bai nao vay

\(A=1.\left(-1\right)+3.\left(-1\right)^2+5.\left(-1\right)^3.7+\left(-1\right)^4+9.\left(-1\right)^5\)

\(A=1.\left(-1\right)+3.1+5.\left(-1\right).7+1+9.\left(-1\right)\)

\(A=\left(-1\right)+3+\left(-5\right).7+1+\left(-9\right)\)

\(A=-1+3-35+1-9\)

\(A=-41\)

= 0 nha

1 - 2 - 3 + 4 + 5 - 6 - 7 + 8 + 9 - 10 - 11 + 12 + ... + 61 - 62 - 63 + 64 ( 64 số )

= ( 1 - 2 - 3 + 4 ) + ( 5 - 6 - 7 + 8 ) + ( 9 - 10 - 11 + 12 ) + ... + ( 61 - 62 - 63 + 64 )   ( 16 nhóm )

= 0 + 0 + 0 + ... + 0         ( 16 số 0 )

= 0 . 16

= 0 

a ) bằng 0.

b ) bằng 65.

a. (45-63+18) x (1+2+3+4+5+6+7+8+9)

= 0 x (1+2+3+4+5+6+7+8+9) = 0

b. 60-61+62-63+64-65+66-67+68-69+70

= 60 + (-61-69)+(62+68)+(-63-67)+(64+66)-65+70

= 60 + (-130)+130+(-130)+130-65-70

= 60 + (-130+130) + (-130+130)-65+70

= 60 - 65 + 70 = 65

(sai rồi)

3/4 - 5/7 = 21/28 - 20/28 = 1/28

64 - 61 = 3

Đs:

^.^

3/4 - 5/7 = 1/28

64 - 61 = 3.

3/4 - 5/7 = 1/28

64 - 61 = 3

1: \(2^x=64\)

=>\(x=log_264=6\)

2: \(2^x\cdot3^x\cdot5^x=7\)

=>\(\left(2\cdot3\cdot5\right)^x=7\)

=>\(30^x=7\)

=>\(x=log_{30}7\)

3: \(4^x+2\cdot2^x-3=0\)

=>\(\left(2^x\right)^2+2\cdot2^x-3=0\)

=>\(\left(2^x\right)^2+3\cdot2^x-2^x-3=0\)

=>\(\left(2^x+3\right)\left(2^x-1\right)=0\)

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

=>\(2^x=1\)

=>x=0

4: \(9^x-4\cdot3^x+3=0\)

=>\(\left(3^x\right)^2-4\cdot3^x+3=0\)

Đặt \(a=3^x\left(a>0\right)\)

Phương trình sẽ trở thành:

\(a^2-4a+3=0\)

=>(a-1)(a-3)=0

=>\(\left[{}\begin{matrix}a-1=0\\a-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}a=1\left(nhận\right)\\a=3\left(nhận\right)\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}3^x=1\\3^x=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=0\end{matrix}\right.\)

5: \(3^{2\left(x+1\right)}+3^{x+1}=6\)

=>\(\left[3^{x+1}\right]^2+3^{x+1}-6=0\)

=>\(\left(3^{x+1}\right)^2+3\cdot3^{x+1}-2\cdot3^{x+1}-6=0\)

=>\(3^{x+1}\left(3^{x+1}+3\right)-2\left(3^{x+1}+3\right)=0\)

=>\(\left(3^{x+1}+3\right)\left(3^{x+1}-2\right)=0\)

=>\(3^{x+1}-2=0\)

=>\(3^{x+1}=2\)

=>\(x+1=log_32\)

=>\(x=-1+log_32\)

6: \(\left(2-\sqrt{3}\right)^x+\left(2+\sqrt{3}\right)^x=2\)
=>\(\left(\dfrac{1}{2+\sqrt{3}}\right)^x+\left(2+\sqrt{3}\right)^x=2\) 

=>\(\dfrac{1}{\left(2+\sqrt{3}\right)^x}+\left(2+\sqrt{3}\right)^x=2\)

Đặt \(b=\left(2+\sqrt{3}\right)^x\left(b>0\right)\)

Phương trình sẽ trở thành:

\(\dfrac{1}{b}+b=2\)

=>\(b^2+1=2b\)

=>\(b^2-2b+1=0\)

=>(b-1)²=0

=>b-1=0

=>b=1

=>\(\left(2+\sqrt{3}\right)^x=1\)

=>x=0

7: ĐKXĐ: \(x^2+3x>0\)

=>x(x+3)>0

=>\(\left[{}\begin{matrix}x>0\\x< -3\end{matrix}\right.\)
\(log_4\left(x^2+3x\right)=1\)

=>\(x^2+3x=4^1=4\)

=>\(x^2+3x-4=0\)

=>(x+4)(x-1)=0

=>\(\left[{}\begin{matrix}x+4=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\left(nhận\right)\\x=-4\left(nhận\right)\end{matrix}\right.\)

ko xem dc de ban oi :/

2

\(\dfrac{1}{4}.\dfrac{2}{6}.\dfrac{3}{8}.\dfrac{4}{10}.\dfrac{5}{12}.....\dfrac{30}{62}.\dfrac{31}{64}=2x\)

\(\dfrac{1.2.3.4.5.....30.31}{\left(2.2\right)\left(2.3\right)\left(2.4\right)\left(2.5\right)\left(2.6\right).....\left(2.31\right)\left(2.32\right)}=2x\)

\(\dfrac{1\left(2.3.4.5....30.31\right)}{32\left(2.3.4.5.....31\right).2^{31}}=2x\)

\(\dfrac{1}{2^5.2^{31}}=2x\Rightarrow2x=\dfrac{1}{2^{36}}\Rightarrow x=\dfrac{1}{2^{36}}\div2=\dfrac{1}{2^{37}}\)

Vậy x = \(\dfrac{1}{2^{37}}\)

\(\dfrac{1}{4}\cdot\dfrac{2}{6}\cdot\dfrac{3}{8}\cdot\dfrac{4}{10}\cdot...\cdot\dfrac{30}{62}\cdot\dfrac{31}{64}=2x\\ \dfrac{1}{2\cdot2}\cdot\dfrac{2}{2\cdot3}\cdot\dfrac{3}{2\cdot4}\cdot\dfrac{4}{2\cdot5}\cdot...\cdot\dfrac{30}{2\cdot31}\cdot\dfrac{31}{2\cdot32}=2x\\ \dfrac{1}{2}\cdot\left(\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot\dfrac{3}{4}\cdot\dfrac{4}{5}\cdot...\cdot\dfrac{30}{31}\cdot\dfrac{31}{32}\right)=2x\\ \dfrac{1}{2}\cdot\left(\dfrac{1\cdot2\cdot3\cdot4\cdot...\cdot30\cdot31}{2\cdot3\cdot4\cdot5\cdot...\cdot31\cdot32}\right)=2x\\ \dfrac{1}{2}\cdot\dfrac{1}{32}=2x\\ 2x=\dfrac{1}{64}\\ x=\dfrac{1}{64}:2\\ x=\dfrac{1}{128}\)