dk \(x\ge-\frac{4}{3}\)

\(x^2-5x+4=8\sqrt{3x+4}-32\)

\(\Leftrightarrow\left(x-1\right)\left(x-4\right)=8\left(\sqrt{3x+4}-4\right)\)

\(\Leftrightarrow\left(x-1\right)\left(x-4\right)-8\frac{\left(\sqrt{3x+4}-4\right)\left(\sqrt{3x+4}+4\right)}{\sqrt{3x+4}+4}=0\)

\(\left(x-1\right)\left(x-4\right)-8.\frac{3\left(x-4\right)}{\sqrt{3x+4}+4}=0\)

\(\Leftrightarrow\left(x-4\right)\left(x-1-\frac{24}{\sqrt{3x+4}+4}=0\right)\)

đến đây để rồi tự làm nhé ^^

bài toán của mk k có -23 ở vế sau ạ

\(a,PT\Leftrightarrow\left|x+3\right|=3x-6\\ \Leftrightarrow\left[{}\begin{matrix}x+3=3x-6\left(x\ge-3\right)\\x+3=6-3x\left(x< -3\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{9}{2}\left(tm\right)\\x=\dfrac{3}{4}\left(ktm\right)\end{matrix}\right.\\ \Leftrightarrow x=\dfrac{9}{2}\\ b,PT\Leftrightarrow\left|x-1\right|=\left|2x-1\right|\\ \Leftrightarrow\left[{}\begin{matrix}x-1=2x-1\\1-x=2x-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{2}{3}\end{matrix}\right.\)

\(c,ĐK:x\le\dfrac{2}{5}\\ PT\Leftrightarrow4-5x=25x^2-20x+4\\ \Leftrightarrow25x^2-15x=0\\ \Leftrightarrow5x\left(5x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\left(tm\right)\\x=\dfrac{3}{5}\left(ktm\right)\end{matrix}\right.\Leftrightarrow x=0\\ d,ĐK:x\le\dfrac{2}{5}\\ PT\Leftrightarrow4-5x=2-5x\\ \Leftrightarrow x\in\varnothing\)

a) 7x - 35 = 0

<=> 7x = 0 + 35

<=> 7x = 35

<=> x = 5

b) 4x - x - 18 = 0

<=> 3x - 18 = 0

<=> 3x = 0 + 18

<=> 3x = 18

<=> x = 5

c) x - 6 = 8 - x

<=> x - 6 + x = 8

<=> 2x - 6 = 8

<=> 2x = 8 + 6

<=> 2x = 14

<=> x = 7

d) 48 - 5x = 39 - 2x

<=> 48 - 5x + 2x = 39

<=> 48 - 3x = 39

<=> -3x = 39 - 48

<=> -3x = -9

<=> x = 3

có bị viết nhầm thì thông cảm nha!

la`thu'hai nga`y 19 nhe

a.

ĐKXĐ: \(x\ge2\)

\(\left(x+\sqrt{x}+1\right)\sqrt{x-2}=\left(x+1\right)^2-x\)

\(\Leftrightarrow\left(x+\sqrt{x}+1\right)\sqrt{x-2}=\left(x+\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)\)

\(\Leftrightarrow\sqrt{x-2}=x-\sqrt{x}+1\)

\(\Leftrightarrow\sqrt{x-2}+\sqrt{x}=x+1\)

\(\Leftrightarrow2x-2+2\sqrt{x^2-2x}=x^2+2x+1\)

\(\Leftrightarrow x^2-2\sqrt{x^2-2x}+3=0\)

\(\Leftrightarrow\left(\sqrt{x^2-2x}-1\right)^2+2x+2=0\) (vô nghiệm do \(2x+2>0\))

Vậy pt đã cho vô nghiệm

b. ĐKXĐ: \(\left[{}\begin{matrix}x\ge1\\x\le\dfrac{1}{2}\end{matrix}\right.\)

\(\Leftrightarrow2x^2-3x+1+2\left(x-1\right)\sqrt{2x^2-3x+1}+x^2-2x-3=0\)

Đặt \(\sqrt{2x^2-3x+1}=t\ge0\)

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

\(\Delta'=\left(x-1\right)^2-\left(x^2-2x-3\right)=4\)

\(\Rightarrow\left[{}\begin{matrix}t=1-x-2=-x-1\\t=1-x+2=3-x\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}\sqrt{2x^2-3x+1}=-x-1\left(x\le-1\right)\\\sqrt{2x^2-3x+1}=3-x\left(x\le3\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2-5x=0\left(vn\right)\\x^2+3x-8=0\left(x\le3\right)\end{matrix}\right.\)

\(\Rightarrow x=\dfrac{-3\pm\sqrt{41}}{2}\)

a: =>3x^2-3x-2x+2=0

=>(x-1)(3x-2)=0

=>x=2/3 hoặc x=1

b: =>2x^2=11

=>x^2=11/2

=>\(x=\pm\dfrac{\sqrt{22}}{2}\)

c: Δ=5^2-4*1*7=25-28=-3<0

=>PTVN

f: =>6x^4-6x^2-x^2+1=0

=>(x^2-1)(6x^2-1)=0

=>x^2=1 hoặc x^2=1/6

=>\(\left[{}\begin{matrix}x=\pm1\\x=\pm\dfrac{\sqrt{6}}{6}\end{matrix}\right.\)

d: =>(5-2x)(5+2x)=0

=>x=5/2 hoặc x=-5/2

e: =>4x^2+4x+1=x^2-x+9 và x>=-1/2

=>3x^2+5x-8=0 và x>=-1/2

=>3x^2+8x-3x-8=0 và x>=-1/2

=>(3x+8)(x-1)=0 và x>=-1/2

=>x=1

Bài 1: 

a: \(\Leftrightarrow x^2-5x+6< =0\)

=>(x-2)(x-3)<=0

=>2<=x<=3

b: \(\Leftrightarrow\left(x-6\right)^2< =0\)

=>x=6

c: \(\Leftrightarrow x^2-2x+1>=0\)

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

hay \(x\in R\)