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a) (x2 + 2xy + y2) : (x + y) = (x + y)2 : (x + y) = x + y.
b) (125x3 + 1) : (5x + 1) = [(5x)3 + 1] : (5x + 1)
= (5x)2 – 5x + 1 = 25x2 – 5x + 1.
c) (x2 – 2xy + y2) : (y – x) = (x – y)2 : [-(x – y)] = - (x – y) = y – x
Hoặc (x2 – 2xy + y2) : (y – x) = (y2 – 2xy + x2) : (y – x)
= (y – x)2 : (y – x) = y - x.
Bài giải:
a) (x2 + 2xy + y2) : (x + y) = (x + y)2 : (x + y) = x + y.
b) (125x3 + 1) : (5x + 1) = [(5x)3 + 1] : (5x + 1)
= (5x)2 – 5x + 1 = 25x2 – 5x + 1.
c) (x2 – 2xy + y2) : (y – x) = (x – y)2 : [-(x – y)] = - (x – y) = y – x
Hoặc (x2 – 2xy + y2) : (y – x) = (y2 – 2xy + x2) : (y – x)
= (y – x)2 : (y – x) = y - x.
dẽ mà giúp mình bài này đi
cho tam giac abc . co canh bc=12cm, duong cao ah=8cm
a> tinh s tam giac abc
b> tren canh bc lay diem e sao cho be=3/4bc. tinh s tam giac abe va s tam giac ace ( bằng nhiều cách
c> lay diem chinh giua cua canh ac va m . tinh s tam giac ame
\(\left(m+n\right)\left(m^2-mn+n^2\right)\)
\(=m^3-m^2n+mn^2-m^2n-mn^2+n^3\)
\(=m^3-n^3\)
a) \(\left(1+x\right)^2+\left(1-x\right)^2\)
\(=1+2x+x^2+1-2x+x^2\)
\(=2x^2+2\)
b) \(\left(x+2\right)^2+\left(1+x\right)\left(1-x\right)\)
\(=x^2+4x+4+1-x^2\)
\(=4x+5\)
c) \(\left(x-3\right)^2+3\left(x+1\right)^2\)
\(=x^2-6x+9+3x^2+6x+3\)
\(=4x^2+12\)
d)\(\left(2+3x\right)\left(3x-2\right)-\left(3x+1\right)^2\)
\(=9x^2-4-9x^2-6x-1\)
\(=-6x-5\)
e) \(\left(x+5\right)\left(x-2\right)-\left(x+2\right)^2\)
\(=x^2-2x+5x-10-x^2-4x-4\)
\(=-x-14\)
f) \(\left(x+3\right)\left(2x-5\right)-2\left(1+x\right)^2\)
\(=2x^2-5x+6x-15-2-4x-2x^2\)
\(=-3x-17\)
g) \(\left(4x-1\right)\left(4x+1\right)-4\left(1-2x\right)^2\)
\(=16x^2-1-4+16x-16x^2\)
\(=16x-5\)
#Học tốt!
a, (y-3)(y+3)=y2-32=y2-9 (hằng đẳng thức)
b, (a-b-c)2 - (a-b+c)2= ((a-b-c)-(a-b+c)).((a-b-c)+(a-b+c))
=(a-b-c-a+b-c).(a-b-c+a-b+c)=-2c+2a-2b
c, (m+n)(m2 -mn+n2)=m3+n3(hằng đẳng thức)
d
a) \(x\left(x-1\right)\left(x+1\right)-\left(x+1\right)\left(x^2-x+1\right)\)
\(=\left(x+1\right)\cdot\left[x\cdot\left(x-1\right)-\left(x^2-x+1\right)\right]\)
\(=\left(x+1\right)\left(x^2-x-x^2+x-1\right)\)
\(=\left(x+1\right)\cdot\left(-1\right)\)
\(=-1\left(x+1\right)\)
b) \(\left(x-1\right)^3-\left(x+2\right)\left(x^2-2x+4\right)+3\left(x+4\right)\left(x-4\right)\)
\(=x^3-3x^2+3x-1-\left(x^3+8\right)+\left(3x+12\right)\left(x-1\right)\)
\(=x^3-3x^2+3x-1-\left(x^3+8\right)+3x^2-3x+12x-12\)
\(=x^3-1-x^3-8+12x-12\)
\(=-21+12x\)
c) \(3x^2\left(x+1\right)\left(x-1\right)+\left(x^2-1\right)^3-\left(x^2-1\right)\left(x^4+x^2+1\right)\)
\(=3x^2\left(x^2-1\right)+x^6-3x^4+3x^2-1-\left(x^6-1\right)\)
\(=3x^4-3x^2+x^6-3x^4+3x^2-1-x^6+1\)
\(=0\)
câu 2:
a(b-c)-b(a+c)+c(a-b)=-2bc
ta có:
a( b-c ) - b ( a +c )+ c(a-b)
=ab-ac-(ba+bc)+(ca-cb)
=ab-ac-ba-bc+ca-cb
=ab-ba-ac+ca-bc-cb
=0-0-bc-cb
=bc+(-cb)
=-2cb hay -2bc
b)a(1-b)+a(a^2-1)=a(a^2-b)
Ta có:
a(1-b) + a(a^2-1)
=a-ab+(a^3-a)
=a-ab+a^3-a
=a-a-ab+a^3
=0-ab+a^3
=-ab+a^3
=a(-b +a^2) hay a(a^2-b)
1. a) $(5-2x)^2-16=0$
$=>(5-2x)^2-4^2=0$
$=>(5-2x-4)(5-2x+4)=0$
$=>(1-2x)(9-2x)=0$
\(=>\left[{}\begin{matrix}1-2x=0=>x=0,5\\9-2x=0=>x=4,5\end{matrix}\right.\)
b) $x^2-4x=29$
$=>x^2-4x-29=0$
$=>(x^2-4x+4)-33=0$
$=>(x-2)^2-(\sqrt{33})^2=0$
$=>(x-2-\sqrt{33})(x-2+\sqrt{33})=0$
\(=>\left[{}\begin{matrix}x-2-\sqrt{33}=0=>x=\sqrt{33}+2\\x-2+\sqrt{33}=0=>x=2-\sqrt{33}\end{matrix}\right.\)
Bài 1:
a) \(\left(5-2x\right)^2-16=0\) (1)
\(\Leftrightarrow\left(5-2x\right)^2=16\)
\(\Leftrightarrow5-2x=\pm4\)
\(\Leftrightarrow\left[{}\begin{matrix}5-2x=4\\5-2x=-4\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{9}{2}\end{matrix}\right.\)
Vậy tập nghiệm phương trình (1) là \(S=\left\{\dfrac{1}{2};\dfrac{9}{2}\right\}\)
b) \(x^2-4x=29\) (2)
\(\Leftrightarrow x^2-4x-29=0\)
\(\Leftrightarrow x=\dfrac{4\pm2\sqrt{33}}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{4+2\sqrt{33}}{2}\\x=\dfrac{4-2\sqrt{33}}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2+\sqrt{33}\\x=2-\sqrt{33}\end{matrix}\right.\)
Vậy tập nghiệm phương trình (2) là \(S=\left\{2-\sqrt{33};2+\sqrt{33}\right\}\)
c) \(\left(x-3\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+9\left(x+1\right)^2=15\) (3)
\(\Leftrightarrow x^3-9x^2+27x-27-\left(x^3-27\right)+9\left(x^2+2x+1\right)=15\)
\(\Leftrightarrow x^3-9x^2+27x-27-\left(x^3-27\right)+9x^2+18x+9=15\)
\(\Leftrightarrow x^3+27x-27-x^3+27+18x+9=15\)
\(\Leftrightarrow45x+9=15\)
\(\Leftrightarrow45x=15-9\)
\(\Leftrightarrow45x=6\)
\(\Leftrightarrow x=\dfrac{2}{15}\)
Vậy tập nghiệm phương trình (3) là \(S=\left\{\dfrac{2}{15}\right\}\)
d) \(2\left(x-5\right)\left(x+5\right)-\left(x+2\right)\left(2x-3\right)+x\left(x^2+8\right)=\left(x+1\right)\left(x^2-x+1\right)\)(4)
\(\Leftrightarrow2\left(x^2-25\right)-\left(2x^2-3x+4x-6\right)+x^3-8x=x^3+1\)
\(\Leftrightarrow2x^2-50-\left(2x^2+x-6\right)+x^3-8x=x^3+1\)
\(\Leftrightarrow2x^2-50-2x^2-x+6-8x=1\)
\(\Leftrightarrow-44-9x=1\)
\(\Leftrightarrow-9x=1+45\)
\(\Leftrightarrow-9x=45\)
\(\Leftrightarrow x=-5\)
Vậy tập nghiệm phương trình (4) là \(S=\left\{-5\right\}\)
a/ \(\left(m+n\right)\left(m^3-mn+n^2\right)=m^3+n^3\)
b/ \(\left(a-b-c\right)^2-\left(a-b+c\right)^2=\left(a-b-c-a+b-c\right)\left(a-b-c+a-b+c\right)=-2c\left(2a-2b\right)=-4c\left(a-b\right)\)c/
\(\left(1+x+x^2\right)\left(1-x\right)\left(1+x\right)\left(1-x+x^2\right)=\left(\left(1+x+x^2\right)\left(1-x\right)\right)\left(\left(1-x+x^2\right)\left(1+x\right)\right)=\left(1-x^3\right)\left(1+x^3\right)=1-x^6\)
a) m3+n3
b) (a -b-c+a-b+c)(a-b-c-a+b-c)
= -4c(a-b)
c) (1-x3)(1+x3)
=1-x6