đi ạ
Em hứa sẽ vote nhiệt tình luôn huhuuu
a, (x2 - 3)(2 + 4x)
= 2x2 + 4x3 - 6 - 12x
b, 6(3x + 5)(x - 4)
= (18x + 30)(x - 4)
= 18x2 - 72 + 30x - 120
= 18x2 + 30x - 192