a) \(x+\dfrac{4}{9}=\dfrac{5}{27}\)  

    \(x=\dfrac{5}{27}-\dfrac{4}{9}\)

   \(x=-\dfrac{7}{27}\)

b) \(x-\dfrac{4}{11}=\dfrac{7}{33}\)

   \(x=\dfrac{7}{33}+\dfrac{4}{11}\)

   \(x=\dfrac{19}{33}\)

c) \(\dfrac{8}{5}-x=\dfrac{1}{3}\times\dfrac{2}{5}\)

  \(\dfrac{8}{5}-x=\dfrac{2}{15}\)

          \(x=\dfrac{8}{5}-\dfrac{2}{15}\)

          \(x=\dfrac{22}{15}\)

d) \(x-\dfrac{3}{4}=\dfrac{1}{2}+\dfrac{2}{6}\)

   \(x-\dfrac{3}{4}=\dfrac{5}{6}\)

   \(x=\dfrac{5}{6}+\dfrac{3}{4}\)

   \(z=\dfrac{19}{12}\)

Ví dụ : 5/2 = 5 phần 2

a) \(\left(x+3\right)\left(x^2-3x+9\right)-x\left(x-2\right)^2=27.\)

\(\Leftrightarrow x^3+27-x\left(x^2-4x+4\right)-27=0.\)

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

\(\Leftrightarrow4x\left(x-1\right)=0.\)

\(\Leftrightarrow\left[{}\begin{matrix}4x=0.\\x-1=0.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0.\\x=1.\end{matrix}\right.\)

Vậy \(S=\left\{0;1\right\}.\)

Câu a) -3 phần 1/2 

Câu a) 2 mũ 2

a: =>-7/2:(4/5-1/2x)=4

=>4/5-1/2x=-7/2:4=-7/8

=>1/2x=4/5+7/8=67/40

=>x=67/20

b: =>5x=5

=>x=1

c: =>x(-2/3-1/3)=-2

=>-x=-2

=>x=2

d: =>-2/3(x+1)=-1/3+1/2=1/6

=>x+1=-1/6:2/3=-1/6*3/2=-3/12=-1/4

=>x=-1/4-1=-5/4

a) \(\sqrt{x-1}+\sqrt{x+3}+2\sqrt{\left(x+3\right)\left(x-1\right)}=-\left(x+3+x-1-6\right)\)\(\left(Đk:x\ge1\right)\)

\(\left(\sqrt{x-1}+\sqrt{x+3}\right)^2+\sqrt{x-1}+\sqrt{x-3}-6=0\)

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

Đến đây em xét các trường hợp rồi bình phương lên là được nha

b) \(\sqrt{3x-2}+\sqrt{x-1}=3x-2+x-1-6+2\sqrt{\left(3x-2\right)\left(x-1\right)}\left(Đk:x\ge1\right)\)

\(\left(\sqrt{3x-2}+\sqrt{x-1}\right)^2-\left(\sqrt{3x-2}+\sqrt{x-1}\right)-6=0\)

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

Đến đây em xét các trường hợp rồi bình phương lên là được nha

a/ ĐKXĐ: $x\geq 1$

Đặt $\sqrt{x-1}=a; \sqrt{x+3}=b$ thì pt trở thành:

$a+b+2ab=6-(a^2+b^2)$

$\Leftrightarrow a^2+b^2+2ab+a+b-6=0$

$\Leftrightarrow (a+b)^2+(a+b)-6=0$

$\Leftrightarrow (a+b-2)(a+b+3)=0$

Hiển nhiên do $a\geq 0; b\geq 0$ nên $a+b+3>0$. Do đó $a+b-2=0$

$\Leftrightarrow a+b=2$

Mà $b^2-a^2=(x+3)-(x-1)=4$

$\Leftrightarrow (b-a)(b+a)=4\Leftrightarrow (b-a).2=4\Leftrightarrow b-a=2$

$\Rightarrow \sqrt{x+3}=b=(a+b+b-a):2=(2+2):2=2$

$\Leftrightarrow x=1$ (tm)

b/

ĐKXĐ: $x\geq 1$

Đặt $\sqrt{3x-2}=a; \sqrt{x-1}=b(a,b\geq 0)$. Khi đó pt đã cho trở thành:

$a+b=a^2+b^2-6+2ab$

$\Leftrightarrow a^2+b^2+2ab-(a+b)-6=0$

$\Leftrightarrow (a+b)^2-(a+b)-6=0$

$\Leftrightarrow (a+b+2)(a+b-3)=0$

Hiển nhiên $a+b+2>0$ với mọi $a,b\geq 0$

Do đó $a+b-3=0\Leftrightarrow a+b=3$

$\Leftrightarrow b=3-a$.

Ta thấy $a^2-3b^2=1$. Thay $b=3-a$ vô thì:

$a^2-3(3-a)^2=1$

$\Leftrightarrow (a-2)(a-7)=0$

$\Leftrightarrow a=2$ hoặc $a=7$

Vì $a+b=3$ mà $a,b>0$ nên $a,b<3$. Do đó $a=2$

$\Leftrightarrow \sqrt{3x-2}=2$ 

$\Leftrightarrow x=2$ 

a: ĐKXD: x<>0

\(\dfrac{14x^3+12x^2-14x}{2x}=\left(x+2\right)\left(3x-4\right)\)

=>\(\dfrac{2x\left(7x^2+6x-7\right)}{2x}=\left(x+2\right)\left(3x-4\right)\)

=>\(7x^2+6x-7=3x^2-4x+6x-8\)

=>\(7x^2+6x-7=3x^2+2x-8\)

=>\(4x^2+4x+1=0\)

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

=>2x+1=0

=>x=-1/2(nhận)

b: \(\left(4x-5\right)\left(6x+1\right)-\left(8x+3\right)\left(3x-4\right)=15\)

=>\(24x^2+4x-30x-5-\left(24x^2-32x+9x-12\right)=15\)

=>\(24x^2-26x-5-24x^2+23x+12=15\)

=>-3x+7=15

=>-3x=8

=>\(x=-\dfrac{8}{3}\)

a: =>x^2-25-x^2-3x=10

=>-3x=35

=>x=-35/3

b: =>4x^2-9-4(x^2+4x+4)=5

=>4x^2-9-4x^2-16x-16-5=0

=>-16x-30=0

=>x=-15/8

c: =>9x^2+45x-9x^2+4=7

=>45x=3

=>x=1/15

d: =>x^3+3x^2+3x+1-x^3-3x^2+5x=8

=>8x=7

=>x=7/8

\(a,\Leftrightarrow\left(5x+1\right)\left(x-4\right)-\left(x-4\right)=0\\ \Leftrightarrow\left(x-4\right)\left(5x+1-x\right)=0\\ \Leftrightarrow5x\left(x-4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\\ b,\Leftrightarrow2x^2-10x-2x^2-3x=26\\ \Leftrightarrow-13x=26\\ \Leftrightarrow x=-2\\ c,\Leftrightarrow x^3+1-x^3+3x=15\\ \Leftrightarrow3x=14\\ \Leftrightarrow x=\dfrac{14}{3}\)

\(d,\Leftrightarrow x^3-5x+2x^2-10+5x-2x^2-17=0\\ \Leftrightarrow x^3-27=0\\ \Leftrightarrow x^3=27\\ \Leftrightarrow x=3\)

a: Ta có: \(\sqrt{\sqrt{x}+3}=4\)

\(\Leftrightarrow\sqrt{x}+3=16\)

\(\Leftrightarrow\sqrt{x}=13\)

hay x=169

b: Ta có: \(\sqrt{x+3}=\sqrt{1-5x}\)

\(\Leftrightarrow x+3=1-5x\)

\(\Leftrightarrow6x=-2\)

hay \(x=-\dfrac{1}{3}\left(nhận\right)\)

a) \(\sqrt{3+\sqrt{x}}=4\left(đk:x\ge0\right)\)

\(\Leftrightarrow3+\sqrt{x}=16\Leftrightarrow\sqrt{x}=13\Leftrightarrow x=169\left(tm\right)\)

b) \(\sqrt{x+3}=\sqrt{1-5x}\left(đk:\dfrac{1}{5}\ge x\ge-3\right)\)

\(\Leftrightarrow x+3=1-5x\Leftrightarrow6x=-2\Leftrightarrow x=-\dfrac{1}{3}\left(ktm\right)\)

Vậy \(S=\varnothing\)

c) \(\sqrt{x^2+6x+9}=3x-1\left(đk:x\ge\dfrac{1}{3}\right)\)

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

\(\Leftrightarrow\left|x+3\right|=3x-1\)

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

a. \(\sqrt{3+\sqrt{x}}=4\)      ĐKXĐ: \(x\ge0\)

<=> 3 + \(\sqrt{x}\) = 4²

<=> \(3+\sqrt{x}=16\)

<=> \(\sqrt{x}=16-3\)

<=> \(\sqrt{x}=13\)

<=> x = 13²

<=> x = 169 (TM)

b. \(\sqrt{x+3}=\sqrt{1-5x}\)           ĐKXĐ: \(x\ge\dfrac{1}{5}\)

<=> \(\left(\sqrt{x+3}\right)^2=\left(\sqrt{1-5x}\right)^2\)

<=> \(|x+3|=|1-5x|\)

<=> \(\left[{}\begin{matrix}x+3=1-5x\\-\left(x+3\right)=-\left(1-5x\right)\\x+3=-\left(1-5x\right)\\-\left(x+3\right)=1-5x\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}x=\dfrac{-1}{3}\\x=\dfrac{-1}{3}\\x=1\\x=1\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}x=\dfrac{-1}{3}\\x=1\end{matrix}\right.\)

c. \(\sqrt{x^2+6x+9}=3x-1\)

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

<=> \(|x+3|=3x-1\)

<=> \(\left[{}\begin{matrix}x+3=-\left(3x-1\right)\\x+3=3x-1\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}x+3=-3x=1\\-2x=-4\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}4x=-2\\x=2\end{matrix}\right.\)

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

a) \(\left(x+1\right)^3-\left(x-1\right)^3-6\cdot\left(x-1\right)^2=10\)

\(\Rightarrow x^3+3x^2+3x+1-x^3+3x^2-3x+1-6\cdot\left(x^2-2x+1\right)=10\)

\(\Rightarrow6x^2+2-6x^2+12x-6=10\)

\(\Rightarrow12x-4=10\)

\(\Rightarrow12x=14\)

\(\Rightarrow x=\dfrac{7}{6}\)

b) \(x\left(x+5\right)\left(x-5\right)-\left(x+2\right)\left(x^2-2x+4\right)=42\)

\(\Rightarrow x\left(x^2-25\right)-\left(x^3+8\right)=42\)

\(\Rightarrow x^3-25x-x^3-8=42\)

\(\Rightarrow-25x-8=42\)

\(\Rightarrow-25x=50\)

\(\Rightarrow x=\dfrac{50}{-25}=-2\)

c) \(\left(x-2\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+6\left(x+1\right)^2=49\)

\(\Rightarrow x^3-6x^2+12x-8-\left(x^3-27\right)+6\left(x^2+2x+1\right)=49\)

\(\Rightarrow x^3-6x^2+12x-8-x^3+27+6x^2+12x+6=49\)

\(\Rightarrow24x+25=49\)

\(\Rightarrow24x=24\)

\(\Rightarrow x=\dfrac{24}{24}=1\)