Làm hộ mk , mk tích cho:)))
Phân tích thành nhân tử:
\(a\cdot\left(b^2+c^2+bc\right)+b\cdot\left(c^2+a^2+ac\right)+c\cdot\left(a^2+b^2+ab\right)\)
help me,please!!
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Bài 1:Phân tích đa thức thành nhân tử
a) 2x4+3x3-9x2-3x2+2
b) \(a\cdot\left(b+c\right)\cdot\left(b^2-c^2\right)+b\cdot\left(a+c\right)\cdot\left(c^2-b^2\right)+c\cdot\left(a+b\right)\cdot\left(a^2-b^2\right)\)
Bài 2: Cho x-y=12. Tính A=x3-y3-36xy
Giúp mình nhanh nhé
\(x^3-y^3-36xy\)
\(=\left(x-y\right)^3+3xy\left(x-y\right)-36xy\)
\(=12^3+36xy-36xy\)
\(=1728\)
Phân tích đa thức sau thành nhân tử:
a) \(8\cdot\left(x+y+z\right)^3-\left(x+y\right)^3-\left(y+z\right)^3-\left(z-x\right)^3\)
b) \(\left[4abcd+\left(a^2+b^2\right)\cdot\left(c^2+d^2\right)\right]^2-4\cdot\left[cd\cdot\left(a^2+b^2\right)+ab\cdot\left(c^2+d^2\right)\right]^2\)
Các bạn giúp mk giải bài tập này nhá.mk cảm ơn nhìu
Phân tích đa thức thành nhân tử
a)\(x\cdot\left(x+1\right)\cdot\left(x+2\right)\cdot\left(x+3\right)+1\)
b)\(\left(x^2-x+2\right)^2+4\cdot x^2-4\cdot x-4\)
c)\(\left(x+2\right)\cdot\left(x+4\right)\cdot\left(x+6\right)\cdot\left(x+8\right)+16\)
a)\(x\left(x+1\right)\left(x+2\right)\left(x+3\right)+1=\left(x^2+3x\right)\left(x^2+3x+2\right)+1\)
Đặt \(t=x^2+3x\) thì biểu thức có dạng \(t\left(t+2\right)+1=t^2+2t+1=\left(t+1\right)^2=\left(x^2+3x+1\right)^2\)
b)\(\left(x^2-x+2\right)^2+4x^2-4x-4=\left(x^2-x+2\right)^2+4\left(x^2-x-1\right)\)
Đặt \(k=x^2-x+2\) thì biểu thức có dạng
k2+4(k-3)=k2+4k-12=k2-2k+6k-12=k(k-2)+6(k-2)=(k-2)(k+6)=(x2-x)(x2-x+8)=(x-1)x(x2-x+8)
c)làm tương tự câu a
rút gọn phân thức\(\frac{a^2\cdot\left(b-c\right)+b^2\cdot\left(c-a\right)+c^2\cdot\left(a-b\right)}{a^4\cdot\left(b^2-c^2\right)+b^4\cdot\left(c^2-a^2\right)+c^4\cdot\left(a^2-b^2\right)}\)
Rút gọn các phân thức sau
a) \(A=\frac{a^2\cdot\left(b-c\right)+b^2\cdot\left(c-a\right)+c^2\cdot\left(a-b\right)}{a\cdot b^2-a\cdot c^2-b^3+b\cdot c^2}\)
b) \(B=\frac{x^3+y^3+z^3-3\cdot x\cdot y\cdot z}{\left(x+y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2}\)
a. Ta có:
\(a^2\left(b-c\right)+b^2\left(c-a\right)+c^2\left(a-b\right)=a^2\left(b-c\right)-b^2\left(b-c+a-b\right)+c^2\left(a-b\right)=a^2\left(b-c\right)-b^2\left(b-c\right)-b^2\left(a-b\right)+c^2\left(a-b\right)\)
\(=\left(a-b\right)\left(c-a\right)\left(c-b\right)\)
và \(ab^2-ac^2-b^3+bc^2=a\left(b^2-c^2\right)-b\left(b^2-c^2\right)=\left(a-b\right)\left(b-c\right)\left(b+c\right)\)
Vậy, \(A=\frac{\left(a-b\right)\left(c-a\right)\left(c-b\right)}{\left(a-b\right)\left(b-c\right)\left(b+c\right)}=\frac{c-a}{-c-b}=\frac{a-c}{c+b}\)
Tính:
B = \(\dfrac{\text{(a^2 +b^2 +c^2)*(a+b+c)^2+(a*b+b*c+c*a)^2}}{\left(a+b+c\right)^2-\left(a\cdot b+b\cdot c+c\cdot a\right)}\)
C = \(\dfrac{\left(b-c\right)^3+\left(c-a\right)^3+\left(a-b\right)^3}{a^2\cdot\left(b-c\right)+b^2\cdot\left(c-a\right)+c^2\cdot\left(a-b\right)}\)
\(C=\dfrac{\left(b-c+c-a\right)^3+3\left(b-c\right)\left(c-a\right)\left(b-c+c-a\right)+\left(a-b\right)^3}{a^2b-a^2c+b^2c-b^2a+c^2a-c^2b}\)
\(=\dfrac{3\left(b-c\right)\left(c-a\right)\left(b-a\right)}{a^2b-b^2a-a^2c+b^2c+c^2a-c^2b}\)
\(=\dfrac{3\left(b-c\right)\left(c-a\right)\left(b-a\right)}{\left(a-b\right)\cdot ab-c\left(a-b\right)\left(a+b\right)+c^2\left(a-b\right)}\)
\(=\dfrac{3\left(b-c\right)\left(a-c\right)\left(a-b\right)}{\left(a-b\right)\left(ab-ac-bc+c^2\right)}\)
\(=\dfrac{3\left(b-c\right)\left(a-c\right)}{a\left(b-c\right)-c\left(b-c\right)}=3\)
help me!!!
a) tính giá trị nhỏ nhất: H=5.\(\left|3\cdot x-6\right|\)+100
b)cho \(\dfrac{a}{b}\)=\(\dfrac{c}{d}\) c/m \(\dfrac{a\cdot c}{b\cdot d}\)=\(\dfrac{\left(a+2018\cdot c\right)^2}{\left(b+2018\cdot d\right)^2}\)(các tỉ lệ thức đều có nghĩa)
giúp mk nhé mai mk kiểm tra học kì rồi
a: H=5|3x-6|+100>=100
Dấu = xảy ra khi x=2
b: Đặt a/b=c/d=k
=>a=bk; c=dk
\(\dfrac{ac}{bd}=\dfrac{bk\cdot dk}{bd}=k^2\)
\(\left(\dfrac{a+2018c}{b+2018d}\right)^2=\left(\dfrac{bk+2018dk}{b+2018d}\right)^2=k^2\)
=>ĐPCM
Phân tích thành nhân tử:
a)\(a^2\left(a-b\right)-b^2\left(a-c\right)-c^2\left(b-a\right)\)
b)\(a\left(b-c\right)^3+b\left(c-a\right)^3+c\left(a-b\right)^3\)
c) \(abc-\left(ab+ac+bc\right)+\left(a+b+c\right)-1\)
Giups mk vs!! Làm đc nhiều và đúng mk sẽ tick
a) a2(a-b)-b2(a-c)-c2(b-a)
=a2(a-b)-b2(a-c)+c2(a-b)
=(a-b)(a2-c2)-b2(a-c)
=(a-b)(a-c)(a+c)-b2(a-c)
=(a-c)[(a-b)(a+c)-b2]
b)a(b-c)3+b(c-a)3+c(a-b)3
=a(b-c)3-b[(a-b)+(b-c)]+c(a-b)3
=a(b-c)3-b[(a-b)3+3(a-b)2(b-c)+3(a-b)(b-c)2+(b-c)3]+c(a-b)3
=a(b-c)3-b(a-b)3+3b(a-b)2(b-c)+3b(a-b)(b-c)2+b(b-c)3+c(a-b)3
=(b-c)3(a-b)-(a-b)3(b-c)-3b(a-b)(b-c)(a-b+b-c)
=(b-c)3(a-b)-(a-b)3(b-c)-3b(a-b)(b-c)(a-c)
=(a-b)(b-c)[(b-c)2-(a-b)2-3b(a-c)]
=(a-b)(b-c)[(b-c-a+b)(b-c+a-b)-3b(a-c)]
=(a-b)(b-c)[(2b-a-c)(a-c)-3b(a-c)]
=(a-b)(b-c)(a-c)(2b-a-c-3b)
=-(a-b)(b-c)(a-c)(a+b+c)
=(a-b)(b-c)(c-a)(a+b+c)
c)abc-(ab+ac+bc)+(a+b+c)-1
=abc-ab-ac-bc+a+b+c-1
=abc-bc-ab+b-ac+c+a-1
=bc(a-1)-b(a-1)-c(a-1)+a-1
=(a-1)(bc-b-c+1)
=(a-1)[b(c-1)-(c-1)]
=(a-1)(c-1)(b-1)
=(a-1)(b-1)(c-1)
cho tỉ lệ thức \(\frac{a}{b}=\frac{c}{d}\)CM
a)\(\frac{a\cdot c}{b\cdot d}=\frac{a^2+c^2}{b^2+d^2}\)
b)\(\frac{ab}{cd}=\frac{\left(a+b\right)^2}{\left(c+d\right)^2}\)
c)\(\left(a+2c\right)\cdot\left(b+d\right)=\left(a+c\right)\cdot\left(b+2d\right)\)
giúp mk vs