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Bài 1:
a) \(\left(a-b^2\right)\left(a+b^2\right)=a^2-b^4\)
b) \(\left(a^2+2a-3\right)\left(a^2+2a+3\right)=\left(a^2+2a\right)^2-9\)
c) \(\left(a^2+2a+3\right)\left(a^2-2a-3\right)=a^2-\left(2a+3\right)^2\)
d) \(\left(a^2-2a+3\right)\left(a^2+2a+3\right)=9-\left(a^2-2a\right)^2\)
e) \(\left(-a^2-2a+3\right)\left(-a^2-2a+3\right)=\left(-a^2-2a+3\right)^2\)
g) \(\left(a^2+2a+3\right)\left(a^2-2a+3\right)=\left(a^2+3\right)^2-4a^2\)
f) \(\left(a^2+2a\right)\left(2a-a^2\right)=4a^2-a^4\)
Bài 2 :
a) \(\left(x+1\right)\left(x^2-x+1\right)=x^3+1\)
b) \(\left(x+y+z\right)^2=\left(x+y+z\right)\left(x+y+z\right)=x^2+xy+xz+yx+y^2+yz+zx+zy+z^2=x^2+2xy+2yz+2xz+y^2+z^2\)
c) \(\left(x-y+z\right)^2=\left(x-y+z\right)\left(x-y+z\right)=x^2-xy+xz-xy+y^2-yz+xz-yz+z^2=x^2+y^2+z^2-2xy+2xz-2yz\)d) \(\left(x-2y\right)\left(x^2+2xy+4y^2\right)=\left(x-2y\right)^3\)
e) \(\left(x-y-z\right)^2=\left(x-y-z\right)\left(x-y-z\right)=x^2-xy-xz-xy+y^2+yz-xz+yz+z^2=x^2-2xy-2xz+2yz+y^2+z^2\)
a)Ta có:
\(\left(x-2\right)^2-\left(x-3\right)\left(x-1\right)\\ =x^2-4x+4-x^2+4x-3\\ =1\)
Vậy biểu thức \(\left(x-2\right)^2-\left(x-3\right)\left(x-1\right)\)không phụ thuộc vào biến
b) Ta có:
\(\left(x-1\right)^3-\left(x+1\right)^3+6\left(x+1\right)\left(x-1\right)\\ =x^3-3x^2+3x-1-x^3-3x^2-3x-1+6x^2-6\\ =-8\)
Vậy.....
c) Ta có:
\(\left(x-3\right)\left(x+3\right)\left(x^2+9\right)-\left(x^2-2\right)\left(x^2+2\right)\\ =\left(x^2-9\right)\left(x^2+9\right)-x^4+4\\ =x^4-81-x^4+4=-77\)
Vậy....
d) Ta có: \(\left(3x+1\right)^2-2\left(3x+1\right)\left(3x-5\right)+\left(3x-5\right)^2\\ =\left(3x+1-3x+5\right)^2\\ =6^2=36\)
Vậy....
Ta có:
\(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=4\left(a^2+b^2+c^2-ab-bc-ac\right)\)
\(\Leftrightarrow a^2-2ab+b^2+b^2-2bc+c^2+c^2-2ac+a^2=4a^2+4b^2+4c^2-4ab-4bc-4ac\)
\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2bc-2ac=4a^2+4b^2+4c^2-4ab-4bc-4ac\)
\(\Leftrightarrow0=2a^2+2b^2+2c^2-2ab-2bc-2ac\)
\(\Leftrightarrow0=a^2-2ab+b^2+b^2-2bc+c^2+c^2-2ac+a^2\)
\(\Leftrightarrow0=\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\)
Mà \(\left\{\begin{matrix}\left(a-b\right)^2\ge0\\\left(b-c\right)^2\ge0\\\left(c-a\right)^2\ge0\end{matrix}\right.\)
\(\Rightarrow\left\{\begin{matrix}\left(a-b\right)^2=0\\\left(b-c\right)^2=0\\\left(c-a\right)^2=0\end{matrix}\right.\)
\(\Rightarrow\left\{\begin{matrix}a=b\\b=c\\c=a\end{matrix}\right.\)
\(\Rightarrow a=b=c\) ( đpcm )
\(a\left(b+c\right)\left(b^2-c^2\right)+b\left(a+c\right)\left(c^2-a^2\right)+c\left(a+b\right)\left(a^2-b^2\right)\)
\(=\left(a-b\right)\left(b-c\right)\left(c-a\right)\left(a+b+c\right)\)
a)\(=5c\left(a^2-9b^2\right)+5c^2.\left(3b-a\right)\))\
\(=5c\left(a-3b\right)\left(a+3b\right)-5c^2.\left(a-3b\right)\)
= \(5c.\left(a-3b\right)\left(a+3b-c\right)\)
b)\(=xy.\left(3x-y\right)-xyz.\left(3x-y\right)\)
\(=xy.\left(3x-y\right)\left(1-z\right)\)
c) \(=a^2m^2+2ambp+b^2p^2-a^2p^2-2ambp-b^2m^2\)
\(=a^2m^2-a^2p^2+b^2p^2-b^2m^2\)
\(=a^2.\left(m^2-p^2\right)+b^2\left(p^2-m^2\right)\)
\(=a^2.\left(m^2-p^2\right)-b^2\left(m^2-p^2\right)\)
\(=\left(a^2-b^2\right)\left(m^2-p^2\right)\)
\(=\left(a-b\right)\left(a+b\right)\left(m-p\right)\left(m+p\right)\)
a) \(\left(9x-1\right)^2+\left(1-5x\right)^2+2\left(9x-1\right)\left(1-5x\right)\)
\(=\left(9x-1\right)^2+2\left(9x-1\right)\left(1-5x\right)+\left(1-5x\right)^2\)
\(=\left(9x-1+1-5x\right)^2=\left(4x\right)^2\)
b) \(x^2\left(x+4\right)\left(x-4\right)-\left(x^2+1\right)\left(x^2-1\right)\)
\(=x^2\left(x^2-16\right)-\left(x^4-1\right)\)
\(=x^4-16x^2-x^4+1=-16x^2+1\)
a)\(\left(5x-1\right)^2-\left(5x-4\right)\left(5x+4\right)=7\)
\(\Leftrightarrow25x^2-10x+1-25x^2+16=7\)
\(\Leftrightarrow-10x=-10\)
\(\Leftrightarrow x=1\)
b) k hiểu đề
a: Xét ΔAEB vuông tại E và ΔAFC vuông tại F có
góc BAE chung
Do đó: ΔAEB\(\sim\)ΔAFC
SUy ra:AE/AF=AB/AC
hay \(AE\cdot AC=AB\cdot AF\)
b: Xét ΔBDH vuông tại D và ΔBEC vuông tại E có
góc EBC chung
Do đó; ΔDBH\(\sim\)ΔEBC
SUy ra: BD/BE=BH/BC
hay \(BD\cdot BC=BH\cdot BE\)
c: Xét ΔHFA vuông tại F và ΔHDC vuông tại D có
\(\widehat{AHF}=\widehat{CHD}\)
Do đó: ΔHFA\(\sim\)ΔHDC
Suy ra: HF/HD=HA/HC
hay \(HF\cdot HC=HD\cdot HA\left(1\right)\)
Xét ΔHFB vuông tại F và ΔHEC vuông tạiE có
\(\widehat{FHB}=\widehat{EHC}\)
Do đó: ΔHFB\(\sim\)ΔHEC
Suy ra: HF/HE=HB/HC
hay \(HF\cdot HC=HB\cdot HE\left(2\right)\)
Từ (1) và (2) suy ra \(HA\cdot HD=HB\cdot HE=HC\cdot HF\)
3a2c2 + bd + 3abc + acd
= 3ac(ac + b) + d(ac + b)
= (ac + b)(3ac + d)
ab(a + b) - bc(a + c) + abc
= b(a2 + ab - ac - c2 + ac)
= b(a2 + ab - c2)
a(b2 + c2) + b(c2 + a2) + c(a2 + b2) + 2abc
= ab2 + ac2 + bc2 + a2b + c(a2 + 2ab + b2)
= c2(a + b) + ab(a + b) + c(a + b)2
= (a + b)(c2 + ab + ac + bc)
= (a + b)[c(b + c) + a(b + c)]
= (a + b)(a + c)(b + c)
bc(b + c) + ac(c - a) - ab(a + b)
= bc(b + c) + ac[(b + c) - (a + b)] - ab(a + b)
= bc(b + c) + ac(b + c) - ac(a + b) - ab(a + b)
= c(b + c)(a + b) - a(a + b)(b + c)
= (a + b)(b + c)(c - a)