Những hằng đẳng thức đáng nhớ - Hỏi đáp

Ta có: \(a+b=1\)

\(\Rightarrow a^2+2ab+b^2=1\)

\(\Rightarrow a^2+b^2=1-2ab\)

Ta có: \(B=4\left(a^3+b^3\right)-6\left(a^2+b^2\right)\)

\(=4\left(a+b\right)\left(a^2+b^2-ab\right)-6\left(1-2ab\right)\)

\(=4a^2+4b^2+8ab-6\)

\(=4\left(a^2+2ab+b^2\right)-6\)

\(=4\left(a+b\right)^2-6\)

\(=-2\)

Bạn tham khảo nhé!

Tham khảo :3

Ta có :

\(VT=\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)....\left(2^{1024}+1\right)\)

\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)...\left(2^{1024}+1\right)\)

\(=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)....\left(2^{1024}+1\right)\)

\(=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right).....\left(2^{1024}+1\right)\)

\(=\left(2^{16}-1\right)\left(2^{16}+1\right)....\left(2^{1024}+1\right)\)

\(=\left(2^{1024}-1\right)\left(2^{1024}+1\right)\)

\(=2^{2048}-1\)

Đặt VP = A = \(1+2+2^2+....+2^{2047}\)

⇒ 2A = \(2+2^2+....+2^{2048}\)

⇒ 2A - A = \(2^{2048}-1\)

⇒ VP = A \(=2^{2048}-1\)

Do đó VT = VP ( \(=2^{2048}-1\))

Vậy....

hình như phải là : \(1+\frac{1}{2^n}\) mới đúng nhể

ĐKXĐ : \(x\ne0\)

Ta có : \(x-\frac{1}{x}=4\)

=> \(x^2-1=4x\)

=> \(x^2-4x-1=0\)

=> \(x^2-2.2x+4-5=0\)

=> \(\left(x-2\right)^2=5\)

=> \(x=2\pm\sqrt{5}\left(TM\right)\)

Ta có : \(\left\{{}\begin{matrix}\left\{{}\begin{matrix}F=\frac{\left(2-\sqrt{5}\right)^2}{\left(2-\sqrt{5}\right)^4+1}\approx0,06\\F=\frac{\left(2+\sqrt{5}\right)^2}{\left(2+\sqrt{5}\right)^4+1}=\frac{1}{18}\end{matrix}\right.\\\left\{{}\begin{matrix}G=\left(2-\sqrt{5}\right)^4+\frac{1}{\left(2-\sqrt{5}\right)^4}=322\\G=\left(2+\sqrt{5}\right)^4+\frac{1}{\left(2+\sqrt{5}\right)^4}=322\end{matrix}\right.\end{matrix}\right.\)

Vậy ...

\(x-\frac{1}{x}=4\Rightarrow\left(x-\frac{1}{x}\right)^2=16\Rightarrow x^2+\frac{1}{x^2}=18\)

\(\frac{1}{F}=\frac{x^4+1}{x^2}=x^2+\frac{1}{x^2}=18\Rightarrow F=\frac{1}{18}\)

\(\left(x^2+\frac{1}{x^2}\right)^2=324\Rightarrow x^4+\frac{1}{x^4}=322\)

\(\Rightarrow G=322\)

Lời giải:

Ta có:

$C=x^2-y^2=(x-y)(x+y)=7(x+y)=7\sqrt{(x+y)^2}$

$=7\sqrt{(x-y)^2+4xy}=7\sqrt{7^2+4.60}=119$

$D=x^4+y^4=(x^2-y^2)^2+2(xy)^2=C^2+2(xy)^2=119^2+2.60^2=21361$

\(2\left(a^2+b^2\right)=\left(a-b\right)^2\) \(\Leftrightarrow2a^2+2b^2=a^2-2ab+b^2\)

\(\Leftrightarrow2a^2-a^2+2b^2-b^2+2ab=0\)

\(\Leftrightarrow a^2+2ab+b^2=0\Leftrightarrow\left(a+b\right)^2=0\)

Để a+b = 0 thì a = b = 0 hoặc a và b là hai số đối nhau thì (a+b)² = 0 (Đpcm)

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

\(\Leftrightarrow2a^2+2b^2=a^2-2ab+b^2\)

\(\Leftrightarrow2a^2+2b^2-a^2+2ab-b^2=0\)

\(\Leftrightarrow a^2+2ab+b^2=0\Leftrightarrow\left(a+b\right)^2=0\)

\(\Leftrightarrow a+b=0\Leftrightarrow a=-b\)

Tức là \(a;b\) là 2 số đối nhau

a) \(A=25x^2+3y^2-10x+11\)

\(A=\left(5x-1\right)^2+3y^2+11\ge11\forall x;y\)

Dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{1}{5}\\y=0\end{matrix}\right.\)

b) \(B=\left(x-3\right)^2+\left(x-11\right)^2\)

\(B=2\left(x^2-14x+65\right)\)

\(B=2\left[\left(x-7\right)^2+16\right]\)

\(B=2\left(x-7\right)^2+32\ge32\forall x\)

Dấu "=" xảy ra \(\Leftrightarrow x=7\)

c) \(C=\left(x+1\right)\left(x-2\right)\left(x-3\right)\left(x-6\right)\)

\(C=\left(x^2-5x-6\right)\left(x^2-5x+6\right)\)

Đặt \(x^2-5x-6=a\)

\(C=a\left(a+12\right)\)

\(C=a^2+12a+36-36\)

\(C=\left(a+6\right)^2-36\ge-36\)

Dấu "=" xảy ra \(\Leftrightarrow a=-6\Leftrightarrow\left[{}\begin{matrix}x=0\\x=5\end{matrix}\right.\)

\(C=\left(x+1\right)\left(x-2\right)\left(x-3\right)\left(x-6\right)\\ C=\left(x+1\right)\left(x-6\right)\left(x-2\right)\left(x-3\right)\\ C=\left(x^2-5x-6\right)\left(x^2-5x+6\right)\\ C=\left(x^2-5x\right)^2-6^2\\ C=\left(x^2-5x\right)^2-36\)

Ta có:

\(\left(x^2-5x\right)^2\ge0\\ \Rightarrow C=\left(x^2-5x\right)^2-36\ge-36\)

Dấu "=" xảy ra khi và chỉ khi:

(x² - 5x)² = 0 => x² - 5x = 0 => x(x - 5) = 0

=> x = 5 hoặc x = 0

Vậy MinC = -36 <=> x = 5; x = 0

d) Ta có: \(\left(x+3y\right)^2+6x+18y+9\)

\(=\left(x+3y\right)^2+2\cdot\left(x+3y\right)\cdot3+3^2\)

\(=\left(x+3y+3\right)^2\)

a) \(\left(x+1\right)^2-\left(x-1\right)^2-2\left(x+1\right)\left(x-1\right)\)

\(=\left(x^2+2x+1^2\right)-\left(x^2-2x+1^2\right)-2.\left(x^2-1^2\right)\)

\(=x^2+2x+1-x^2+2x-1-2x^2+2\)

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

b) Check lại đề nha! Thiếu rồi

c) \(\left(3x+1\right)^2-2\left(3x+1\right)\left(3x+5\right)+\left(3x+5\right)^2\)

\(=\left(9x^2+2.3x.1+1^2\right)-2.\left[3x\left(3x+5\right)+1\left(3x+5\right)\right]+\left(9x^2+2.3x.5+5^2\right)\)

\(=9x^2+6x+1-2\left(9x^2+15x+3x+5\right)+9x^2+30x+25\)

\(=9x^2+6x+1-18x^2-30x-6x-10+9x^2+30x+25\)

\(=16\)

P/s: Ko chắc!

Đề là gì vậy ạ?

B = x(x + 2) + y(y - 2) - 2xy + 100

Thay x = y + 6 vaof B ta co

\(B=\left(y+6\right)\left(y+2+6\right)+y\left(y-2\right)-2\left(y+6\right)y+100\)

⇔ B = \(\left(y+6\right)\left(y+8\right)+y\left(y-2\right)-y\left(2y+12\right)+100\)

⇔ B = \(y^2+6y+8y+48+y^2-2y-2y^2-12y+100\)

⇔ B = 148

VayaJ B = 148 tai x = y + 6

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

⇔ \(A=16x^2-16x+9x^2+4-6x+1-\left(12x-3-12x^2+3x\right)\)

⇔ \(A=25x^2-22x+5-12x+3+12x^2-3x\)

⇔ \(A=37x^2-37x+8\)

Thay x = - 3 vào A ta có

A = \(\left(-3\right)^2.37-37.\left(-3\right)+8=9.37+37.3+8=111.\left(3+1\right)+8=111.4+8=444+8=452\)Vậy A = 452 tại x = -3

Ta có: \(A=\left(4x-2\right)^2+\left(-3x+1\right)^2-\left(4x-1\right)\left(3-3x\right)\)

\(=16x^2-16x+4+9x^2-6x+1-\left(12x-12x^2-3+3x\right)\)

\(=25x^2-22x+5-15x+12x^2+3\)

\(=37x^2-37x+8\)

Thay x=-3 vào biểu thức \(A=37x^2-37x+8\), ta được:

\(A=37\cdot\left(-3\right)^2-37\cdot\left(-3\right)+8\)

\(=37\cdot9+111+8\)

\(=333+111+8\)

\(=452\)

Vậy: 452 là giá trị của biểu thức \(A=\left(4x-2\right)^2+\left(-3x+1\right)^2-\left(4x-1\right)\left(3-3x\right)\) tại x=-3

a) Ta có: \(x^2-8x+16\)

\(=x^2-2\cdot x\cdot4+4^2\)

\(=\left(x-4\right)^2\)

b) Ta có: \(16x^2+y^2-8xy\)

\(=\left(4x\right)^2-2\cdot4x\cdot y+y^2\)

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

c) Ta có: \(49a^2+4b^2+28ab\)

\(=\left(7a\right)^2+2\cdot7a\cdot2b+\left(2b\right)^2\)

\(=\left(7a+2b\right)^2\)

e) Ta có: \(\left(3x-2\right)^2-\left(3x+2\right)^2+4x^2+36\)

\(=\left[\left(3x-2\right)-\left(3x+2\right)\right]\cdot\left[\left(3x-2\right)+\left(3x+2\right)\right]+4\left(x^2+9\right)\)

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

\(=-4\cdot6x+4\left(x^2+9\right)\)

\(=4\left(-6x+x^2+9\right)\)

\(=4\left(x^2-6x+9\right)\)

\(=4\left(x-3\right)^2\)

\(=\left(2x-6\right)^2\)

a, (x-4)²

b, (4x-y)²

c, (7a+2b)²

Hằng đẳng thức r cần cm à ? :((

a, (a+b)³

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

\(=\left[a\left(a+b\right)+b\left(a+b\right)\right]\left(a+b\right)\)

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

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

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

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

b, (a-b)³

\(=\left(a-b\right)\left(a-b\right)\left(a-b\right)\)

\(=\left[a\left(a-b\right)-b\left(a-b\right)\right]\left(a-b\right)\)

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

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

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

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

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

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

\(=\left[\left(a+b\right)^3+3.\left(a+b\right)^2.c+3.\left(a+b\right).c^2+c^3\right]-\left[\left(b+c\right)^3-3.\left(b+c\right)^2.a+3.\left(b+c\right).a^2-a^3\right]-\left[\left(a+c\right)^3-3.\left(a+c\right)^2.b+3.\left(a+c\right).b^2-b^3\right]-\left[\left(a+b\right)^3-3.\left(a+b\right)^2.c+3.\left(a+b\right).c^2-c^3\right]\)\(=\left[\left(a^3+3a^2b+3ab^2+b^3\right)+3\left(a^2+2ab+b^2\right).c+3c^2a+3c^2b+c^3\right]-\left[\left(b^3+3b^2c+3bc^2+c^3\right)-3.\left(b^2+2bc+c^2\right).a+3a^2b+3a^2c-a^3\right]-\left[\left(a^3+3a^2c+3ac^2+c^3\right)-3\left(a^2+2ab+b^2\right).c+3c^2a+3c^2b-c^3\right]\)\(=\left(a^3+3a^2b+3ab^2+b^3+3ca^2+6abc+3b^2c+3c^2a+3c^2b+c^3\right)-\left(b^3+3b^2c+3bc^2+c^3-3ab^2-6abc-3ac^2+3a^2b+3a^2c-a^3\right)-\left(a^3+3a^2c+3ac^2+c^3-3a^2c-6abc-3b^2c+3c^2a+3c^2b-c^3\right)\)\(=a^3+3a^2b+3ab^2+b^3+3ca^2+6abc+3b^2c+3c^2a+3c^2b+c^3-b^3-3b^2c-3bc^2-c^3+3ab^2+6abc+3ac^2-3a^2b-3a^2c+a^3-a^3-3a^2c-3ac^2-c^3+3a^2c+6abc+3b^2c-3c^2a-3c^2b+c^3\)\(=3ab^2+6abc+3ab^2+6abc+a^3+6abc+3b^2c-3c^2b\)

\(=6ab^2+18abc+a^3+3b^2c-3bc^2\)

P/s: Ko chắc! Đây là kết quả của hơn 50 phút !