Ta có (\(^{x^{2^{ }}^{ }+3x}\)) (\(^{x^{2^{ }}+3x+4}\))

Đặt \(x^{2^{ }^{ }}+3x\) là a ta có

a.(a+4)=-4

4a+\(a^2\) -4=0

\(^{ }\left(a-2\right)^2\)=0

Suy ra a=2

hay \(x^{2^{ }^{ }^{ }}+3x=2\)

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

𝑥=−3±17√/2

a) 5-(x-6)=4.(3-2x)

<=>5-x+6=12-8x

<=>-x+8x=-5-6+12

<=>7x=1

<=>x=1/7

vậy nghiệm của phương trình là 1/7

b) 7-(2x+4)=-(x+4)

<=>7-2x-4=-x-4

<=>-2x+x=-7+4-4

<=>-x=-7

<=>x=7

vậy nghiệm của phương trình là 7

7:

a: =>0,5x-5=2 hoặc 0,5x-5=-2

=>0,5x=3 hoặc 0,5x=7

=>x=6 hoặc x=14

b: |5x-2|=-3

mà |5x-2|>=0

nên ptvn

c: =>1/4x+3=0

=>1/4x=-3

=>x=-12

a, \(\left(x-5\right)\left(x-5+3\right)=0\Leftrightarrow x=5;x=2\)

b, \(-4x=\dfrac{274}{21}\Leftrightarrow x=-\dfrac{137}{42}\)

c, đk x khác - 2 ; 2 

\(x^2-3x+2-x^2-2x=6-7x\Leftrightarrow-5x+2=6-7x\)

\(\Leftrightarrow2x-4=0\Leftrightarrow x=2\left(ktm\right)\)

Vậy pt vô nghiệm 

a: \(\Leftrightarrow\dfrac{x-51}{9}-1+\dfrac{x-52}{8}-1=\dfrac{x-53}{7}-1+\dfrac{x-54}{6}-1\)

=>x-60=0

hay x=60

b: \(\Leftrightarrow\left(x-2\right)^2-3\left(x+2\right)=x-14\)

\(\Leftrightarrow x^2-4x+4-3x-6-x+14=0\)

\(\Leftrightarrow x^2-8x+12=0\)

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

=>x=2(loại) hoặc x=6(nhận)

\(a,ĐK:-9\le x\le16\\ PT\Leftrightarrow\left(\sqrt{16-x}-3\right)+\left(\sqrt{x+9}-4\right)=0\\ \Leftrightarrow\dfrac{7-x}{\sqrt{16-x}+3}+\dfrac{x-7}{\sqrt{x+9}+4}=0\\ \Leftrightarrow\left(x-7\right)\left(\dfrac{1}{\sqrt{x+9}+4}-\dfrac{1}{\sqrt{16-x}+3}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=7\left(tm\right)\\\dfrac{1}{\sqrt{x+9}+4}-\dfrac{1}{\sqrt{16-x}+3}=0\end{matrix}\right.\)

Với \(x\ge-9\) thì \(\dfrac{1}{\sqrt{x+9}+4}-\dfrac{1}{\sqrt{16-x}+3}>0\)

Do đó PT có nghiệm duy nhất \(x=7\)

\(b,ĐK:-\sqrt{2}\le x\le\sqrt{2}\\ PT\Leftrightarrow\left(\sqrt{2-x^2}-1\right)+\left(\sqrt{x^2+8}-3\right)=0\\ \Leftrightarrow\dfrac{1-x^2}{\sqrt{2-x^2}+1}+\dfrac{x^2-1}{\sqrt{x^2+8}+3}=0\\ \Leftrightarrow\left(x^2-1\right)\left(\dfrac{1}{\sqrt{x^2+8}+3}-\dfrac{1}{\sqrt{2-x^2}+1}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\\\dfrac{1}{\sqrt{x^2+8}+3}-\dfrac{1}{\sqrt{2-x^2}+1}=0\end{matrix}\right.\)

Với \(x\ge-\sqrt{2}\) thì \(\dfrac{1}{\sqrt{x^2+8}+3}-\dfrac{1}{\sqrt{2-x^2}+1}>0\)

Vậy pt có tập nghiệm \(x=\pm1\)

a) Đk: \(\left\{{}\begin{matrix}16-x\ge0\\x+9\ge0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x\le16\\x\ge-9\end{matrix}\right.\) \(\Rightarrow x\in\left[-9;16\right]\)

    Pt: \(\Rightarrow\left(\sqrt{16-x}+\sqrt{x+9}\right)^2=7^2\)

         \(\Rightarrow16-x+x+9+2\sqrt{144+7x-x^2}=49\)

         \(\Rightarrow\sqrt{144+7x-x^2}=12\)

         \(\Rightarrow144+7x-x^2=144\)

 Bạn tự tìm x nhé rồi đối chiếu đk ta đc \(x=0\) hoặc \(x=7\)

1:

a: =>(|x|+4)(|x|-1)=0

=>|x|-1=0

=>x=1; x=-1

b: =>x^2-4>=0

=>x>=2 hoặc x<=-2

d: =>|2x+5|=2x-5

=>x>=5/2 và (2x+5-2x+5)(2x+5+2x-5)=0

=>x=0(loại)

a,\(x\in\left\{5;1,5;\dfrac{-4}{3}\right\}\)

a) (x - 7)(2x + 8) = 0

\(\Leftrightarrow\left[{}\begin{matrix}x-7=0\\2x+8=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=7\\2x=-8\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=7\\x=-4\end{matrix}\right.\)

Vậy: S = {7; -4}

b) Tương tự câu a

c)  (x - 1)(2x + 7)(x² + 2) = 0

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\2x+7=0\\x^2+2=0\end{matrix}\right.\)

Mà: x² + 2 > 0 với mọi x

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\2x+7=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\2x=-7\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{7}{2}\end{matrix}\right.\)

Vậy: \(S=\left\{1;-\dfrac{7}{2}\right\}\)

d) (2x - 1)(x + 8)(x - 5) = 0

\(\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\x+8=0\\x-5=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x=1\\x=-8\\x=5\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-8\\x=5\end{matrix}\right.\)

Vậy \(S=\left\{\dfrac{1}{2};-8;5\right\}\)

a/ Pt \(\Leftrightarrow\left[{}\begin{matrix}x-7=0\\2x+8=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=7\\x=-4\end{matrix}\right.\)

Vậy \(S=\left\{7;-4\right\}\)

b/ pt \(\Leftrightarrow\left[{}\begin{matrix}3x+1=0\\5x-2=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{3}\\x=\dfrac{2}{5}\end{matrix}\right.\)

c/ pt \(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\2x+7=0\end{matrix}\right.\) (\(x^2+2>0\forall x\))\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{7}{2}\end{matrix}\right.\)

d/ pt \(\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\x+8=0\\x-5=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-8\\x=5\end{matrix}\right.\)

a)(x-7)(2x+8)=0

⇔x-7=0 hoặc 2x+8=0

1.x-7=0⇔x=7

2.2x+8=0⇔2x=-8⇔x=-4

phương trình có 1 nghiệm x=7 và x=-4

b)(3x+1)(5x-2)=0

⇔3x+1=0 hoặc 5x-2=0

1.3x+1=0⇔3x=-1⇔x=-1/3

2.5x-2=0⇔5x=2⇔x=5/2

phương trình có 2 nghiệm x=-1/3 và x=5/2

\(\dfrac{2x}{x-1}+\dfrac{4}{x^2+2x-3}=\dfrac{2x-5}{x+3}\)

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

\(ĐK:x\ne1;-3\)

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

\(\Leftrightarrow2x\left(x+3\right)+4=\left(2x-5\right)\left(x-1\right)\)

\(\Leftrightarrow2x^2+6x+4=2x^2-2x-5x+5\)

\(\Leftrightarrow13x=1\)

\(\Leftrightarrow x=\dfrac{1}{13}\left(tm\right)\)