\(\dfrac{1}{2}-3x+\left|x-1\right|=0\\ \Rightarrow3x+\left|x-1\right|=\dfrac{1}{2}-0\\ \Rightarrow3x+\left|x-1\right|=\dfrac{1}{2}\\ \Rightarrow\left|x-1\right|=\dfrac{1}{2}-3x\\ \Rightarrow\left[{}\begin{matrix}x-1=\dfrac{1}{2}-3x\\x-1=-\dfrac{1}{2}+3x\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x+3x=\dfrac{1}{2}+1\\x-3x=-\dfrac{1}{2}+1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}4x=\dfrac{3}{2}\\2x=\dfrac{1}{2}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{8}\\x=\dfrac{1}{4}\end{matrix}\right.\)

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\(\dfrac{1}{2}\left|2x-1\right|+\left|2x-1\right|=x+1\\ \Rightarrow\left|2x-1\right|\cdot\left(\dfrac{1}{2}+1\right)=x+1\\ \Rightarrow\left|2x-1\right|\cdot\dfrac{3}{2}=x+1\\ \Rightarrow\left|2x-1\right|=x+1:\dfrac{3}{2}\\ \Rightarrow\left|2x-1\right|=x+\dfrac{2}{3}\\ \Rightarrow\left[{}\begin{matrix}2x-1=x+\dfrac{2}{3}\\2x-1=-x-\dfrac{2}{3}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}2x-x=\dfrac{2}{3}+1\\2x+x=-\dfrac{2}{3}+1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\\3x=\dfrac{1}{3}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\\x=\dfrac{1}{9}\end{matrix}\right.\)

a/

\(\left(x-1\right)^2-\left(x+1\right)^2=2x-6\\ x^2-2x+1-\left(x^2+2x+1\right)=2x-6\\ \)

\(\Leftrightarrow x^2-2x+1-x^2-2x-1-2x+6=0\)

\(\Leftrightarrow6-6x=0\)

=> x=1

b, \(4x^2+12x+9-3\left(x^2-16\right)=x^2-4x+4\)

\(\Leftrightarrow12x+9+48=-4x+4\Leftrightarrow16x=-53\Leftrightarrow x=-\dfrac{53}{16}\)

c, \(\left(x-3\right)\left(x+3\right)-\left(x-3\right)\left(5x+2\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(x+3-5x-2\right)=0\Leftrightarrow\left(x-3\right)\left(-4x+1\right)=0\Leftrightarrow x=3;x=\dfrac{1}{4}\)

d, \(x^2\left(x+4\right)-9\left(x+4\right)=0\Leftrightarrow\left(x-3\right)\left(x+3\right)\left(x+4\right)=0\Leftrightarrow x=-3;3;-4\)

Ta có ( x + 2 )( -x - 5 )( x² + 9 ) = 0

Suy ra x + 2 = 0; -x - 5 = 0 hoặc x² + 9 = 0

Nếu x + 2 = 0 ⇒ x = -2

Nếu -x - 5 = 0 ⇒ -x = 5 ( loại )

Nếu x² + 9 = 0 ⇒ x² = -9 ( loại )

Vậy x = -2 thì ( x + 2 )( -x - 5 )( x² + 9 ) = 0

x=-5 sao lại loại z ?

(x^2+9)(x+1)(x^2+4)=0

<=> x²+9=0 hoặc x+1=0 hoặc x²+4=0

<=> x=-1 

Vậy x=-1

Xét từng TH nha bạn

Th1 : x^2 + 9 = 0 => x^2 = -9 < Loại >

th2 : x^2 + 4 = 0 => x^2 = -4 < Loại >

Th3 : x + 1 = 0 => x = -1 

(x²+9)(x+1)(x²+4)=0

\(\Rightarrow\hept{\begin{cases}x^2+9=0\\x+1=0\\x^2+4=0\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}x^2=9\left(vl\right)\\x=-1\\x^2=-4\left(vl\right)\end{cases}}\)

Vậy x=-1

\(A=4\sqrt{x}-\frac{\left(\sqrt{x}+3\right)^2\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)

\(=4\sqrt{x}-\left(\sqrt{x}+3\right)\)

\(=3\sqrt{x}-3\)

\(B=\frac{\sqrt{\left(3x+2\right)^2}}{3x+2}=\frac{|3x+2|}{3x+2}\)

\(TH1:3x+2>0\Rightarrow B=1\)

\(TH2:3x+2< 0\Rightarrow B=-1\)

A <=> 4√x - [ ( (√x )^2 + 2√x3+ 3^2)*( √x -3)]/ (x-9)

<=> 4√x - [(√x+3)^2×(√x-3)]/( x-9)

<=> 4√x - [(√x+3)*(x-9)]/(x-9)

<=> 4√x - √x -3

<=> 3√x -3

b, <=> √[(3*x) ^2+2*3x*2+2^2]/(3x+2)

<=> √[( 3x+2)^2] /(3x+2) 

<=> (3x+2)/(3x+2) = 1

bai tap nay lop may day

Điên à 

a)

(2x-1)²=9

<=> 2x-1=\(\sqrt{9}\)

<=> 2x-1=3

<=> 2x=4

<=> x=2

b)

2x(x-3)=0

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x-3=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=0\\x=3\end{cases}}\)

Vậy x=0 hoặc x=3

Tìm số nguyên x,biết:

a,(2x-1)² = 9

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

\(\Rightarrow2x-1=3\)

\(\Rightarrow2x=4\Rightarrow x=2\)

b,2x(x-3) = 0

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

giúp mik với!!

b.x=3

a. x=2

Cách 1:

\(\left(\dfrac{1}{3}+\dfrac{2}{9}\right)\times\dfrac{4}{9}\)

\(=\dfrac{5}{9}\times\dfrac{4}{9}\)

\(=\dfrac{20}{81}\)

Cách 2:

\(\left(\dfrac{1}{3}+\dfrac{2}{9}\right)\times\dfrac{4}{9}\)

\(=\dfrac{1}{3}\times\dfrac{4}{9}+\dfrac{2}{9}\times\dfrac{4}{9}\)

\(=\dfrac{4}{27}+\dfrac{8}{81}\)

\(=\dfrac{20}{81}\)

C1: \(\left(\dfrac{1}{3}+\dfrac{2}{9}\right)\)x\(\dfrac{4}{9}\)=\(\left(\dfrac{3}{9}+\dfrac{2}{9}\right)\text{x}\dfrac{4}{9}\)=\(\dfrac{5}{9}\text{x}\dfrac{4}{9}=\dfrac{20}{81}\)

C2: \(\left(\dfrac{1}{3}+\dfrac{2}{9}\right)\)x\(\dfrac{4}{9}\)=\(\dfrac{1}{3}\text{ x}\dfrac{4}{9}+\dfrac{2}{9}\text{ x}\dfrac{4}{9}\)=\(\dfrac{4}{27}+\dfrac{8}{81}=\dfrac{12}{81}+\dfrac{8}{81}=\dfrac{20}{81}\)

 \(C1:\left(\dfrac{1}{3}+\dfrac{2}{9}\right)\times\dfrac{4}{9}\\ =\dfrac{5}{9}\times\dfrac{4}{9}\\ =\dfrac{20}{81}\)

\(C2:\left(\dfrac{1}{3}+\dfrac{2}{9}\right)\times\dfrac{4}{9}\\ =\dfrac{1}{3}\times\dfrac{4}{9}+\dfrac{2}{9}\times\dfrac{4}{9}\\ =\dfrac{4}{27}+\dfrac{8}{81}\\ =\dfrac{20}{81}\)