2410 là 24¹⁰

A = 2⁰+2¹+2²+2³+...+2⁴⁰

-> 2A = 2¹+2²+2³+...+2⁴⁰+2⁴¹

=> A = 2A - A = 2⁴¹-1

mà B = 2⁴¹

=> A < B (2⁴¹-1 < 2⁴¹)

Ta có:
\(A=2^0+2^1+2^2+...+2^{40}\)

\(\Rightarrow A=1+2+2^2+...+2^{40}\)

\(\Rightarrow2A=2+2^2+2^3+...+2^{41}\)

\(\Rightarrow2A-A=\left(2+2^2+2^3+...+2^{41}\right)-\left(1+2+2^2+...+2^{40}\right)\)

\(\Rightarrow A=2^{41}-1\)

Vì \(2^{41}-1< 2^{41}\) nên A < B

Vậy A < B

\(\dfrac{4^5\cdot9^4-2\cdot6^9}{2^{10}\cdot3^8+6^8\cdot20}\)=\(\dfrac{\left(2^2\right)^5\cdot\left(3^2\right)^4-2\cdot6^9}{2^{10}\cdot3^8+6^8\cdot2\cdot10}=\dfrac{2^{10}\cdot3^8-2\cdot6^9}{2^{10}\cdot3^8+6^8\cdot2\cdot10}=\dfrac{6}{10}=\dfrac{3}{5}\)

\(3x^2y^4\)-\(5xy^3\)-\(\dfrac{3}{2}x^2y^4\)+\(3xy^3\)+\(2xy^3\)+1=1,5\(x^2y^4\)+1>0

3x² y⁴ - 5xy³ -3/2 x2y⁴ + 3xy³ + 2xy³ +1

=\(\dfrac{3}{2}x^2y^4+1\)

Ta có:x²\(\ge0\forall x\in Q\)

y^{4\(\ge0\forall x\in Q\)}

\(\Rightarrow x^2y^4\)\(\ge0\forall x\in Q\)

\(\Rightarrow\dfrac{3}{2}\)\(x^2y^4\) \(\ge0\forall x\in Q\)

\(\Rightarrow\dfrac{3}{2}\)\(x^2y^4\) \(\ge1\forall x\in Q\) (đpcm)

1. a) (x-2)² =1

=> x - 2 = \(\pm\sqrt{1}\)

=> x - 2 = 1 hoặc -1

=> x = 3 hoặc 1

b) 2x - 1= -8

=> 2x = -7

=>x = \(\dfrac{-7}{2}\)

c)thiếu đề

d) (x-1)^x+2 = (x-1)^x+4

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

(x-1)^x+2 = (x-1)^x+2. (x-1)²

(x-1)^x+2 - (x-1)^x+2. (x-1)² = 0

=> (x-1)^x+2. [1 - (x-1)²] = 0

\(\left[{}\begin{matrix}\left(x-1\right)^{x+2}=0\\1-\left(x-1\right)^2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x-1=0\\x-1=1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)

2a) \(\dfrac{45^{10}.5^{10}}{75^{10}}\) = \(\dfrac{\left(3.3.5\right)^{10}.5^{10}}{\left(5.5.3\right)^{10}}\) = \(\dfrac{3^{10}.3^{10}.5^{10}.5^{10}}{5^{10}.5^{10}.3^{10}}\) = \(3^{10}\)

b) \(\dfrac{2^{15}.9^4}{6^6.8^3}\)=\(\dfrac{2^{15}.\left(3^2\right)^4}{\left(2.3\right)^6.\left(2^3\right)^3}\)=\(\dfrac{2^{15}.3^8}{2^6.3^6.2^9}\)=\(3^2\)

Ta có:

A =2¹⁰⁰-2⁹⁹+2⁹⁸-2⁹⁷+.....+2²-2¹

=>2A=2¹⁰¹-2¹⁰⁰+2⁹⁹-2⁹⁸+.....+2³-2²

=>2A+A=(2¹⁰¹-2¹⁰⁰+2⁹⁹-2⁹⁸+.....+2³-2²) + (2¹⁰⁰-2⁹⁹+2⁹⁸-2⁹⁷+....+2²-2¹⁾

^=>3A=2¹⁰¹-2

=>A=\(\frac{2^{101}-2}{3}\)

Vậy A=\(\frac{2^{101}-2}{3}\).

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

\(\Rightarrow2A=2^{101}-2^{100}+2^{99}-2^{98}+...+2^3-2^2\)

\(\Rightarrow2A+A=\left(2^{101}-2^{100}+2^{99}-2^{98}+...+2^3-2^2\right)+\left(2^{100}-2^{99}+2^{98}-2^{97}+...+2^2-2\right)\)

\(\Rightarrow3A=2^{101}-2\)

\(\Rightarrow A=\frac{2^{101}-2}{3}\)

a.

\(\frac{x}{y}=\frac{7}{3}\Rightarrow\frac{x}{7}=\frac{y}{3}\Rightarrow\frac{5x}{35}=\frac{2y}{6}\)

Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:

\(\frac{5x}{35}=\frac{2y}{6}=\frac{5x-2y}{35-6}=\frac{87}{29}=3\)

\(\frac{5x}{35}=3\Rightarrow x=\frac{35\times3}{5}=21\)

\(\frac{2y}{6}=3\Rightarrow y=\frac{6\times3}{2}=9\)

Vậy \(x=21\) và \(y=9\)

b.

\(\frac{x}{19}=\frac{y}{21}\Rightarrow\frac{2x}{38}=\frac{y}{21}\)

Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:

\(\frac{2x}{38}=\frac{y}{21}=\frac{34}{17}=2\)

\(\frac{2x}{38}=2\Rightarrow x=\frac{38\times2}{2}=38\)

\(\frac{y}{21}=2\Rightarrow y=2\times21=42\)

Vậy \(x=38\) và \(y=42\)

c.

\(\frac{x^3}{8}=\frac{y^3}{64}=\frac{z^3}{216}\Rightarrow\frac{x^3}{2^3}=\frac{y^3}{4^3}=\frac{z^3}{6^3}\Rightarrow\left(\frac{x}{2}\right)^3=\left(\frac{y}{4}\right)^3=\left(\frac{z}{6}\right)^3\Rightarrow\frac{x}{2}=\frac{y}{4}=\frac{z}{6}\Rightarrow\frac{x^2}{2^2}=\frac{y^2}{4^2}=\frac{z^2}{6^2}\Rightarrow\frac{x^2}{4}=\frac{y^2}{16}=\frac{z^2}{36}\)

Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:

\(\frac{x^2}{4}=\frac{y^2}{16}=\frac{z^2}{36}=\frac{x^2+y^2+z^2}{4+16+36}=\frac{14}{56}=\frac{1}{4}\)

\(\frac{x^2}{4}=\frac{1}{4}\Rightarrow x=\sqrt{1}=\pm1\)

\(\frac{y^2}{16}=\frac{1}{4}\Rightarrow y=\sqrt{\frac{16}{4}}=\sqrt{4}=\pm2\)

\(\frac{z^2}{36}=\frac{1}{4}\Rightarrow z=\sqrt{\frac{36}{4}}=\sqrt{9}=\pm3\)

Vậy \(x=1;y=2;z=3\) hoặc \(x=-1;y=-2;z=-3\)

d.

Cách 1:

Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:

\(\frac{2x+1}{5}=\frac{3y-2}{7}=\frac{2x+3y-1}{6x}=\frac{2x+1+3y-2}{5+7}=\frac{2x+3y-1}{12}\)

\(6x=12\Rightarrow x=\frac{12}{6}=2\Rightarrow y=3\)

Vậy \(x=2\) và \(y=3\)

Cách 2:

Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:

\(\frac{2x+1}{5}=\frac{3y-2}{7}=\frac{2x+3y-1}{6x}=\frac{\left(2x+3y-1\right)-\left(2x+3y-1\right)}{5+7-6x}=0\)

\(2x+1=0\Rightarrow x=-\frac{1}{2}\)

\(3y-2=0\Rightarrow y=\frac{2}{3}\)

Vậy \(x=-\frac{1}{2}\) và \(y=\frac{2}{3}\)

Chúc bạn học tốt ^^

mk trả lời ở dưới rồi nhé

Giải:

Đặt \(\frac{a}{b}=\frac{c}{d}=k\)

\(\Rightarrow a=bk,c=dk\)

a) Ta có:
\(\frac{ab}{cd}=\frac{bkb}{dkd}=\frac{b^2}{d^2}\) (1)

\(\frac{\left(a+b\right)^2}{\left(c+d\right)^2}=\frac{\left(bk+b\right)^2}{\left(dk+d\right)^2}=\frac{\left[b\left(k+1\right)\right]^2}{\left[d\left(k+1\right)\right]^2}=\frac{b^2}{d^2}\) (2)

Từ (1) và (2) suy ra \(\frac{ab}{cd}=\frac{\left(a+b\right)^2}{\left(c+d\right)^2}\)

Vậy \(\frac{ab}{cd}=\frac{\left(a+b\right)^2}{\left(c+d\right)^2}\)

b) Ta có:
\(\frac{a^4+b^4}{c^4+d^4}=\frac{\left(bk\right)^4+b^4}{\left(dk\right)^4+d^4}=\frac{b^4.k^4+b^4}{d^4.k^4+d^4}=\frac{b^4.\left(k^4+1\right)}{d^4.\left(k^4+1\right)}=\frac{b^4}{d^4}\) (1)

\(\frac{\left(a+b\right)^4}{\left(c+d\right)^4}=\frac{\left(bk+b\right)^4}{\left(dk+d\right)^4}=\frac{\left[b\left(k+1\right)\right]^4}{\left[d\left(k+1\right)\right]^4}=\frac{b^4}{d^4}\) (2)

Từ (1) và (2) suy ra \(\frac{a^4+b^4}{c^4+d^4}=\frac{\left(a+b\right)^4}{\left(c+d\right)^4}\)

Ta có:\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}=\frac{a+b}{c+d}\)

\(\Rightarrow\left(\frac{a+b}{c+d}\right)^2=\frac{a}{c}\cdot\frac{b}{d}=\frac{ab}{cd}=\frac{\left(a+b\right)^2}{\left(c+d\right)^2}\)(đpcm)