mk ko biet

1) 4⁶

2) 4⁶⁰ 

3) 8¹⁶

4) 16¹⁰

5) 32⁸

\(\dfrac{2^{4-x}}{16^5}=32^6\)

\(\Leftrightarrow2^{4-x}=16^5.32^6\)

\(\Rightarrow2^{4-x}=\left(2^4\right)^5.\left(2^5\right)^6\)

\(\Rightarrow2^{4-x}=2^{20}.2^{30}\)

\(\Rightarrow2^{4-x}=2^{50}\)

\(\Rightarrow4-x=50\)

\(\Rightarrow x=-46\)

Vậy...

\(\dfrac{2^{4-x}}{16^5}=32^6\Rightarrow\dfrac{2^{4-x}}{2^{20}}=2^{30}\Rightarrow2^{4-x}=2^{50}\Rightarrow x=-46\)

\(32^{-x}\times16^x=1024\\ 16^{-x}\times2^{-x}\times16^x=2^{10}\\ \left(16^{-x}\times16^x\right)\times2^{-x}=2^{10}\\ 2^{-x}=2^{10}\\ \Rightarrow-x=10\\ x=-10\) \(\left(\dfrac{4}{5}\right)^{2x+7}=\dfrac{625}{256}\\ \left(\dfrac{4}{5}\right)^{2x+7}=\left(\dfrac{4}{5}\right)^{-4}\\ \Rightarrow2x+7=4\\ 2x=-3\\ x=-1,5\)

Ta có: 

 2²+4²+6²+...+20²

= 2.1²+2.2²+...+2.10²

=2.(1²+...+10²)=2.385=770

Ta có:

 1²+3.1²+3.2²+3.3²+...+3.10²

=1²+3.(1²+2²+...+10²)

=1+3.385=1156

\(4^x=32^{24}\)

\(2^{2x}=2^{120}\)

\(2x=120\)

\(x=60\)

\(8^x=16^{12}\)

\(2^{3x}=2^{48}\)

\(3x=48\)

\(x=16\)

a.

 (25)-n. (24)n = 1024
 2-5n. 24n = 210
 2-n = 210
 n = -10

1) x^​2019 ​-1 =0 => x^​2019 ​=1 => x=1

​

2) x⁴ -16 =0 => x^​4 ​=16 => x⁴ = 4⁴ hoặc (-4)⁴ => x = 4 hoặc -4

\(\frac{625}{5^n}=5^3\)

\(\Leftrightarrow5^3\cdot5^n=625\)

\(\Leftrightarrow5^{3+n}=625\)

\(\Leftrightarrow5^{3+n}=5^4\)

\(\Leftrightarrow3+n=4\Leftrightarrow n=1\)

\(32< 2^x< 512\)

\(\Leftrightarrow2^5< 2^x< 2^9\)

\(\Leftrightarrow5< x< 9\)

\(\Leftrightarrow x\in\left\{6;7;8\right\}\)

Giải:

a) Đặt \(\frac{x}{10}=\frac{y}{6}=k\)

\(\Rightarrow x=10k,y=6k\)

Mà \(xy=60\)

\(\Rightarrow10k6k=60\)

\(\Rightarrow60k^2=60\)

\(\Rightarrow k^2=1\)

\(\Rightarrow k=\pm1\)

+) \(k=1\Rightarrow x=10;y=6\)

+) \(k=-1\Rightarrow x=-10;y=-6\)

Vậy cặp số \(\left(x;y\right)\) là \(\left(10;6\right);\left(-10;-6\right)\)

b) Hình như đề sai !!!

c) Giải:

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

\(\frac{x^2}{9}=\frac{y^2}{16}=\frac{x^2+y^2}{9+16}=\frac{100}{25}=4\)

+) \(\frac{x^2}{9}=4\Rightarrow x^2=36\Rightarrow x=\pm6\)

+) \(\frac{y^2}{16}=4\Rightarrow y^2=64\Rightarrow y=\pm8\)

( x, y cùng dấu )

Vậy cặp số ( x; y ) là ( 6; 8 ) ; ( -6; -8 )
 

b) $$\dfrac{x-1}2 = \dfrac{y-2}3 = \dfrac{z-3}3$$

$$\iff \dfrac{x-1}2 = \dfrac{2y-4}{6} = \dfrac{3z - 9}9$$

Áp dụng tính chất dãy tỉ số bằng nhau

$$\dfrac{x-1}2 = \dfrac{2y-4}{6} = \dfrac{3z - 9}9 = \dfrac{(x-1) - (2y-4) + (3z - 9)}{2 - 6 + 9} = \dfrac{(x - 2y + 3z) - 6}5 = \dfrac{16 - 6}5 = 2$$

+) $\dfrac{x-1}2 = 2 \iff x = 5$

+) $\dfrac{2y-4}6 = 2 \iff y = 8$

+) $\dfrac{3z-9}9 = 2 \iff z = 9$