(0.25-0.3x)x1/3-1/4=-5/16

(0.25-0.3x)x1/3=-1/16  (làm tắt)

0.25-0.3x=-1/16:1/3

0.25-0.3x=-3/16

0.3x=0.4375 =>x=35/24

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 \(x=\dfrac{1}{128}\)

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

\(\Leftrightarrow2^x=\dfrac{1\cdot2\cdot3\cdot4\cdot...\cdot31}{2\cdot\left(2\cdot3\cdot4\cdot...\cdot31\right)\cdot64}\)

\(\Leftrightarrow2^x=\dfrac{1}{2}\cdot\dfrac{1}{64}=\dfrac{1}{128}\)
\(\Leftrightarrow2^x=\dfrac{1}{2^6}\)

\(\Leftrightarrow2^{x+6}=1\)

\(\Leftrightarrow x+6=0\)

hay x=-6

Vậy: x=-6

`1/4 . 2/6 . 3/8 ... . 30/62 .31/64 =2^x`

`-> (1.2.3....30.31)/(4.6.8....62.64)=2^x`

`-> (1.(2.3...31))/(2.(2.3.4...31).32)=2^x`

`-> 1/(2.32)=2^x`

`-> 1/64=2^x`

`-> 1/(2^6)=2^x`

`-> x=-6`.

\(\dfrac{3}{2}x+\dfrac{3}{7}=-\dfrac{4}{5}\)

\(\Leftrightarrow\dfrac{3}{2}x=-\dfrac{43}{35}\)

\(\Leftrightarrow x=-\dfrac{86}{105}\)

Vậy \(x=-\dfrac{86}{105}\)

\(-\dfrac{11}{12}x+0,25=\dfrac{5}{6}\)

\(\Leftrightarrow-\dfrac{11}{12}x+\dfrac{1}{4}=\dfrac{5}{6}\)

\(\Leftrightarrow-\dfrac{11}{12}x=\dfrac{7}{12}\)

\(\Leftrightarrow x=-\dfrac{7}{11}\)

Vậy \(x=-\dfrac{7}{11}\)

\(\left(x-2\right)^2=1\)

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

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

\(\Leftrightarrow\left(x-3\right)\left(x-1\right)=0\)

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

Vậy x = {3; 1}\(\left(2x-1\right)^3=-8\)

\(\Leftrightarrow2x-1=-2\)

\(\Leftrightarrow2x=-1\)

\(\Leftrightarrow x=-\dfrac{1}{2}\)

Vậy \(x=-\dfrac{1}{2}\)

       (x - 2)² = 1

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

Vậy x = 3; 1

(2x - 1)³ = -8

<=> 2x - 1 = -2

<=> 2x = -1

<=> x = \(\dfrac{-1}{2}\)

Vậy x = \(\dfrac{-1}{2}\)

\(|\) 2,5 - x\(|\) = 1,3

1,6- \(|\)x - 0,2 \(|\)= 0

13^x = 169

\(\dfrac{-2}{x}\)=\(\dfrac{-x}{8\dfrac{ }{25}}\)

\(\dfrac{x}{-15}\)=\(\dfrac{-60}{x}\)

\(|\)x\(|\)+0,573=2

TÌM X:

a) 2x - 3 = \(\frac{1}{2}\)

2x = \(\frac{1}{2}+3\)

2x = \(\frac{7}{2}\)

x = 2 : \(\frac{7}{2}\)

x = 2 . \(\frac{2}{7}\)

x = \(\frac{4}{7}\)

b) /x+1/ = 0.25

/x+1/ = \(\frac{1}{4}\)

\(\orbr{\begin{cases}x+1=\frac{1}{4}\\x+1=-\frac{1}{4}\end{cases}}\)

\(\orbr{\begin{cases}x=\frac{1}{4}-1\\x=-\frac{1}{4}-1\end{cases}}\)

\(\orbr{\begin{cases}x=-\frac{3}{4}\\x=-\frac{5}{4}\end{cases}}\)

c) 32 : 2x = 2

\(2x=32:2\)

\(2x=16\)

\(x=16:2\)

\(x=8\)

~GOOD STUDY~

a) 2x-3=1/2

=> 2x=1/2+3

=> 2x=7/2

=> x=7/2:2

=> x=7/4

b) |x+1|=0.25

=> \(\orbr{\begin{cases}x+1=0,25\\x+1=-0,25\end{cases}}\)=>\(\orbr{\begin{cases}x=0,25-1\\x=-0,25-1\end{cases}}\)=>\(\orbr{\begin{cases}x=-0,75\\x=-1,25\end{cases}}\)

b) Ta có: \(3+\left(x-5\right)=2\left(3x-2\right)\)

\(\Leftrightarrow3+x-5=6x-4\)

\(\Leftrightarrow x-2-6x+4=0\)

\(\Leftrightarrow-5x+2=0\)

\(\Leftrightarrow-5x=-2\)

\(\Leftrightarrow x=\dfrac{2}{5}\)

Vậy: \(S=\left\{\dfrac{2}{5}\right\}\)

c) Ta có: \(2\left(x-0.5\right)+3=0.25\left(4x-1\right)\)

\(\Leftrightarrow2x-1+3=x-\dfrac{1}{4}\)

\(\Leftrightarrow2x+2-x+\dfrac{1}{4}=0\)

\(\Leftrightarrow x+\dfrac{9}{4}=0\)

\(\Leftrightarrow x=-\dfrac{9}{4}\)

Vậy: \(S=\left\{-\dfrac{9}{4}\right\}\)

d) Ta có: \(2\left(x-\dfrac{1}{4}\right)-4=-6\left(-\dfrac{1}{3}x+0.5\right)+2\)

\(\Leftrightarrow2x-\dfrac{1}{2}-4=2x-3+2\)

\(\Leftrightarrow2x-\dfrac{9}{2}=2x-1\)

\(\Leftrightarrow2x-2x=-1+\dfrac{9}{2}\)

\(\Leftrightarrow0x=\dfrac{7}{2}\)(vô lý)

Vậy: \(S=\varnothing\)

b. 

→ -2 + x = 6x - 4

→ -2 + 4 = 6x - x

→ 2 = 5x

→ x = \(\dfrac{2}{5}\) 

Vậy, phương trình có tập nghiệm S = {\(\dfrac{2}{5}\)}

`1/4. 2/6. 3/8. 4/10........ 30/62. 31/64=2^x`

`=>\underbrace{1/2. 1/2. 1/2. 1/2..............1/2}_{\text{30 số 2}}. 1/64=2^x`

`=>(1/2)^{30}.(1/2)^{6}=2^x`

`=>(1/2)^{36}=2^x`

`=>2^{-36}=2^x`

`=>x=-36`

Vậy `x=-36`

các bạn thêm bằng 2^x hộ mk nha , mk quên

Có: \(\frac{1}{4}.\frac{2}{6}.\frac{3}{8}.\frac{4}{10}...\frac{30}{62}.\frac{31}{64}=\frac{1}{2.2}.\frac{2}{2.3}.\frac{3}{2.4}...\frac{30}{2.31}.\frac{31}{2.32}=\frac{1}{2}.\frac{1}{2}.\frac{1}{2}...\frac{1}{2}.\frac{1}{2}.\frac{1}{32}\)

\(=\frac{1}{2^{31}.2^5}=\frac{1}{2^{36}}=2^x\)\(\Rightarrow1=2^x.2^{36}=2^{36+x}\)\(\Rightarrow2^{36+x}=2^0\Rightarrow36+x=0\Rightarrow x=-36\)