Phân tích đa thức thành nhân tử
1).a(b+c)^2(b-c)+b(c+a)^2(c-a)+c(a+b)^2(a-b)
2).x^8+x^4+1
3).x^12+1
trình bày cách lm nhaq
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M
31 tháng 10 2015
\(ab\left(a-b\right)+bc\left(b-c\right)+ca\left(c-a\right)\)
\(=ab\left(a-b\right)+bc\left(b-c\right)-ca\left(a-c\right)\)
\(=ab\left(a-b\right)+bc\left(b-c\right)-ca\left(a-b+b-c\right)\)
\(=ab\left(a-b\right)+bc\left(b-c\right)-ca\left(a-b\right)-ca\left(b-c\right)\)
\(=\left(a-b\right)\left(ab-ca\right)+\left(b-c\right)\left(bc-ca\right)\)
\(=\left(a-b\right)a\left(b-c\right)+\left(b-c\right)c\left(b-a\right)\)
\(=\left(a-b\right)a\left(b-c\right)-\left(b-c\right)c\left(a-b\right)\)
\(=\left(a-b\right)\left(b-c\right)\left(a-c\right)\)
mình làm vội, có chỗ nào sai bạn thông cảm nha
8 tháng 9 2019
\(x^2-y^2+4x+4\)
\(=\left(x+2\right)^2-y^2\)
\(=\left(x+2+y\right)\left(x+2-y\right)\)
\(4x^2-y^2+8\left(y-2\right)\)
\(=4x^2-\left(y^2-8y+16\right)\)
\(=4x^2-\left(y-4\right)^2\)
\(=\left(2x+y-4\right)\left(2x-y+4\right)\)
AN
1
NH
15 tháng 7 2017
\(D=a^3\left(b-c\right)+b^3\left(c-a\right)+c^3\left(a-b\right)\)
\(D=a^3\left(b-c\right)+\left[b^3\left(c-a\right)+c^3\left(a-b\right)\right]\)
\(D=a^3\left(b-c\right)\left(b^3c-ab^3+ac^3-bc^3\right)\)
\(D=a^3\left(b-c\right)\left[\left(b^3c-bc^3\right)-\left(ab^3-ac^3\right)\right]\)
\(D=a^3\left(b-c\right)\left[bc\left(b^2-c^2\right)-a\left(b^3-c^3\right)\right]\)
\(D=a^3\left(b-c\right)\left[bc\left(b-c\right)\left(b+c\right)-a\left(b-c\right)\left(b^2+bc+c^2\right)\right]\)
\(D=\left(b-c\right)\left[a^3+bc\left(b+c\right)-a\left(b^2+bc+c^2\right)\right]\)
\(D=\left(b-c\right)\left(a^3+b^2c+bc^2-ab^2-abc-ac^2\right)\)
\(D=\left(b-c\right)\left[\left(b^2c-ab^2\right)+\left(bc^2-abc\right)-\left(ac^2-a^3\right)\right]\)
\(D=\left(b-c\right)\left[b^2\left(c-a\right)+bc\left(c-a\right)-a\left(c^2-a^2\right)\right]\)
\(D=\left(b-c\right)\left[b^2\left(c-a\right)+bc\left(c-a\right)-a\left(c-a\right)\left(c+a\right)\right]\)
\(D=\left(b-c\right)\left(c-a\right)\left[b^2+bc-a\left(c+a\right)\right]\)
\(D=\left(b-c\right)\left(c-a\right)\left(b^2+bc-ac-a^2\right)\)
\(D=\left(b-c\right)\left(c-a\right)\left[\left(bc-ac\right)+\left(b^2-a^2\right)\right]\)
\(D=\left(b-c\right)\left(c-a\right)\left[c\left(b-a\right)+\left(b-a\right)\left(b+a\right)\right]\)
\(D=\left(b-c\right)\left(c-a\right)\left(b-a\right)\left(c+b+a\right)\)
\(D=\left(a-b\right)\left(b-c\right)\left(a-c\right)\left(a+b+c\right)\)
Chúc bạn học tốt.
T
8 tháng 10 2023
Bài 1.
\(a\Big) 9(4x+3)^2=16(3x-5)^2\\\Leftrightarrow 9[(4x)^2+2\cdot 4x\cdot3+3^2]=16[(3x)^2-2\cdot3x\cdot5+5^2]\\\Leftrightarrow9(16x^2+24x+9)=16(9x^2-30x+25)\\\Leftrightarrow 144x^2+216x+81=144x^2-480x+400\\\Leftrightarrow (144x^2-144x^2)+(216x+480x)=400-81\\\Leftrightarrow 696x=319\\\Leftrightarrow x=\dfrac{11}{24}\\Vậy:x=\dfrac{11}{24}\\---\)
\(b\Big)(x-3)^2=4x^2-20x+25\\\Leftrightarrow(x-3)^2=(2x)^2-2\cdot2x\cdot5+5^2\\\Leftrightarrow(x-3)^2=(2x-5)^2\\\Leftrightarrow (x-3)^2-(2x-5)^2=0\\\Leftrightarrow (x-3-2x+5)(x-3+2x-5)=0\\\Leftrightarrow (-x+2)(3x-8)=0\\\Leftrightarrow \left[\begin{array}{} -x+2=0\\ 3x-8=0 \end{array} \right.\\\Leftrightarrow \left[\begin{array}{} -x=-2\\ 3x=8 \end{array} \right.\\\Leftrightarrow \left[\begin{array}{} x=2\\ x=\dfrac{8}{3} \end{array} \right.\\Vậy:...\)