Bài 4: Những hằng đẳng thức đáng nhớ (Tiếp)

a, \(VP=-\left(b-a\right)^3=-\left(b^3-3a^2b+3b^2a-a^3\right)\)

\(=a^3-3b^2a+3a^2b-b^3=\left(a-b\right)^3=VT\left(đpcm\right)\)

b, \(VT=\left(-a-b\right)^2=\left(-a\right)^2-2\left(-a\right)b+b^2\)

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

cách nhân của tú ko thông minh lắm

a)\(-\left(b-a\right)^3=\left(-1\right)^3\left(b-a\right)^3=\left(-\left(b-a\right)\right)^3=\left(-b+a\right)^3=\left(a-b\right)^3\)

b)\(\left(-a-b\right)^2=\left(-\left(a+b\right)\right)^2=\left(-1\right)^2\left(a+b\right)^2=\left(a+b\right)^2\)

Tham khảo nha bạn !

Luyện tập - Bài 38 Sgk tập 1 - trang 17 - Toán lớp 8 | Học trực tuyến

a,

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

b,

\(2\left(x-y\right)\left(x+y\right)+\left(x+y\right)^2+\left(x-y\right)^2\\ =2\left(x^2-y^2\right)+x^2+2xy+y^2+x^2-2xy+y^2\\ =2x^2-2y^2+2x^2+2y^2\\ =4x^2\)

c,

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

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

\(\left(2^x-8\right)^3+\left(4^x+13\right)^3=\left(4^x+2^x+5\right)^3\)

Đặt \(\left\{{}\begin{matrix}2^x-8=a\\4^x+13=b\end{matrix}\right.\) thì ta có:

\(a^3+b^3=\left(a+b\right)^3\)

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

\(\Leftrightarrow ab\left(a+b\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}a=0\\b=0\\a+b=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}2^x-8=0\\4^x+13=0\left(l\right)\\4^x+2^x+5=0\left(l\right)\end{matrix}\right.\)

\(\Rightarrow x=3\)

Đề sai

a) \(27x^3+27x^2+9x+1\\ =\left(3x\right)^3+3.\left(3x\right)^2.1+3.3x.1^2+1^3\\ =\left(3x+1\right)^3\)

Thay x=13 vào biểu thức, ta có:

\(\left(3x+1\right)^3=\left(3.13+1\right)^3=40^3=64000\)

b) \(x^3+12x^2+48x+65\\ =\left(x^3+3.x^2.4+3.x.4^2+4^3\right)+1\\ =\left(x+4\right)^3+1\)

Thay x=6 vào biểu thức, ta có:

\(\left(x+4\right)^3+1=\left(6+4\right)^3+1=10^3+1=1000+1=1001\)

a) \(27x^3+27x^2+9x+1\)

= \(\left(3x\right)^3+3.\left(3x\right)^2.1+3.1^2.3x+1^3\)

= \(\left(3x+1\right)^3\) (1)

Thay x=13 vào pt(1),ta được:

\(\left(3x+1\right)^3=\left(3.13+1\right)^3=40^3=64000\)

b) \(x^3+12x^2+48x+65\)

= \(x^3+3.x^2.4+3.4^2.x+4^3+1\)

= \(\left(x+4\right)^3+1\) (2)

Thay x=6 vào pt(2),ta được:

\(\left(x+4\right)^3+1=\left(6+4\right)^3+1=10^3+1=1001\)

Ta có :

\(a+b+c\Rightarrow a+b=-c\Rightarrow\left(a+b\right)^3=\left(-c\right)^3\)\(=a^3+3a^2b+3ab^2+b^3+c^3=0\Rightarrow a^3+b^3+c^3=-3ab\left(a+b\right)\)\(=a^3+b^3+c^3=-3ab.-c\)

\(=a^3+b^3+c^3=3abc\Rightarrowđpcm\)

Ta cm \(a^3+b^3+c^3=3abc\) là đúng khi \(a+b+c=0\)

Ta có: \(a^3+b^3+c^3=3abc\)

\(\Leftrightarrow\) \(a^3+b^3+c^3-3abc=0\)

\(\Leftrightarrow\) \(\left(a+b\right)^3-3ab\left(a+b\right)+c^3-3abc=0\)

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

\(\Leftrightarrow\) \(\left(a+b+c\right)\left[\left(a+b+c\right)^2-3\left(a+b\right)c-3ab\right]=0\)(điều này đúng vì a+b+c=0)

\(\Rightarrow\) \(a^3+b^3+c^3=3abc\)

ko ai làm đc àĐoàn Đức HiếuNguyễn Huy TúAce Legona

Hồng Phúc NguyễnTrương Hồng Hạnhsoyeon_Tiểubàng giảiTuấn Anh Phan Nguyễnsoyeon_Tiểubàng giải

\(a,\left(x+2\right)\left(x-2\right)+4x=4+x^2\)

\(\Leftrightarrow x^2-4+4x-4-x^2=0\)

\(\Leftrightarrow4x=16\)

\(\Rightarrow x=4\)

\(b,2\left(x+5\right)\left(x-5\right)+5x=8+x^2\)

\(\Leftrightarrow2x^2-50+5x-8-x^2=0\)

\(\Leftrightarrow x^2-5x-42=0\Rightarrow\left(x^2-5x+\dfrac{25}{4}\right)-\dfrac{193}{4}=0\Rightarrow\left(x-\dfrac{5}{2}\right)^2=\dfrac{193}{4}\Rightarrow\left[{}\begin{matrix}x-\dfrac{5}{2}=\sqrt{\dfrac{193}{4}}\\x-\dfrac{5}{2}=-\sqrt{\dfrac{193}{4}}\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=\sqrt{\dfrac{193}{4}}+\dfrac{5}{2}\\x=-\sqrt{\dfrac{193}{4}}+\dfrac{5}{2}\end{matrix}\right.\)

\(c,\left(x+4\right)^2-\left(x+1\right)\left(x-1\right)=16\)

\(\Leftrightarrow x^2+8x+16-x^2-1-16=0\)

\(\Leftrightarrow8x-1=0\Leftrightarrow8x=1\Rightarrow x=\dfrac{1}{8}\)

\(d,\left(2x-1\right)^2+\left(x+3\right)^2-5\left(x+7\right)\left(x-7\right)=0\)\(\Leftrightarrow4x^2-4x+1+x^2+6x+9-5x^2-245=0\)\(\Leftrightarrow2x=235\Leftrightarrow x=117,5\)

Ta có :

A= X^3 - 3X^2 + 3X - 1

<=> A= x^3-3*x^2*1+3*x*1^2-1

<=> A=(x-1)^3

thay x=0 vào biểu thức trên ta có

A=(x-1)^3=(0-1)^3=-1

B= X^3 + 3X^2 +3X + 1 VỚI X=1

( tương tự hằng đẳng thức trên)
C= X^3 + 9X^2 + 27X + 27 VỚI X=5

( tương tự)
D= (X+2)^2 - (X-2)^2 VỚI X=-2

<=> D= (x+2-x+2)(x+2+x-2)

<=> D=8x

thay x=-2 ta có D=-16