(a + b)3 – (a – b)3 – 2b3

= (a3 + 3a2b + 3ab2 + b3) – (a3 – 3a2b + 3ab2 – b3) – 2b3 (Áp dụng HĐT (4) và (5))

= a3 + 3a2b + 3ab2 + b3 – a3 + 3a2b – 3ab2 + b3 – 2b3

= (a3 – a3) + (3a2b + 3a2b) + (3ab2 – 3ab2) + (b3 + b3 – 2b3)

= 6a2b

Rút gọn biểu thức: (a+b)\(^3\) – (a–b)\(^3\)

Ta có: \(\left(a+b\right)^3-\left(a-b\right)^3\)

= \(a^3+3a^2b+3ab^2+b^3-a^3-3a^2b+3ab^2-b^3\)

= \(6ab^2\)

(a+b)³-(a-b)³ = a³+3a²b+3ab²+b³-a³+3a²b-3ab²+b³

=6a²b+2b³ =2b(3a² +b²)

\(1.\\ A=\sqrt{\left(2+\sqrt{3}\right)^2}+\sqrt{\left(2-\sqrt{3}\right)^2}\\ =\left|2+\sqrt{3}\right|+\left|2-\sqrt{3}\right|\\ =2+\sqrt{3}+2-\sqrt{3}=4\)

\(2.\\a.\\ P=3x-\sqrt{\left(x-5\right)^2}=3x-\left|x-5\right|\\ b.\\ x=2\Rightarrow P=3\)

\(3.\\ M=\dfrac{\sqrt{\left(x-1\right)^2}}{x-1}=\dfrac{\left|x-1\right|}{x-1}\)

\(\cdot x>1\Rightarrow M=1\\ \cdot x=1\Rightarrow M=0\\\cdot x< 1\Rightarrow M=-1\)

B1.

Ta có:A\(=\sqrt{3+4\sqrt{3}+4}+\sqrt{3-4\sqrt{3}+4}\)

            \(=\sqrt{\left(\sqrt{3}+2\right)^2}+\sqrt{\left(\sqrt{3}-2\right)^2}\)

           \(=\sqrt{3}+2+\sqrt{3}-2=2\sqrt{3}\)

Bài 1 : 

\(A=\sqrt{\left(\sqrt{3}+2\right)^2}+\sqrt{\left(\sqrt{3}-2\right)^2}\\ =\sqrt{3}+2+2-\sqrt{3}=4\)

Bài 2 : 

a) \(P=3x-\sqrt{\left(x-5\right)^2}=3x-\left|x-5\right|\)

b) khi x = 2 thì \(P=3.2-\left|2-5\right|=3\)

Bài 3 : 

\(M=\dfrac{\sqrt{\left(\sqrt{x}-1\right)^2}}{x-1}=\dfrac{\left|\sqrt{x}-1\right|}{x-1}\)

Áp dụng hằng đẳng thức dưới dạng 

\(x^3+y^3=\left(x+y\right)^3-3xy\left(x+y\right)\)

\(\left(a+b+c\right)^3+\left(a-b-c\right)^3=\left(2a\right)^3-3\left(a+b+c\right)\left(a-b-c\right).2a\)

\(\left(b-c-a\right)^3+\left(c-a-b\right)^3=\left(-2a\right)^3-3\left(b-c-a\right)\left(c-a-b\right).\left(-2a\right)\)

\(\Rightarrow\left(a+b+c\right)^3+\left(a-b-c\right)^3+\left(b-c-a\right)^3+\left(c-a-b\right)^3\)

\(=\left(2\right)^3+\left(-2a\right)^3-6a\left[a+\left(b+c\right)\right]\left[a-\left(b+c\right)\right]+6a\left[-a+\left(b-c\right)\right]\left[-a-\left(b-c\right)\right]\)

\(=-6a\left\{a^2-\left(b+c\right)^2-\left[\left(-a\right)^2-\left(b-c\right)^2\right]\right\}\)

\(=-6a\left\{a^2-a^2+\left(b-c\right)^2-\left(b+c\right)^2\right\}\)

\(=-6a\left[b-c+b+c\right]\left[b-c-\left(b+c\right)\right]=-6a.2b.\left(-2c\right)\)

\(=24abc\)

=\(\dfrac{\left(a-b\right)\left(\sqrt{a}-\sqrt{b}\right)}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}-\dfrac{\sqrt{a^3}-\sqrt{b^3}}{a-b}\)

=\(\dfrac{\sqrt{a^3}-\sqrt{b^3}}{a-b}-\dfrac{\sqrt{a^3}-\sqrt{b^3}}{a-b}\)

=\(0\)

ừ chie cần k vaod chữ đúng thôi

a,Đặt a+b-c=x, c+a-b=y, b+c-a=z

=>x+y+z=a+b-c+c+a-b+b+c-a=a+b+c

Ta có hằng đẳng thức:

(x+y+z)^3-3x-3y-3z=3(x+y)(x+z)(y+z)

=>(a+b+c)^3-(b+c-a)^3-(a+c-b)^3-(a+b-c)^3=(x+y+z)^3-x^3-y^3-z^3

=3(x+y)(x+z)(y+z)

=3(a+b-c+c+a-b)(c+a-b+b+c-a)(b+c-a+a+b-c)

=3.2a.2b.2c

=24abc

mình mới có tài khoản,vậy k cho bn chỉ cần k đúng thôi đk ^^? 

Đặt \(x=a+b;y=b+c;z=c+a\) ta có:

\(x^3+y^3+z^3-3xyz\)

\(=\left(x+y\right)^3-3xy\left(x-y\right)+z^3-3xyz\)

\(=\left[\left(x+y\right)^3+z^3\right]-3xy\left(x+y+z\right)\)

\(=\left(x+y+z\right)^3-3z\left(x+y\right)\left(x+y+z\right)-3xy\left(x-y-z\right)\)

\(=\left(x+y+z\right)\left[\left(x+y+z\right)^2-3z\left(x+y\right)-3xy\right]\)

\(=\left(x+y+z\right)\left(x^2+y^2+z^2+2xy+2xz+2yz-3xz-3yz-3xy\right)\)

\(=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-yx\right)\)

Thay vào ta có:\(\left(a+b+b+c+c+a\right)\left[\left(a+b\right)^2+\left(b+c\right)^2+\left(c+a\right)^2-\left(a+b\right)\left(b+c\right)-\left(b+c\right)\left(c+a\right)-\left(c+a\right)\left(a+b\right)\right]\)

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

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

Câu hỏi của Nhàn Nguyễn - Toán lớp 8 - Học toán với OnlineMath