(x-1)(x-2)(x+2)-(x-3)\(^3\)

=(x-1)(x\(^2\)-4)-(x-3)\(^3\)

(xy-1)(xy-2)-(xy-2)\(^2\)

=(xy-2)(xy-1-xy+2)

=xy-2

Giải 2 bài luôn

Rút gọn:

\(Y=\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+....+\frac{1}{\sqrt{99}-\sqrt{100}}\)

\(Y=\sqrt{2}-\sqrt{1}+\sqrt{2}-\sqrt{3}+....+\sqrt{100}-\sqrt{99}\)

\(Y=\sqrt{10}-1\)

\(Y=9\)

Tính:

\(Y=\frac{2014}{\sqrt{1}+\sqrt{2}}+\frac{2014}{\sqrt{2}+\sqrt{3}}+....+\frac{2014}{\sqrt{99}+\sqrt{100}}\)

\(Y=\sqrt{2}-\sqrt{1}+\sqrt{2}-\sqrt{3}+...+\sqrt{100}-\sqrt{99}\)

\(Y=\sqrt{10}-1\)

\(Y=9\)

\(Y=2014.9\)

\(Y=18126\)

Y=\(\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+...+\frac{1}{\sqrt{99}+\sqrt{100}}\)

\(=\sqrt{2}-1+\sqrt{2}-\sqrt{3}+...+\sqrt{100}-\sqrt{99}\)

\(=-1+\sqrt{100}=\sqrt{100}-1=10-1=9\)

y = 18126 nha anh k em nha

\(Q=\frac{2-\frac{1}{3}+\frac{1}{4}}{2+\frac{1}{6}-\frac{1}{4}}\)

\(Q=\frac{2-\frac{1}{3}}{2+\frac{1}{6}}\)

Còn lại dễ mà, bn tự làm nhé!

B=1/2²+1/2³+...+1/2¹⁰⁰

2B=1/2¹+1/2²+...1/2⁹⁹

2B-B=(1/2¹+1/2²+...+1/2⁹⁹)-(1/2²+1/2³+...+1/2¹⁰⁰)

B=1/2¹-1/2¹⁰⁰=2⁹⁹/2¹⁰⁰-1/2¹⁰⁰=2⁹⁹-1/2¹⁰⁰

Lời giải:

Áp dụng hằng đẳng thức \((a-1)(a+1)=a^2-1\) ta có:

\(A=3(2^2+1)(2^4+1)(2^8+1)(2^{16}+1)\)

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

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

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

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

\(D=\dfrac{5}{2x^2+6x}-\dfrac{4-3x^2}{x^2-9}-3\) (đk:\(x\ne3;x\ne-3\))

\(=\dfrac{5}{2x\left(x+3\right)}-\dfrac{4-3x^2}{\left(x-3\right)\left(x+3\right)}-3\)

\(=\dfrac{5\left(x-3\right)}{2x\left(x-3\right)\left(x+3\right)}-\dfrac{\left(4-3x^2\right).2x}{2x\left(x-3\right)\left(x+3\right)}-\dfrac{3.2x\left(x-3\right)\left(x+3\right)}{2x\left(x-3\right)\left(x+3\right)}\)

\(=\dfrac{5x-15-8x+6x^3-6x\left(x^2-9\right)}{2x\left(x-3\right)\left(x+3\right)}\)

\(=\dfrac{51x-15}{2x\left(x-3\right)\left(x+3\right)}\)

\(A=100^2-99^2+98^2-97^2+....+2^2-1^2\)

\(=\left(100-99\right).\left(100+99\right)+\left(98-97\right).\left(98+97\right)+....+\left(2-1\right).\left(2+1\right)\)

\(=1+2+....+97+98+99+100=\frac{100.\left(100+1\right)}{2}=5050\)

\(B=3\left(2^2+1\right)\left(2^4+1\right)....\left(2^{64}+1\right)+1=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)...\left(2^{64}+1\right)+1\)

\(=\left(2^4-1\right)\left(2^4+1\right)......\left(2^{64}+1\right)+1=\left(2^8-1\right).....\left(2^{64}+1\right)+1\)

Tiếp tục rút gọn như vậy,ta đc \(B=\left(2^{64}-1\right)\left(2^{64}+1\right)=2^{128}-1+1=2^{128}\)

a: ĐKXĐ: \(a\notin\left\{0;1;-1\right\}\)

\(A=\dfrac{a^2}{\left(a-1\right)\left(a+1\right)}-\dfrac{a^2}{a^2+1}\cdot\dfrac{a^2+1}{a\left(a+1\right)}\)

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

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

b: Để A=3 thì \(3a^2-3=a\)

\(\Leftrightarrow2a^2=3\)

hay \(a\in\left\{\dfrac{\sqrt{6}}{2};-\dfrac{\sqrt{6}}{2}\right\}\)