a) \(x.\left(x+4\right)\left(x-4\right)-\left(x^2+1\right)\left(x^2-1\right)=x.\left(x^2-16\right)-\left(x^4-1\right)=x^3-16x-x^4+1\)

ý này ko rút gọn được hết đâu.

b) \(\left(y-3\right)\left(y+3\right)\left(y^2+9\right)-\left(y^2+2\right)\left(y^2-2\right)=\left(y^2-9\right)\left(y^2+9\right)-\left(y^4-4\right)\)

\(=y^4-81-y^4+4=-77\)

c)  \(\left(a+b-c\right)^2-\left(a-c\right)^2-2ab+2bc=a^2+b^2+c^2+2ab-2bc-2ac-a^2+2ac-c^2-2ab+2bc=b^2\)

Trần Anh: Cảm ơn pạn nhiều nhé ~~!! ;) ;) ;) 

Từ \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\)

=> \(\frac{ab+bc+ac}{abc}=0\)

=> \(ab+bc+ac=0\)

=> \(\hept{\begin{cases}ab=-bc-ac\\bc=-ab-ac\\ac=-ab-bc\end{cases}}\)

a) \(N=\frac{bc}{a^2+2bc}+\frac{ca}{b^2+2ac}+\frac{ab}{c^2+2ab}\)

\(=\frac{bc}{a^2-ab-ac+bc}+\frac{ca}{b^2-ab-bc+ac}+\frac{ab}{c^2-ac-bc+ab}\)

\(=\frac{bc}{a\left(a-b\right)-c\left(a-b\right)}+\frac{ca}{b\left(b-a\right)-c\left(b-a\right)}+\frac{ab}{c\left(c-a\right)-b\left(c-a\right)}\)

\(=\frac{bc}{\left(a-b\right)\left(a-c\right)}+\frac{ca}{\left(b-a\right)\left(b-c\right)}+\frac{ab}{\left(c-a\right)\left(c-b\right)}\)

\(=\frac{bc}{\left(a-b\right)\left(a-c\right)}-\frac{ca}{\left(a-b\right)\left(b-c\right)}+\frac{ab}{\left(a-c\right)\left(b-c\right)}\)

\(=\frac{bc\left(b-c\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}-\frac{ca\left(a-c\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}+\frac{ab\left(a-b\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)

\(=\frac{b^2c-bc^2}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}-\frac{ca^2-c^2a}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}+\frac{ab\left(a-b\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)

\(=\frac{b^2c-bc^2-ca^2+c^2a+ab\left(a-b\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)

\(=\frac{\left(c^2a-bc^2\right)-\left(ca^2-b^2c\right)+ab\left(a-b\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)

\(=\frac{c^2\left(a-b\right)-c\left(a-b\right)\left(a+b\right)+ab\left(a-b\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)

\(=\frac{\left(a-b\right)\left(c^2-ac-bc+ab\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)

\(=\frac{\left(a-b\right)\left[\left(ab-bc\right)-\left(ac-c^2\right)\right]}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}=\frac{\left(a-b\right)\left[b\left(a-c\right)-c\left(a-c\right)\right]}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)

\(=\frac{\left(a-b\right)\left(b-c\right)\left(a-c\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}=1\)

b) \(P=\frac{a^2}{a^2+2bc}+\frac{b^2}{b^2+2ac}+\frac{c^2}{c^2+2ab}\)

\(=\frac{a^2}{a^2-ab-ac+bc}+\frac{b^2}{b^2-ab-bc+ac}+\frac{c^2}{c^2-bc-ac+ab}\)

\(=\frac{a^2}{a\left(a-b\right)-c\left(a-b\right)}+\frac{b^2}{b\left(b-a\right)-c\left(b-a\right)}+\frac{c^2}{c\left(c-b\right)-a\left(c-b\right)}\)

\(=\frac{a^2}{\left(a-b\right)\left(a-c\right)}+\frac{b^2}{\left(b-a\right)\left(b-c\right)}+\frac{c^2}{\left(c-b\right)\left(c-a\right)}\)

\(=\frac{a^2}{\left(a-b\right)\left(a-c\right)}-\frac{b^2}{\left(a-b\right)\left(b-c\right)}+\frac{c^2}{\left(b-c\right)\left(a-c\right)}\)

\(=\frac{a^2\left(b-c\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}-\frac{b^2\left(a-c\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}+\frac{c^2\left(a-b\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)

\(=\frac{a^2b-a^2c}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}-\frac{b^2a-b^2c}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}+\frac{c^2\left(a-b\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)

\(=\frac{a^2b-a^2c-b^2a+b^2c+c^2\left(a-b\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)

\(=\frac{ab\left(a-b\right)-c\left(a^2-b^2\right)+c^2\left(a-b\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}=\frac{ab\left(a-b\right)-c\left(a-b\right)\left(a+b\right)+c^2\left(a-b\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)

\(=\frac{\left(a-b\right)\left(ab-ac-bc+c^2\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}=\frac{\left(a-b\right)\left[a\left(b-c\right)-c\left(b-c\right)\right]}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)

\(=\frac{\left(a-b\right)\left(b-c\right)\left(a-c\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}=1\)

4a) \(\left(a+b\right)^2=a^2+2ab+b^2\)

\(\left(a-b\right)^2+4ab=a^2-2ab+b^2+4ab=a^2+b^2+2ab\)

=> (a+b)^2=(a-b)^2+4ab

(x – 3)(2x + 1) = 0

x = 3 hay x = -1/2

a, 1,5 +|2x - 2/3| = 3/2

            |2x - 2/3| = 3/2 - 1,5 

            |2x - 2/3| = 0

<=> 2x - 2/3 = 0 

<=> 2x = 0 + 2/3

<=> 2x = 2/3

<=> x = 2/3 : 2

<=> x = 1/3 

Vậy x = 1/3

b, 3/4 - |1/4 - x| = 5/8

            |1/4 - x| = 3/4 - 5/8

            |1/4 - x| = 1/8

<=> 1/4 - x = 1/8

       1/4 - x = /1/8

<=> x = 1/4 - 1/8

       x = 1/4 - ( -1/8)

<=> x = 1/8

       x = 3/8

Vậy x thuộc { 1/8 ; 3/8 }

c, 3 + | x+2 | = 2

          | x+2 | = 2 - 3 

          | x+2 | = -1 

<=> x + 2 = -1

       x + 2 = 1

<=> x = -1 - 2

       x = 1 -2 

<=> x = -3

       x = -1

Vậy x thuộc { -3 ; -1 }

a) \(\sqrt{x^2-10+25}\)=lx-5l=2

=>x=7 hoặc x=3

b) bình phường lên ta đc x^2-2x=25

từ đây bạn giải bình thường là đc chúc hk tốt

(x+1)+(x+2)+(x+3)=4x

x+1+x+2+x+3=4x

(x+x+x)+(1+2+3)=4x

x*3+6=4x

6=1*x(bớt cả hai vế đi 3*x)

x=6/1(Tìm thừa số)

x=6

a) x=0 hoặc x=1

b)x=-1 hoặc x=2

Bài làm

a) x( x - 1) = 0

=> x = 0 hoặc x - 1 = 0

=> x = 0 hoặc x = 1

Vậy .....

b) ( x + 1 )( x - 2 ) = 0

=> x + 1 = 0 hoặc x - 2 = 0

=> x = -1 hoặc x = 2

Vậy ...

Trl:

a) x.(x - 1) = 0

=> x = 0 hoặc x - 1 = 0 

=> x = 0 hoặc x = 0 + 1

=> x = 0 hoặc x = 1

Vậy \(x\in\left\{0;1\right\}\)

b) (x + 1).(x - 2) = 0

=> x + 1 = 0 hoặc x - 2 = 0

=> x = 0 - 1 hoặc x = 0 + 2

=> x = -1 hoặc x = 2

Vậy \(x\in\left\{-1;2\right\}\)