a) \(\left|x\right|-\frac{7}{6}=\frac{9}{15}\)

=> \(\left|x\right|=\frac{9}{15}+\frac{7}{6}=\frac{53}{30}\)

=> \(\orbr{\begin{cases}x=\frac{53}{30}\\x=-\frac{53}{30}\end{cases}}\)

b) \(\left|x-\frac{4}{3}\right|=\frac{1}{6}\)

=> \(\orbr{\begin{cases}x-\frac{4}{3}=\frac{1}{6}\\x-\frac{4}{3}=-\frac{1}{6}\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{3}{2}\\x=\frac{7}{6}\end{cases}}\)

c) \(\left|x-\frac{4}{3}\right|-\frac{1}{3}=\frac{1}{2}\)

=> \(\left|x-\frac{4}{3}\right|=\frac{1}{2}+\frac{1}{3}\)

=> \(\left|x-\frac{4}{3}\right|=\frac{5}{6}\)

=> \(\orbr{\begin{cases}x-\frac{4}{3}=\frac{5}{6}\\x-\frac{4}{3}=-\frac{5}{6}\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{13}{6}\\x=\frac{1}{2}\end{cases}}\)

d) \(\frac{8}{3}-\left|\frac{7}{9}-x\right|=-\frac{1}{5}\)

=> \(\left|\frac{7}{9}-x\right|=\frac{43}{15}\)

=> \(\orbr{\begin{cases}\frac{7}{9}-x=\frac{43}{15}\\\frac{7}{9}-x=-\frac{43}{15}\end{cases}}\Rightarrow\orbr{\begin{cases}x=-\frac{94}{45}\\x=\frac{164}{45}\end{cases}}\)

e) \(\left|x-\left(\frac{1}{4}\right)^2\right|-\frac{25}{64}=0\)

=> \(\left|x-\frac{1}{16}\right|=\frac{25}{64}\)

=> \(\orbr{\begin{cases}x-\frac{1}{16}=\frac{25}{64}\\x-\frac{1}{16}=-\frac{25}{64}\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{29}{64}\\x=-\frac{21}{64}\end{cases}}\)

f) \(\left(x-\frac{1}{4}\right)^2+\frac{17}{64}=\frac{21}{32}\)

=> \(\left(x-\frac{1}{4}\right)^2=\frac{25}{64}\)

=> \(\left(x-\frac{1}{4}\right)^2=\left(\frac{5}{8}\right)^2\)

=> \(\orbr{\begin{cases}x-\frac{1}{4}=\frac{5}{8}\\x-\frac{1}{4}=-\frac{5}{8}\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{7}{8}\\x=-\frac{3}{8}\end{cases}}\)

\(A=1.\left(x+y\right)\left(x^2+y^2\right)...\left(x^{64}+y^{64}\right)\)

\(=\left(x-y\right)\left(x+y\right)\left(x^2+y^2\right)...\left(x^{64}+y^{64}\right)\)

\(=\left(x^2-y^2\right)\left(x^2+y^2\right)\left(x^4+y^4\right)...\left(x^{64}+y^{64}\right)\)

\(=\left(x^4-y^4\right)...\left(x^{64}+y^{64}\right)\)

\(=...=\left(x^{64}-y^{64}\right)\left(x^{64}+y^{64}\right)=x^{128}-y^{128}\)

\(1,\Rightarrow4^{x+1}=4^{3x}\\ \Rightarrow x+1=3x\\ \Rightarrow2x=1\\ \Rightarrow x=\dfrac{1}{2}\\ 2,\Rightarrow5x-2x=10+10x\\ \Rightarrow7x=-10\\ \Rightarrow x=-\dfrac{10}{7}\)

1. 4^{x + 1} = 64^x

<=> 4^{x + 1} = 4^3x

<=> x + 1 = 3x

<=> 1 = 2x

<=> \(x=\dfrac{1}{2}\)

2. \(\dfrac{x}{2}-\dfrac{x}{5}=1+x\)

<=> \(\dfrac{5x}{10}-\dfrac{2x}{10}=\dfrac{10}{10}+\dfrac{10x}{10}\)

<=> 5x - 2x = 10 + 10x

<=> 5x - 2x - 10x = 10

<=> -7x = 10

<=> \(x=\dfrac{-10}{7}\)

5 + ( x + 27 ) = 64

a) \(\Rightarrow x+27=59\Rightarrow x=32\)

b) \(\Rightarrow x-2=39\Rightarrow x=41\)

c) \(\Rightarrow x+5=-322\Rightarrow x=-327\)

d) \(\Rightarrow5x=35\Rightarrow x=7\)

e) \(\Rightarrow4\left(x-5\right)=56\Rightarrow x-5=14\Rightarrow x=19\)

f) \(\Rightarrow15+x=37\Rightarrow x=22\)

g) \(\Rightarrow7\left(13-x\right)=35\Rightarrow13-x=5\Rightarrow x=8\)

h) \(\Rightarrow10\left(x+1\right)=100\Rightarrow x+1=10\Rightarrow x=9\)

`a,`\(2^x -15= 2^4+1\)

`-> 2^x-15=17`

`-> 2^x=17+15`

`-> 2^x=32`

`-> 2^x=2^5`

`-> x=5`

`b,` Có phải đề là \(\dfrac{x+1}{65}+\dfrac{x+2}{64}=\dfrac{x+3}{63}+\dfrac{x+4}{62}\) ?

`=>`\(\dfrac{x+1}{65}+1+\dfrac{x+2}{64}+1=\dfrac{x+3}{63}+1+\dfrac{x+4}{62}+1\)

`=>`\(\dfrac{x+1+65}{65}+\dfrac{x+2+64}{64}-\dfrac{x+3+63}{63}-\dfrac{x+4+62}{62}=0\)

`=>`\(\dfrac{x+66}{65}+\dfrac{x+66}{64}-\dfrac{x+66}{63}-\dfrac{x+66}{62}=0\)

`=>`\(\left(x+66\right)\left(\dfrac{1}{65}+\dfrac{1}{64}-\dfrac{1}{63}-\dfrac{1}{62}\right)=0\)

Mà `1/65+1/64-1/63-1/62 \ne 0`

`-> x+66=0`

`-> x=-66`

a: =>2^x=2^4+16=32

=>x=5

b: Sửa đề: \(\dfrac{x+1}{65}+\dfrac{x+2}{64}=\dfrac{x+3}{63}+\dfrac{x+4}{62}\)

=>\(\left(\dfrac{x+1}{65}+1\right)+\left(\dfrac{x+2}{64}+1\right)=\left(\dfrac{x+3}{63}+1\right)+\left(\dfrac{x+4}{62}+1\right)\)

=>x+66=0

=>x=-66

a) Ta có: \(\left(x-1\right)^{x+2}-\left(x-1\right)^{x+4}=0\)

\(\Leftrightarrow\left(x-1\right)^x\cdot\left(x-1\right)^2-\left(x-1\right)^x\cdot\left(x-1\right)^4=0\)

\(\Leftrightarrow\left(x-1\right)^{x+2}\cdot\left[1-\left(x-1\right)^2\right]=0\)

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

b) Ta có: \(\dfrac{1}{4}\cdot\dfrac{2}{6}\cdot\dfrac{3}{8}\cdot\dfrac{4}{10}\cdot\dfrac{5}{15}\cdot...\cdot\dfrac{30}{62}\cdot\dfrac{31}{64}=2x\)

\(\Leftrightarrow2x=\dfrac{1}{64}\)

hay \(x=\dfrac{1}{128}\)

Cài vào điện thoại phần mềm photo math, nó sẽ hỗ trợ bạn các tính nhanh luôn, cách dùng thid seach bác google hen.

bn ơi hãy lấy máy tính

mà tính đó bna j

tính thế này thì mất

mấy tiếng

Câu 1:2x31x12+4x46x42+8x27x3=9120

Câu 2 :........................................=12280

k and kb nha