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

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

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

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

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

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

Tỉ số của 2 số lúc đầu là 1/6 hay số bé bằng 1/5 hiệu 2 số.

Khi thêm vào mỗi số 18,4 đơn vị thì hiệu vẫn không đổi và số bé bằng 0,25 số lớn hay bằng 1/4 số lớn. Nên số bé bằng 1/3 hiệu hai số.

Tỷ số giữa số bé lúc đầu và số bé lúc sau là: 1/5 : 1/3 = 3/5

Hiệu số phần bằng nhau là: 6 - 3 = 2 phần

Số bé lúc đầu là: 18,4 : 2 x 3 = 27,6

Số lớn lúc đầu là: 27,6 x 6 = 165,6

ĐS: Số lớn 165,6; Số bé 27,6

\(2x=\frac{y}{3}\Rightarrow\frac{x}{\frac{1}{2}}=\frac{y}{3}\Rightarrow\frac{x}{\frac{1}{2}}=\frac{3y}{9}\)

Theo t/c dãy tỉ số = nhau ta có:

\(\frac{x}{\frac{1}{2}}=\frac{3y}{9}=\frac{x-3y}{\frac{1}{2}-9}=\frac{34}{-\frac{17}{2}}=34.\left(-\frac{2}{17}\right)=-4\)

=> \(\frac{x}{\frac{1}{2}}=-4\Rightarrow x=-4.\frac{1}{2}=-2\)

=>\(\frac{y}{3}=-4\Rightarrow y=-4.3=-12\)

Vậy x=-2; y=-12.

\(x^2-0,25=0\)

\(\Leftrightarrow x^2-0,5^2=0\)

\(\Leftrightarrow x-0,5=0\)

\(\Leftrightarrow x=0+0,5\)

\(\Leftrightarrow x=0,5\)

b.

\(\dfrac{8}{2x}=2\Leftrightarrow2x=\dfrac{8}{2}\Rightarrow x=4:2=2\)

c.

\(\left(2x-1\right)^3=8\)

\(\Rightarrow\left(2x-1\right)^3=2^3\)

\(\Rightarrow2x-1=2\)

\(\Rightarrow2x=2+1\)

\(\Rightarrow2x=3\)

\(\Rightarrow x=\dfrac{3}{2}\)

d.

\(\left(x-2\right)^2=16\)
\(\Rightarrow\left(x-2\right)^2=4^2\)

\(\Rightarrow x-2=4\)

\(\Rightarrow x=4+2\)

\(\Rightarrow x=6\)

Chúc bạn học tốt!!!!

\(x^2-0,25=0\)

\(\Rightarrow x^2=0,25\)

\(\Rightarrow x=0,5\)

b. \(\dfrac{8}{2x}=2\)

\(\Rightarrow2x=4\)

\(\Rightarrow x=2\)

c. \(\left(2x-1\right)^3=8\)

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

\(\Rightarrow2x-1=2\)

\(\Rightarrow x=\dfrac{3}{2}\)

d. \(\left(x-2\right)^2=16\)

\(\left(x-2\right)^2=\pm4^2\)

\(\Rightarrow\left[{}\begin{matrix}x-2=4\\x-2=-4\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=6\\x=-2\end{matrix}\right.\)

\(x^2-0,25=0\Rightarrow x^2=0,25\)

\(\Rightarrow\left[{}\begin{matrix}x=0,5\\x=-0,5\end{matrix}\right.\)

\(\dfrac{8}{2x}=2\Rightarrow2x=4\Rightarrow x=2\)

\(\left(2x-1\right)^3=8\)

\(\Rightarrow2x-1=2\Rightarrow2x=3\Rightarrow x=\dfrac{3}{2}\)

\(\left(x-2\right)^2=16\)

\(\Rightarrow\left[{}\begin{matrix}x-2=4\\x-2=-4\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=6\\x=-2\end{matrix}\right.\)

Lời giải:
PT $\Leftrightarrow \frac{11}{17}x+\frac{3}{17}=\frac{78}{17}x-\frac{22}{17}$

$\Leftrightarrow \frac{25}{17}=\frac{67}{17}x$

$x=\frac{25}{67}$

\(x-\dfrac{3}{17}\left(2x-1\right)=\dfrac{7}{34}\left(1-2x\right)+\dfrac{10x-3}{2}\)

<=>\(x-\dfrac{3}{17}\left(2x-1\right)-\dfrac{7}{34}\left(1-2x\right)=\dfrac{10x-3}{2}\)

<=>\(x-\dfrac{3}{17}\left(2x-1\right)+\dfrac{7}{34}\left(2x-1\right)=\dfrac{10x-3}{2}\)

<=>\(x+\left(-\dfrac{3}{17}+\dfrac{7}{34}\right)\left(2x-1\right)-\dfrac{10x-3}{2}=0\)

<=>\(x+\dfrac{1}{34}\left(2x-1\right)-\dfrac{10x-3}{2}=0\)

<=>\(x+\dfrac{1}{17}x-\dfrac{1}{34}-\dfrac{10x-3}{2}=0\)

<=>\(\dfrac{18}{17}x-\dfrac{85x-25.5}{17}=\dfrac{0.5}{17}\)

=> \(18x-85x+25.5=0.5\)
<=>\(-67x=0.5-25.5\)
<=>\(-67x=-25\)
<=> \(x=\dfrac{25}{67}\)
đây là cách làm khác (dù dài hơn) nếu bạn không hiểu cách bên dưới 
chúc bạn học tốt!

a)

\(\left(1-2x\right)\cdot\left(1+2x\right)+\left(2x+3\right)^2=34\\ \Leftrightarrow1-4x^2+4x^2+12x+9=34\\ \Leftrightarrow12x+10=34\\ \Rightarrow x=\frac{34-10}{12}=2\)

b)

\(\left(2x-3\right)^2+\left(3-2x\right)\cdot\left(x-1\right)=0\\ \Leftrightarrow4x^2-12x+9+3x-3-2x^2+2x=0\\ \Leftrightarrow2x^2-7x+6=0\\ \Leftrightarrow6-7x+2x^2=0\\ \Leftrightarrow6-3x-4x+2x^2=0\\ \Leftrightarrow3\cdot\left(2-x\right)-2x\cdot\left(2-x\right)=0\\ \Leftrightarrow\left(3-2x\right)\cdot\left(2-x\right)=0\\\Rightarrow\left[{}\begin{matrix}3-2x=0\\2-x=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\frac{3}{2}\\x=2\end{matrix}\right.\)

\(a,\left(1-2x\right)\left(1+2x\right)+\left(2x+3\right)^2=34\)

\(1-4x^2+4x^2+12x+9=34\)

\(10+12x=34\)

\(12x=24\)

\(x=2\)

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

\(4x^2+12x+9+3x-3-2x^2+2x=0\)

\(2x^2-7x+6=0\)

\(2x^2-3x-4x+6=0\)

\(x\left(2x-3\right)-2\left(2x-3\right)=0\)

\(\left(x-2\right)\left(2x-3\right)=0\)

\(\left[{}\begin{matrix}x-2=0\\2x-3=0\end{matrix}\right.\)

\(\left[{}\begin{matrix}x=2\\x=\frac{3}{2}\end{matrix}\right.\)

\(\left(1-2x\right)\left(1+2x\right)+\left(2x+3\right)^2=34\)

\(1-4x^2+4x^2+12x+9=34\)

\(12x+10=34\)

\(12x=24\Rightarrow x=2\)

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

\(4x^2-12x+9+3x-3-2x^2_{ }+2x=0\)

\(2x^2-7x+6=0\)

\(\left(x-2\right)\left(2x-3\right)=0\)

\(\left[{}\begin{matrix}x-2=0\\2x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\frac{3}{2}\end{matrix}\right.\)