-16 + 23 + x = -16

          7 + x = -16 

               x = -16 - 7

               x = -23

2x + 35 = -15

        2x = -15 - 35

         2x = -50

           x = -25

-13 x |x | = -26

         x = -26 : (-13)

          x = 2

Vậy x = 2 hoặc -2

|2x - 5| = 13

      2x = 13 + 5

       2x = 18

         x = 9

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

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

2)2x+35=-15

2x      =-15-35

2x      = -50

x        =-25

1)-16+23+x=-16

7+x          =-16

x             =-16-7

x             =-23

3)-13 * lxl =-26

lxl         =  -26:-13

lxl           =2

<=>x=-2 hoac x=2

5) (x-3).(x+2) =0

<=>\(\orbr{\begin{cases}x-3=0\\x+2=0\end{cases}}\) <=>\(\orbr{\begin{cases}x=3\\x=-2\end{cases}}\)

\(a,ĐKXĐ:x\ge1\\ 13-\sqrt{x-1}=10\\ \Leftrightarrow\sqrt{x-1}=3\\ \Leftrightarrow x-1=9\\ \Leftrightarrow x=10\\ b,ĐKXĐ:x\in R\\ \sqrt{\left(2x-1\right)^2}-1=3\\ \Leftrightarrow\left|2x-1\right|=4\\ \Leftrightarrow\left[{}\begin{matrix}2x-1=-4\\2x-1=4\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{5}{2}\end{matrix}\right.\)

a: ĐKXĐ: x>=2/3

\(\dfrac{x-2}{\sqrt{3x-2}+2}=9\)

=>\(x-2=9\sqrt{3x-2}+18\)

=>\(9\sqrt{3x-2}=x-2-18=x-20\)

=>\(\Leftrightarrow\left\{{}\begin{matrix}x>=20\\81\left(3x-2\right)=x^2-40x+400\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x>=20\\x^2-40x+400-243x+162=0\end{matrix}\right.\)

=>x>=20 và x^2-283x+562=0

=>x=281(nhận) hoặc x=2(loại)

b: ĐKXĐ: x>=2/5

\(\sqrt{5x-2}=9\)

=>5x-2=81

=>5x=83

=>x=83/5

c: ĐKXĐ: x>=-1; x<>8

\(\dfrac{2x-16}{\sqrt{x+1}-3}=5\)

=>\(2x-16=5\sqrt{x+1}-15\)

=>\(\sqrt{25x+25}=2x-16+15=2x-1\)

=>\(\left\{{}\begin{matrix}x>=\dfrac{1}{2}\\4x^2-4x+1=25x+25\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>=\dfrac{1}{2}\\4x^2-29x-24=0\end{matrix}\right.\)

=>x=8(nhận) hoặc x=-3/4(loại)

a) 

\(13x+3=16\\ 13x=16-3\\ 13x=13\\ x=13:13\\ x=1\)

b) 

\(2x-138=24:23\\ 2x-138=\dfrac{24}{23}\\ 2x=\dfrac{24}{23}+138\\ 2x=\dfrac{3198}{23}\\ x=\dfrac{3198}{23}:2\\ x=\dfrac{1599}{23}\)

@ Nguyễn Quốc Đạt em bị nhầm đề bài nhé

2\(x\) - 138 = 2⁴ : 2³

2\(x\) - 138 = 2

2\(x\)           = 2 + 138

2\(x\)            = 140

  \(x\)            = 140 : 2

  \(x\)            = 70 

\(ĐKXĐ:...\)

Đặt \(\sqrt{2x+3}+\sqrt{x+1}=a>0\)

\(\Rightarrow a^2-4=3x+2\sqrt{2x^2+5x+3}\left(1\right)\)

Phương trình trở thành :

\(a=a^2-4-16\Leftrightarrow a^2-a-20=0\Rightarrow\orbr{\begin{cases}a=5\\a=-4\left(l\right)\end{cases}}\)

Thay vào (1)

\(\sqrt{2x+3}+\sqrt{x+1}=5\)

\(\Leftrightarrow\sqrt{2x+3}-3+\sqrt{x+1}-2=0\)

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

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

\(\Rightarrow x=3\)

1) \(ĐK:x\in R\)

2) \(ĐK:x< 0\)

3) \(ĐK:x\in\varnothing\)

4) \(=\sqrt{\left(x+1\right)^2+2}\) 

\(ĐK:x\in R\)

5) \(=\sqrt{-\left(a-4\right)^2}\)

\(ĐK:x\in\varnothing\)

a) x = 4

b) x = 3

c) x = 2

d) x = 1

e) x = 3

f) x = 2

g) x = 4

h) x = 3

1. \(A=\frac{1}{\left(\sqrt{x}+\frac{2016}{\sqrt{x}}\right)^2}\)

Áp dụng bất đẳng thức Côsi cho mẫu số.

2. Thế y theo x từ pt đầu xuống pt sau rồi quy đồng, giải pt bậc 4.

C2: \(pt\left(1\right)-2pt\left(2\right)\Leftrightarrow\left(x-y+5\right)\left(x-y-13\right)=0\)

3. a.

\(\text{ĐK: }2x^2-x=x\left(2x-1\right)\ge0\Leftrightarrow x\le0\text{ hoặc }x\ge\frac{1}{2}\)

Để pt có nghiệm thì \(2x-x^2\ge0\Leftrightarrow x\left(2-x\right)\ge0\Leftrightarrow0\le x\le2\)

Vậy \(\frac{1}{2}\le x\le2\)

\(pt\Leftrightarrow\sqrt{x\left(2x-1\right)}=x\left(2-x\right)\Leftrightarrow\sqrt{2x-1}=\sqrt{x}\left(2-x\right)\text{ (do }x>0\text{)}\)

\(\Leftrightarrow2x-1=x\left(2-x\right)^2\Leftrightarrow\left(x-1\right)\left(x^2-3x-1\right)=0\)

b.

\(\text{ĐK: }......\)

\(\sqrt{2x+1}=a;\text{ }\sqrt[3]{4-3x}=b\text{ }\left(a\ge0\right)\)

\(pt\Leftrightarrow3a-2b=13\Leftrightarrow a=\frac{2b+13}{3}\)

Lại có: \(3a^2+2b^3=3\left(2x+1\right)+2\left(4-3x\right)=11\)

Thay vào: \(3\left(\frac{2b+13}{3}\right)^2+2b^3=11\Leftrightarrow6b^3+4b^2+52b+136=0\)

\(\Leftrightarrow\left(b+2\right)\left(6b^2-8b+68\right)=0\)