\(\dfrac{3x+2}{5x+7}=\dfrac{3x-1}{5x-3}\)

\(\Leftrightarrow\left(3x+2\right)\left(5x-3\right)=\left(5x+7\right)\left(3x-1\right)\)

\(\Leftrightarrow\left(3x+2\right)\left(5x-3\right)-\left(5x+7\right)\left(3x-1\right)=0\)

\(\Leftrightarrow\left(15x^2-9x+10x-6\right)-\left(15x^2-5x+21x-7\right)=0\)

\(\Leftrightarrow15x^2-9x+10x-6-15x^2+5x-21x+7=0\)

\(\Leftrightarrow-15x+1=0\)

\(\Leftrightarrow-15x=-1\)

\(\Leftrightarrow x=\dfrac{-1}{-15}=\dfrac{1}{15}\)

Vậy \(x=\dfrac{1}{15}\)

-1+3+(-5)+7+...+x=100

=2+2+...+2=100

2.50=100

Co 50 cap suy ra co 100 so

So thu 100(x) la 1+(100-1).2=......

a) 3\(^x\) . 3 = 243

3\(^x\) = 81

3\(^x\) = 3\(^4\)

=> = 4 

b) 64 . 4\(^x\) = 16\(^8\)

4\(^3\) . 4\(^x\) = 4\(^{16}\)

4\(^{3+x}\) = 4\(^{16}\)

=> 3 + x = 16

x = 13

Học tốt

Đúng thì t cho mk

a) 3^x . 3 = 243 

243 = 3^5 

3^4 . 3 = 234 

nên x = 4 

b) 64 . 4^x = 16^8

4^3 . 4^x = ( 4^2)^8 

4^3 . 4^x = 4^16 

4^(4+x) = 4^16 

x = 4^16 : 4^4 

x = 4^12 

x = 12

a) 3^x . 3 = 243

3^x = 243 : 3

3^x = 81

3^x = 3⁴

\(\Rightarrow\)x = 4

Vậy x = 4

b ) 64 . 4^x = 16⁸

4^x = 16⁸ : 64

4^x = 4¹⁶ : 4 ³

4^x = 4^{16 - 3}

4^x = 4¹³

\(\Rightarrow\)x = 13

Vậy x = 13

#THPT - TQY

A

mk nghĩ zậy

mình nghĩ là D

đúng thì k tớ nhé

Ừ thank

\(=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{7\cdot8}\)

\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-...-\frac{1}{8}\)

\(=\frac{1}{1}-\frac{1}{8}=\frac{7}{8}\)

\(=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{3.7}+\frac{1}{7.8}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{7}-\frac{1}{8}\)

\(=1-\frac{1}{8}+0+0+...+0\)

\(=\frac{7}{8}\)

= \(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{6.7}+\frac{1}{7.8}\)
= \(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}\)
= \(1-\frac{1}{8}=\frac{7}{8}\)

\(\frac{x}{5}=\frac{y}{6};\frac{y}{8}=\frac{z}{7}\Rightarrow\frac{x}{40}=\frac{y}{48}=\frac{z}{42}\)

Áp dụng tính chất của DTSBN , ta có :

( bn tự lm )

ta có : \(\frac{x}{5}=\frac{y}{6}\Rightarrow\frac{x}{20}=\frac{y}{24}\left(1\right)\)

\(\frac{y}{8}=\frac{z}{7}\Rightarrow\frac{y}{24}=\frac{z}{21}\left(2\right)\)

từ (1);(2) ta có : \(\frac{x}{20}=\frac{y}{24}=\frac{z}{21}=\frac{x+y+z}{20+24+21}=\frac{69}{65}\)( AD t/c của dãy tỉ số = nhau)

\(\frac{x}{20}=\frac{69}{65}\Rightarrow x=\frac{60}{65}.20=\frac{240}{13}\)

\(\frac{y}{24}=\frac{69}{65}\Rightarrow y=\frac{69}{65}.24=\frac{1656}{65}\)

\(\frac{z}{21}=\frac{69}{65}\Rightarrow z=\frac{69}{65}.21=\frac{1449}{65}\)

vậy (x,y,z)= \(\left(\frac{240}{13},\frac{1656}{65},\frac{1449}{65}\right)\)

Áp dụng tính chất của DTSBN ta có :\(\frac{x}{40}=\frac{y}{48}=\frac{z}{42}=\frac{x+y+z}{40+48+42}=\frac{69}{120}=\frac{23}{40}\)

\(\Rightarrow\)\(\frac{x}{40}=\frac{23}{40}\Rightarrow x=23\)

\(\frac{y}{48}=\frac{23}{40}\Rightarrow\frac{23}{40}.48=\frac{138}{5}\)

\(\frac{z}{42}=\frac{23}{40}\Rightarrow\frac{23}{40}.42=\frac{483}{20}\)

Vậy \(x=23\); \(y=\frac{138}{5}\); \(z=\frac{483}{20}\)

\(\dfrac{7}{9}< \dfrac{x}{y}< \dfrac{7}{8}\)

\(\dfrac{56}{72}< \dfrac{x}{y}< \dfrac{63}{72}\)

\(\dfrac{x}{y}=\left\{\dfrac{57}{72};\dfrac{58}{72};\dfrac{59}{72};\dfrac{60}{72};\dfrac{61}{72};\dfrac{62}{72}\right\}\)

\(\left(8-5x\right)\left(x+2\right)+4\left(x-2\right)\left(x+1\right)=\left(x-2\right)\left(x+2\right)\)

\(\Rightarrow8x+16-5x^2-10x+4x^2+4x-8x-8=x^2-4\)

\(\Rightarrow-6x-x^2-8-x^2+4=0\)

\(\Rightarrow-6x-2x^2-4=0\)

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

\(\Rightarrow\left(x+1,5\right)^2-0,25=0\)

\(\Rightarrow\left(x+2\right)\left(x+1\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x+2=0\\x+1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-2\\x=-1\end{cases}}}\)

a)-5/2

b)1

a)-5/2

b)1

a)-5/2

b)1