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A) 1/2 x(x^2-4)+4(x+2)
=1/2x(x-2)(x+2)+4(x+2)
=(x+2)(1/2x^2-x+4)
b) 21(x-y)^2-7(x-y)^3
= (x-y)^2(21-7x+7y)
=(x-y)^2.7(3-x+y)
c) 1/8x^3-3/4x^2+3/2x-1
=(1/2x)^3-3.(1/2x)^2.1+3.1/2x.1^2-1
=(1/2x-1)^3
a,x4-4x3+8x2-16x+16
=(x4-4x3+4x2)+(4x2-16x+16)
=(x^2-2x)^2+(2x-4)^2
=x^2(x-2)^2+4(x-2)^2
=(x-2)^2(x^2+4)
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 = \(\frac{5}{2}\)
\(A=x^3+4x^2-8x-8=\left(x^3-8\right)+4x\left(x-2\right)=\left(x^3-2^3\right)+4x\left(x-2\right)\)
\(=\left(x-2\right)\left(x^2+2x+4\right)+4x\left(x-2\right)=\left(x-2\right)\left(x^2+2x+4+4x\right)=\left(x-2\right)\left(x^2+6x+4\right)\)
\(B=a^2+b^2-a^2b^2+ab-a-b=\left(ab-a\right)-\left(a^2b^2-a^2\right)+\left(b^2-b\right)\)
\(=a\left(b-1\right)-a^2\left(b^2-1\right)+b\left(b-1\right)=a\left(b-1\right)-a^2\left(b-1\right)\left(b+1\right)+b\left(b-1\right)\)
\(=\left(b-1\right)\left(a-a^2b-a^2+b\right)\)
\(C=x^4-x^3-x+1=x^3\left(x-1\right)-\left(x-1\right)=\left(x-1\right)\left(x^3-1\right)\)
Đoàn Thị Huyền Đoan: Hình như câu A bạn chép xuống bị sai đề rồi!
bằng phương pháp nào zậy bn????
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a: \(x^4-4x^3-8x^2+8x\)
\(=x\left(x^3-4x^2-8x+8\right)\)
\(=x\left[\left(x+2\right)\left(x^2-2x+4\right)-4x\left(x+2\right)\right]\)
\(=x\left(x+2\right)\left(x^2-6x+4\right)\)
b: \(x^2-1-xy+y\)
\(=\left(x-1\right)\left(x+1\right)-y\left(x-1\right)\)
\(=\left(x-1\right)\left(x-y+1\right)\)
c: Ta có: \(\left(x-1\right)\left(x-2\right)\left(x-3\right)+\left(x-1\right)^2\cdot\left(x-2\right)\)
\(=\left(x-1\right)\cdot\left(x-2\right)\cdot\left(x-3-x-1\right)\)
\(=2\cdot\left(x-1\right)\cdot\left(x-2\right)^2\)