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\(\left(a+b\right).\left(b+c\right).\left(c-a\right)+\left(b+c\right).\left(c+a\right).\left(a-b\right)+\left(c+a\right).\left(a+b\right).\left(b-c\right)\)
\(=\left(a+b\right).\left[\left(b+c\right).\left(c-a\right)+\left(c+a\right).\left(a-b\right)\right]+\left(c+a\right).\left(a+b\right).\left(b-c\right)\)
\(=\left(a+b\right).\left(ac-a^2+bc-ab+a^2-ab+ac-bc\right)+\left(c+a\right).\left(a+b\right).\left(b-c\right)\)
\(=-\left(a+b\right).2a.\left(b-c\right)+\left(c+a\right).\left(a+b\right).\left(b-c\right)\)
\(=\left(a+b\right).\left(b-c\right).\left(-2a+c+a\right)=\left(a+b\right).\left(b-c\right).\left(c-a\right)\)
giai lai:
\(\left(b+c\right).\left[\left(a+b\right).\left(c-a\right)+\left(c+a\right).\left(a-b\right)\right]+\left(c+a\right).\left(a+b\right).\left(b-c\right)\)
\(=-\left(b+c\right).2a.\left(b-c\right)+\left(b-c\right).\left(ac+bc+a^2+ab\right)\)
\(=\left(b-c\right).\left(-2ab-2ac+ac+bc+a^2+ab\right)\)
\(=\left(b-c\right).\left(-ab-ac+bc+a^2\right)\)
\(=\left(b-c\right).\left(a+b\right).\left(a-c\right)\)
Ta có : ( a - b )2 + 4ab
= a2 - 2ab + b2 + 4ab
= a2 + 2ab + b2
= ( a + b )2 ( Vế trái )
Do đó : ( a + b )2 = ( a - b )2 + 4ab
+) Biến đổi vế phải ta có :
\(\left(A-B\right)^2+4AB\)
\(=A^2-2AB+B^2+4AB\)
\(=A^2+2AB+B^2=\left(A+B\right)^2=VT\left(đpcm\right)\)
\(\dfrac{3x^2+ax^2+x+a}{x+1}\)
\(=\dfrac{3x^2+3x+ax^2+ax-\left(a+2\right)x-\left(a+2\right)+a+2}{x+1}\)
\(=3x+ax-a-2+\dfrac{a+2}{x+1}\)
Để đây là phép chia hết thì a+2=0
hay a=-2
Bài 2:
Diện tích khu vườn là:
\(\left(14+x\right)\left(18-x\right)\)
\(=252-14x+18x-x^2\)
\(=-x^2+4x+252\)
\(=-\left(x^2-4x+4-256\right)\)
\(=-\left(x-2\right)^2+256\le256\forall x\)
Dấu '=' xảy ra khi x=2
Chu vi hình chữ nhật là:
\(C=2\left[14+x+18-x\right]=2\cdot32=64\left(cm\right)\)
Bài 1:
\(a+b=15\)
\(\Rightarrow\left(a+b\right)^2=225\)
\(\Leftrightarrow a^2+2ab+b^2=225\)
\(\Leftrightarrow a^2+4+b^2=225\)
\(\Leftrightarrow a^2+b^2=221\)
Ta có: \(\left(a-b\right)^2=a^2-2ab+b^2\)
\(=221-4\)
\(217\)
Bài 2:
Vì \(x:7\)dư 6
\(\Rightarrow x\equiv-1\left(mod7\right)\)
\(\Rightarrow x^2\equiv1\left(mod7\right)\)
Vậy \(x^2:7\)dư 1
a: =>(3x+1)(3x-1)-(3x+1)(2x-3)=0
=>(3x+1)(3x-1-2x+3)=0
=>(3x+1)(x+2)=0
=>x=-1/3 hoặc x=-2
b: =>(3x+1)(6x+2)-(3x+1)(x-2)=0
=>(3x+1)(6x+2-x+2)=0
=>(3x+1)(5x+4)=0
=>x=-1/3 hoặc x=-4/5