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phân tích đa thức thành nhân tử
a/x2(x+1)-2x(x+1)+(x+1)=(x+1)(x^2-2x+1)=(x+1)(x-1)^2
b/a2+b2+2a-2b-2ab=(a^2-ab)+(b^2-ab)+2(a-b)=a(a-b)-b(a-b)+2(a-b)=(a-b)(a-b+2)
c/ 4x2-8x+3=(2x-2)^2-1=(2x-2-1)(2x-2+1)=(2x-3)(2x-1)
d/25-16x2=5^2-(4x)^2=(5-4x)(5+4x)
a) 2x3 + 8x2 - 8x
= 2x(x2 + 4x - 4)
= 2x(x2 + 4x + 4 - 8)
= 2x[(x + 2)2 - 8]
= \(2x\left(x+2-\sqrt{8}\right)\left(x+2+\sqrt{8}\right)\)
b) a2 - b2 + 4a + 4b
= (a - b)(a + b) + 4(a + b)
= (a + b)(a - b + 4)
c) x2 - 2x - 3
= x2 + x - 3x - 3
= x(x + 1) - 3(x + 1)
= (x + 1)(x - 3)
d) x2 - 4x - 3
= x2 - 4x + 4 - 7
= (x + 2)2 - 7
= \(\left(x+2-\sqrt{7}\right)\left(x+2+\sqrt{7}\right)\)
\(a,3x^2-11x+6=3x^2-9x-2x+6=3x\left(x-3\right)-2\left(x-3\right)=\left(3x-2\right)\left(x-3\right)\)
\(b,8x^2+10x-3=8x^2+12x-2x-3=4x\left(2x+3\right)-\left(2x+3\right)=\left(4x-1\right)\left(2x+3\right)\)
\(c,8x^2-2x-1=9x^2-x^2-2x-1=9x^2-\left(x+1\right)^2=\left(3x-x-1\right)\left(3x+x+1\right)\)
\(=\left(2x-1\right)\left(4x+1\right)\)
Đề đúng: \(M=\left(a^2+b^2-c^2\right)^2-4a^2b^2\)
a) Ta có:
\(M=\left(a^2+b^2-c^2\right)^2-4a^2b^2\)
\(M=\left(a^2+b^2-c^2-2ab\right)\left(a^2+b^2-c^2+2ab\right)\)
\(M=\left[\left(a^2-2ab+b^2\right)-c^2\right]\left[\left(a^2+2ab+b^2\right)-c^2\right]\)
\(M=\left[\left(a-b\right)^2-c^2\right]\left[\left(a+b\right)^2-c^2\right]\)
\(M=\left(a-b-c\right)\left(a-b+c\right)\left(a+b-c\right)\left(a+b+c\right)\)
b) Nếu a,b,c là độ dài 3 cạnh của tam giác thì:
\(\hept{\begin{cases}a+b>c\\c+a>b\\b+c>a\end{cases}}\Leftrightarrow\hept{\begin{cases}a+b-c>0\\a-b+c>0\\a-b-c< 0\end{cases}}\) , mà a + b + c > 0
=> \(M< 0\)
a.\(3x^2-11x+6\)
= \(3x^2-9x-2x+6\)
=\(3x\left(x-3\right)-2\left(x-3\right)\)
=\(\left(x-3\right)\left(3x-2\right)\)
b\(8x^2+10x-3\)
=.\(8x^2-2x+12x-3\)
=\(2x\left(4x-1\right)+3\left(4x-1\right)\)
=\(\left(4x-1\right)\left(2x+3\right)\)
d.\(x^2-y^2+10x-6y+16\)
=\(\left(x^2+10x+25\right)-\left(y^2+6y+9\right)\)
=\(\left(x+5\right)^2-\left(y+3\right)^2\)
=\(\left(x+5-y-3\right)\left(x+5+y+3\right)\)
=\(\left(x-y+2\right)\left(x+y+8\right)\)
e.\(x^4+x^2y^2+y^4\)
=\(x^4+2x^2y^2+y^4-x^2+y^2\)
=\(\left(x^2+y^2\right)^2-x^2y^2\)
=\(\left(x^2+y^2-xy\right)\left(x^2+y^2+xy\right)\)
a)
\(=3x^2-9x-2x+6=3x\left(x-3\right)-2\left(x-3\right)=\left(x-3\right)\left(3x-2\right)\)
3a) x2 (x-1) - 4x2 + 8x - 4
= x2(x-1) - ( 2x - 2)2
= (x\(\sqrt{x-1}\))2 -( 2x - 2)2
= (x\(\sqrt{x-1}\)- 2x+2) ( x\(\sqrt{x-1}\)+ 2x - 2)
3b) = x3 +33 + (x+3) (x-9)
= (x + 3)( x2 - 3x + 9) + (x+3)(x-9)
= (x+3)(x2 -2x) = (x + 3)(x - 2)x
phân tích đa thức ->nhân tử:
a)2x2+4x-70
b)x3-5x2+8x-4
c)x2-10+16
rút gọn:
(8x-8x3-10x2+3x4-5):(3x2-2x+1)
Bài 1:
a)2x2+4x-70
=2(x2+2x-35)
=2(x2+7x-5x-35)
=2[x(x+7)-5(x+7)]
=2(x-5)(x+7)
b)x3-5x2+8x-4
=x3-4x2+4x-x2+4x-4
=x(x2-4x+4)-(x2-4x+4)
=(x2-4x+4)(x-1)
=(x-2)2(x-1)
c)x2-10x+16
=x2-2x-8x+16
=x(x-2)-8(x-2)
=(x-8)(x-2)
Bài 2:
\(\frac{8x-8x^3-10x^2+3x^4-5}{3x^2-2x+1}=\frac{\left(x^2-2x-5\right)\left(3x^2-2x+1\right)}{3x^2-2x+1}=x^2-2x-5\)
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/ phân tích thành nhân tử ;
= C2-( a +b )2=( c-a -b ) . ( c+a +b )