\(2x+\frac{3}{7}=\frac{17}{7}\Leftrightarrow2x=\frac{14}{7}\Leftrightarrow x=1\)

\(\left|2x+1\right|=3\)

TH1 : \(2x+1=3\Leftrightarrow x=1\)

TH2 : \(2x+1=-3\Leftrightarrow x=-2\)

\(2x+\frac{3}{7}=\frac{17}{7}\Leftrightarrow2x=\frac{17}{7}-\frac{3}{7}=2\Leftrightarrow x=2\)

\(\left|2x+1\right|=3\Leftrightarrow\orbr{\begin{cases}2x+1=3\\2x+1=-3\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=1\\x=-2\end{cases}}\)

Trả lời :

a, (x + 1)³ = 8² 

=> (x + 1)³ = 64 = 4³

=> x + 1 = 4

=> x = 3

b, 9^{2x - 3} = 27^{3x + 1}

=> 3^{4x - 6} = 3^{9x + 3}

=> 4x - 6 = 9x + 3

=> 4x - 9x = 3 + 6

=> - 5x = 9

=> x = \(\frac{-9}{5}\)

Bài làm :

\(A=8^{12}+8^{4x+2}\)

\(=8^{12}+8^{4x}.8^2\)

\(=8^2.\left(8^{10}.8^{4x}\right)\)

\(=64.\left(8^{10}.8^{4x}\right)⋮64\)

=> đpcm 

Học tốt nhé

Sửa lại bài làm chút

Bài làm :

\(A=8^{12}+8^{4x+2}\)

\(=8^{12}+8^{4x}.8^2\)

\(=8^2.\left(8^{10}+8^{4x}\right)\)

\(=64.\left(8^{10}+8^{4x}\right)⋮64\)

=> đpcm

Học tốt nhé

\(\left(0,75\right)^3\cdot1024=\left(0,75\right)^3\cdot8^3\cdot2=\left(0,75\cdot8\right)^3\cdot2=6^3\cdot2=216\cdot2=432\)

Ta có : 6x² = 11x - 3

=> 6x² - 11x + 3 = 0

=> 6x² - 2x - 9x + 3 = 0

=> 2x(3x - 1) - 3(3x - 1) = 0

=> (2x - 3)(3x - 1) = 0

=> \(\orbr{\begin{cases}2x-3=0\\3x-1=0\end{cases}}\Rightarrow\orbr{\begin{cases}2x=3\\3x=1\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{3}{2}\\x=\frac{1}{3}\end{cases}}\)

Vậy \(x\in\left\{\frac{3}{2};\frac{1}{3}\right\}\)

Ta có : 6x² = 11x - 3

\(\Leftrightarrow\)6x² - 11x + 3 = 0

\(\Leftrightarrow\) 6x² - 9x - 2x + 3 = 0

\(\Leftrightarrow\) 3x(2x - 3) - (2x - 3) = 0

\(\Leftrightarrow\)(3x - 1)(2x - 3) = 0

\(\Leftrightarrow\) x \(\in\){ 1/3 ; 3/2}

+) \(40\div\left\{\left[11+\left(26-27\right)\right]\times2\right\}\)

\(=40\div\left(10\times2\right)=2\)

+) \(\left|\frac{3}{5}\right|\times4\frac{1}{7}+\left|\frac{-6}{7}\right|\times\frac{3}{5}\)

\(=\frac{3}{5}\times\frac{29}{7}+\frac{6}{7}\times\frac{3}{5}=\frac{3}{5}\times\left(\frac{29}{7}+\frac{6}{7}\right)=\frac{3}{5}\times5=3\)

a. \(\left(\frac{1}{3}x\right):\frac{2}{3}=1\frac{3}{4}:\frac{2}{5}\)

\(\left(\frac{1}{3}x\right):\frac{2}{3}=\frac{35}{8}\)

\(\frac{1}{3}x=\frac{35}{8}.\frac{2}{3}=\frac{35}{12}\)

\(x=\frac{35}{12}.3=\frac{35}{4}\)

b) 4,5:0,3=2,25:(0,1.x)

\(\frac{45}{10}:\frac{10}{3}=\frac{225}{100}:\left(\frac{1}{10}x\right)\)

\(15=\frac{9}{4}:\left(\frac{1}{10}x\right)\)

\(\frac{9}{4}:\left(\frac{1}{10}x\right)=15\)

\(\frac{1}{10}x=\frac{9}{4}.\frac{1}{15}=\frac{3}{20}\)

\(x=\frac{3}{20}.10=\frac{3}{2}\)

c) 8:\(\left(\frac{1}{4}x\right)=2:0,02\)

8:\(\left(\frac{1}{4}x\right)\)=2:\(\frac{2}{100}\)

8:\(\left(\frac{1}{4}x\right)=2.\frac{100}{2}=100\)

\(\frac{1}{4}x=\frac{8}{100}\)

\(x=\frac{8}{100}.4=\frac{8}{25}\)

d)\(3:2\frac{1}{4}=\frac{3}{4}:\left(6x\right)\)

\(3.\frac{9}{4}=\frac{3}{4}:\left(6x\right)\)

\(\frac{3}{4}:\left(6x\right)=\frac{27}{4}\)

\(6x=\frac{3}{4}.\frac{4}{27}=\frac{1}{9}\)

\(x=\frac{1}{9}.\frac{1}{6}=\frac{1}{54}\)

 a.=> 1/2 . x = 35/8

 => x = 31/8

 b. => 15 = 22,5 . x

=> x = 2/3

c. => 32 . x =  100

=>x=25/8 

d. 4/3= 1/8 . x 

x = 32/3

Ta có: -1/3=-4/12=-8/24=-16/48

           -1/4=-3/12=-6/24=-12/48

Ta gọi x là tập hợp các số hữu tỉ cần tìm

       =>-16/48<x<-12/48

Suy ra x={-15/48;-14/48;-13/48}

Vậy các số cần tìm là -5/16;-7/24;-13/48

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

Đặt \(\frac{a}{c}=\frac{b}{d}=k\Rightarrow\hept{\begin{cases}a=ck\\b=dk\end{cases}}\)(*)

Khi đó \(\frac{a+2c}{a-c}=\frac{ck+2c}{ck-c}=\frac{c\left(k+2\right)}{c\left(k-1\right)}=\frac{k+2}{k-1}\)(1) ; 

Lại có \(\frac{b+2d}{b-d}=\frac{dk+2d}{dk-d}=\frac{d\left(k+2\right)}{d\left(k-1\right)}=\frac{k+2}{k-1}\)(2)

Từ (1)(2) = > \(\frac{a+2c}{a-c}=\frac{b+2d}{b-d}\left(\text{đpcm}\right)\)

Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow\hept{\begin{cases}a=kb\\c=kd\end{cases}}\)

\(\Rightarrow VT=\frac{a+2c}{a-c}=\frac{kb+2kd}{kb-kd}=\frac{k\left(b+2d\right)}{k\left(b-d\right)}=\frac{b+2d}{b-d}=VP\)

=> đpcm

a) \(\hept{\begin{cases}\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}=\frac{5a}{5c}=\frac{3b}{3d}=\frac{5a-3b}{5c-3d}\\\frac{a}{c}=\frac{b}{d}\Rightarrow\frac{3a}{3c}=\frac{2b}{2d}=\frac{3a+2b}{3c+2d}\end{cases}}\)

\(\Rightarrow\frac{5a-3b}{5c-3d}=\frac{3a+2b}{3c+2d}\)

\(\Rightarrow\frac{5a-3b}{3a+2b}=\frac{5c-3d}{3c+2d}\)

b) Chứng minh tương tự 

ko biet nghen