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

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

Do đó :\(A=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)+\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\)\(A=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\)

\(A=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\)

\(A=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\)

\(A=\left(2^{16}-1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\)

\(A=\left(2^{32}-1\right)\left(2^{32}+1\right)=2^{64}-1\)

Đây là bài giải pt chứ có phải biểu thức đâu mà thu gọn hả bạn?

Lời giải:

a. ĐKXĐ: $x\neq 1$

PT $\Leftrightarrow \frac{x^2+x+1}{(x-1)(x^2+x+1)}+\frac{2x(x-1)}{(x-1)(x^2+x+1)}=\frac{3x^2}{(x-1)(x^2+x+1)}$

$\Leftrightarrow x^2+x+1+2x(x-1)=3x^2$

$\Leftrightarrow 3x^2-x+1=3x^2$

$\Leftrightarrow x=1$ (không thỏa đkxđ)

Vậy pt vô nghiệm.

b. ĐKXĐ: $x\neq \pm 3$

PT $\Leftrightarrow \frac{(x+2)(x+3)}{(x-3)(x+3)}=\frac{x^2+3x}{(x-3)(x+3)}$

$\Leftrightarrow (x+2)(x+3)=x^2+3x$

$\Leftrightarrow x^2+5x+6=x^2+3x$

$\Leftrightarrow 2x+6=0$

$\Leftrightarrow x=-3$ (không thỏa mãn đkxđ)

Do đó pt vô nghiệm.

c. ĐKXĐ: $x\neq \pm 2$

PT $\Leftrightarrow \frac{(x-2)^2-(x+2)^2}{(x+2)(x-2)}=\frac{-16}{(x-2)(x+2)}$

$\Leftrightarrow (x-2)^2-(x+2)^2=-16$

$\Leftrightarrow -8x=-16$

$\Leftrightarrow x=2$ (vi phạm đkxđ)

Do đó pt vô nghiệm.

`a)sqrt{8-2sqrt7}+sqrt{16-6sqrt7}`

`=sqrt{(sqrt7-1)^2}+sqrt{(3-sqrt7)^2}`

`=sqrt7-1+3-sqrt7=2`

`b)sqrt{(sqrt7-1)^2}-sqrt{11+4sqrt7}`

`=sqrt7-1-sqrt{(2+sqrt7)^2}`

`=sqrt7-1-2-sqrt7=-3`

a, \(=\sqrt{7-2\sqrt{7}+1}+\sqrt{7-2.3\sqrt{7}+9}\)

\(=\sqrt{\left(\sqrt{7}-1\right)^2}+\sqrt{\left(3-\sqrt{7}\right)^2}=\left|\sqrt{7}-1\right|+\left|3-\sqrt{7}\right|\)

\(=\sqrt{7}-1+3-\sqrt{7}=2\)

\(b,=\left|\sqrt{7}-1\right|-\sqrt{7+2.2\sqrt{7}+4}\)

\(=\left|\sqrt{7}-1\right|-\sqrt{\left(\sqrt{7}+2\right)^2}=\left|\sqrt{7}-1\right|-\left|\sqrt{7}+2\right|\)

\(=\sqrt{7}-1-\sqrt{7}-2=-3\)

a \(\Rightarrow\sqrt{7-2\sqrt{7}+1}+\sqrt{9-6\sqrt{7}+7}=\sqrt{\left(\sqrt{7}-1\right)^2}+\sqrt{\left(3-\sqrt{7}\right)^2}=\sqrt{7}-1+3-\sqrt{7}=2\) b \(\Rightarrow\sqrt{7}-1-\sqrt{7-4\sqrt{7}+4}=\sqrt{7}-1-\sqrt{\left(\sqrt{7}-2\right)^2}=\sqrt{7}-1-\sqrt{7}+2=1\)

a) Ta có: \(2\sqrt{8}-3\sqrt{18}+\sqrt{32}\)

\(=4\sqrt{2}-6\sqrt{2}+4\sqrt{2}\)

\(=2\sqrt{2}\)

b) Ta có: \(\sqrt{\left(1-\sqrt{2}\right)^2}+\sqrt{\left(1+\sqrt{2}\right)^2}\)

\(=\sqrt{2}-1+\sqrt{2}+1\)

\(=2\sqrt{2}\)

c) Ta có: \(\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{4-2\sqrt{3}}\)

\(=2-\sqrt{3}+\sqrt{3}-1\)

=1

a) Ta có: 

b) Ta có: 

= 36 đó bn

tich nha!!

1*2*4+2*4*8+4*8*16+8*16*32/1*3*4+2*6*8+4*12*16+8*24*32 = 56744

1*2*4+2*4*8+4*8*16+8*16*32/1*3*4+2*6*8+4*12*16+8*24*32 = 56744

#nhớ tk 

\(\dfrac{1}{1-x}+\dfrac{1}{1+x}+\dfrac{2}{1+x^2}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)

\(=\dfrac{1+x+1-x}{1-x^2}+\dfrac{2}{1+x^2}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)

\(=\dfrac{2+2x^2+2-2x^2}{1-x^4}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)

\(=\dfrac{4+4x^4+4-4x^4}{1-x^8}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)

\(=\dfrac{8+8x^8+8-8x^8}{1-x^{16}}+\dfrac{16}{1+x^{16}}\)

\(=\dfrac{16+16x^{16}+16-16x^{16}}{1-x^{32}}=\dfrac{32}{1-x^{32}}\)

VT=[(2-1)(2+1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)]/2

=[(2^2-1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)]/2

=[(2^4-1(2^4+1)(2^8+1)(2^16+1)]/2

=[(2^8-1)(2^8+1)(2^16+1)]/2

=[(2^16-1)(2^16+1)]/2

=(2^32-1)/2

cau nau de sai roi

(2+1)(2^2+1)(2^4 +1)(2^8+1)(2^16+1) - 2^32

=1.(2+1)(2²+1)(2⁴ +1)(2⁸+1)(2¹⁶+1) - 2³²

=(2-1).(2+1)(2²+1)(2⁴ +1)(2⁸+1)(2¹⁶+1) - 2³²

=(2²-1)(2²+1)(2⁴ +1)(2⁸+1)(2¹⁶+1) - 2³²

=(2⁴-1)(2⁴ +1)(2⁸+1)(2¹⁶+1) - 2³²

=(2⁸-1)(2⁸+1)(2¹⁶+1) - 2³²

=(2¹⁶-1)(2¹⁶+1) - 2³²

=2³²-1-2³²

=-1

    (2+1 ) ( 2^2 + 1) ... (2^16 + 1) - 2^32

=  3 ( 2^2 + 1) ....( 2^16 + 1) -2^32

= ( 2^2 - 1)( 2^2 +1)....(2^16  + 1 ) - 2^32

= (2^4 - 1)( 2^4 + 1)( 2^8 + 1)( 2^16 + 1) - 2^32

= ( 2^8 - 1) ( 2^8 + 1) ( 2^16  - 1 ) - 2^32

= ( 2^ 16 - 1) (2^16 + 1) - 2^32

= 2^32 - 1 - 2^32

=-1

tìm x biết: 

câu a: (2x+1)-4(x+2)^2=9

câu b: (x+3)^2-(x-4)(x+8)=1

câu c: (x+1)^3 - (x-1)^3 - 6(x-1)^2=-19