\(a.\left(3x+2\right)\left(x^2-1\right)=\left(9x^2-4\right)\left(x+1\right)\)

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

\(\Leftrightarrow x-1=3x-2\)

\(\Leftrightarrow2x=1\)

\(\Leftrightarrow x=\dfrac{1}{2}\)

c: =>x-3=0

hay x=3

d: \(\Leftrightarrow\left(3x-1\right)\cdot\left(x^2+2-7x+10\right)=0\)

\(\Leftrightarrow\left(3x-1\right)\left(x-3\right)\left(x-4\right)=0\)

hay \(x\in\left\{\dfrac{1}{3};3;4\right\}\)

 \(\left(3x+2\right)\left(x^2-1\right)=\left(9x^2-4\right)\left(x+1\right).\)

\(\Leftrightarrow\left(3x+2\right)\left(x-1\right)\left(x+1\right)-\left(3x-2\right)\left(3x+2\right)\left(x+1\right)=0.\)

\(\Leftrightarrow\left(3x+2\right)\left(x+1\right)\left(x-1-3x+2\right)=0.\)

\(\Leftrightarrow\left(3x+2\right)\left(x+1\right)\left(-2x+1\right)=0.\)

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

c: =>(x-3)(x²+3x+5)=0

=>x-3=0

hay x=3

d: =>(3x-1)(x²+2-7x+10)=0

=>(3x-1)(x-3)(x-4)=0

hay \(x\in\left\{\dfrac{1}{3};3;4\right\}\)

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

Ta có: \(3 - 2x \ge 0 \Leftrightarrow x \le \frac{3}{2}\)

Bình phương hai vế của (1) ta được:

\(\begin{array}{l}2 - x = {\left( {3 - 2x} \right)^2}\\ \Rightarrow 2 - x = 9 - 12x + 4{x^2}\\ \Leftrightarrow 4{x^2} - 11x + 7 = 0\\ \Leftrightarrow \left[ \begin{array}{l}x = 1\left( {TM} \right)\\x = \frac{7}{4}\left( {KTM} \right)\end{array} \right.\end{array}\)

Vậy tập nghiệm của phương trình là \(S = \left\{ 1 \right\}\)

b) \(\sqrt { - {x^2} + 7x - 6}  + x = 4\)\( \Leftrightarrow \sqrt { - {x^2} + 7x - 6}  = 4 - x\)  (2)

Ta có: \(4 - x \ge 0 \Leftrightarrow x \le 4\)

Bình phương hai vế của (2) ta được:

\(\begin{array}{l} - {x^2} + 7x - 6 = {\left( {4 - x} \right)^2}\\ \Leftrightarrow  - {x^2} + 7x - 6 = 16 - 8x + {x^2}\\ \Leftrightarrow 2{x^2} - 15x + 22 = 0\\ \Leftrightarrow \left[ \begin{array}{l}x = 2\left( {TM} \right)\\x = \frac{{11}}{2}\left( {KTM} \right)\end{array} \right.\end{array}\)

Vậy tập nghiệm của phương trình là \(S = \left\{ 2 \right\}\)

Bài 1: Giải các phương trình sau:

a) 3(2,2-0,3x)=2,6 + (0,1x-4)

<=> 6.6 - 0.9x = 2,6 + 0,1x - 4

<=> - 0.9x - 0,1x = -6.6 -1,4

<=> -x = -8

<=> x = 8

Vậy x = 8

b) 3,6 -0,5 (2x+1) = x - 0,25(22-4x)

<=> 3,6 - x - 0,5 = x - 5,5 + x

<=> - x - 3,1 = -5,5

<=> - x = -2.4

<=> x = 2.4

Vậy  x = 2.4

a)2x + 3 = 7x - 7
(=)2x-7x=-7-3
(=)-5x=-10
(=)x=-2
Vậy S={2}

a. 3(x-2)-10=5(2x + 1)

<=> 3x - 6 - 10 = 10x + 5

<=> 3x - 10x = 5 + 6 + 10

<=> -7x = 21

<=> x = -3

b. 3x + 2=8 -2(x-7)

<=> 3x + 2 = 8 - 2x + 14

<=> 3x + 2x = 8 + 14 - 2

<=> 5x = 20

<=> x = 4

c. 2x-(2+5x)= 4(x + 3)

<=> 2x - 2 - 5x = 4x + 12

<=> 2x - 5x - 4x = 12 + 2

<=> -7x = 14

<=> x = -2

d. 5-(x +8)=3x + 3(x-9)

<=> 5 - x - 8 = 3x + 3x - 27

<=> -x - 3x - 3x = -27 + 8 - 5

<=> -7x = -24

<=> x = 24/7

e. 3x - 18 + x= 12-(5x + 3)

<=> 3x - 18 + x = 12 - 5x - 3

<=> 3x + x - 5x = 12 - 3 + 18

<=> -x = 27

<=> x = - 27

a. 3(x-2)-10=5(2x + 1)

<=> 3x - 6 - 10 = 10x + 5

<=> 3x - 10x = 5 + 6 + 10

<=> -7x = 21

<=> x = -3

b. 3x + 2=8 -2(x-7)

<=> 3x + 2 = 8 - 2x + 14

<=> 3x + 2x = 8 + 14 - 2

<=> 5x = 20

<=> x = 4

c. 2x-(2+5x)= 4(x + 3)

<=> 2x - 2 - 5x = 4x + 12

<=> 2x - 5x - 4x = 12 + 2

<=> -7x = 14

<=> x = -2

d. 5-(x +8)=3x + 3(x-9)

<=> 5 - x - 8 = 3x + 3x - 27

<=> -x - 3x - 3x = -27 + 8 - 5

<=> -7x = -24

<=> x = 24/7

e. 3x - 18 + x= 12-(5x + 3)

<=> 3x - 18 + x = 12 - 5x - 3

<=> 3x + x - 5x = 12 - 3 + 18

<=> -x = 27

<=> x = - 27

\(\sqrt{x^{ }2-6x+9}=4-x\)
\(\sqrt{\left(x-3\right)^{ }2}=4-x\)
x-3=4-x
x+x=4+3
2x=7
x=\(\dfrac{7}{2}\)

Lời giải:
a.

PT \(\Leftrightarrow \left\{\begin{matrix} 4-x\geq 0\\ x^2-6x+9=(4-x)^2=x^2-8x+16\end{matrix}\right.\)

\(\Leftrightarrow \left\{\begin{matrix} x\leq 4\\ 2x=7\end{matrix}\right.\Leftrightarrow x=\frac{7}{2}\)

b.

ĐKXĐ: $x\geq \frac{3}{2}$

PT \(\Leftrightarrow \sqrt{(2x-3)+2\sqrt{2x-3}+1}+\sqrt{(2x-3)+8\sqrt{2x-3}+16}=5\)

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

\(\Leftrightarrow |\sqrt{2x-3}+1|+|\sqrt{2x-3}+4|=5\)

\(\Leftrightarrow \sqrt{2x-3}+1+\sqrt{2x-3}+4=2\sqrt{2x-3}+5=5\)

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

\(a,=>x^3-2x^2+4x+2x^2-4x+8-x^3+2x-15=0\)

\(< =>2x-7=0< =>x=\dfrac{7}{2}\)

b,\(=>x\left(x^2-25\right)-\left(x+2\right)\left(x^2-2x+4\right)-3=0\)

\(< =>x^3-25x-x^3+2x^2-4x-2x^2+4x-8-3=0\)

\(< =>-25x-11=0\)

\(< =>x=-0,44\)

cảm ơn bạn nhiều nha!

\(a,\Leftrightarrow x^2+2x+1+2x+3-2\sqrt{2x+3}+1=0\\ \Leftrightarrow\left(x+1\right)^2+\left(\sqrt{2x+3}-1\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=-1\\2x+3=1\end{matrix}\right.\Leftrightarrow x=-1\left(N\right)\)

\(b,\Leftrightarrow3x^2+3x-2\sqrt{x^2+x}=0\left(x\le-1;x\ge0\right)\\ \Leftrightarrow3x\left(x-1\right)-2\sqrt{x\left(x+1\right)}=0\\ \Leftrightarrow\sqrt{x\left(x+1\right)}\left(3\sqrt{x\left(x-1\right)}-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x\left(x-1\right)=0\\\sqrt{x\left(x-1\right)}=\dfrac{2}{3}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x^2-x-\dfrac{4}{9}=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\9x^2-9x-4=0\left(1\right)\end{matrix}\right.\)

\(\Delta\left(1\right)=81-4\left(-4\right)\cdot9=225\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=\dfrac{9-15}{18}\\x=\dfrac{9+15}{18}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(N\right)\\x=1\left(N\right)\\x=-\dfrac{1}{3}\left(L\right)\\x=\dfrac{4}{3}\left(N\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=\dfrac{4}{3}\end{matrix}\right.\)