Lời giải:

a. ĐKXĐ: $x\geq -9$

PT $\Leftrightarrow x+9=7^2=49$

$\Leftrightarrow x=40$ (tm)

b. ĐKXĐ: $x\geq \frac{-3}{2}$

PT $\Leftrightarrow 4\sqrt{2x+3}-\sqrt{4(2x+3)}+\frac{1}{3}\sqrt{9(2x+3)}=15$

$\Leftrightarrow 4\sqrt{2x+3}-2\sqrt{2x+3}+\sqrt{2x+3}=15$

$\Leftrgihtarrow 3\sqrt{2x+3}=15$

$\Leftrightarrow \sqrt{2x+3}=5$

$\Leftrightarrow 2x+3=25$

$\Leftrightarrow x=11$ (tm)

c.

PT \(\Leftrightarrow \left\{\begin{matrix} 2x+1\geq 0\\ x^2-6x+9=(2x+1)^2\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\geq \frac{-1}{2}\\ 3x^2+10x-8=0\end{matrix}\right.\)

\(\Leftrightarrow \left\{\begin{matrix} x\geq \frac{-1}{2}\\ (3x-2)(x+4)=0\end{matrix}\right.\)

\(\Leftrightarrow x=\frac{2}{3}\)

d. ĐKXĐ: $x\geq 1$

PT \(\Leftrightarrow \sqrt{(x-1)+4\sqrt{x-1}+4}-\sqrt{(x-1)+6\sqrt{x-1}+9}=9\)

\(\Leftrightarrow \sqrt{(\sqrt{x-1}+2)^2}-\sqrt{(\sqrt{x-1}+3)^2}=9\)

\(\Leftrightarrow \sqrt{x-1}+2-(\sqrt{x-1}+3)=9\)

\(\Leftrightarrow -1=9\) (vô lý)

Vậy pt vô nghiệm.

a) \(\sqrt{x+9}=7\left(x\ge-9\right)\Rightarrow x+9=49\Rightarrow x=40\)

b) \(4\sqrt{2x+3}-\sqrt{8x+12}+\dfrac{1}{3}\sqrt{18x+27}=15\left(x\ge-\dfrac{3}{2}\right)\)

\(\Rightarrow4\sqrt{2x+3}-\sqrt{4\left(2x+3\right)}+\dfrac{1}{3}\sqrt{9\left(2x+3\right)}=15\)

\(\Rightarrow4\sqrt{2x+3}-2\sqrt{2x+3}+\sqrt{2x+3}=15\)

\(\Rightarrow3\sqrt{2x+3}=15\Rightarrow\sqrt{2x+3}=5\Rightarrow2x+3=25\Rightarrow x=11\)

c) \(\sqrt{x^2-6x+9}=2x+1\)

Vì \(VT\ge0\Rightarrow VP\ge0\Rightarrow x\ge-\dfrac{1}{2}\)

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

\(\Rightarrow\left[{}\begin{matrix}x=-4\left(l\right)\\x=\dfrac{2}{3}\end{matrix}\right.\)

d) \(\sqrt{x+3+4\sqrt{x-1}}-\sqrt{x+8+6\sqrt{x-1}}=9\left(x\ge1\right)\)

\(\Rightarrow\sqrt{x-1+4\sqrt{x-1}+4}-\sqrt{x-1+6\sqrt{x-1}+9}=9\)

\(\Rightarrow\sqrt{\left(\sqrt{x-1}+2\right)^2}-\sqrt{\left(\sqrt{x-1}+3\right)^2}=9\)

\(\Rightarrow\left|\sqrt{x-1}+2\right|-\left|\sqrt{x-1}+3\right|=9\)

\(\Rightarrow\sqrt{x-1}+2-\sqrt{x-1}-3=9\Rightarrow-1=9\) (vô lý)

1) /7x+14/=63

    /7x/     =63-14

   /x/        =49:7

   x          =7

2) /18+9x/=81

         /9x/=81-18

         /x/=63:9

          x=7

a) -5 . (2 - x) + 4(x - 3) = 10x - 15

-10 + 5x + 4x -12 = 10x - 15

5x + 4x - 10x = -15 + 10 + 12

-x = 7

x = -7

b) 5 . (3 - 2x) + 5 . (x - 4) = 6 - 4x

15 - 10x + 5x - 20 = 6 - 4x

-10x + 5x + 4x = 6 - 15 + 20

-x = 11

x = -11

c) - 7 . (3x - 5) + 2 . (7x - 14) = 28

-21x + 35 + 14x - 28 = 28

-21x + 14x = 28 - 35 + 28

-7x = 21

x = 21 : (-7)

x = -3

d) 4 . (x - 5) - 3 . (x + 7) = 5 . (-4)

4x - 20 - 3x - 21 = -20

4x - 3x = -20 + 20 + 21

x = 21

e) 5 . (4 - x) - 7. (-x + 2) = 4 - 9 + 3

20 - 5x + 7x - 14 = -2

-5x + 7x = -2 - 20 + 14

2x = -8

x = -8 : 2

x = -4

Đúng 100%

câu c

- 7 ( 3x - 5 ) + 2 ( 7x - 14 ) = 28

- 21x + 35 + 14x - 28 = 28

 21x - 14x = 35 - 28 - 28

7x = - 21

x = ( - 21) : 7

x = - 3 

a, -5 .(2- x) +4 (x-3) =10x -15
<=> -19 +5x +4x-12 -10x +15=0
<=> -x=16
<=>x=16
b, 5(3-2x) +5.(x-4)=6-4x
<=> 15-10x +5x-20 -6+4x=0
<=> -x=-11
<=> x=11
c, -7.(3x-5) +7.(7x -14)=28
<=> -21x +35+49x -98-28=0
<=>28x=-91
<=>x= -\(\frac{13}{4}\)
d, 4.(x-5) -3.(x+7)=5.(-4)
<=>4x-20-3x-21=-20
<=>x=21
e,5.(4-x)-7.(-x+2)=4-9+3
<=>20x -5x +7x -14= -2
<=>22x=12
<=>x=\(\frac{6}{11}\)
P tự kết luận nhé k hiểu hay mk sai chỗ nào p pảo mk nhé

a: \(81x^5-x^3\)

\(=x^3\left(81x^2-1\right)\)

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

b: \(9x^2y-12xy+4y\)

\(=y\left(9x^2-12x+4\right)\)

\(=y\left(3x-2\right)^2\)

c: \(\left(5-x\right)^2-16\left(x-2\right)^2\)

\(=\left(x-5\right)^2-\left(4x-8\right)^2\)

\(=\left(x-5-4x+8\right)\left(x-5+4x-8\right)\)

\(=-3\left(x-1\right)\left(5x-13\right)\)

d: Ta có: \(9x^2-y^2-21x-7y\)

\(=\left(3x-y\right)\left(3x+y\right)-7\left(3x+y\right)\)

\(=\left(3x+y\right)\left(3x-y-7\right)\)

e: Ta có: \(-y^2+8y-16+9x^2\)

\(=-\left(y^2-8y+16-9x^2\right)\)

\(=-\left(y-4-3x\right)\left(y-4+3x\right)\)

f: Ta có: \(5x^2-4x-1\)

\(=5x^2-5x+x-1\)

\(=5x\left(x-1\right)+\left(x-1\right)\)

\(=\left(x-1\right)\left(5x+1\right)\)

e: Ta có: \(\left(x+2\right)\left(x^2-2x+4\right)-x\left(x^2+2\right)=15\)

\(\Leftrightarrow x^3+8-x^3-2x=15\)

\(\Leftrightarrow2x=-7\)

hay \(x=-\dfrac{7}{2}\)

f: Ta có: \(x^3-6x^2+12x-19=0\)

\(\Leftrightarrow x^3-6x^2+12x-8-11=0\)

\(\Leftrightarrow\left(x-2\right)^3=11\)

hay \(x=\sqrt[3]{11}+2\)

a, =>-2x+12+6x-60=8

=>4x+48=8

=>4x=56

=>x=14

b,=>-8x-36-8x-3-x-13=0

=>-17x-52=0

=>-17x=52

=>x=-52/17

c,=>14x+7x^2-7x^2-21x=14

=>-7x=14

=>x=-2

a <=> -2x+12 +6x-60-8=0

<=>4x-56=0

<=>x=56/4=14

Vậy S=(4)

\(a.\dfrac{3}{2}+\dfrac{-1}{3}< \dfrac{x}{6}< \dfrac{1}{9}+\dfrac{31}{18}\)

\(\Leftrightarrow\dfrac{7}{6}< \dfrac{x}{6}< \dfrac{11}{6}\)

\(\Leftrightarrow7< x< 11\)

\(\Leftrightarrow x\in\left\{8;9;10\right\}\)

\(b.\dfrac{-5}{12}+\dfrac{7}{12}+\dfrac{-1}{12}< \dfrac{x}{12}< \dfrac{2}{15}+\dfrac{1}{5}\)

\(\Leftrightarrow\dfrac{1}{12}< \dfrac{x}{12}< \dfrac{1}{3}\)

\(\Leftrightarrow\dfrac{1}{12}< \dfrac{x}{12}< \dfrac{4}{12}\)

\(\Leftrightarrow1< x< 4\)

\(\Leftrightarrow x\in\left\{2;3\right\}\)

Bn cộng 2 phân số đó lại

Bài 1: 

a) Ta có: \(\dfrac{17}{6}-x\left(x-\dfrac{7}{6}\right)=\dfrac{7}{4}\)

\(\Leftrightarrow\dfrac{17}{6}-x^2+\dfrac{7}{6}x-\dfrac{7}{4}=0\)

\(\Leftrightarrow-x^2+\dfrac{7}{6}x+\dfrac{13}{12}=0\)

\(\Leftrightarrow-12x^2+14x+13=0\)

\(\Delta=14^2-4\cdot\left(-12\right)\cdot13=196+624=820\)

Vì Δ>0 nên phương trình có hai nghiệm phân biệt là:

\(\left\{{}\begin{matrix}x_1=\dfrac{14-2\sqrt{205}}{-24}=\dfrac{-7+\sqrt{205}}{12}\\x_2=\dfrac{14+2\sqrt{2015}}{-24}=\dfrac{-7-\sqrt{205}}{12}\end{matrix}\right.\)

b) Ta có: \(\dfrac{3}{35}-\left(\dfrac{3}{5}-x\right)=\dfrac{2}{7}\)

\(\Leftrightarrow\dfrac{3}{5}-x=\dfrac{3}{35}-\dfrac{10}{35}=\dfrac{-7}{35}=\dfrac{-1}{5}\)

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

ai giúp mik vs

Bài 3: 

a) Ta có: \(2x-3=x+\dfrac{1}{2}\)

\(\Leftrightarrow2x-x=\dfrac{1}{2}+3\)

\(\Leftrightarrow x=\dfrac{7}{2}\)

b) Ta có: \(4x-\left(x+\dfrac{1}{2}\right)=2x-\left(\dfrac{1}{2}-5\right)\)

\(\Leftrightarrow3x-\dfrac{1}{2}-2x+\dfrac{1}{2}-5=0\)

\(\Leftrightarrow x=5\)