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\(a,\dfrac{3}{2}\cdot x-1=\dfrac{1}{2}x-\dfrac{3}{5}\)

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

\(\Rightarrow\left(\dfrac{3}{2}-\dfrac{1}{2}\right)x=-\dfrac{3}{5}+\dfrac{5}{5}\)

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

\(b,\dfrac{1}{2}x+\dfrac{1}{2}\left(x-2\right)=\dfrac{3}{4}-2x\)

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

\(\Rightarrow\left(\dfrac{1}{2}+\dfrac{1}{2}+2\right)x=\dfrac{3}{4}+1\)

\(\Rightarrow3x=\dfrac{7}{4}\)

\(\Rightarrow x=\dfrac{7}{4}:3\)

\(\Rightarrow x=\dfrac{7}{12}\)

\(c,\left(x-\dfrac{1}{2}\right)-\dfrac{1}{4}=0\)

\(\Rightarrow x-\dfrac{1}{2}=\dfrac{1}{4}\)

\(\Rightarrow x=\dfrac{1}{4}+\dfrac{1}{2}\)

\(\Rightarrow x=\dfrac{1}{4}+\dfrac{2}{4}\)

\(\Rightarrow x=\dfrac{3}{4}\)

\(d,4^{x-3}+1=17\)

\(\Rightarrow4^{x-3}=17-1\)

\(\Rightarrow4^{x-3}=16\)

\(\Rightarrow4^{x-3}=4^2\)

\(\Rightarrow x-3=2\)

\(\Rightarrow x=2+3\)

\(\Rightarrow x=5\)

#Toru

`3/2 x -1 =1/2x -3/5`

`=> 3/2x -1/2x = -3/5 +1`

`=> 2/2x= -3/5 + 5/5`

`=> x= 2/5`

__

`1/2x +1/2(x-2) = 3/4 -2x`

`=> 1/2x + 1/2x - 2/2 = 3/4 -2x`

`=> 1/2x +1/2x +2x = 3/4 + 1`

`=> 1/2x +1/2x + 4/2x = 3/4 +4/4`

`=> 6/2x = 7/4`

`=> x= 7/4 : 3`

`=>x=7/12`

__

`(x-1/2) -1/4=0`

`=> x-1/2=1/4`

`=> x=1/4 +1/2`

`=> x= 1/4 +2/4`

`=>x=3/4`

__

`4^(x-3) +1=17`

`=> 4^(x-3) =17-1`

`=> 4^(x-3)=16`

`=> 4^(x-3)=4^2`

`=> x-3=2`

`=>x=2+3`

`=>x=5`

Để olm.vn  giúp em nhá:

(\(x-5\))²⁰⁰² + (2\(x\) + 1)²⁰⁰⁰ = 0

vì (\(x\) - )²⁰²² ≥ 0 ∀ \(x\) 

    (2\(x\) + 1)²⁰⁰⁰ \(\ge\) 0 ∀ \(x\)

⇒ (\(x\) - 5)²⁰⁰² + (2\(x\) + 1)²⁰⁰⁰ = 0 

   ⇔ \(\left\{{}\begin{matrix}\left(x-5\right)^{2002}=0\\\left(2x+1\right)^{2000}=0\end{matrix}\right.\)

    \(\Rightarrow\) \(\left\{{}\begin{matrix}x-5=0\\2x+1=0\end{matrix}\right.\)

     \(\Rightarrow\) \(\left\{{}\begin{matrix}x=5\\2x=-1\end{matrix}\right.\)

     ⇒   \(\left\{{}\begin{matrix}x=5\\x=-\dfrac{1}{2}\end{matrix}\right.\)

vì - \(\dfrac{1}{2}\) \(\ne\) 5 vậy \(x\in\) \(\varnothing\)

Không thể tìm ''x'' trong bài này. 

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

\(2x^3-2x^2+x^2-x+5x+x^2-2x^3=3x+7\)

\(5x-x=3x+7\)

\(4x-3x=7\)

\(x=7\)

(2x+1)(x^2-x)+x(-2x^2+x+5)=3x+7

=>2x^3-2x^2+x^2-x-2x^3+x^2+5x=3x+7

=>-x^2-x+x^2+5x=3x+7

=>4x=3x+7

=>x=7

=>\(\dfrac{7}{2^x}=\dfrac{7}{2^9}\)

=>2^x=2^9

=>x=9

Có: \(\dfrac{x}{3}=\dfrac{y}{5}\Rightarrow\dfrac{2x^2}{18}=\dfrac{y^2}{25}\)

Áp dụng t/c của dãy tỉ số = nhau ta có:

\(\dfrac{2x^2}{18}=\dfrac{y^2}{25}=\dfrac{2x^2-y^2}{18-25}=\dfrac{-28}{-7}=4\)

\(\Rightarrow\left\{{}\begin{matrix}2x^2=72\\y^2=100\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=6;x=-6\\y=10;y=-10\end{matrix}\right.\)

Vậy................

\(a,\frac{1}{2}x+\frac{5}{2}=\frac{7}{2}x-\frac{3}{4}\)

\(\Leftrightarrow\frac{1}{2}x+\frac{5}{2}-\frac{7}{2}x=-\frac{3}{4}\)

\(\Leftrightarrow\frac{1}{2}x-\frac{7}{2}x+\frac{5}{2}=-\frac{3}{4}\)

\(\Leftrightarrow-3x+\frac{5}{2}=-\frac{3}{4}\)

\(\Leftrightarrow-3x=-\frac{13}{4}\)

\(\Leftrightarrow x=-\frac{13}{4}:(-3)=-\frac{13}{4}:\frac{-3}{1}=-\frac{13}{4}\cdot\frac{-1}{3}=\frac{13}{12}\)

\(b,\frac{2}{3}x-\frac{2}{5}=\frac{1}{2}x-\frac{1}{3}\)

\(\Leftrightarrow\frac{2}{3}x-\frac{2}{5}-\frac{1}{2}x=-\frac{1}{3}\)

\(\Leftrightarrow\frac{2}{3}x-\frac{1}{2}x-\frac{2}{5}=-\frac{1}{3}\)

\(\Leftrightarrow\frac{1}{6}x-\frac{2}{5}=-\frac{1}{3}\)

\(\Leftrightarrow\frac{1}{6}x=\frac{1}{15}\)

\(\Leftrightarrow x=\frac{1}{15}:\frac{1}{6}=\frac{1}{15}\cdot6=\frac{6}{15}=\frac{2}{5}\)

\(c,\frac{1}{3}x+\frac{2}{5}(x+1)=0\)

\(\Leftrightarrow\frac{1}{3}x+\frac{2}{5}x+\frac{2}{5}=0\)

\(\Leftrightarrow\frac{11}{15}x=-\frac{2}{5}\)

\(\Leftrightarrow x=-\frac{6}{11}\)

d,e,f Tương tự

\(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}=\frac{2\left(x-1\right)}{4}=\frac{3\left(y-2\right)}{9}=\frac{2x-2+3y-6-z+3}{4+9-4}=\frac{2x+3y-z-5}{9}=\frac{90}{9}=10\)

=> x-1 = 10.2 = 20 => x= 21

    y-2 = 10.3 = 30 => y = 32

    z-3 = 10.4 =40 => z = 43