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a,\(6x^2-11x+3\)
\(\Leftrightarrow6x^2-2x-9x+3\)
\(\Leftrightarrow2x\left(3x-1\right)-3\left(3x-1\right)\)
\(\Leftrightarrow\left(3x-1\right)\left(2x-3\right)\)
b,\(2x^2+3x-27\)
\(\Leftrightarrow2x^2+9x-6x-27\)
\(\Leftrightarrow x\left(2x+9\right)-3\left(2x+9\right)\)
\(\Leftrightarrow\left(2x+9\right)\left(x-3\right)\)
c,\(\left(x^2+x+2\right)\left(x^2+x+5\right)-4\)
\(\Leftrightarrow x^4+x^3+5x^2+x^3+x^2+5x+2x^2+2x+10-4\)
\(\Leftrightarrow x^4+2x^3+8x^2+7x+6\)
\(\left(25-m^2\right)=\left(5-m\right)\left(5+m\right)\)
\(10a-25-a^2=-\left(a^2-10a+25\right)=-\left(a^2-2.a.5+5^2\right)=-\left(a-5\right)^2\)
\(4a^2b^2-\left(a^2+b^2-c^2\right)^2\)
\(=4a^2b^2-2ab\left(a^2+b^2-c^2\right)+2ab\left(a^2+b^2-c^2\right)-\left(a^2+b^2-c^2\right)^2\)
\(=2ab\left[2ab-\left(a^2+b^2-c^2\right)\right]+\left(a^2+b^2-c^2\right)\left[2ab-\left(a^2+b^2-c^2\right)\right]\)
\(=\left(2ab+a^2+b^2-c^2\right)\left(2ab-a^2-b^2+c^2\right)\)
\(=\left(a^2+ab+ab+b^2-c^2\right)\left[c^2-\left(a^2-ab-ab+b^2\right)\right]\)
\(=\left[a\left(a+b\right)+b\left(a+b\right)-c^2\right]\left[c^2-\left(a\left(a-b\right)-b\left(a-b\right)\right)\right]\)
\(=\left[\left(a+b\right)^2-c^2\right]\left[c^2-\left(a-b\right)^2\right]\)
\(=\left[\left(a+b\right)^2-c\left(a+b\right)+c\left(a+b\right)-c^2\right]\left[c^2+c\left(a-b\right)-c\left(a-b\right)-\left(a-b\right)^2\right]\)
\(=\left[\left(a+b\right)\left(a+b-c\right)+c\left(a+b-c\right)\right]\left[c\left(c+a-b\right)-\left(a-b\right)\left(c+a-b\right)\right]\)
\(=\left(a+b+c\right)\left(a+b-c\right)\left(c+a-b\right)\left(c-a+b\right)\)
\(x^3-x^2y+3x-3y\)
\(=x^2\left(x-y\right)+3\left(x-y\right)\)
\(=\left(x-y\right)\left(x^2+3\right)\)
\(=x^2\left(x-y\right)+3\left(x-y\right)=\left(x^2+3\right)\left(x-y\right)\)
mình tìm không tháy bạn ơi ~ chủ yếu là mình nhờ mấy bạn từng học qua rồi chỉ giúp những dạng chủ yếu,mẹo vặt các loại đấy bạn !! không phải mình tìm đề đâu ~~`
a) x12 + 4 = x12 + 4x6 + 4 - 4x6 = (x6 + 2)2 - (2x3)2
= (x6 - 2x3 + 2)(x6 + 2x3 + 2)
b) 4x8 + 1 = 4x8 + 4x4 + 1 - 4x4 = (2x4 + 1)2 - (2x2)2
= (2x4 + 2x2 + 1)(2x4 - 2x2 + 1)
c) x7 + x5 - 1 = x7 - x + x5 + x2 - (x2 - x + 1) = x(x6 - 1) + x2(x3 + 1) - (x2 - x + 1)
= x(x3 - 1)(x3 + 1) + x2(x + 1)(x2 - x + 1) - (x2 - x + 1)
= (x4 - x)(x + 1)(x2 - x + 1) + (x3 + x2)(x2 - x + 1) - (x2 - x + 1)
= (x5 + x4 - x2 - x + x3 + x2 - 1)(x2 -x + 1)
= (x5 + x4 + x3 - x - 1)(x2 - x + 1)
d) x7 + x5 + 1 = x7 - x + x5 - x2 + (x2 + x + 1)
= x(x3 - 1)((x3 + 1) + x2(x3 - 1) + (x2 + x + 1)
= (x4 + x)(x - 1)(x2 + x + 1) + x2(x - 1)((x2 + x + 1) + (x2 + x + 1)
= (x2 + x + 1)(x5 - x4 + x2 - x + x3 - x2 + 1)
= (x2 + x + 1)(x5 - x4 + x3 - x + 1)
Ta có:
(2 - 3x)(x + 8) = (3x - 2)(3 - 5x)
⇔ (2 - 3x)(x + 8) - (3x - 2)(3 - 5x) = 0
⇔ (2 - 3x)(x + 8) + (2 - 3x)(3 - 5x) = 0
⇔ (2 - 3x)(x + 8 + 3 - 5x) = 0
⇔ (2 - 3x)(11 - 4x) = 0
⇔ 2 - 3x = 0 hay 11 - 4x = 0
⇔ 2 = 3x hay 11 = 4x
⇔ x = \(\dfrac{2}{3}\) hay x = \(\dfrac{11}{4}\)
Vậy tập nghiệm của pt S = \(\left\{\dfrac{2}{3};\dfrac{11}{4}\right\}\)
<=> (2-3x ) (x+8) + (2-3x ) (3-5x)=0
<=> (2-3x ) ( x+8 + 3-5x ) =0
<=> (2-3x ) ( 11 - 4x ) = 0
=> 2-3x =0 hoặc 11-4x =0
3x = 2 4x =11
x = 2/3 x = 11/4
a,\(8y^2-\dfrac{1}{8}\)
=\(\dfrac{1}{8}\left(64y^2-1\right)\)
=\(\dfrac{1}{8}\left(8y-1\right)\left(8y+1\right)\)
b,\(\dfrac{1}{25}y^2-64m^2\)
=\(\left(\dfrac{1}{5}y-8m\right)\left(\dfrac{1}{5}y+8m\right)\)
a) \(6x^2-11x+3\)
\(=6x^2-9x-2x+3\)
\(=3x\left(2x-3\right)-\left(2x-3\right)\)
\(=\left(3x-1\right)\left(2x-3\right)\)
b) \(2x^2+3x-27\)
\(=2x^2-6x+9x-27\)
\(=2x\left(x-3\right)+9\left(x-3\right)\)
\(=\left(2x+9\right)\left(x-3\right)\)