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\(A=x^3+4x^2-8x-8=\left(x^3-8\right)+4x\left(x-2\right)=\left(x^3-2^3\right)+4x\left(x-2\right)\)
\(=\left(x-2\right)\left(x^2+2x+4\right)+4x\left(x-2\right)=\left(x-2\right)\left(x^2+2x+4+4x\right)=\left(x-2\right)\left(x^2+6x+4\right)\)
\(B=a^2+b^2-a^2b^2+ab-a-b=\left(ab-a\right)-\left(a^2b^2-a^2\right)+\left(b^2-b\right)\)
\(=a\left(b-1\right)-a^2\left(b^2-1\right)+b\left(b-1\right)=a\left(b-1\right)-a^2\left(b-1\right)\left(b+1\right)+b\left(b-1\right)\)
\(=\left(b-1\right)\left(a-a^2b-a^2+b\right)\)
\(C=x^4-x^3-x+1=x^3\left(x-1\right)-\left(x-1\right)=\left(x-1\right)\left(x^3-1\right)\)
Đoàn Thị Huyền Đoan: Hình như câu A bạn chép xuống bị sai đề rồi!
a
4x2--25=0
=> (2x)22 --52 =0
=> (2x-5)(2x+5)=0
\(\orbr{\begin{cases}2x-5=0\\2x+5=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}X=\frac{5}{2}\\X=\frac{-5\:\:. \:\:\:\:\:\:\:\:\:\:TT}{2}\end{cases}Mình\:}\)
\(4x^2=25\Rightarrow x^2=\frac{25}{4}\Rightarrow x=\sqrt{\frac{25}{4}}\) \(=\frac{5}{2}\)
\(\left(x^3-x^2\right)^2-\left(4x^2-8x+4\right)=0\)
= \(\left(x^3-x^2\right)^2-\left(2x-2\right)^2=0\)
=(\(\left(x^3-x^2-2x+2\right)\left(x^3-x^2+2x-2\right)=0\)
=\(\left[x^2\left(x-1\right)-2\left(x-1\right)\right]\) \(\left[x^2\left(x-1\right)+2\left(x-1\right)\right]\)=0
=\(\left(x-1\right)\left(x^2-2\right)\left(x-1\right)\left(x^2+2\right)\) = 0
= \(\left(x-1\right)\left(x^2-2\right)\left(x^2+2\right)=0\)
=\(\left(x-1\right)\left(x^4-4\right)\) = 0
=> \(x-1=0\) hoặc \(x^4-4=0\)
=> \(x=1\) hoặc \(x=\pm\sqrt{2}\)
câu 2
a)\(\left(3x^2\right)^3-\left(2x\right)^3\)
= \(\left(3x^2-2x\right)\left(9x^4-54x^5+36x^4-4x^2\right)\)
= \(x\left(3x-2\right)\left(9x^4-54x^5+36x^4-4x^2\right)\)
may be wrong , but chawsc k nhiều , chỗ nào k hiểu ib hỏi mk sai nha <3
1,
a, \(\left(2x-5\right)\cdot\left(2x+5\right)=0\)
\(x=\frac{5}{2}\)
x\(=-\frac{5}{2}\)
b \(\left(x^3-x^2\right)^2-\left(2x-2\right)^2\)=0
(x-2x+2)(x+2x-2)=0
x=2
x=2/3
2,
a (3x^2)^3-(2x)^3
(3x^2-2x)(9x^4+6x^3+4x^2)
\(4x^2-25=0\)
\(\left(2x-5\right)\left(2x+5\right)=0\)
\(\Rightarrow\orbr{\begin{cases}2x-5=0\\2x+5=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{5}{2}\\x=-\frac{5}{2}\end{cases}}\)
Vậy \(\orbr{\begin{cases}x=\frac{5}{2}\\x=-\frac{5}{2}\end{cases}}\)
\(27x^6-8x^3=\left(3x^2\right)^3-\left(2x\right)^3=\left(3x^2-2x\right)\left[\left(3x^2\right)^2+3x^2.2x+\left(2x\right)^2\right]=x^3.\left(3x-2\right).\left(3x^2+6x+4\right)\)
1a) 4x2 - 25 = 0 => 4x2 = 25 => x2 = \(\frac{25}{4}\)= \(\left(\frac{5}{2}\right)^2\)=> x = \(\frac{5}{2}\)
\(\left(a^2+a+4\right)^2+8a\left(a^2+a+4\right)+15a^2=\left(a^2+a+4\right)^2+8a\left(a^2+a+4\right)+16a^2-a^2=\left(a^2+a+4+4a\right)^2-a^2\)
\(\left(a^2+5a+4\right)^2-a^2=\left(a^2+5a+a+4\right)\left(a^2+5a-a+4\right)=\left(a^2+6a+4\right)\left(a^2+4a+4\right)=\left(a^2+6a+4\right)\left(a+2\right)^2\)
a,9a^3-13a+6
=9a^3-6a^2+6a^2-4a-9a+6
=3a^2(3a-2)+2a(3a-2)-3(3a-2)
=(3a^2+2a-2)(3a-2)
\(x^2-y^2+4x+4\)
\(=\left(x+2\right)^2-y^2\)
\(=\left(x+2+y\right)\left(x+2-y\right)\)
\(4x^2-y^2+8\left(y-2\right)\)
\(=4x^2-\left(y^2-8y+16\right)\)
\(=4x^2-\left(y-4\right)^2\)
\(=\left(2x+y-4\right)\left(2x-y+4\right)\)
a/ 9a^3 - 13a + 6 = 9a^3 - 6a^2 + 6a^2 - 4a - 9a + 6 = (9a^3 - 6a^2) + (6a^2 - 4a) - (9a - 6) = 3a^2(3a - 2) + 2a(3a - 2) - 3(3a - 2) = (3a^2 + 2a - 3)(3a - 2) Mình gửi luôn cho nóng^^Được câu nào hay câu đó. Yên tâm mình sẽ cố nghĩ &gửi nốt :)))
b/x^4 - 4x^3 + 8x + 3 = x^4 - 3x^3 - x^3 + 3x^2 - 3x^2 + 9x - x + 3 = (x^4 - 3x^3) - (x^3 - 3x^2) - (3x^2 - 9x) - (x - 3) = x^3(x - 3) - x^2(x - 3) - 3x(x - 3) - (x - 3) = (x^3 - x^2 - 3x - 1)(x - 3) Mình đang cố nghĩ nốt con c đây, có vẻ khó^^